Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity

Bihar Board 12th Physics Objective Questions and Answers

Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity

Question 1.
1 ampere current is equivalent to
(a) 6.25 × 1018 electrons s-1
(b) 2.25 × 1018 electrons s-1
(c) 6.25 × 1014 electrons s-1
(d) 2.25 × 1014 electrons s-1
Answer:
(a) 6.25 × 1018 electrons s-1

Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity

Question 2.
In an atom electrons revolves around the nucleus along a path of radius 0.72 Å making 9.4 x 1018 revolution per second. The equivalent current is
(e = 1.6 × 10-19 C)
(a) 1.2 A
(b) 1.5 A
(c) 1.4 A
(d) 1.8 A
Answer:
(b) 1.5 A

Question 3.
The direction of the flow of current through electric circuit is
(a) from low potential to high potential.
(b) from high potential to low potential.
(c) does not depend upon potential value.
(d) current cannot flow through circuit.
Answer:
(b) from high potential to low potential.

Question 4.
The electrical resistance of a conductor depends upon
(a) size of conductor
(b) temperature of conductor
(c) geometry of conductor
(d) all of these
Answer:
(d) all of these

Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity

Question 5.
A wire of resistance 4Ω is used to wind a coil of radius 7 cm. The wire has a diameter of 1.4 mm and the specific resistance of its material is 2 × 10-7 Ω m. The number of turns in the coil is
(a) 50
(b) 40
(c) 60
(d) 70
Answer:
(d) 70
Solution:
(d) Let n be the number of turns in the coil.
Then total length of wire used,
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 9

Question 6.
Space between two concentric spheres of radii r1 and r2, such that r1 < r2, is filled with a material of resistivity ρ . Find the resistance between inner and outer surface of the material.
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 1
Answer:
(b) \(\frac{r_{2}-r_{1}}{r_{1} r_{2}} \frac{\rho}{4 \pi}\)
Solution:
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 10

Question 7.
The I-V characteristic shown in figure represents
(a) ohmic conductors
(b) non-ohmic conductors
(c) insulators
(d) superconductors
Answer:
(b) non-ohmic conductors

Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity

Question 8.
Range of resistivity for metals is
(a) 10-6 Ω m to 10-4 Ω m
(b) 10-7 Ω m to 10-5 Ω m
(c) 10-8 Ω m to 10-6 Ω m
(d) 10-9 Ω m to 10-7 Ω m
Answer:
(c) 10-8 Ω m to 10-6 Ω m

Question 9.
Wire bound resistors are made by
(a) winding the wires of an alloy viz, Cu, Al, Ag
(b) winding the wires of an alloy viz, Si, Tu, Fe
(c) winding the wires of an alloy viz, Ge, Au, Gr
(d) winding the wires of an alloy viz, manganin, constantan, nichrome.
Answer:
(d) winding the wires of an alloy viz, manganin, constantan, nichrome.

Question 10.
The resistivity of alloy manganin is
(a) Nearly independent of temperature
(b) Increases rapidly with increase in temperature
(c) Decreases with increase in temperature
(d) Increases rapidly with decrease in temperature
Answer:
(a) Nearly independent of temperature

Question 11.
What is the order of magnitude of the resistance of a dry human body ?
(a) 10 Ω
(b) 10 KΩ
(c) 10 MΩ
(d) 10 μΩ
Answer:
(b) 10 KΩ

Question 12.
Two copper wires of length l and 21 have radii, r and 2r respectively. What is the ratio of their specific resistances ?
(a) 1 : 2
(b) 2 : 1
(c) 1 : 1
(d) 1 : 3
Answer:
(c) 1:1

Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity

Question 13.
Nichrome or Manganin is widely used in wire bound standard resistors because of their
(a) temperature independent resistivity
(b) very weak temperature dependent resistivity
(c) strong dependence of resistivity with temperature
(d) mechanical strength
Answer:
(b) very weak temperature dependent resistivity

Question 14.
An electric heater is connected to the voltage supply. After few seconds, current get its steady value then its initial current will be
(a) equal to its steady current
(b) slightly higher than its steady current
(c) slightly less than its steady current
(d) zero
Answer:
(b) slightly higher than its steady current

Question 15.
In the circuit shown in figure heat developed across 2 Ω,4 Ω and 3 Ωresistances are in the ratio of
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 2
(a) 2 : 4 : 3
(b) 8 : 4 : 12
(c) 4 : 8 : 27
(d) 8 : 4 : 27
Answer:
(d) 8 : 4 : 27

Question 16.
The total resistance in the parallel combination of three resistances 9 Ω,7 Ω, and 5Ω is
(a) 1.22 Ω
(b) 2.29 Ω
(c) 4.22 Ω
(d) 2.02 Ω
Answer:
(d) 2.02 Ω
Solution:
(d) In the parallel combination of three resistances, the equivalent resistance is
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 11
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 12

Question 17.
Five equal resistances of 10 £2 are connected between A and B as shown in figure. The resultant resistance
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 3
(a) 10Ω
(b) 5Ω
(c) 15Ω
(d) 6Ω
Answre:
(b) 5Ω
Solution:
(b) According to the given circuit 10 Ω and 10 Ω resistances are connected in series.
∴ R’= 10 + 10 = 20Ω
Again 10 Ω and 10Ω resistances are connected in series
∴R”= 10 + 10 = 20 Ω
R’, R” and 10 Ω all connected in parallel than
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 13

Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity

Question 18.
A copper cylindrical tube has inner radius a and outer radius b. The resistivity is P. The resistance of the cylinder between the two ends is
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 4
Answer:
(c) \(\frac{\rho^{f}}{\pi\left(b^{2}-a^{2}\right)}\)

Question 19.
A and B are two points on a uniform ring of resistance 15Ω . The ∠AOB = 45°. The equivalent resistance between A and B is
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 5
(a) 1.64 Ω
(b) 2.84 Ω
(c) 4.57 Ω
(d) 2.64 Ω
Answer:
(a) 1.64 Ω

Question 20.
The reading of ammeter shown in figure is
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 6
(a) 6.56 A
(b) 3.28 A
(c) 2.18 A
(d) 1.09 A
Answer:
(d) 1.09 A

Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity

Question 21.
In a circuit a cell with internal resistance r is connected to an external resistance R. The condition for the maximum current that drawn from the cell is
(a) R = r
(b) R < r
(c) R > r
(d) R = 0
Answer:
(d) R = 0

Question 22.
In parallel combination of n cells, we obtain
(a) more voltage
(b) more current
(c) less voltage
(d) less current
Answer:
(b) more current

Question 23.
If n cells each of emf e and internal resistance r are connected in parallel, then the total emf and internal resistances will be
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 7
Answer:
(a) \(\varepsilon, \frac{\mathrm{r}}{\mathrm{n}}\)

Question 24.
The potential difference between A and B as shown in figure is
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 8
(a) 1 V
(b) 2 V
(c) 3 V
(d) 4 V
Answer:
(a) 1 V
Solution:
(a) Resistance of the upper arm CAD = 2 Ω +3 Ω = 5 Ω
Resistance of the lower arm CBD = 3 Ω +2 Ω =5 Ω
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 14
As the resistance of both arms are equal, therefore same amount of current flows in both the arms. Current through each arm CAD or CBD = 1A
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 15
Potential difference across C and A is
VC – VB = (2Ω) (1 A)
= 2 V …..(i)
Potential difference across C and B is
VC – VB = (3Ω) (1 A)
= 3 V ………..(ii)
Subtracting (i) from (ii), we get VA – VB = 3V – 2V = 1V

Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity

Question 25.
In a Wheatstone bridge if the battery and galvanometer are interchanged then the deflection in galvanometer will
(a) change in previous direction
(b) not change
(c) change in opposite direction
(d) none of these
Answer:
(b) not change

Question 26.
In a Wheatstone’s network, P = 2Ω, Q = 2Ω, R = 2Ω, and S = 3Ω The resistance with which S is to be shunted in order that the bridge may be balanced is
(a) 1Ω
(b) 2Ω
(c) 4Ω
(d) 6Ω
Answer:
(d) 6Ω
Solution:
(d) Let x be the resistance shunted with S for the bridge to be balanced.
For a balance Wheatone’s bridge
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 15 - 1
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 16

Question 27.
When a metal conductor connected to left gap of a meter bridge is heated, the balancing point
(a) shifts towards right
(b) shifts towards left
(c) remains unchanged
(d) remains at zero
Answer:
(a) shifts towards right

Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity

Question 28.
In a meter bridge experiment, the ratio of the left gap resistance to right gap resistance is 2 : 3, the balance point from left is
(a) 60 cm
(b) 50 cm
(c) 40 cm
(d) 20 cm
Answer:
(c) 40 cm
Solution:
Bihar Board 12th Physics Objective Answers Chapter 3 Current Electricity - 17

Question 29.
Which of the following characteristics of electrons determines the current in a conductor ?
(a) Drift velocity alone
(b) Thermal velocity alone
(c) Both drift velocity and thermal velocity
(d) Neither drift not thermal velocity
Answer:
(a) Drift velocity alone

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Bihar Board 12th Chemistry Objective Questions and Answers

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 1.
Which of the following is the correct cell representation for the given cell reaction ?
Zn + H2SO4→ ZnSO4 + H2
(a) Zn|Zn2+||H + |H2
(b) Zn|Zn2+||H + ,H2|Pt
(c) Zn |ZnSO4| |H2SO4| Zn
(d) Zn |H2SO4||ZnSO4| H2
Answer:
(b) Zn|Zn2+||H + ,H2|Pt

Question 2.
The cell reaction of the galvanic cell:
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 1
(a) Hg + Cu2+→ Hg2++Cu
(b) Hg + Cu2+→ Cu+Hg+
(c) Cu + Hg → CuHg
(d) Cu + Hg2+→ Cu2+ Hg
Answer:
(d) Cu + Hg2+→ Cu2+ Hg

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 3.
Which of the following reaction is possible at anode ?
(a) 2Cr3+ +7H2O → Cr2O72- + 14H+
(b) F2 → 2F
(c) (1/2)O2 +2H+ → H2O
(d) None of these.
Answer:
(a) 2Cr3+ +7H2O → Cr2O72- + 14H+

Question 4.
In the cell, |Zn2+| |Cu2+|Cu, the negative terminals ………..
(a) Cu
(b) Cu2+
(c) Zn
(d) Zn2+
Answer:
(c) Zn

Question 5.
In a cell reaction,
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 2
If the concentration of Cu2+ ions is doubled then EºCell will be
(a) doubled
(b) halved
(c) increased by four times
(d) unchanged
Answer:
(d) unchanged

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 6.
A standard hydrogen electrode that a zero potential because ……………..
(a) hydrogen can be most’easily oxidised
(b) hydrogen has only one electron
(c) the electrode potential is assumed to be zero
(d) hydrogen is the lightest element.
Answer:
(c) the electrode potential is assumed to be zero

Question 7.
Which of the following is the correct order in which metals displace each other from the salt solution of their salts  ?
(a) Zn, Al, Mg, Fe, Cu
(b) Cu, Fe, Mg, Al, Zn
(c) Mg, Al, Zn, Fe, Cu
(d) Al, Mg, Fe, Cu, Zn
Answer:
(d) Al, Mg, Fe, Cu, Zn

Question 8.
Fluorine is the best oxidising agent because it has ………..
(a) highest electron affinity
(b) highest reduction potential
(c) highest oxidation potential
(d) lowest electron affinity.
Answer:
(b) highest reduction potential

Question 9.
At 25°C, Nemst equation is …………..
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 3
Answer:
(a)

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 10.
The correct Nernst equation for the given cell
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 4
Answer:
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 5
Question 11.
What will be the emf of the following concentration cell at 25°C ?
Ag(s)| AgNO3(0.01M)| | AgNO3(0.05M)| Ag(s)
(a) 0.828 V
(b) 0.0413 V
(c) -0.0413 V
(d) -0.828 V
Answer:
(b) 0.0413 V
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 6

Question 12.
What will be the reduction potential for the following half-cell reaction at 298 K ? [Given: [Ag+] = 0.1 M and
e„ = +0.80 V)] ………….
(a) 0.741V
(b) 0.80 V
(c) -0.80 V
(d) -0.741V
Answer:
(a) 0.741V
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 7
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 8

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 13.
For the cell reaction: 2Cu(aq) → Cu(s) + Cu2+ the standard ceil potential is 0.36 V. The equilibrium constant for the reaction is ………….
(a) 1.2 x 106
(b) 7.4 x 1012
(c) 2.4 x 106
(d) 5.5 x 108
Answer:
(a) 1.2 x 106

Question 14.
cell for the reaction, 2H2O+ H3O+ + OH” at 25°C is -0.8277 V. The equilibrium constant for the reaction is ………….
(a) 10-14
(b) 10-23
(c) 10-7
(d) 10-21
Answer:
(a) 10-14

Question 15.
Cell reaction is spontaneous, when …………….
(a) E°d is negative
(b) ΔG° is negative
(c) E°oxid is positive
(d) ΔG° is positive
Answer:
(b) ΔG° is negative

Question 16.
ΔG° f°r the reaction, Cu2++ Fe → Fe2+ + Cu is
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 9
(a) 11.44 kJ
(b) 180.8 kJ
(c) 150.5 kJ
(d) 28.5 kJ
Answer:
(c) 150.5 kJ

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 10

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 17.
Units of the properties measured are given below. Which of the properties has not been matched correctly ?
(a) Molar conductance = S m2 mol-1
(b) Cell constant = m-1
(c) Specific conductance = S m2
(d) Equivalent conductance = S m2 (g eq)-1
Answer:
(c) Specific conductance = S m2

Question 18.
Molar conductivity of 0.15 M solution of KCl at 298 K, if its conductivity is 0.0152 S cm-1 will be
(a) 124 Ω-1 cm2    mol-1
(b) 204  Ω-1 cm2 mol-1
(c) 101 Ω-1 cm2 mol-1
(d) 300  Ω-1 cm2 mol-1
Answer:
(c) 101 Ω-1 cm2   mol-1

Question 19.
The molar conductivity is maximum for the solution of concentration
(a) 0.004 M
(b) 0.002 M
(c) 0.005 M
(d) 0.001 M
Answer:
(d) 0.001 M

Question 20.
Specific conductance of 0.1 M NaCl solution is 1.01 x 10-2 ohm-1 cm-1. Its molar conductance in ohm-1 cm2 mol-1
(a) 1.01 x 102
(b) 1.01 x 103
(c) 1.01 x 104
(d) 1.01
Answer:
(a) 1.01 x 102

Question 21.
When water is added to an aqueous solution of an electrolyte, what is the change in specific conductivity of the electrolyte ?
(a) Conductivity decreases
(b) Conductivity increases
(c) Conductivity remain same
(d) Conductivity does not depend on number of ions
Answer:
(a) Conductivity decreases

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 22.
Mark the correct choice of electrolytes represented in the graph.
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 11
(a) A → 4 NH4OH, B → NaCl
(b) A →NH4OH,B → NH4Cl
(c) A → CH3COOH, B → 4CH3COONa
(d) A → KCl, B → 4 NH4OH
Answer:
(d) A → KCl, B → 4 NH4OH

Question 23.
Molar conductivity of NH4OH can be calculated by the equation,
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 12
Answer:
(c)

Question 24.
Limiting molar conductivity of NaBr is
(a) Δ°mNaBr = Δ°mNaCl + Δ°mKBr
(b) Δ°mNaBr = Δ°mNaCl + Δ°mKBr – Δ°mKCl
(c) Δ°mNaBr = Δ°mNaOH + Δ°mNaBr – Δ°mNaCl
(d) Δ°mNaBr = Δ°mNaCl – Δ°mNaBr
Answer:
(b) Δ°mNaBr = Δ°mNaCl + Δ°mKBr – Δ°mKCl

Question 25.
Δ°m(NH4OH) is equal to …………..
(a) Δ°m (NH4OH) + Δ°m (NH4a) –  -Δ°m(HCl)
(b) Δ°m(NH4Cl) + Δ°m (NaOH) – Δ°m(Nao)
(c) Δ°m(NH4Cl) + Δ°m(NaCl) –  Δ°m(NaOH)
(d) Δ°m(NaOH) +Δ°m(NaCI) – Δ°m(NH4Cl)
Answer:
(b) Δ°m(NH4Cl) + Δ°m (NaOH) – Δ°m(Naa)

Question 26.
In an electrolytic cell, the flow of electrons is
(a) from cathode to anode in the solution
(b) from cathode to anode through external supply
(c) from cathode to anode through internal supply
(d) from anode to cathode through internal supply.
Answer:
(c) from cathode to anode through internal supply

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 27.
How long would it take to deposit 50 g of Ai from an electrolytic cell containing Al2O3 using a current of 105 ampere
(a) 54 h
(b) 1.42 h
(c) 32 h
(d) 2.15 h
Answer:
(b) 1.42 h

Question 28.
The charge required for reducing 1 mole of MnO4 to Mn2+ is ……………….
(a) 93 x 105 C
(b) 2.895 x  105 C
(c) 4.28 x  105 C
(d) 4.825 x  105 C
Answer:
(d) 4.825 x  105 C

Question 29.
How much electricity in terms of Faraday is required to  produce 100 g of Ca from molten CaCl2 ?
(a) IF
(b) 2F
(c) 3 F
(d) 5F
Answer:
(a) IF

Question 30.
How many coulombs of electricity is required to reduce 1 mole of Cr2O2 in acidic medium ?
(a) 4 x  96500C
(b) 6 x  96500 C
(c) 2 x 96500 C
(d) 1 x  96500 C
Answer:
(b) 6 x  96500 C

Question 31.
How much metal will be deposited when a current of 12 ampere with 75% efficiency is passed through the cell for 3 h ? (Given : Z = 4 x 10-4)
(a) 32.4 g
(b) 38.8 g
(c) 0 g
(d) 22.4 g
Answer:
(b) 38.8 g

Question 32.
How many moles of Pt may be deposited on the cathode when 0.80 F of electricity is passed through a 1.0 M solution of Pt4+ ?
(a) 0.1 mol
(b) 0.2  mol
(c) 4 mol
(d) 0.6  mol
Answer:
(b) 0.2  mol
(b) Pt4+ + 4e → Pt
4 moles of electricity or 4 F of electricity is required to deposit 1 mole of Pt.
0. 80 F of electricity will deposit
\(\frac{1}{4}\) 0.80 = 0.20 mol

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 33.
If 54 g of silver is deposited during an electrolysis reaction, how much aluminium will be deposited by the same amount of electric current ?
(a) 2.7 g
(b) 4.5  g
(c) 27 g
(d) 5.4  g
Answer:
(b) 4.5g
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 13

Question 34.
During the electrolysis of dilute sulphuric acid, the following process is possible at anode.
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 14
Answer:
(a)

Question 35.
When a lead storage battery is discharged,
(a) lead sulphate is consumed
(b) oxygen gas is evolved
(c) lead sulphate is formed
(d) lead sulphide is formed.
Answer:
(c) lead sulphate is formed

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 36.
The reaction which is taking place in nickel-cadmium battery can be represented by which of the following equation
(a) Cd + NiO2 + 2H2O → 4 Cd(OH)2 + Ni(OH)2
(b) Cd + NiO2 + 2 OH+ → Ni + Cd(OH)2
(c) Ni + Cd(OH)2 → Cd + Ni(OH)2
(d) Ni(OH)2 +Cd(OH)2 → Ni + Cd + 2H2O
Answer:
(a) Cd + NiO2 + 2H2O → 4 Cd(OH)2 + Ni(OH)2

Question 37.
Which of the following reactions does not take place during rusting ?
(a) 42 CO3 2H+ + CO+
(b) 4Fe2+ + O2(dry) → Fe2O3
(c) 4Fe2+ +O2 +4H2O → 2Fe2O3 +8H+
(d) Fe2O3 + xH2O → Fe2O3. x H2O
Answer:
(a) 42 CO3 2H+ + CO+

Question 38.
Which of the following reactions does not take place during rusting ?
(a)
Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry 15
(b) 4Fe2+ + O2(dry) → Fe2O3
(c) 4Fe2+ + O2 +4H2O → 2Fe2O3 +8H+
(d) Fe2O3 + xH2O → Fe2O3. xH2O
Answer:
(b) 4Fe2+ + O2(dry) → Fe2O3

Question 39.
The difference between the electrode potentials of two electrodes when no current is drawn through the cell is called
(a) cell potential
(b) cell emf
(c) potential difference
(d) cell voltage
Answer:
(b) cell emf

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 40.
An electrochemical cell can behave like an electrolytic cell when ___
(a) Ecell = 0
(b) Ecell > Eext
(c) Eextl > Ecell
(d) Ecell = Eext
Answer:
(c) Eextl > Ecell

Question 41.
Use the data given in Q. 61 and find out the most stable ion in its reduced form.
(a) Cl-
(b) Cr3+
(c) Cr
(d) Mn2+
Answer:
(d) Mn2+

Question 42.
Use the data of Q. 61 and find out the most stable oxidised species…….
(a) Cr3+
(b) MnO4
(c) Cr2O2-
(d) Mn2+
Answer:
(a) Cr3+

Question 43.
The quantity of charge required to obtain one mole of aluminium from Al2O3 is____________
(a) 1F
(b) 6F
(c) 3F
(d) 2F
Answer:
(c) 3F

Question 44.
The cell constant of a conductivity cell_____________
(a) changes with change of electrolyte
(b) changes with change of concentration of electrolyte
(c) changes with temperature of electrolyte
(d) remains constant for a cell
Answer:
(d) remains constant for a cell

Bihar Board 12th Chemistry Objective Answers Chapter 3 Electrochemistry

Question 45.
While charging the lead storage battery___________
(a) PbSO4 anode is reduced to Pb
(b) PbSO4 cathode is reduced to Pb
(c) PbSO4 cathode is oxidised to Pb
(d) PbSO4 anode is oxidised to PbO2
Answer:
(a) PbSO4 anode is reduced to Pb

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Bihar Board 12th Physics Objective Questions and Answers

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 1.
Who invented world wide web?
(a) J.C.R. Licklider
(b) Tim Berners-Lee
(c) Alexander Graham Bell
(d) Samuel F.B. Morse
Answer:
(b) Tim Berners-Lee

Question 2.
Essential elements of a communication system are
(a) transmitter and receiver
(b) receiver and communication channel
(c) transmitter and communication channel
(d) transmitter, communication channel and receiver
Answer:
(d) transmitter, communication channel and receiver

Question 3.
Which of the following is an example of point to point communication mode?
(a) Radio
(b) Television
(c) Telephony
(d) All of these
Answer:
(c) Telephony

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 4.
Which among the following is not a basic terminology used in electronic communication systems?
(a) Transducer
(b) Transmitter
(c) Telegraph
(d) Attenuation
Answer:
(c) Telegraph

Question 5.
Which of the following is not a transducer?
(a) Loudspeaker
(b) Amplifier
(c) Microphone
(d) All of these
Answer:
(b) Amplifier

Question 6.
The loss of strength of a signal while propagating through a medium is known as
(a) reception
(b) absorption
(c) transmission
(d) attenuation
Answer:
(d) attenuation

Question 7.
The process of increasing the strenth of a signal using an electronic circuit is called
(a) amplification
(b) modulation
(c) demodulation
(d) transmission
Answer:
(a) amplification

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 8.
Modem is a device which performs
(a) modulation
(b) demodulation
(c) rectification
(d) modulation and demodulation
Answer:
(d) modulation and demodulation

Question 9.
Modulation is the process of superposing
(a) low frequency audio signal on high frequency radio waves
(b) low frequency radio signals on low frequency audio . wave
(c) high frequency radio signal on low frequency audio signal
(d) high frequency audio signal on low frequency radio waves
Answer:
(a) low frequency audio signal on high frequency radio waves

Question 10.
The device which is a combination of a receiver and a transmitter is
(a) Amplifier
(b) Repeater
(c) Transducer
(d) Modulator
Answer:
(b) Repeater

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 11.
Large bandwidth for higher data rate is achieved by using
(a) high frequency carrier wave
(b) high frequency audio wave
(c) low frequency carrier wave
(d) low frequency audio wave
Answer:
(a) high frequency carrier wave

Question 12.
In a video signal for transmission of picture, what value of bandwidth used in communication system?
(a) 2.4 MHz
(b) 4.2 MHz
(c) 24 MHz
(d) 42 MHz
Answer:
(b) 4.2 MHz

Question 13.
Which of the following is an example of broadcast mode of communication?
(a) Radio
(b) Television
(c) Mobile
(d) Both (a) & (b)
Answer:
(d) Both (a) & (b)

Question 14.
Which of the following devices is full duplex?
(a) Mobile phone
(b) Walky-talky
(c) Loud speaker
(d) Radio
Answer:
(a) Mobile phone

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 15.
FM broadcast is preferred over AM broadcast becuase
(a) it is less noisy
(b) reproduction is of much better quality
(c) it is more noisy
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 16.
The frequency band used in the downlink of satellite communication is
(a) 9.5 to 2.5 GHz
(b) 896 to 901 MHz
(c) 3.7 to 4.2 GHz
(d) 840 to 935 MHz
Answer:
(c) 3.7 to 4.2 GHz

Question 17.
The radio waves of frequency 30 MHz to 300 MHz belong to
(a) high frequency band
(b) very high frequency band
(c) ultra high frequency band
(d) super high frequency band
Answer:
(b) very high frequency band

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 18.
Ground waves have wavelength
(a) less than that of sky waves
(b) greater than that of sky waves
(c) less than that of space waves
(d) equal to that of space waves
Answer:
(b) greater than that of sky waves

Question 19.
The mode of propagation used by short wave broadcast services is
(a) space wave
(b) sky wave
(c) ground wave
(d) both (a) and (c)
Answer:
(b) sky wave

Question 20.
The skip zone in radio wave transmission is that range where
(a) there is no reception of either ground wave or sky wave
(b) the reception of ground-wave is maximum but that of sky wave is minimum
(c) the reception of ground wave is minimum, but that of sky wave is maximum
(d) the reception of both ground and sky wave is maximum
Answer:
(a) there is no reception of either ground wave or sky wave

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 21.
The waves that are bent down by the ionosphere are
(a) ground waves
(b) surface waves
(c) space waves
(d) sky waves
Answer:
(d) sky waves

Question 22.
The maximum line-of-sight distance dM between two antennas having heights hT and hR above the earth is
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 1
Answer:
(d)

Question 23.
Frequencies in the UHF range normally propagate by means of
(a) ground waves
(b) sky waves
(c) surface waves
(d) space waves
Answer:
(d) space waves

Question 24.
A ground reciver in line-of-sight communication cannot receive direct waves due to
(a) its low frequency
(b) curvature of earth
(c) its high intensity
(d) smaller antenna
Answer:
(b) curvature of earth

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 25.
Which of the following modes is used for line of sight communication as well as satellite communication?
(a) Ground wave
(b) Sky wave
(c) Space wave
(d) All of these
Answer:
(c) Space wave

Question 26.
Through which mode of propagation, the radio waves can be sent from one place to another
(a) ground wave propagation
(b) sky wave propagation
(c) space wave propagation
(d) all of them
Answer:
(d) all of them

Question 27.
Which of the following frequency will be suitable for beyond the horizon communication using sky waves?
(a) 10 kHz
(b) 10 MHz
(c) 1 GHz
(d) 1000 GHz
Answer:
(b) 10 MHz

Question 28.
What should be the length of the dipole antenna for a carrier wave of frequency 3 x 108 Hz?
(a) 1 m
(b) 0.5 m
(c) 2 m
(d) 2.5 m
Answer:
(b) 0.5 m

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 29.
A radio can tune to any station in 7.5 MHz to 12 MHz band. The corresponding wavelength band is
(a) 25 m-40m
(b) 10 m-30m
(c) 20-40 m
(d) 25 m – 35 m
Answer:
(a) 25 m-40m

Question 30.
In frequency modulation
(a) the amplitude of modulated wave varies as frequency of carrier wave
(b) the frequency of modulated wave varies as amplitude of modulating wave
(c) the amplitude of modulated wave varies as amplitude of carrier wave
(d) the frequency of modulated wave varies as frequency of modulating wave
Answer:
(d) the frequency of modulated wave varies as frequency of modulating wave

Question 31.
Audio signal cannot be transmitted because
(a) the signal has more noise.
(b) the signal cannot be amplified for distance communication.
(c) the transmitting antenna length is very small to design.
(d) the transmitting antenna length is very large and impracticable.
Answer:
(d) the transmitting antenna length is very large and impracticable.

Question 32.
The modulation in which pulse duration varies in accordance with the modulating signal is called
(a) PAM
(b) PPM
(c) PWM
(d) PCM
Answer:
(c) PWM

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 33.
A 300 W carrier is modulated to a depth 75%. The total power in the modulated wave is
(a) 200 W
(b) 284 W
(c) 320 W
(d) 384 W
Answer:
(d) 384 W

Question 34.
A signal wave of frequency 12 kHz is modulated with a carrier wave of frequency 2.51 MHz. The upper and lower side band frequencies are respectively
(a) 2512 kHz and 2508 kHz
(b) 2522 kHz and 2488 kHz
(c) 2502 kHz and 2498 kHz
(d) 2522 kHz and 2498 kHz
Answer:
(d) 2522 kHz and 2498 kHz

Question 35.
The maximum amplitude of an amplitude modulated wave is found to be 15 V while its minimum amplitude is found to be 3 V. The modulation index is
\((a) \frac{3}{2}
(b) \frac{2}{3}
(c) \frac{1}{2}
(d) \frac{1}{3}\)
Answer:
\((b) \frac{2}{3}\)

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 36.
The LC product of a tuned amplifier circuit require to generate a carrrier wave of 1 MHz for amplitude modulation is
(a) 1.5 x 10-14 s
(b) 1.2 x 10-12 s
(c) 3.2 x 10-12 s
(d) 2.5 x 10-14 s
Answer:
(d) 2.5 x 10-14 s

Question 37.
The essential condition for demodulation is
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 2
Answer:
(b)

Question 38.
Figure shows a communication system. What is the output power when input signal is 1.01 m W?
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 3
(a) 90 mW
(b) 101 mW
(c) 112 m W
(d) 120 mW
Answer:
(b) 101 mW

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 39.
A 100 m long antenna is mounted on a 500 m tall building. The complex can become a transmission tower for waves with X
(a) ~ 400 m
(b)~25m
(c)~150m
(d)~ 2400 m
Answer:
(a) ~ 400 m

Question 40.
A male voice after modulation-transmission sounds like that of a female to the receiver. The problem is due to
(a) poor selection of modulation index (selected 0<w<l)
(b) poor bandwidth selection of amplifiers
(c) poor selection of carrier frequency
(d) loss of energy in transmission
Answer:
(b) poor bandwidth selection of amplifiers

Question 41.
A basic communication system consists of
(A) transmitter
(B) information source
(C) user of information
(D) channel
(E) receiver

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Choose the correct sequence in which these are arranged in a basic communication system.
(a) ABCDE
(b) BADEC
(c) BDACE
(d) BEADC
Answer:
(b) BADEC

Question 42.
Identify the mathematical expression for amplitude modulated wave.
\( (a) A_{c} \sin \left[\left\{\omega_{c}+k_{1} v_{m}(t)\right\} t+\phi\right]\)
\( (b) A_{c} \sin \left\{\omega_{c} t+\phi+k_{2} v_{m}(t)\right\}\)
\( (c) \left\{A_{c}+k_{2} v_{m}(t)\right\} \sin \left(\omega_{n} t+\phi\right)\)
\( (d) A_{c} v_{m}(t) \sin \left(\omega_{c} t+\phi\right)\)
Answer:
\((c) \left\{A_{c}+k_{2} v_{m}(t)\right\} \sin \left(\omega_{n} t+\phi\right)\)

Hints And Explanations

Question 29.
(a) Wavelength corresponding to 7.5 MHz frequency
Answer:
\(\lambda_{1}=\frac{c}{v}=\frac{3 \times 10^{8}}{7.5 \times 10^{6}}=40 \mathrm{m}\)
Wavelength corresponding to 12 MHz frequency
\(\lambda_{2}=\frac{3 \times 10^{8}}{12 \times 10^{6}}=25 \mathrm{m}\)

Question 33.
Pc = 300 W, μ, = 75% = 0.75
Answer:
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 4

Question35.
(b) Here, Ac + Am = 15 V …………………. (i)
Ac – Am = 3V ………………………… (ii)
Answer:
Solving (I) and (ii), we get, Ac =9 V, Am = 6 V
∴ Modulation index, \(\mu=\frac{A_{m}}{A_{c}}=\frac{6}{9}=\frac{2}{3}\)

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 36.
(d) Frequency of tuned amplifier is \(v=\frac{1}{2 \pi \sqrt{L C}}\)
Answer:
As per question, u = 1 MHz = 106 Hz
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 5

Question 38.
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 6
Answer:
Here, transmission path = 5 km
Loss suffered in transmission path
= – 2 dB km-1 x 5 km = – 10 dB
Total amplifier gain = 10 dB + 20 dB = 30 dB
Overall gain of signal = 30 dB – 10 dB = 20 dB
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 7

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 39.
(a) Length of antenna, ¡ = 100 m,
As l = \(\frac{\lambda}{4}\) or λ = 4l = 4 x 100 = 400m.

Very Important Questions

Question 1.
The electric Held at a point near an infinite thin sheet of charged conductor is :
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 8
Answer:
(c)

Question 2.
Two capacitors C1 = 2μF and C2 = 4μF are connected in series and a potential difference (p.d.) of 1200 V is applied across it. The Potential difference across 2μF will be :
(a) 400 V
(b) 600 V
(c) 800V
(d) 900V
Answer:
(c) 800V

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 3.
In the figure, ifnet force on Q is zero then value of \(\frac{Q}{q}\) is:
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 9
Answer:
(b)

Question 4.
When a body is charged, its mass
(a) increases
(b) decreases
(c) remains same
(d) may increase or decrease
Answer:
(d) may increase or decrease

Question 5.
The electric potential due to a small electric dipole at a large distance r from the center of the dipole is proportional to :
\((a) r
(b) \frac{1}{r}
(c) \frac{1}{r^{2}}
(d) \frac{1}{r^{3}}\)
Answer:
\((c) \frac{1}{r^{2}}\)

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 6.
Minimum number of capacitors of 2pF each required to obtain a capacitance of 5fxF will be
(a) 4
(b) 3
(c) 5
(d) 6
Answer:
(a) 4

Question 7.
The specific resistance of conductor increases with :
(a) increase of temperature
(b) increase of cross-sectional area
(c) decrease in length
(d) decrease of cross-sectional area
Answer:
(a) increase of temperature

Question 8.
The Drift velocity (vd) and applied electric field (E) of a conductor are related as :
\((a) \mathrm{V}_{\mathrm{d}} \propto \sqrt{\mathrm{E}}
(b) \mathrm{V}_{\mathrm{d}} \propto \mathrm{E}
(c) \mathrm{V}_{\mathrm{d}} \propto \mathrm{E}^{2}
(d) v_{d}= Constant\)
Answer:
\((b) \mathrm{V}_{\mathrm{d}} \propto \mathrm{E}\)

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 9.
A charge ‘q’ moves in a region where electric field ‘E’ and magnetic field ‘B’ both exist, then force on it is :
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 10
Answer:
(b)

Question 10.
A bar magnet of magnetic moment M is cut into two parts of equal length. The magnetic moment of either part is :
(a) M
(b) M/2
(c) 2M
(d) Zero
Answer:
(b) M/2

Question 11.
The dimensional formula for \(\frac{1}{2} \varepsilon_{0} \mathbf{E}^{2}\) is identical to that of:
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 11
Answer:
(a)

Question 12.
A circular coil of radius R carries a current I. The magnetic field at its center is B. At what distance from the center, on the axis of the coil, the magnetic field will be B/8.
(a) \(\sqrt{2} \mathrm{R}\)
(b) 2R
(c) \(\sqrt{3} \mathrm{R}\)
(d) 3R
Answer:
(c) \(\sqrt{3} \mathrm{R}\)

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 13.
Magnetic dipole moment is a vector quantity directed from:
(a) South to North Pole
(b) North to South Pole
(c) East to West direction
(d) West to East direction
Answer:
(a) South to North Pole

Question 14.
A wire of magnetic dipole moment M and L is bent into shape of a semicircle of radius r. What will be its new dipole moment?
\((a) \mathrm{M}
(b) \frac{M}{2 \pi}
(c) \frac{M}{\pi}
(d) \frac{2 M}{\pi}\)
Answer:
\((d) \frac{2 M}{\pi}\)

Question 15.
In a closed circuit of resistance 10W, the linked flux varies with time according to relation Φ = 6t2 – 5t +1. At t = 0.25 second, the current (in Ampere) flowing through the circuit is :
(a) 0.4
(b) 0.2
(c) 2.0
(d) 4.0
Answer:
(b) 0.2

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 16.
In A.C. circuit, the current and voltage are given by i = 5 cos wt, and V = 200 sin wt respectively. Power loss in the circuit is :
(a) 20 W
(b) 40W
(c) 1000W
(d) zero
Answer:
(d) zero

Question 17.
The energy of an electromagnetic radiation is 13.2 keV.
This radiation is related to which region of spectrum?
(a) Visible
(b) X-rays
(c) Ultra violet
(d) Infrared
Answer:
(b) X-rays

Question 18.
The image formed by objective lens of a compound Microscope is :
(a) Virtual and diminished
(b) Real and diminished
(c) Real and large
(d) Virtual and Large
Answer:
(c) Real and large

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 19.
A convex lens is dipped in a liquid, whose refractive index is equal to refractive index of material of lens. Then its focal length will:
(a) become zero
(b) become infinite
(c) decrease
(d) increase
Answer:
(b) become infinite

Question 20.
Critical angle for light moving from medium I to medium II is 6. The speed of light in medium I is v. Then speed in medium II is :
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 12
Answer:
(b)

Question 21.
A magnifying glass is to be used at the fixed object distance of 1 inch. If it is to produce an erect image 5 items magnified, its focal length should be :
(a) 0.2″
(b) 0.8″
(c) 1.5″
(d) 5″
Answer:
(c) 1.5″

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 22.
To remove hyper metropia lens used is :
(a) Convex
(b) Concave
(c) Cylindrical
(d) Plano-convex
Answer:
(a) Convex

Question 23.
Which of the following phenomena taken place when a monochromatic light is incident on a prism?
(a) Dispersion
(b) Deviation
(c) Interference
(d) All of the above
Answer:
(a) Dispersion

Question 24.
Optical fiber communication is based on which of the following phenomena:
(a) Total Internal reflection
(b) Scattering
(c) Reflection
(d) Interference
Answer:
(a) Total Internal reflection

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 25.
The value of maximum amplitude produced due to interference of two waves is given by- y1 = 4 sin wt and y2 = 3 cos wt
(a) 7
(b) 5
(c) 1
(d) 25
Answer:
(b) 5

Question 26.
The phase difference 0 is related to Path difference Δx by:
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 13
Answer:
(c)

Question 27.
The resolving power of human eye (in minute) is :
(a) \(\frac{1}{60}\)
(b) 1
(c) 10
(d) \(\frac{1}{2}\)
Answer:
(a) \(\frac{1}{60}\)

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 28.
A particle of mass m and charged q is accelerated through a potential V. The De-Broglie wavelength of the particle will be :
Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems in english medium 14
Answer:
(c)

Question 29.
The number of Photons of frequency 1014 Hz in radiation of 6.62J will be :
(a) 1010
(b) 1015
(c) 1020
(d) 1025
Answer:
(c) 1020

Question 30.
The minimum angular momentum of electron in Hydrogen atom will be :
\((a) \frac{h}{\pi} J s
(b) \frac{h}{2 \pi} J s
(c) \mathrm{h} \pi \mathrm{Js}
(d) 2 \pi h J s\)
Answer:
\((b) \frac{h}{2 \pi} J s\)

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Question 31.
The atomic number and mass number for a specimen are Z and A respectively. The number of neutrons in the atom will be :
(a) A
(b) Z
(c) A + Z
(d) A-Z
Answer:
(d) A-Z

Question 32.
The quantities, which remain conserved in a nuclear reaction:
(a) Total Charge
(b) Angular momentum
(c) Linear momentum
(d) All the above
Answer:
(d) All the above

Question 33.
Meaning of “FAX” is :
(a) Full Access Transmission
(b) Facsimile Telegraphy
(c) Factual Auto Access
(d) Feed Auto Exchange
Answer:
(b) Facsimile Telegraphy

Question 34.
A semiconductor is cooled from T1K to T2K, then its resistance will:
(a) increase
(b) decrease
(c) remain constant
(d) First decrease then increase
Answer:
(a) increase

Question 35.
If the current constant for a transistor are α & β then :
(a) αβ = l
(b) β > l, α < l
(c) α = β
(d) β < l, α > l
Answer:
(b) β > l, α < l

Bihar Board 12th Physics Objective Answers Chapter 15 Communication Systems

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Bihar Board 12th Physics Objective Questions and Answers

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 1.
Which of the following statement is not true ?
(a) Electrostatic force is a conservative force.
(b) Potential at a point is the work done per unit charge in bringing a charge from infinity to that point.
(c) Electrostatic force is non-conservative.
(d) Potential is the product of charge and work.
Answer:
(b) Potential at a point is the work done per unit charge in bringing a charge from infinity to that point.

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 2.
I volt is equivalent to
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 1
Answer:
\(\frac{\text { joule }}{\text { coulomb }}\)

Question 3.
The potential at a point due to a charge of 5 × 10-7 C located 10 cm away is
(a) 3.5 × 105 V
(b) 3.5 × 104 V
(c) 4.5 × 104 V
(d) 4.5 × 105 V
Answer:
(c) 4.5 × 104 V
Solution:
(c) Here, q = 5 × 10-7C,r = 10cm = 0.1 m

Potential,
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 10

Question 4.
A charge +q is placed at the origin O of x-y axes as shown in the figure. The work done in taking a charge Q from A to B along the straight line AB is
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 2
Answer:
(a) \(\frac{\mathrm{qQ}}{4 \pi \varepsilon_{0}}\left(\frac{\mathrm{a}-\mathrm{b}}{\mathrm{ab}}\right)\)

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 5.
The electric field and the potential of an electric dipole vary with distance r as
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 3
Answer:
(d) \(\frac{1}{r^{3}} \text { and } \frac{1}{r^{2}}\)

Question 6.
An electric dipole is placed at the centre of a hollow conducting sphere. Where of the following is correct ?
(a) Electric field is zero at every point of the sphere
(b) Electric field is not zero anywhere on the sphere
(c) The flux of electric field is not zero through the sphere
(d) All of these
Answer:
(b) Electric field is not zero anywhere on the sphere

Question 7.
The distance between H+ and Cl ions in HCl molecules is 1.38 Å. The potential due to this dipole at a distance of 10 Å on the axis of dipole is
(a) 2.1 V
(b) 1.8 V
(c) 0.2 V
(d) 1.2 V
Answer:
(c) 0.2 V

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 8.
A cube of side x has a charge q at each of its vertices. The potential due to this charge array at the centre of the cube is
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 4
Answer:
(b) \(\frac{4 q}{\sqrt{3} \pi \varepsilon_{0} x}\)

Question 9.
A hexagon of side 8 cm has a charge 4 µC at each of its vertices. The potential at the centre of the hexagon is
(a) 2.7 × 106 V
(b) 7.2 × 1011 V
(c) 2.5 × 1012 V
(d) 3.4 × 104 V
Answer:
(a) 2.7 × 106 V
Solution:
As shown in the figure, O is the centre of hexagon ABCDEF of each side 8 cm. As it is a regular hexagon OAB, OBC, etc are equilateral triangles,

∴OA = OB = OC = OD = OE = OF = 8 cm = 8 × 10-2 m The potential at O is
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 11

Question 10.
Equipotential surfaces
(a) are closer in regions of large electric fields compared to regions of lower electric fields
(b) will be more crowded near sharp edges of a conductor
(c) will always be equally spaced
(d) both (a) and (b) are correct
Answer:
(d) both (a) and (b) are correct

Question 11.
What is the angle between electric field and equipotential surface ?
(a) 90° always
(b) 0° always
(c) 0°to90°
(d) 0° to 180°
Answer:
(a) 90° always

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 12.
The work done to move a unit charge along an equipotential surface from P to Q
(a) must be defined as \(-\int_{\mathrm{P}}^{Q} \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{d} \ell}\)
(b) is zero
(c) can have a non-zero value
(d) both (a) and (b) are correct
Answer:
(d) both (a) and (b) are correct

Question 13.
The work done in carrying a charge q once round a circle of radius a with a charge Q at its centre is
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 5
Answer:
(d) zero

Question 14.
A test charge is moved from lower potential point to a higher potential point. The potential energy of test charge will
(a) remains the same
(b) increase
(c) decrease
(d) becomes zero
Answer:
(c) decrease

Question 15.
Which among the following is an example of polar molecule ?
(a) O2
(b) H2
(c) N2
(d) HCl
Answer:
(d) HCl

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 16.
Dielectric constant for a metal is
(a) zero
(b) infinite
(c) 1
(d) 10
Answer:
(b) infinite

Question 17.
Metallic sphere of radius R is charged to potential V. Then charge q is proportional to
(a) V
(b) R
(c) both V and R
(d) none of these
Answer:
(c) both V and R

Question 18.
In a parallel plate capacitor, the capacity increases if
(a) area of the plate is decreased
(b) distance between the plates increases
(c) area of the plate is increased
(d) dielectric constant decreases.
Answer:
(c) area of the plate is increased

Question 19.
A parallel plate capacitor having area A and separated by distance d is filled by copper plate of thickness b. The new capacity is
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 6
Answer:
(c) \(\frac{\varepsilon_{0} A}{d-b}\)

Question 20.
Figure shows the Held lines of a positive point charge. The work done by the field in moving a small positive charge from Q to P is
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 7
(a) zero
(b) positive
(c) negative
(d) data insufficient
Answer:
(c) negative

Question 21.
The charge on 3 μF capacitor shown in the figure is
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 8
Answer:
(a) 2 μC
(b) 10 μC
(c) 6 μC
(d) 8 μC
Answer:
(b) 10 μC

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 22.
Minimum number of capacitors each of 8 μF and 250 V used to make a composite capacitor of 16 μF and 1000 V are
(a) 8
(b) 32
(c) 16
(d) 24
Answer:
(b) 32
Solution:
(b) Minimum number of capacitors in each row

= \(\frac{1000}{250}=4\)
Therefore, 4 capacitors connected in series.
If there are m such rows, then total capacity = m x 2 = 16
∴ m = 16/2 = 8
∴ minimum number of capacitors = 4 x 8 = 32

Question 23.
A parallel plate capacitor is made by placing n equally spaced plates connected alternatively. If the capacitance between any two adjacent plates is C then the resultant capacitance is
(a) nC
(b) C/n
(c) (n + 1)C
(d) (n – 1)C
Answer:
(d) (n – 1)C

Question 24.
A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be
(a) 400%
(b) 66.6%
(c) 33.3%
(d) 200%
Answer:
(b) 66.6%

Question 25.
A capacitor of capacitance 700 pF is charged by 100 V battery. The electrostatic energy stored by the capacitor is
(a) 2.5 × 10-8 J
(b) 3.5 × 10-6 J
(c) 2.5 × 10-4J
(d) 3.5 × 10-4J
Answer:
(b) 3.5 × 10-6 J
Solution:
(b) Here, C = 700 pF = 700 × 10-12 F, V = 100 V

Energy stored
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 12

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 26.
A 16 pF capacitor is connected to 70 V supply. The amount of electric energy stored in the capacitor is
(a) 4.5 × 10-12 J
(b) 5.1 × 10-8 J
(c) 2.5 × 10-12 J
(d) 3.2 × 10-8 J
Answer:
(b) 5.1 × 10-8 J

Question 27.
A metallic sphere of radius 18 cm has been given a charge of 5 × 10-6 C. The energy of the charged conductor is
(a) 0.2 J
(b) 0.6 J
(c) 1.2 J
(d) 2.4 J
Answer:
(b) 0.6 J
Solution:
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 13

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 28.
Energy stored in a capacitor and dissipated during charging a capacitor bear a ratio
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 3
Answer:
(c) 2 : 1
Solution:
(c) Half of the energy is dissipated during charging a capacitor.
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 14

Question 29.
Van deGraaff generator is used to
(a) store electrical energy
(b) build up high voltages of few million volts
(c) decelerate charged particle like electrons
(d) both (a) and (b) are correct
Answer:
(b) build up high voltages of few million volts

Question 30.
Who established the fact of animal electricity ?
(a) Van deGraaff
(b) Count Alessandro Volta
(c) Gustav Robert Kirchhoff
(d) Hans Christian Oersted
Answer:
(b) Count Alessandro Volta

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 31.
In case of a Van de Graaff generator, the breakdown field of air is
(a) 2 × 108 V m-1
(b) 3 × 106 V m-1
(c) 2 × 1CT8 V m-1
(d) 3 × 104 V m-1
Answer:
(b) 3 × 106 V m-1

Question 32.
A capacitor of 4 μF is connected as shown in the circuit. The internal resistance of the battery is 0.5 Ω . The amount of charge on the capacitor plates will be
Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance - 9
(a) 0
(b) 4 μC
(c) 16 μC
(d) 8 μC
Answer:
(d) 8 μC

Bihar Board 12th Physics Objective Answers Chapter 2 Electrostatic Potential and Capacitance

Question 33.
Equipotentials at a great distance from a collection of charges whose total sum is not zero are approximately
(a) spheres
(b) planes
(c) paraboloids
(d) ellipsoids
Answer:
(a) spheres

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Bihar Board 12th Biology Objective Questions and Answers

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 1.
RCH stands for
(a) routine check-up of health
(b) reproduction cum hygiene
(c) reversible contraceptive hazards
(d) reproductive and child health care.
Answer:
(d) reproductive and child health care.

Question 2.
Amminocentesis is a technique used to
(a) determine errors in amino acid metabolism in embryo
(b) pin point specific cardiac ailments in embryo
(c) determine any hereditary genetic abnormality in embryo
(d) all of these.
Answer:
(a) determine errors in amino acid metabolism in embryo

Question 3.
Which of the following cannot be detected in a developing foetus by amniocentesis ?
(a) Jaundice
(b) Down’s syndrome
(c) Cystic fibrosis
(d) Colourblindness
Answer:
(a) Jaundice

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 4.
In which of the following weeks of pregnancy CVS is done ?
(a) 12th – 14th week
(b) 8th – 10th week
(c) 5th – 7th week
(d) None of these
Answer:
(b) 8th – 10th week

Question 5.
Causes for increased population growth in india is/are
(a) increase in birth rate
(b) decrease in death rate
(c) lack of education
(d) all of these.
Answer:
(d) all of these.

Question 6.
Which of the following pairs contributes to an increase in population ?
(a) Natality and immigration
(b) Mortality and emigration
(c) Natality and emigration
(d) Mortality and immigration
Answer:
(a) Natality and immigration

Question 7.
Which of the following is correct regarding the consequences of lower population ?
(a) It increases the poverty of a country.
(b) It leads to shortage of food supply.
(c) It results in unemployment
(d) All of these
Answer:
(d) All of these

Question 8.
The best way to decrease population of a country is
(a) to educate people
(b) to have better houses
(c) to kill people on a large scale
(d) to practice and implement familay planning.
Answer:
(d) to practice and implement familay planning.

Question 9.
Which of the following correctly describes the measures
that can be used to control over-population ?
(a) Educating people about the advantages of a small family
(b) Raising the age of marriage
(c) Encouraging family planning programme
(d) All of these
Answer:
(d) All of these

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 10.
Which of the following is not a characteristic of an ideal contraceptive ?
(a) User-friendly
(b) Irreversible
(c) Easily available
(d) Least side-effects
Answer:
(b) Irreversible

Question 11.
On which of the following facts does the method of periodic abstinence is based ?
(a) Ovulation occurs on about the 14th day of menstruation.
(b) Ovum remains alive for about 1-2 days.
(c) Sperms survive for about 3 days.
(d) All of these
Answer:
(d) All of these

Question 12.
Which of the following is a non-medicated intrauterine device (IUD) ?
(a) CuT
(b) Lippes Loop
(c) Cu7
(d) LNG-20
Answer:
(b) Lippes Loop

Question 13.
Multiload device contains
(a) manganese
(b) iron
(c) copper
(d) calcium
Answer:
(c) copper

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 14.
Which of the following is not an intrauterine device ?
(a) Progestasert
(b) Multiload-375
(c) Norplant
(d) Lippes loop
Answer:
(c) Norplant

Question 15.
What is the work of copper-T ?
(a) To inhibit ovulation
(b) To inhibit fertilisation
(c) To inhibit implantation of blastocyst
(d) To inhibit gametogenesis
Answer:
(b) To inhibit fertilisation

Question 16.
IUDs prevent pregnancy by
(a) inhibiting physiological and morphological uterine changes required for implantation.
(b) increasing phagocytosis of spermatozoa within uterus
(c) suppressing motility of sperms as well as their fertilising capacity
(d) all of these.
Answer:
(d) all of these.

Question 17.
Cu ions released from copper releasing intrauterine devices (IUDs)
(a) prevent ovulation
(b) make uterus unsuitable for implantation
(c) decrease phagocytosis of sperms
(d) suppress sperm motility.
Answer:
(d) suppress sperm motility.

Question 18.
The birth control device used by women is
(a) diaphragm
(b) vault
(c) copper T
(d) all of these.
Answer:
(d) all of these.

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 19.
The most important component of oral contraceptive pills is
(a) progesterone-estrogen
(b) growth hormore
(c) thyroxine
(d) luteinising hormone.
Answer:
(a) progesterone-estrogen

Question 20.
Progesterone pill helps in preventing pregnancy by not allowing
(a) ova formation
(b) fertilisation
(c) implantation
(d) none of these.
Answer:
(a) ova formation

Question 21.
What is true about “Saheli” ?
(i) Developed at the CDRI, Lucknow
(ii) Contains a steroidal preparation
(iii) “Once-a-weak” pill
(iv) Many side effects
(v) High contraceptive value
(vi) Very few side effects value
(vii) Low contraceptive value
(a) (i), (ii), (iii), (v), (vi)
(b) (i), (iii), (v), (vi), (vii)
(c) (i), (ii), (iii), (iv), (v)
(d) (i), (iii), (iv), (v)
Answer:
(d) (i), (iii), (iv), (v)

Question 22.
What is the figure given below showing in particular ?
Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health
(a) Ovarian cancer
(b) Uterine cancer
(c) Tubectomy
(d) Vasectomy
Answer:
(c) Tubectomy

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 23.
Which of the following birth control measures can be considered as the safest ?
(a) The rhythm method
(b) The use of physical barriers
(c) Termination of unwanted pregnancy
(d) Sterilisation techniques
Answer:
(d) Sterilisation techniques

Question 24.
Which of the following is the most widely accepted method of contraception in India at present ?
(a) Cervical caps
(b) Tubectomy
(c) Diaphragms
(d) IUDs (Intra uterine devices)
Answer:
(d) IUDs (Intra uterine devices)

Question 25.
Which of the following is a full proof method contraception ?
(a) Implantation
(b) Lactational amenorrhea
(c) Condoms
(d) Sterilisation
Answer:
(d) Sterilisation

Question 26.
Which of the following statements is correct regarding vasectomy ?
(a) It prevents the production of sperm in the testes.
(b) It prevents the production of semen.
(c) It prevents the movement of sperm into the urethra.
(d) It prevents a man from having an erection.
Answer:
(c) It prevents the movement of sperm into the urethra.

Question 27.
Which of the following statements is incorrect regarding the medical termination of pregnancy (MTP) ?
(a) These help in getting rid of unwanted pregnancies.
(b) These help in aborting the pregnancies which may be harmful to either mother of foetus or both.
(c) These contribute in decreasing the human population.
(d) None of these.
Answer:
(d) None of these.

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 28.
Which of the following represents the correct match of a sexually transmitted disease with its pathogen ?
(a) Syphilis-Treponema pallidum
(b) Gonorrhoea-Entamoeba histolytica
(c) Urethritis-Bacillus anthracis
(d) Softsore-Bacillus brevis
Answer:
(a) Syphilis-Treponema pallidum

Question 29.
The sexually transmitted disease, that can affect both the male and the female genitals and may damage the eyes of babies born of infected mothers is
(a) AIDS
(b) syphilis
(c) gonorrhoea
(d) hepatitis
Answer:
(c) gonorrhoea

Question 30.
A sexually transmitted disease symptomised by the development of chancre on the genitals is caused by the infection of
(a) Treponema pallidum
(b) Neisseria gonorrhoeae
(c) human immunodeficiency virus
(d) hepatitis B virus
Answer:
(a) Treponema pallidum

Question 31.
Which of the following is correct regarding HIV, hepatitis B, gonorrhoea, trichomoniasis ?
(a) Trichomoniasis is an STD whereas others are not.
(b) Gonorrhoea is a viral disease whereas others are bacterial.
(c) HIV is a pathogen whereas others are diseases.
(d) Hepatitis B is eradicated completely whereas others are not.
Answer:
(c) HIV is a pathogen whereas others are diseases.

Question 32.
Which one of the following groups includes sexually transmitted diseases caused by bacteria only ?
(a) Syphilis, gonorrhoea, chancroid
(b) Syphilis, chlamydiasis, chancroid
(c) Syphilis, gonorrhoea, scabies
(d) Syphilis, scabies, pediculosis
Answer:
(a) Syphilis, gonorrhoea, chancroid

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 33.
Confirmatory test for STDs is
(a) ELISA
(b) PCR
(c) DNA hybridisation
(d) all of these.
Answer:
(d) all of these.

Question 34.
Hepatitis B is transmitted through
(a) blood transfusion
(b) intimate physical contact
(c) sexual contact
(d) all of these
Answer:
(d) all of these

Question 35.
The common means of transmission of AIDS is
(a) sexual intercourse
(b) blood transfusion
(c) placental transfer
(d) all of these.
Answer:
(d) all of these.

Question 36.
Which of the following STDs are not completely curable ?
(a) Chlamydiasis, gonorrhoea, trichomoniasis
(b) Chancroid syphilis, genital warts
(c) AIDS, syphilis, hepatitis B
(d) AIDS, genital herpes, hepatitis B
Answer:
(d) AIDS, genital herpes, hepatitis B

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 37.
World AIDS day is
(a) December 21
(b) December 1
(c) November 1
(d) June 11
Answer:
(b) December 1

Question 38.
In vitro fertilisation is a technique that involves transfer of which one of the following into the fallopian tube ?
(a) Embryo only, upto 8 cell stage
(b) Either zygote or early embryo upto 8 cell stage
(c) Embro of 32 cell stage
(d) Zygote only
Answer:
(b) Either zygote or early embryo upto 8 cell stage

Question 39.
The test-tube baby programme employs which one of the following techniques ?
(a) Zygote intra Fallopian transfer (ZIFT)
(b) Intra uterine insemination (IUI)
(c) Gamete intra Fallopian transfer
(d) All of these
Answer:
(a) Zygote intra Fallopian transfer (ZIFT)

Question 40.
The first case of IVF-ET technique success, was reported by
(a) Louis joy Brown and Banting Best
(b) Patrick Steptoe and Rebert Edwards
(c) Robert Steptoe and Gilbert Brown
(d) Baylis and Starling Taylor.
Answer:
(b) Patrick Steptoe and Rebert Edwards

Question 41.
Which of the following is ART ?
(a) IUDs
(b) GIFT
(c) ZIFT
(d) Both (b) & (c)
Answer:
(d) Both (b) & (c)

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 42.
The technique called Gamete Intra Fallopian Transfer (GIFT) is recommended for those females
(a) who cannot produce an ovum
(b) who cannot retain the foetus inside uterus
(c) who cannot provide suitable environment for fertilisation
(d) all of these
Answer:
(a) who cannot produce an ovum

Question 43.
Which method can be used for women that cannot produce ovum but can provide suitable environment ?
(a) IUD
(b) GIFT
(c) IUI
(d) ICSI
Answer:
(b) GIFT

Question 44.
Which of these can be used to cure infertility in couples where male partner has very low sperm count ?
(a) IUD
(b) GIFT
(c) IUI
(d) None of these
Answer:
(c) IUI

Question 45.
The method of directly injecting a sperm into ovum in assisted reproductive technology is called
(a) GIFT
(b) ZIFT
(c) ICSI
(d) ET
Answer:
(c) ICSI

Question 46.
Increased IMR and decreased MMR in a population will
(a) cause rapid increase in growth rate
(b) result in decline in growth rate
(c) not cause significant change in growth rate
(d) result in an explosive population.
Answer:
(b) result in decline in growth rate

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Question 47.
Intensely lactating mothers do not generally conceive due to the
(a) suppression of gonadotropins
(b) hypersecretion of gonadotropins .
(c) suppression of gametic transport
(d) suppression of fertilisation.
Answer:
(a) suppression of gonadotropins

Question 48.
A national level approach to build up a reproductively healthy society was taken up in our country in
(a) 1950s
(b) 1960s
(c) 1980s
(d) 1990s
Answer:
(a) 1950s

Question 49.
Emergency contraceptives are effective if used within,
(a) 72 hrs of coitus
(b) 72 hrs of ovulation
(c) 72 hrs of menstruation
(d) 72 hrs of implantation.
Answer:
(a) 72 hrs of coitus

Question 50.
From the sexually transmitted diseases mentioned below, identify the one which does not specifically affect the sex organs.
(a) Syphilis
(b) AIDS
(c) Gonorrhea
(d) Genital warts
Answer:
(b) AIDS

Question 51.
Condoms are one of the most popular contraceptives because of the following reasons.
(a) These are effective barriers for insemination.
(b) They do not interfere with coital act.
(c) These help in reducing the risk of STDs.
(d) All of the above
Answer:
(d) All of the above

Question 52.
The correct surgical procedure as a contraceptive method is
(a) ovariectomy
(b) hysterectomy
(c) vasectomy
(d) castration.
Answer:
(c) vasectomy

Bihar Board 12th Biology Objective Answers Chapter 4 Reproductive Health

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Bihar Board 12th Biology Objective Questions and Answers

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 1.
Which of the following facrtors affect human health ?
(i) Infections
(ii) Silent mutation
(iii) Life style
(iv) Genetic disorders
(a) (i), (ii)and(iv)
(b) (i) and (ii)
(c) (i), (iii) and (iv)
(d) (i), (ii), (iii) and (iv)
Answer:
(c) (i), (iii) and (iv)

Question 2.
Which one of the following disease is non – communicable ?
(a) Diphtheria
(b) Flu
(c) Cancer
(d) Malaria
Answer:
(c) Cancer

Question 3.
Which of the following pairs contains an infectious and a non-infectious disease respectively ?
(a) Typhoid and AIDS
(b) AIDS and cancer
(c) Pneumonia and malaria
(d) Cancer and malaria
Answer:
(b) AIDS and cancer

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 4.
Typhoid fever in human beings is caused by
(a) Plasmodium vivax
(b) Trichophyton
(c) Salmonella typhi
(d) Rhinoviruses.
Answer:
(c) Salmonella typhi

Question 5.
Which of the following is the bacterial disease in humans ?
(a) Dysentery
(b) Malaria
(c) Plague
(d) Both (a) & (c)
Answer:
(d) Both (a) & (c)

Question 6.
Which of the following pathogens causes whooping coough ?
(a) Legionella spp.
(b) Bordetella pertussis
(c) Vibrio cholerae
(d) Burcella melitensis
Answer:
(b) Bordetella pertussis

Question 7.
Which one of the following sets includes bacterial diseases ?
(a) Tetanus, tuberculosis, measles
(b) Diptheria, leprosy, plague
(c) Cholera, typhoid, mumps
(d) Malaria, mumps, poliomyelits
Answer:
(b) Diptheria, leprosy, plague

Question 8.
The common cold is caused by
(a) Rhino viruses
(b) Streptococcus pnemoniae
(c) Salmonella typhimurium
(d) Plasmodium vivax.
Answer:
(a) Rhino viruses

Question 9.
Hepatitis B is transmitted through
(a) sneezing
(b) female Anopheles
(c) coughing
(d) blood transfusion.
Answer:
(b) female Anopheles

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 10.
A toxic substance, responsible for the chills and high fever recurring every three to four days in malarial fever, is
(a) interferon
(b) haemozoin
(c) hirudin
(d) colostrum
Answer:
(b) haemozoin

Question 11.
During the life cycle of Plasmodium, sexual reproduction takes place in which of the following hosts ?
(a) Human
(b) Female Anopheles mosquito
(c) Male Anopheles mosquito
(d) Both (a) and (b)
Answer:
(b) Female Anopheles mosquito

Question 12.
Amoebic dysentery (amoebiasis) is caused by
(a) Entamoeba histolytica
(b) E. coli
(c) Streptococcus pneumoniae
(d) Trichophyton.
Answer:
(a) Entamoeba histolytica

Question 13.
Which one of the following diseases cannot be cured by taking antibiotics ?
(a) Plague
(b) Amoebiasis
(c) Leprosy
(d) Whooping cough
Answer:
(b) Amoebiasis

Question 14.
An intestinal parasite which causes blockage of the intestinal passage and whose eggs are excreted along with the faeces of infected person is ______.
(a) Wuchereria bancrofti
(b) Ascaris
(c) Epidermophyton
(d) Microsporum
Answer:
(b) Ascaris

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 15.
Elephantiasis, a chronic inflammation that results in gross deformities is caused by
(a) Ascaris
(b) E.coli
(c) Wuchereria
(d) Trichophyton
Answer:
(c) Wuchereria

Question 16.
Which of the following is affected by the infection of Wuchereria bancrofti ?
(a) Lymphatic vessels
(b) Respiratory system
(c) Nervous system
(d) Blood circulation
Answer:
(a) Lymphatic vessels

Question 17.
Which of the following diseases is transmitted by the bite of the female mosquito vector ?
(a) Filariasis
(b) Amoebiasis
(c) Typhoid
(d) Pneumonia
Answer:
(a) Filariasis

Question 18.
Which of the following pairs correctly matches a disease and a pathogen causing it ?
(a) Typhoid – Salmonella typhi
(b) Pneumonia – Haemophilus pneumoniae
(c) Malaria – Ascaris lumbricoides
(d) Ringworm – Entamoeba histolytica
Answer:
(a) Typhoid – Salmonella typhi

Question 19.
The pathogen Microsporum responsible for ringworm disease in humans belongs to the same kingdom as that of
(a) Taenia, a tapeworm
(b) Ascaris, a roundworm
(c) Rhizopus, a mould
(d) Wuchereria, a filarial worm,
Answer:
(c) Rhizopus, a mould

Question 20.
Appearance of dry, scaly lesions with itching on various parts of the body are the symptoms of _______.
(a) elephantiasis
(b) ringworm
(c) ascariasis
(d) amoebiasis
Answer:
(b) ringworm

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 21.
Gambusia is a fish which is being introduced into the ponds in order to check the vector-borne diseases such as
(a) dengue
(b) malaria
(c) chikungunya
(d) all of these.
Answer:
(d) all of these.

Question 22.
Which one of the following pairs of diseases is viral as well as transmitted by mosquitoes ?
(a) Encephalitis and sleeping sickness
(b) Yellow fever and sleeping sickness
(c) Elephantiasis and dengue
(d) Yellow fever and dengue
Answer:
(d) Yellow fever and dengue

Question 23.
Which one of the following pairs is not correctly matched ?
(a) Dengue fever – Fiavi-ribo virus
(b) Syphilis – Trichu ris inch jura
(c) Plague – Yersinia pesris
(d) Filariasis – Wuchereria banchrofri
Answer:
(b) Syphilis – Trichu ris inch jura

Question 24.
The term ‘immunity’ refers to
(a) mutalism between host and parasite
(b) ability of the host to fight the disease-causing organism
(c) ability of the parasite to survive within a host
(d) a fatal disease.
Answer:
(b) ability of the host to fight the disease causing organism

Question 25.
A person has developed interferons in his body. He seems to carry an infection of
(a) tetanus
(b) malaria
(c) measles
(d) typhoid
Answer:
(c) measles

Question 26.
The first line of defence in the immune system is provided by
(a) skin and mucous membrane
(b) inflammatory response
(c) the complement system
(d) none of these.
Answer:
(a) skin and mucous membrane

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 27.
Primary response produced due to first time encounter with a pathogen is of
(a) high intensity
(b) low intenisty
(c) intermediate intensity
(d) no intensity.
Answer:
(b) low intenisty

Question 28.
Which of the following components does not participate in innate immunity ?
(a) Neutrophils
(b) Macrophages
(c) B-lymphocytes
(d) Natural killer cells
Answer:
(c) B-lymphocytes

Question 29.
Antibodies are secreted by
(a) T-lymphocytes
(b) B-lymphocytes
(c) both (a) and (b)
(d) natural killer cell.
Answer:
(b) B-lymphocytes

Question 30.
The antigen binding site of an antibody is present at
(a) the constant region
(b) the C-terminal
(c) the variable region
(d) between constant and variable region.
Answer:
(c) the variable region

Question 31.
Humoral immunity is associated with
(a) T-cells
(b) B-cells
(c) macrophages
(d) both (a) and (b)
Answer:
(b) B-cells

Question 32.
The antibody which can cross placental barrier is.
(a) IgA
(b) JgE
(c) IgM
(d) IgG.
Answer:
(d) IgG.

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 33.
The most abundant class of immunoglobulins (Igs) in the body is
(a) IgA
(b) IgG
(c) IgE
(d) IgM
Answer:
(b) IgG

Question 34.
A protein or polysaccharide molecule that stimulates antibody formation.
(a) antigen
(b) antibiotics
(c) exotoxin
(d) endotoxins.
Answer:
(a) antigen

Question 35.
Passive immunity can be conferred directly by
(a) vaccines
(b) antitoxins
(c) colostrum
(d) both (b) & (c)
Answer:
(d) both (b) & (c)

Question 36.
Which form of pathogen is used in vaccination ?
(a) Activated and strong pathogenic antigens
(b) Inactivated and weakened pathogenic antigens
(c) Hyperactive and strong pathogen
(d) Preformed antibodies
Answer:
(b) Inactivated and weakened pathogenic antigens

Question 37.
Injection of antitoxin in tetanus confers which type of immunisation ?
(a) Active immunisation
(b) Passive immunisation
(c) Auto-immunisation
(d) Humoral immunisation
Answer:
(b) Passive immunisation

Question 38.
The term ‘antitoxin’ refers to a preparation containing
(a) B-lymphoctyes and T-lymphocytes
(b) antibodies to the toxin
(c) weakend pathogen
(d) inactivated T-lymphocytes.
Answer:
(b) antibodies to the toxin

Question 39.
The injection given against the snake venom contains
(a) antigenic proteins
(b) preformed antibodies
(c) attenuated pathogen
(d) all of these.
Answer:
(b) preformed antibodies

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 40.
Vaccine against polio viruses is an example of
(a) auto-immunisation
(b) passive immunisation
(c) active immunisation
(d) simple immunisation
Answer:
(c) active immunisation

Question 41.
Hepatitis B vaccine is produced from
(a) inactivated viruses
(b) yeast
(c) Haemophilus influenzae
(d) Salmonella typhimurium.
Answer:
(b) yeast

Question 42.
Use of vaccines and immunisation programmes have controlled which of the following infectious diseases ?
(a) Polio and tetanus
(b) Diphtheria and pheumonia
(c) Cancer and AIDS
(d) Both (a) and (b)
Answer:
(d) Both (a) and (b)

Question 43.
The most abundant antibody produced against allergens is
(a) IgE
(b) IgA
(c) IgG
(d) IgM.
Answer:
(a) IgE

Question 44.
Which of the following cells actively participate during allergy ?
(a) B-lymphocytes
(b) Liver cells
(c) Mast cells
(d) Red blood cells
Answer:
(c) Mast cells

Question 45.
The drugs used to quickly reduce the symptoms of allergy are
(a) ati-histamine and adrenaline
(b) histamine and thyroxine
(c) adrenaline and a-interferon
(d) all of these
Answer:
(a) ati-histamine and adrenaline

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 46.
An auto-immune disease is
(a) SCID
(b) rheumatoid arthritis
(c) myasthenia gravis
(d) both (b) and (c).
Answer:
(d) both (b) and (c).

Question 47.
Which out of the following groups represent auto-immune disorders ?
(a) SCID and diphtheria
(b) Diabetes mellitus (type 1) and rheumatic fever
(c) AIDS and cholera
(d) Hepatitis and leukaemia
Answer:
(b) Diabetes mellitus (type 1) and rheumatic fever

Question 48.
The primary lymphoid organs are
(a) spleen and thymus
(b) bone marrow and thymus
(c) bone marrow and lymph node
(d) thymus and MALT.
Answer:
(d) thymus and MALT.

Question 49.
The site where lymphocytes interact with antigens and proliferate to become effector cells are
(a) spleen and lymph nodes
(b) bone marrow and thymus
(c) Peyer’s patches and tonsils
(d) both (a) and (c).
Answer:
(d) both (a) and (c).

Question 50.
Which of these glands is large at the time of birth but in adults, it reduces to a very small size ?
(a) Thyroid
(b) Adrenal
(c) Thymus
(d) Spleen
Answer:
(c) Thymus

Question 51.
MALT is
(a) Muscle Associated Lymphoid Tissues
(b) Mucosal Associated Lymphoid Tissues
(c) Mucosal and Lymphoid Tissue
(d) Memory Association Lymphoid Tissues.
Answer:
(b) Mucosal Associated Lymphoid Tissues

Question 52.
The lymphoid tissue, located within the lining of digestive tract is
(a) spleen
(b) Peyer’s pathes
(c) lymph nodes
(d) MALT
Answer:
(d) MALT

Question 53.
The abbreviation AIDS stands for
(a) Acquired immuno disease syndrome
(b) Acquired immuno deficiency syndrome
(c) Acquired immunity determining syndrome
(d) Acquired immunity delay syndrome.
Answer:
(b) Acquired immunodeficiency syndrome

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 54.
The genetic material of HIV is
(a) dsDNA
(b) dsRNA
(c) ssDNA
(d) ssRNA
Answer:
(d) ssRNA

Question 55.
The human immunodeficiency virus is
(a) an unenveloped, RNA genome containing retrovirus
(b) an enveloped, RNA genome containing retrovirus
(c) an enveloped, DNA genome containing retrovirus
(d) an enveloped, RNA genome containing reovirus.
Answer:
(b) an enveloped, RNA genome containing retrovirus

Question 56.
Which of the following is not a cause of transmission of HIV?
(a) Multiple sexual partners
(b) Sharing infected needles
(c) Mosquito bite
(d) Transfusion of contaminated blood
Answer:
(c) Mosquito bite

Question 57.
Viral DNA after being converted from viral RNA by X, incorporates into host genome to undergo replication. What is‘X’ ?
(a) DNA polymerase
(b) Restriction endonuclease
(c) RNA polymerase
(d) Revers transcriptase
Answer:
(d) Revers transcriptase

Question 58.
The cells called ‘HIV factory’ is
(a) helper T-cells
(b) macrophages
(c) dendritic cells
(d) WBCs.
Answer:
(b) macrophages

Question 59.
HIV is a retrovirus that attacks
(a) helper T-cells
(b) cytotoxin t-cells
(c) B-cells
(d) neutrophils.
Answer:
(a) helper T-cells

Question 60.
AIDS is characterised by
(a) decrease in the number of killer T-cells
(b) decrease in the number of suppressor T-cells
(c) decrease in the number of helper T-cells
(d) increase in the number of helper T-cells
Answer:
(c) decrease in the number of helper T-cells

Question 61.
AIDS is widely diagnosed by
(a) Widal test
(b) ELISA
(c) PC
(d) Chromatography.
Answer:
(b) ELISA

Question 62.
Which of the following day is celebrated as ‘World AIDS Day’ ?
(a) 31st March
(b) 1st March
(c) Ist December
(d) 31st December
Answer:
(c) Ist December

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 63.
Cancer cells do not exhibit the property of
(a) generating tumors
(b) metastasis
(c) contact inhibition
(d) less number of mitochondrial^cristae.
Answer:
(c) contact inhibition

Question 64.
A person suffering from leukaemia has
(a) tumors in adipose tissue
(b) increased number of plasma cells
(c) increased number of melanocytes
(d) increased number of WBCs.
Answer:
(d) increased number of WBCs.

Question 65.
A metastatic cancerous tumour is termed ‘sarcoma’ if the disorder is in
(a) fibroblasts
(b) circulatory system
(c) immune system
(d) epithelial cells.
Answer:
(a) fibroblasts

Question 66.
A chemical carcinogen present in tobacco smoke is responsible for
(a) skin cancer
(b) pancreatic cancer
(c) stomach cancer
(d) lung cancer.
Answer:
(d) lung cancer.

Question 67.
Major factors that cause cancer are
(a) oncogenes and polymorphonuclear leucocytes
(b) oncogenes and tumour suppressor genes
(c) MHC genes
(d) cellular oncogenes and a-interferons.
Answer:
(b) oncogenes and tumour suppressor genes

Question 68.
Which of the following approaches are used for the treatment of cancer ?
(a) Immunotherapy
(b) Surgery
(c) Radiotherapy & chemotherapy
(d) All of these
Answer:
(d) All of these

Question 69.
The substance given to cancer patients in order to activate their immune system and destroy the tumour is
(a) histamine
(b) interleukin
(c) a-interferon
(d) morphine.
Answer:
(c) a-interferon

Question 70.
In humans, receptors for opioids are present in
(a) central nervous system
(b) gastrointestinal tract
(c) respiratory tract
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 71.
Heroin is commonly called as
(a) coke
(b) crack
(c) smack
(d) charas.
Answer:
(c) smack

Question 72.
Which compound is formed by acetylation of morphine ?
(a) Heroin
(b) Cocaine
(C) Tobacco.
(d) Marijuana
Answer:
(a) Heroin

Question 73.
Marijuana is extracted from .
(a) dried leaves and flowers of hemp plant .
(b) ergot fungus
(c) roots of hemp plant
(d) cocoa plant.
Answer:
(a) dried leaves and flowers of hemp plant .

Question 74.
Charas and ganja are the drugs which affect
(a) respiratory system
(b) cardiovascular system
(c) digestive system
(d) nervous system.
Answer:
(b) cardiovascular system

Question 75.
Cocaine is obtained from
(a) Erythroxylon coca
(b) Papaver somniferum
(c) Atropa belladona
(d) Dature stramonium
Answer:
(a) Erythroxylon coca

Question 76.
Cocaine is commonly called as
(a) smack
(b) coke
(c) crack
(d) both(b)&(c)
Answer:
(d) both(b)&(c)

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 77.
_______ is a CNS stimulant as it interferes with the trasport of the neuro-transmitter __________ .
(a) Cocaine, acetylcoline
(b) Barbituarate, glutamate
(c) Cocaine, dopamine
(d) Barbiturate, glycine
Answer:
(c) Cocaine, dopamine

Question 78.
Whihc of the following plants possess hallucinogenic properties ?
(a) Erythroxylon coca
(b) Atropa belladona
(c) Datura stramonium
(d) All of these
Answer:
(d) All of these

Question 79.
Which drug is being excessively taken by some sports persons nowadays ?
(a) Opioids
(b) Barbiturates
(c) Cannabinoids
(d) Lysergic acid diethly amides (LSD)
Answer:
(c) Cannabinoids

Question 80.
Which drug is used as medicine to help patients cope with depression and insomnia ?
(a) Morphine
(b) Amphetamines
(c) Barbiturate
(d) Both (b) and (c)
Answer:
(d) Both (b) and (c)

Question 81.
Which one of the following is an opiate narcotic ?
(a) Barbiturates
(b) Morphine
(c) Amphetamines
(d) LSD
Answer:
(b) Morphine

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 82.
Which one of the following is a mismatched pair of the drug and its effect ?
(a) Amphetamines – CNS stimulants
(b) Lysergic acid Diethlyamide (LSD) – Psychedelic (hallucinogen)
(c) Heroin – Depressant, slows down body functions
(d) Barbiturates – Transquilliser
Answer:
(d) Barbiturates – Transquilliser

Question 83.
The addictive chemical present in tobacco is
(a) caffeine
(b) nicotine
(c) catechol
(d) carbon monoxide.
Answer:
(b) nicotine

Question 84.
Level of which hormones get elevated by the intake of nicotine ?
(a) FSH,LH
(b) Thyroxine, progesterone
(c) Oxytocine, prolactin
(d) Adrenaline, nor-adrenaline
Answer:
(d) Adrenaline, nor-adrenaline

Question 85.
The intravenous drug abusers are more likely to develop
(a) cancer
(b) AIDS
(c) malaria
(d) typhoid.
Answer:
(b) AIDS

Question 86.
The organisms which cause diseases is plants and animals are called
(a) pathogens
(b) vectors
(c) insects
(d) worms.
Answer:
(a) pathogens

Question 87.
The chemical test that is used for diagnosis of tyhpoidis
(a) ELISA-Test
(b) EST-Test
(c) PCR-Test
(d) Widal-Test.
Answer:
(d) Widal-Test.

Question 88.
The sporozoites that cause infection, when a female Anopheles mosquito bites a person are formed in
(a) liver of the person
(b) RBCs of mosquito
(c) salivary glands of mosqiot
(d) intenstine of mosquito.
Answer:
(d) intenstine of mosquito.

Question 89.
The disease chikungunya is transmitted by
(a) house flies
(b) Aedes mosquitoes
(c) cockroach
(d) female Anopheles.
Answer:
(b) Aedes mosquitoes

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 90.
The genes causing cancer are
(a) structural genes
(b) expressor genes
(c) oncogenes
(d) regulatory genes.
Answer:
(c) oncogenes

Question 91.
AIDS is caused by HIV. Among the following, which one is not a mode of transmission of HIV ?
(a) Transfusion of contaminated blood.
(b) Sharing the infected needles.
(c) Shaking hands with infected persons.
(d) Sexual contact with infected persons.
Answer:
(c) Shaking hands with infected persons.

Question 92.
‘Smack’ is a drug obtained from the
(a) latex of Papever somniferum
(b) leaves of Cannabis saliva
(c) flowers of Datura
(d) fruits of Erythroxyl coca.
Answer:
(a) latex of Papever somniferum

Question 93.
The substance produced by a cell in viral infection that can protect other cells from further infection is
(a) serotonin
(b) colostrum
(c) interferon
(d) histamine.
Answer:
(c) interferon

Question 94.
Antibodies present in colostrum which protect the new born from certain diseases is of
(a) IgG type
(b) IgA type
(c) IgD type
(d) IgE type.
Answer:
(b) IgA type

Question 95.
Tobacco consumption is known to stimulate secretion of adrenaline and nor-adrenaline. The component causing this could be
(a) nicotine
(b) tannic acid
(c) curaimin
(d) catechin.
Answer:
(a) nicotine

Question 96.
Antivenom against snake poison contains
(a) antigens
(b) antigen-antibody complexes
(c) antibodies
(d) enzymes
Answer:
(c) antibodies

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Question 97.
Which of the following is not a lymphoid tissue ?
(a) Spleen
(b) Tonsils
(c) Pancreas
(d) Thymus
Answer:
(c) Pancreas

Question 98.
Which of the following glands is large sized at birth but reduces in size with ageing ?
(a) Pineal
(b) Pituitary
(c) Thymus
(d) Thyroid
Answer:
(c) Thymus

Question 99.
Haemozoin is a
(a) precursor of haemoglobin
(b) toxin released from Streptococcus infected cells
(c) toxin released from Plasmodium infected cells
(d) toxin released from Haemophilus infected cells.
Answer:
(c) toxin released from Plasmodium infected cells

Question 100.
One of the following is not the causal organism for ringworm.
(a) Microsporum
(b) Trichophyton
(c) Epidermophyton
(d) Macrosporum
Answer:
(d) Macrosporum

Question 101.
A person with sickle cell anaemia is
(a) more prone to malaria
(b) more prone to typhoid
(c) less prone to malaria
(d) less prone to typhoid.
Answer:
(c) less prone to malaria

Bihar Board 12th Biology Objective Answers Chapter 8 Human Health and Disease

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Bihar Board 12th Biology Objective Questions and Answers

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 1.
One of the possible early sources of energy was/were
(a) CO2
(b) chlorophyll
(c) green plants
(d) UV rays and lightning.
Answer:
(d) UV rays and lightning.

Question 2.
Abiogenesis theory of origin supports
(a) spontaneous generation
(b) origin of life from blue-green algae
(c) origin of life is due to pre-existing organisms
(d) organic evolution is due to chemical reactions.
Answer:
(a) spontaneous generation

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 3.
Who proposed that the first form of the could have come from pre-existing non-living organic molecules ?
(a) S.L. Miller
(b) Oparin and Haldane
(c) Charles Darwin
(d) Alfred Wallace
Answer:
(b) Oparin and Haldane

Question 4.
According to one of the most widely accepted theories, earth’s atmosphere before origin of life was
(a) oxidising
(b) oxidising along with H2
(c) reducing with free O2 in small amount
(d) reducing with oxygen absent in O2 form.
Answer:
(b) oxidising along with H2

Question 5.
According to Oparin, which one of the following was not present in the primitive atmosphere of the earth ?
(a) Methane
(b) Oxygen
(c) Hydrogen
(d) Water vapour
Answer:
(b) Oxygen

Question 6.
The correct sequence for the manufacture of the compounds on the primitive earth is
(a) NH3, CH4, protein and carbohydrate
(b) protein, carbohydrate, water and nucleic acid
(c) NH3, CH4, carbohydrate and nucleic acid
(d) NH3, carbohydrate, protien and nucleic acid.
Answer:
(d) NH3, carbohydrate, protien and nucleic acid.

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 7.
Early at mosphere contained methane and other hydrocarbons. They have been now replaced by
(a) nitrogen
(b) oxygen
(c) carbon dioxide
(d) hydrogen.
Answer:
(c) carbon dioxide

Question 8.
The first life originated
(a) on land
(b) in air
(c) in water
(d) all of these.
Answer:
(c) in water

Question 9.
Coacervates are
(a) colloid droplets
(b) nucleoprotein containing entities
(c) microspheres
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 10.
First life from on earth was a
(a) cyanobacterium
(b) chemoheterotroph
(c) autotroph
(d) photoautotroph.
Answer:
(b) chemoheterotroph

Question 11.
The ship used by Charles Darwin during the sea voyages was
(a) HMS Beagle
(b) HSM Beagle
(c) HMS Eagle
(d) HSM Eagle.
Answer:
(a) HMS Beagle

Question 12.
Fitnes according to Darwin refers to
(a) number of species in a community
(b) useful variation in population
(c) strength of an individual
(d) reproductive fitness of an organism.
Answer:
(d) reproductive fitness of an organism.

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 13.
Alfred Wallace worked in
(a) Galapagos Island
(b) Australian Island Continent
(c) Ma lay Archipelago
(d) none of these.
Answer:
(c) Ma lay Archipelago

Question 14.
The theory of natural selection was given by
(a) Lamarck
(b) Alfred Wallace
(c) Charles Darwin
(d) Oparin and Haldane.
Answer:
(c) Charles Darwin

Question 15.
The preserved fossil remains of Archaeopteryx show that
(a) it was a flying reptile from the Permian period
(b) reptiles gave rise to birds during Jurassic period
(c) it was a flying reptile in the Triassic period
(d) reptiles gave rise to birds during Permian period.
Answer:
(b) reptiles gave rise to birds during Jurassic period

Question 16.
Which of the following isotopes is used for Finding the fossil age maximum about 35,0000 years ?
(a) 238U
(b) 14C
(c) 3H
(d) 206Pb
Answer:
(b) 14C

Question 17.
Which of the following statements is True ?
(a) Wings to birds and insects are homologous organs.
(b) Human hands and bird’s wings are analogous organs.
(c) Human hands and bat’s wings are analogous organs.
(d) Flipper of penguin and dolphin are analogous organs.
Answer:
(d) Flipper of penguin and dolphin are analogous organs.

Question 18.
Replacement of the lighter-coloured variety of peppered moth (Biston betularia) to its darker variety (Biston carbonaria) in England is the example of
(a) natural selection
(b) regeneration
(c) genetic isolation
(d) temporal isolation.
Answer:
(a) natural selection

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 19.
Phenomenon of ‘industrial melanism’ demonstrates
(a) geographical isolation
(b) reproductive isolation
(c) natural selection
(d) induced mutation.
Answer:
(c) natural selection

Question 20.
Which one of the following phenomena supports Darwin’s concept of natural selection in organic evolution ?
(a) Development of transgenic animals
(b) Production of “Dolly’, the sheep by cloning
(c) Prevalence of pesticide resistant insects
(d) Development of organs from ‘stem cells’ for organ transplantation.
Answer:
(c) Prevalence of pesticide resistant insects

Question 21.
The phenomenon ‘ontogeny repeats phylogeny’ is explained by
(a) recapitualtion theory
(b) Inheritcance theory
(c) mutation theory
(d) natural selection theory.
Answer:
(a) recapitualtion theory

Question 22.
The presence of gill slits, in the embryos of vertebrates, supports the theory of
(a) metamorphosis
(b) biogenesis
(c) organic evolution
(d) recapitulation
Answer:
(d) recapitulation

Question 23.
Which is not a vestigial organ in man ?
(a) Nictitating membrane
(b) Tail vertebrae
(c) Vermiform appendix
(d) Nails
Answer:
(d) Nails

Question 24.
Which one is not a vestigia organ ?
(a) Wings of kiwi
(b) Coccyx in man
(c) Pelvic girdle of python
(d) Flipper of seal
Answer:
(d) Flipper of seal

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 25.
By the statement ‘survival of the fittest’, Darwin meant that
(a) the strongest of all species survives
(b) the most intelligent of the species survives
(c) the cleverest of the species survives
(d) the species most adaptable to changes survives.
Answer:
(d) the species most adaptable to changes survives.

Question 26.
Which of the following are the two key concepts of Darwinian theory of evolution ?
(a) Genetic drift and mutation
(b) Adaptive radiation and homology
(c) Mutation and natural selection
(d) Branching descent and natural selection
Answer:
(d) Branching descent and natural selection

Question 27.
According to Lamarckism, long necked giraffes evolved because
(a) nature selected only long necked ones
(b) humans preferred only long necked ones
(c) short necks suddenly changed into long necks
(d) of stretching of necks over many generations by short necked ones.
Answer:
(d) of stretching of necks over many generations by short necked ones.

Question 28.
Which of the following evidences does not favour the Lamarckian concept of inheritance of acquired characters ?
(a) Lack of pigment in cave-dwelling animlas
(b) Melanisation in peppered moth
(c) Absence of limbs in snakes
(d) Presence of webbed toes in aquatic birds
Answer:
(b) Melanisation in peppered moth

Question 29.
“Human population grows in geometric ratio while food materials increase in arithmetic proportion”. It is a statement from
(a) Darwin
(b) Bateson
(c) AmartyaSen
(d) Malthus.
Answer:
(d) Malthus.

Question 30.
Which one of the following scientist’s name is correctly matched with the theory put forth by him ?
(a) de Vries – Theory of natural selection
(b) Darwin – Theory of pangenesis
(c) Weismann – Theory of continuity of germplasm
(d) Pasteur – Theory of inheritance of acquired charcters
Answer:
(c) Weismann-Theory of continuity of germplasm

Question 31.
Single step large mutation leading to speciation is also called
(a) founder effect
(b) saltation
(c) branching descent
(d) natural selection
Answer:
(b) saltation

Question 32.
At a particular locus, frequency of allele A is 0.6 and that of allele a is 0.4. What would be the frequency of heterozygotes in a random mating population at equilibrium ?
(a) 0.36
(b) 0.16
(c) 0.24
(d) 0.48
Answer:
(d) 0.48

Question 33.
Hardy-Weinberg equilibrium is known to be affected by gene flow, genetic drift mutation, genetic recombination and
(a) evolution
(b) limiting facrtors
(c) saltation
(d) natural selection.
Answer:
(d) natural selection.

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 34.
The Hardy-Weinberg principle cannot operate if
(a) a population does not migrate for a longtime to a new habitat.
(b) frequent mutations occur in the population
(c) the population has no chance of interaction with other populations
(d) free interbreeding occurs among all members of the population.
Answer:
(b) frequent mutations occur in the population

Question 35.
Genetic drift operates only in
(a) larger populations
(b) Mendelian populations
(c) island populations
(d) smaller populations.
Answer:
(d) smaller populations.

Question 36.
Which of the following is most important for speciation ?
(a) Seasonal isolation
(b) Reproductive isolation
(c) Behavioural isolation
(d) Tropical isolation
Answer:
(b) Reproductive isolation

Question 37.
The factors involved in the formation of new species are
(a) Isolation and competition
(b) gene flow and competition
(c) competition and mutation
(d) isolation and variation.
Answer:
(d) isolation and variation.

Question 38.
Stabilising selection favours
(a) both extreme forms of a trait
(b) intermediate forms of a trait
(c) environmental differences
(d) one extreme form over the other extreme form and over intermediate forms of a trait.
Answer:
(b) intermediate forms of a trait

Question 39.
The different forms of interbreeding species that live in different geographical regions are called
(a) sibling species
(b) sympatric species
(c) allopatric species
(d) polytypic species.
Answer:
(c) allopatric species

Question 40.
Which of following represents correct order of evolutiton ?
(a) Amoeba → Leucosolenia → Hydra → Ascaris
(b) Leucosolenia → Hydra → Amoeba → Ascaris
(c) Ascaris → Amoeba → Leucosolenia → Hydra
(d) None of these
Answer:
(a) Amoeba → Leucosolenia → Hydra → Ascaris

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 41.
Presence of gills in the tadpole of frog indicated that
(a) fishes were amphibious in the past
(b) fishes evolved from frog-like ancestors
(c) frogs will have gills in future
(d) frogs evolved from gilled ancestors.
Answer:
(d) frogs evolved from gilled ancestors.

Question 42.
The character that proves that frogs have evolved from fishes is
(a) their ability to swim in water
(b) tadpole larva in frogs
(c) similarity in the shape of the head
(d) their feeding on aquatic plants.
Answer:
(b) tadpole larva in frogs

Question 43.
Identify the correct arrangement of periods of Palaeozoic era in ascending order in geological time scale.
(a) Cambrian → Devonian → Ordovician → Silurian → Carboniferous → Permian
(b) Cambrian → Ordovician → Silurian → Devonian→Carboniferous → Permian
(c) Cambrian → Ordovician → Devonian → Silurian → Carboniferous → Permian
(d) Silurian → Devonian → Cambrian → Ordovician → Permian → Carboniferous
Answer:
(b) Cambrian → Ordovician → Silurian → Devonian → Carboniferous → Permian

Question 44.
Which is the correct order of increasing geological time scale for a hypothetical vertebrate evolution ?
(a) Cenozoic, Mesozoic, Palaeozoic, Proterozoic
(b) Cenozoic, Palaeozoic, Mesozoic, Proterozoic
(c) Proterozoic, Cenozoic, Palaezoic, Mesozoic
(d) Proterozoic, Palaeozoic, Mesozoic, Cenozoic
Answer:
(d) Proterozoic, Palaeozoic, Mesozoic, Cenozoic

Question 45.
The ‘Devonian period’ is considered to be as
(a) age of fishes
(b) age of amphibians
(c) age of reptiles
(d) age of mammals.
Answer:
(a) age of fishes

Question 46.
Amphibians were dominant during _______ period.
(a) Carboniferous
(b) Silurian
(c) Ordovician
(d) Cambrian
Answer:
(a) Carboniferous

Question 47.
The primate which existed 15 mya was
(a) Homo habilis
(b) Australopithecus
(c) Ramapithecus
(d) Homo erectus
Answer:
(c) Ramapithecus

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 48.
The extinct stone ancestor, who ate only fruits and hunted with stone weapons was
(a) Ramapithecus
(b) Australophecus
(c) Dryopithecus
(d) Homo erectus
Answer:
(b) Australophecus

Question 49.
One of the oldest, best preserved and most complete hominid fossil commonly known as ‘Lucy’ belongs to the genus
(a) Austraolopithecus
(b) Oreopithecus
(c) Dryopithecus
(d) Pithecanthropus
Answer:
(a) Austraolopithecus

Question 50.
The brain capacity of Homo eractus was about
(a) 650 c.c.
(b) 900 c.c.
(c) 1500 c.c.
(d) 1400 c.c.
Answer:
(b) 900 c.c.

Question 51.
The extinct human who lived 1,00,000 to 40,000 years ago, in East and Central Asia, used hides to protect their bodies and had brain capacity of 1400 c.c. were
(a) Homo habilis
(b) Neanderthal man
(c) Cro-Magnon man
(d) Ramapithecus.
Answer:
(b) Neanderthal man

Question 52.
Which of the following statements is correct ?
(a) Australopithecus has large brain around 900 c.c.
(b) Neanderthal man lived in East Africa and the fruits.
(c) Homo erectus had brain capacity 900 c.c.
(d) Homo sapiens arose in Central Asia and moved to other continuents and developed into distinct races.
Answer:
(c) Homo erectus had brain capacity 900 c.c.

Question 53.
Which of the following statements js correct regarding evolution of mankind ?
(a) Homo erectus is preceded by Homo habilis.
(b) Neanderthal man and Cro-Magnon man were living at the same time.
(c) Australopithecus was living in Australia.
(d) None of these
Answer:
(a) Homo erectus is preceded by Homo habilis.

Question 54.
The cranial capacity was largest among the
(a) Peking man
(b) Java ape man
(c) African man
(d) Neanderthal man.
Answer:
(d) Neanderthal man.

Question 55.
The most apparent change during the evolutionary history oiHomo sapiens is traced in
(a) loss of body hair
(b) walking unright
(c) sortening of the jaws
(d) remarkable, increase in the brain size.
Answer:
(d) remarkable, increase in the brain size.

Question 56.
What kind of evidences suggested that man is more closely related with chimpanzee than with other hominoid apes ?
(a) Evidence from DNA of sex chromosomes only
(b) Comparison of chromosome morphology and number
(c) Evidence from fossil remains, and the fossil mitochondrial DNA alone
(d) Evidence from banding pattern of chromosome 3 and 6
Answer:
(d) Evidence from banding pattern of chromosome 3 and 6

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 57.
Which of the following eras, in geological time scale, corresponds to the period when life had not orginated upon the earth ?
(a) Azoic
(b) Palaeozoic
(c) Mesozoic
(d) Archaeozoic
Answer:
(a) Azoic

Question 58.
Homo sapiens arose during which epoch ?
(a) Pleistocene
(b) Pliocene
(c) Oligocene
(d) Holocene
Answer:
(d) Holocene

Question 59.
Study the human evolution is called
(a) archaeology
(b) anthropology
(c) pedigree analysis
(d) chronobiology.
Answer:
(b) anthropology

Question 60.
Which of the following is used as an atmospheric pollution indicator ?
(a) Lepidoptera
(b) Lichens
(c) Lycopersicon
(d) Lycopodium
Answer:
(b) Lichens

Question 61.
The theory of spontaneous generation stated that
(a) life arose from living forms only
(b) life can arise from both living and non-living
(c) life can arise from non-living things only
(d) life arises spontaneusly, neither from living nor from the non-living.
Answer:
(c) life can arise from non-living things only

Question 62.
Animal husbandry and plant breeding programmes are the examples of
(a) reverse evolution
(b) aritifical selection
(c) mutation
(d) natural selection.
Answer:
(d) natural selection.

Question 63.
Palaentological evidences for evolutaion refer to the
(a) development of embryo
(b) homologous organs
(c) fossils
(d) analogous organs.
Answer:
(c) fossils

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Question 64.
The bones of forelimbs of whale, bat, cheeah and man are similar in structure, because
(a) one organism has given rise to another
(b) they share a common ancestor
(c) they perform the same function.
(d) the have biochemical similarities.
Answer:
(b) they share a common ancestor

Question 65.
Analogous organs arise due to
(a) divergent evolution
(b) artificial selection
(c) genetic drift
(d) convergent evolution.
Answer:
(d) convergent evolution.

Question 66.
(p+q)2 = p2 + 2pq + q2 = 1 represents an equation used in
(a) population genetics
(b) Mendilian genetics
(c) biometircs
(d) molecular genetics.
Answer:
(a) population genetics

Question 67.
Appearnace of antibiotic-resistant bacteria is an example of
(a) adaptive radiation
(b) transduction
(c) pre-existing variation in the population
(d) divergent evolution.
Answer:
(c) pre-existing variation in the population

Question 68.
Fossils are generally found in
(a) sedimentasry rocks
(b) igneous rocks
(c) metamorphic rocks
(d) any type of rock.
Answer:
(a) sedimentasry rocks

Question 69.
Which type of selection is industrial melanism observed in moth, Biston betularia?
(a) Stabilising
(b) Directional
(c) Disruptive
(d) Artificial
Answer:
(b) Directional

Question 70.
Which of the following is an example for link species ?
(a) Lobe fish
(b) Dodo bird
(c) Seaweed
(d) Chimpanzee
Answer:
(a) Lobe fish

Question 71.
Variations during mutations of meiotic recombinations are
(a) random and directionless
(b) random and directional
(c) random and small
(d) random small and directional
Answer:
(a) random and directionless

Bihar Board 12th Biology Objective Answers Chapter 7 Evolution

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Bihar Board 12th Chemistry Model Papers

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Time : 3 Hours 15 Min
Full Marks: 70

Instructions for the candidates

  1. Candidates are required to give answers in their own words as far as practicable.
  2. Figures in the right hand margin indicates full marks.
  3. While answering the questions, candidate should adhere to the words limit as far as practicable.
  4. 15 Minutes of extra time has been allotted for the candidates to read the questions carefully.
  5. This question paper is divided into two sections : Section -A and Section-B
  6. In Section A, there are 35 objective type questions which are compulsory, each carrying 1 mark. Darken the circle with blue/black ball pen against the correct option on OMR Sheet provided to you. Do not use Whitener/Liquid/Blade/ Nail on OMR Sheet otherwise your result will be treated as invalid.
  7. In section-B, there are 18 short answer type questions (each carrying 2 marks), out of which only 10 (ten) questions are to be answered Apart from this there are 06 Long Answer type questions (each carrying 5 marks), out of which 3 questions are to be answered.
  8. Use of any electronic device is prohibited.

Objective Type Questions

In the following questions no. from 1 to 35, there is only one correct answer against each question. For each question, mark (darken) the correct answer on the OMR Sheet provided to you. (1 x 35 = 35)

Question 1.
Which type of crystals contains more than one Bravais lattice ?
(a) Hexagonal
(b) Triclinic
(c) Rhombohedral
(d) Monoclinic
Answer:
(d) Monoclinic

Question 2.
For the structure given below the site marked as S is a …………….
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 1
(a) tetrahedral void
(b) cubic void
(c) octahedral void
(d) none of these
Answer:
(c) octahedral void

Question 3.
If the radius of an octahedral void is r and radius of atoms in close packing is R, the relation between r and R is
(a) r = 0.414 R
(b) R = 0.414 r
(c) r = 2R
(d) r = \(\sqrt{2} R\)
Answer:
(a) r = 0.414 R

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 4.
A metal crystallises into a lattice containing a sequence of layers as AB AB AB………………… What percentage of voids are left in the lattice ?
(a) 72%
(b) 48%
(c) 26%
(d) 32%
Answer:
(c) 26%

Question 5.
In ccp arrangement the pattern of successive layers can be designated as
(a) AB AB AB
(b) ABC ABC ABC
(c) AB ABC AB
(d) ABA ABA ABA
Answer:
(b) ABC ABC ABC

Question 6.
A solution is obtained by mixing 200 g of 30% and 300 g of 20% solution by weight. What is the percentage of solute in the final solution ?
(a) 50%
(b) 28%
(c) 64%
(d) 24%
Answer:
(d) 24%

Question 7.
When 1.04 g of BaCl2 is present in 105 g of solution the concentration of solution is ………….
(a) 104 ppm
(b) 10.4 ppm
(c) 0104 ppm
(d) 104 ppm
Answer:
(b) 10.4 ppm

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 8.
In the cell, |Zn2+| |Cu2+|Cu, the negative terminals
(a) Cu
(b)Cu2+
(c) Zn
(d) Zn2+
Answer:
(d) Zn2+

Question 9.
What will be standard cell potential of galvanic cell with the following reaction ?
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 2
(a) 74 V
(b) 1.14 V
(c) 34 V
(d) – 0.34 V
Answer:
(c) 34 V                         .

Question 10.
In a reaction 2HI → H2 + I2 the concentration of HI decreases from 0.5 mol L-1 in 10 minutes. What is the rate of reaction during this interval ?
(a) 5 x 10-3 M min-1
(b) 5 x 10-3 M min-1
(c) 5 x 10-2 M min-1
(d) 5 x 10-2 M min-1
Answer:
(a) 5 x 10-3 M min-1

Question 11.
The rate of disappearance of SO, in the reaction, 2SO2 + O2  → 2SO3 is 1.28 x 10-5 Ms . The rate of appearance of SO3 is ……………….
(a) 64 x 10-5 M s-1
(b) 0.32 x 10-5 M s-1
(c) 56 x 10-5 Ms1
(d) 1.28 x 10-5  M s-1
Answer:
(d) 1.28 x 10-5  M s-1

Question 12.
Which of the following is a property of physisorption ?
(a) High specificity
(b) Irreversibility
(c) Non-specificity
(d) None of these
Answer:
(c) Non-specificity

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 13.
Which of the following statement is not correct about physisorption ?
(a) It is a reversible process
(b) It requires less heat of adsorption
(c) It requires activation energy
(d) It takes place at low temperature
Answer:
(c) It requires activation energy

Question 14.
The name of the metal which is extracted from the ore is given below. Mark the example which is not correct.
(a) Malachite – Cu
(b) Calamine – Zn
(c) Chromite – Cr
(d) Dolomite – Al
Answer:
(d) Dolomite – Al

Question 15.
Which of the followings is magnetite ?
(a) Fe2CO3
(b) Fe2O3
(c) Fe3O4
(d) Fe2O3.3H2O
Answer:
(c) Fe3O4

Question 16.
Nitrogen can form only one chloride with chlorine which is NCl3 whereas P can form PCl3 and PCl5. This is ……………
(a) due to absence of d-orbitals in nitrogen
(b) due to difference in size of N and P
(c) due to higher reactivity of P towards Cl than N
(d) due to presence of multiple bonding in nitrogen
Answer:
(a) due to absence of d-orbitals in nitrogen

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 17.
On heating a mixture of NH4Cl and KNO2, we get
(a) NH4NO3
(b) KNH4(NO3)2
(c) N2
(d) NO
Answer:
(c) N2

Question 18.
The melting point of copper is higher than that of zinc because
(a) the s, p as well as d-electrons of copper are involved in metallic bonding
(b) the atomic volume of copper is higher
(c) the d-electrons of copper are involved in metallic bonding
(d) the s as well as d-electrons of copper are involved in metallic bonding
Answer:
(c) the d-electrons of copper are involved in metallic bonding

Question 19.
Zr and Hf have almost equal atomic and ionic radii because of ………………….
(a) diagonal relationship
(b) lanthanoid contraction
(c) actinoid contraction
(d) belonging to the same group
Answer:
(b) lanthanoid contraction

Question 20.
Which of the following primary and secondary valencies are not correctly marked against the compounds ?
(a) [Cr(NH3)6]Cl3, p = 3,s = 6
(b) K2[Pt(CI4],p = 2,s = 4
(c) [Pt(NH3)2Cl2],p = 2,s = 4
(d) [Cu(NH3)4]SO4,p = 4,s = 4
Answer:
(d) [Cu(NH3)4]SO4,p = 4,s = 4

Question 21.
When aqueous solution of potassium fluoride is added to the blue coloured aqueous CuSO4 solution, a green precipitate is formed. This observation can be explained as follows.
(a) On adding KF, H2O being weak field ligand is replaced by F ions forming [CuF4 ]2- which is green in colour
(b) Potassium is coordinated to [Cu(H2O)4]2+ ion present in CuSO4 and gives green colour
(c) On adding KF, Cu2+ are replaced by K+ forming a green complex
(d) Blue colour of CuSO4 and yellow colour of KI form green colour on mixingAnswer:
(a) On adding KF, H2O being weak field ligand is replaced by F ions forming [CuF4 ]2- which is green in colour

Question 22.
IUPAC name of CH3)2 CH — CH2 — CH2Br is
(a) 1-bromopentane
(b) 1-bromo-3-methylbutane
(c) 2-methyl-4-bromobutane
(d) 2-methyl-3-bromopropane
Answer:
(b) 1-bromo-3-methylbutane

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 23.
The IUPAC name of the compound is
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 3
(a) 1 -fluoro-4-methyl-2-nitrobenzene
(b) 4-fIu6ro-lmethyl-3-nitrobenzene
(c) 4-methyl-l-fIuoro-2-nitrobenzene
(d) 2-fIuoro-5-methyl-l-nitrobenzene
Answer:
(d) 2-fIuoro-5-methyl-l-nitrobenzene

Question 24.
The C— O — H angle in ether is about
(a) 180°
(b) 190°28′
(c) 110°
(d) 105°
Answer:
(c) 110°

Question 25.
The C— O — H bond angle in alcohols is slightly less than the tetrahedral angle whereas the C— O — C bond angle in ether is slightly greater because ………
(a) of repulsion between the two bulky R groups.
(b) O atom in both alcohols and ethers is sp3-hybridised
(c) lone pair-lone pair repulsion is greater than bond pair­bond pair repulsion
(d) none of these
Answer:
(a) of repulsion between the two bulky R groups.

Question 26.
Propanone can be prepared from ethyne by
(a) passing a mixture of ethyne and steam over a catalyst, magnesium at 420°C
(b) passing a mixture of ethyne and ethanol over a catalyst zinc chromite
(c) boiling ethyne with water and H2SO4
(d) treating ethyne with iodine and NaOH
Answer:
(a) passing a mixture of ethyne and steam over a catalyst, magnesium at 420°C

Question 27.
In the following reaction, product (P) is
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 4
(a) RCHO
(b) RCH3
(c) RCOOH
(d) RCH2OH
Answer:
(a) RCHO

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 28.
IUPAC name of the compound (C2H5)2 NCH is
(a) 2, 2-diethylmethanamine
(b) N, N-diethylmethanamine
(c) N-ethyl-N-methylethanamine
(d) N-methylbutanamine
Answer:
(c) N-ethyl-N-methylethanamine

Question 29.
Reduction of aromatic nitro compounds using Sn and HCI gives
(a) armatic primary amines
(b) aromatic secondary amines
(c) aromatic tertiary amines
(d) aromatic amides
Answer:
(a) armatic primary amines

Question 30.
Which of the fol lowing is an example of an aldopentose ?
(a) D-Ribose
(b) Glyceraldehyde
(c) Fructose
(d) Erythrose
Answer:
(a) D-Ribose

Question 31.
Which of the following treatment will convert starch directly into glucose ?
(a) Heating with dilute H2SO4
(b) Fermentation by diastase
(c) Fermentation by zymase
(d) Heating with dilute NaOH
Answer:
(a) Heating with dilute H2SO4

Question 32.
The correct structure of monomers of buna-S is
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 5
Answer:
(c)

Question 33.
The S in buna-S refers to
(a) sulphur
(b) styrene
(c) sodium
(d) salicylate
Answer:
(b) styrene

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 34.
The main cause of acidity in the stomach is
(a) release of extra gastric acid which decrease the pH level
(b) indidestion and pain in large intestine
(c) increase the pH level in the stomach
(d) release of extra bile juice which increase alkaline medium in stomach.
Answer:
(a) release of extra gastric acid which decrease the pH level

Question 35.
Which of the following will not act as antacid ?
(a) Sodium hydrogen carbonate
(b) Magnesium hydroxide
(c) Sodium carbonate
(d) Aluminium carbonate
Answer:
(c) Sodium carbonate

Non-Objective Type Questions

Short Answer Type Questions

In this Section, there are 18 Short answer type questions (each carrying 2 marks), out of which answer any 10 questions.  2 x 10 = 20

Question 1.
Calculate the osmotic pressure of 5% solution of urea at 272 K . (R = 0.0821 L-atm. deg-1).
Answer:
As we know that \(\pi=\frac{n_{\text {urea }} R T}{V}\)
Number of moles in urea \(=\frac{5}{60}=0.083\)
Volume of Solution \(=\frac{100}{1000}=0.1\) litres
Hence, Osmotic Pressure, \(\pi=\frac{0.083 \times 0.0821 \times 272}{0.1}=18.53 \mathrm{atm}\)
Osmotic Pressure = 18.53atm

Question 2.
Define standard electrode potential.
Answer:
Standard Electrode Potential : The standard electrode potential is the potential difference between the electrode & electrolyte (at 1 M), at standard conditions (1 atm, 298 K). Standard electrode potential is abbreviated by (E or E°).

Question 3.
What is activation energy ? How is the rate constant of a reaction related to its activation energy ?
Answer:
Activation Energy: The minimum extra amount of energy absorbed by the reactant molecules so that their energy becomes equal to threshold value is called activation energy.
Activation energy (Ea)= Threshold energy (ET) – Average energy of the reactant (ER). i.e. Ea = ET – ER
If activation energy will be less, then faster will be the reaction.
i. e . Activation energy = \(\alpha \frac{1}{\text { rate of reaction }}\)
e.g. 2NO + O2 → 2NO2
It is a fast reaction because of low activation energy. 2CO + O2 → 2CO2
It is a slow reaction because of high activation energy.

Question 4.
What is the main difference between physiorption and Chemisorption ?
Answer:
Physisorption:

  • Force of attraction are Vander Waal’s forces.
  • It usually takes place at low temperature and decreases with increasing temperature.
  • It does not require any activation energy.

Chemisorption :

  • Force of attraction are chemical bond forces.
  • It takes place at high temperature.
  • It requires high activation energy.

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 5.
Differentiate the following giving suitable example:
(i) Calcination and Roasting
(ii) Flux and Slag.
Answer:
(i) Calcination :
(a) Calcination is a thermal treatment process in absence of air applied to ores and materials to bring about a thermal decomposition, phase transition or removal of volalatile fraction It is done at temperature below the melting point of the product material.
(b) Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 6

Roasting:
(a) Roasting is reverse process of calcination. It is also thermal process in presence of air applied to ore for removal of volatile fraction under melting point of the product material is called Roasting.
(b) Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 7

Flux :
(a) It is chemical cleaning agent, which remove impurities from metal.
(b) Example – CaO + SiO2 → CaSiO3

Slag :
(a) Slag is partially vitereous by product of the process of smelting ores, which separates the desiredmetal fraction from the unwanted fraction. It is mixture of metal oxide
(b) Example – CaO + SiO2 → CaSiO3

Question 6.
Why do transition elements form coloured compounds ? Explain.
Answer:
The value of electrods potential depends upon the heat of sublimation and ionisation energy
ΔH – ΔHsub + IE + ΔHhyd . Due to the influence of ligand d-orbitals of transition metals divide into two unequal energy containing sets. Unpaired electrons absorb sunlight and jumps from one orbit into another orbit and one clour is reflected.

Question 7.
Predict the geometrical shapes of the following:
(a) sp3
(b)d2sp3
Answer:
(a) sp3 tetra-hedral
(b)d2SP3 – Square tetrahedral.

Question 8.
Specify oxidation numbers of the metals in the following co-ordination compounds:
(a) K4[Fe(CN)6]
(b) [PtCl4]2-
Answer:
(a) Oxidation number of (co-ordination compound)
Let oxidation number of Fe = x
Hence, 1 x 3 + (x -1 x 6 = 0
or, x – 3 = 0
x = 3 = O. N of Fe

(b) Oxidation number of (co-oridnation compaund)
Hence, x + 4 (-1) = – 2
or, x = -2 + 4 = + 2 = ON of Pt

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 9.
Write the structural formulae of the following:
(a) 4,4 dimethyl-2-pentanol
(b) 2-butanol.
Answer:
(a) 4, 4 dimethyl, 2-Pentanol
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 8

Question 10.
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 9
Answer:
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 10

Question 11.
How will you convert the following ?
(a) Ethyl alcohol to ethylamine
(b) Ethylamine to ethyl alcohol.
Answer:
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 11

Question 12.
Explain in which of the following compounds, the chemical bond would have less ionic character: LiCl or KCl.
Answer:
LiCl > KCl – Due to large size of k, KCl compound has less ionic character. LiCl is covalent than Li+ has higher polarisation capacity.

Question 13.
Discuss briefly the structure of CsCl.
Answer:
Structure of CsCl.
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 12

Question 14.
The osmotic pressure of sugar solution is 2.46 atm at 27°C. Calculate the concentration of the solution.
Answer:
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 13

Question 15.
How is molarity of a solution different from molality?
Answer:
Molarity : It is defined as the number of moles of solute dissolved in one liter of solution
\(Molarity (\mathrm{M})=\frac{\text { Moles of solute }}{\text { Volume of Solution in liter }} \)

Molality : It is defined as the number of moles of solute dissolved in one kg of solvent
\(Molality (\mathrm{m})=\frac{\text { Moles of the Solute }}{\text { Mass of the solvent in } \mathrm{kg}}\)

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 16.
Define heat of neutralisation.
Answer:
Defination of heat of neutralization : The heat reaction resulting from the neutralization of an acid or base especially the quantity produced when a gram equivalent of a base or acid is neutralized with a gram equivalent of an acid or base in dilute soliution.

Question 17.
(a) State law of mass action.
(b) What is the effect of temperature on reaction ?
Answer:
(a) Law of mass action: Law stating that the rate of any chemical reaction is proportional to the product of the masses of the reacting substances, with each mass raised to a power equal to the coefficient that occurs in the chemical equation.

(b) The effect of increasing collision frequency on the rate of the reaction is very minor increasing the temperature increases reaction rates because of the disproportionately large increase in the number of high energy collisions.

Question 18.
Discuss the following terms :
(a) Coordination number
(b) Effective atomic number.
Answer:
(a) The coordination number of a central atom in a molecure or crystal is the number of its near neighbours. This number is determined some what differently for molecules than for crystals.

(b) Effective atomic number has two different meaning; one that is the effective nuclear charge of an atom, and other that calculates the average atomic number for a compound or mixture of materials.

Long Answer Type Questions

There are 06 long answer type questions (each carrying 05 marks), out of which answer any there questions.
(5 x 3 = 15)

Question 19.
Define order to reaction and molecularity of relation. Derive a general expression for specific rate constant of first order reaction.
Answer:
It is the numbers of atom, ion, and molecule that collide with one another simultaneously so as to result into a chemical reaction
Rate constant → It is the speed at which the reactions are converted into the products at any moment of time.
Law of Mass Action → At a given temperature, the rate of a chemical reaction is directly proportional to the product of molar concentrations of reacting species with each concentration term raised to the power equal to numerical co-efficient of that species in the chemical reaction.
Thus, A + B → Products
Rate of reaction, r x [A] [B] = K [A] [B]
Where, [A] and [B] are the molar concentrations of the reactants A and B respectively and K is a constant of proportionality.
∴Rate = K [A]a [B]b

First Order Reaction – It is the type of reaction in which the rate of reaction depends only on the first power of the concentration of a single reacting Species.
Let, A + B → Product
If, a be the initial concentration of A and after time t, the concentration of product is x, then the concentration of A left at that instant is (a – x)
A → Products
initial conc.n (a)  ….(i)
after time ‘t’ (a – x)  … (ii)
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 14
Where, C → integration constant, which can be obtained from initial condition of the reaction, when, t = 0, x = 0,
then, from (iii) In (a – 0) = K x 0 + C
or, C = In (a)  ………….. (iv)
from (iii) + (iv) – In (a -x) = Kt – In (a)
or, In (a) – In (a – x) = kt or, In (a) – In (a – x) = kt.
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 15

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 20.
(a) Differentiate between DNA and RNA.
(b) What are Nucleic acids ?
Answer:
(a) The structural difference between DNA and RNA
DNA : DNA is double stranded helix in which two strands are coiled spirally in opposite directions.

  • The sugar molecule is 2-dehydrobase.
  • Nitrogeneous base uracil is not present.
  • DNA molecule are very large their molecules weights may vary from 6 million to 16.
  • DNA has unique property of replication.

RNA :

  • It is single strainded structure.
  • The sugar molecules is ribose.
  • Nitrogeneous base they-mine is not present.
  • RNA molecules are much smaller with molecular weight ranging from 20000 to 40000.
  • RNA usually doesn’t replicate.

(b) Nucleic acid – Nucleic acid are found in all living cells in form of nucleo proteins. They constitute important class of biomolecules. Nucleic acids are genetic material of the cells and are responsible for transmission of herediatry effect from one generation to the other and also carry out the biosynthesis of proteins. Nucleic acid are biopolymers (i.e., polymers present in living system). The genetic information about in nucleic acid controls the structure of all proteins including enzymes and thus governs the entire metabolic activity in the living organism.

Question 21.
(a) Discuss the principle involved in the manufacture of sulphuric acid by contact process.
(b) How will you test for sulphate ion ?
Answer:
From following steps sulphuric acid is manufactured
(i) SO2 gas is produced by burning sulphur of FeS2 in excess of air
S + O2  → SO2
4 FeS2 +11O2 → 2Fe2 O3 + 8SO2

(ii) SO2 is oxidised into SO3 in prensence of catalyst
2 SO2 + O2 → 2 SO3

(iii) SO3 on treatment with cone. H2SO4 oleum (H2S2O7) is obtained.
SO3 + H2SO4  → 2 H2S22O7

(iv) From oleum sulphuric acid is obtained of desired concentration by diluting water.
H2S2O7+ H2O → 2H2OSO4

(b) Test of sulphate radical (of NO2SO4) – We mix up Barium chloride in the salt. We get white precipate of BaSO4 which is insoluble in dil. HCl & HNO3.
BaCl2 4- Na2SO4 → BaSO4 ↓ +2 NaCl
BaCl2 + HNO3 → In soluble

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 22.
What are carbohydrates ? How are they classified ?
Answer:
Carbohydrates are the naturally occuring organic compounds and are a major source of energy to our body. In plants carbohydrates are formed as a result for photosynthesis.
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 16
In animal systems the carbohydrates undergo decomposition to form carbon-dioxide and water accompanied by the release of energy needed for the body.

Composition of carbohydrates – The chemical formula of the carbohydrates suggests that these are the hydrates of carbon. For example, glucose may be represented as C6(H2O)6 while sucrose as C12(H2O)11 . But this definition has certain limitation.

(i) Compounds like formaldehyde (CH2O) and acetic acid (C2H4O2) are the hydrates of carbon do not show characteristics of carbohydrates.

(ii) Similarly, compound like rhammose (C6H12O6) and deoxyribose (C5H10O4) are carbohydrates but not the hydrates of carbon.

The definition of carbohydrates has been modified and they may be defined as—The polyhydroxy aldehydes or polyhdroxy ketones are substances which generally give these on hydrolysis; contain atleast one chiral carbon and are therefore, opitcally active classification of carbohydrates—

Classification of carbohydrates are done in three way

(a) Based on molecular size – on the basis of the molecular size, carbohydrates have been classified into three types. These are

  • Manosaccharides
  • Oligosoccharides
  • Polysaccharides

Based on taste – Carbohydrates with sweet taste are called sugar while those without a sweet taste are called non-sugars. It may be noted that all mono and oligosaccharides are sugars while polysaccharides are non­sugar.

Reducing and non-reducing sugar –  Carbohydrate which reduce Tollen’s reagent and Fehling solution are called reducing sugar while those which don’t reduce, these are called non reducing sugars. For example – glucose and fructose.

Question 23.
Disuss the following :
(a) Kolbe’s reaction
(b) Wurtz’s reaction
(c) Carbylamine reaction.
Answer:
(a) Kolbe’s reaction – The reaction of phenoxide ion with CO2 at the high pressure and temperature followed by acidification to form salicylic acid is called Kolbe’s reaction.
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 17

(b) Wurtz reaction-When two moles of alkyl halide is heated with Na in the presence of ether goes higher alkane.
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 18

(c) Carbylamine reaction-Chloroform when heated with primary amine in presence of alcoholic KOH forms a derivative called isocyanide.
Bihar Board 12th Chemistry Model Question Paper 4 in English Medium 19

Bihar Board 12th Chemistry Model Question Paper 4 in English Medium

Question 24.
Give one example for each of the following :
(a) Synthetic rubber
(b) Naturally occurring amino acid
(c) Condensation polymer
(d) Additional polymer
(e) Artificial sweeteners.
Answer:
(a) Nioprin
(b) Zwitter ion
(c) NYlon-6, 6
(d) Buna-5
(e) Saccharine.

 

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Bihar Board 12th Physics Model Papers

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Objective Type Questions

There are 1 to 35 objective type questions with 4 options. Choose the correct option which is to be answered on OMR Sheet. (35 x 1 = 35)

Question 1.
The electric field intensity at distance on the axis of an electric dipole is.Ej and E2 on the perpendicular bisector of dipole. The angle between E] and E2 is 6, will be
(a) 1 : 1, π
(b) 1 : 2, π/2
(c) 2 : 1, π
(d) 1 : 3 π
Answer:
(c) 2 : 1, π

Question 2.
The metre bridge is shown in figure. The value of x is
(a) 10Ω
(b) 3 ohm
(c) 9 ohm
(d) 10 ohm
Bihar Board 12th Physics Model Question Paper 5 in English Medium 1
Answer:
(a) 10Ω

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 3.
The dimension of electromotive force is
(a) |ML2 T-3]
(b) [ML2T-3I-1]
(c) [MLT-2]
(d) [ML2T-3I-1]
Answer:
(d) [ML2T-3I-1]

Question 4.
The power of electric circuit is
(a) V.R
(b) V2.R
(c) V2/R
(d) V2RI
Answer:
(c) V2/R

Question 5.
If a 60W and 40W bulb are joined in series the
(a) 60W bulb glow more
(b) 40 W bulb glow more
(c) Both bulb glow similar
(d) Only 60W bulb is lighted
Answer:
(b) 40 W bulb glow more

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 6.
Kilowatt hour (kwh) is unit of
(a) Power
(b) Energy
(c) Torque
(d) None of these
Answer:
(b) Energy

Question 7.
A magnet is cut parallel to its length in n equal
parts. Magnetic moment of each part will be
\(\text { (a) } \frac{M}{n}\)
\(\text { (b) } \frac{M}{n^{2}}\)
\(\text { (c) } \frac{M}{2 n}\)
(d) None of these
Answer:
\(\text { (a) } \frac{M}{n}\)

Question 8.
The work done to rotate the magnet with 90° will be
(a) MB
(b) MB cosθ
(c) MB sin θ
(d) MB (1 – sin θ)
Answer:
(b) MB cosθ

Question 9.
S.I. unit of polar strength is
(a) Ampere metre
(b) Tesla
(c) Faraday
(d) Ampere gm2
Answer:
(a) Ampere metre

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 10.
Which is ferromagnetic substance
(a) Mn
(b) Cr
(c) Zn
(d) Alnico
Answer:
(d) Alnico

Question 11.
Which law is based on the principle of conservation of energy?
(a) Lenz’s law
(b) Faraday’s law
(c) Ampere’s law
(d) None of these
Answer:
(a) Lenz’s law

Question 12.
The unit of reactance is
(a) mho
(b) ohm
(c) Faraday
(d) ampere
Answer:
(b) ohm

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 13.
The phase difference between current and voltage in AC circuit is 0. Then power factor will
(a) cos θ
(b) sin θ
(c) tan θ
(d) 1θ
Answer:
(a) cos θ

Question 14.
The velocity of electro magnetic wave in air is
Bihar Board 12th Physics Model Question Paper 5 in English Medium 2
Answer:
(b)

Question 15.
The light waves are transverse in nature which is shown by
(a) Scattering
(b) Diffraction
(c) Interference
(d) Polarisation
Answer:
(d) Polarisation

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 16.
When two converging lens of same total length / are placed in contact then focal length combination is
(a) f
(b) 2f
(c) f2
(d) 3f
Answer:
(c) f2

Question 17.
When the light ray passes from one medium to another medium then it is deviated, is called
(a) Dispersion
(b) Refraction
(c) Diffraction
(d) Reflection
Answer:
(b) Refraction

Question 18.
Convex lens are used for
(a) Myopia
(b) Hypermetropia
(c) Old sightedness
(d) Astigmatisn
Answer:
(b) Hypermetropia

Question 19.
The path difference for destructive interference is
(a) nλ
\(\text { (b) }(2 n+1) \frac{\lambda}{2}\)
(c) Zero
(d) infinity
Answer:
\(\text { (b) }(2 n+1) \frac{\lambda}{2}\)

Question 20.
Brewler’s law is
(a) μ = sin ip
(b) μ = cos ip
(c) μ = tan ip
(d) μ = tan2 ip
Answer:
(c) μ = tan ip

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 21.
Light is made of vibration by
(a) Ether particle
(b) Air particle
(c) Electric and magnetic field
(d) None of these
Answer:
(c) Electric and magnetic field

Question 22.
Photo cell is based on
(a) photo electric effect
(b) chemical effect of current
(c) magnetic effect of current
(d) electromagnetic cell
Answer:
(a) photo electric effect

Question 23.
Which is uncharged
(a) α-particle
(b) β-particle
(c) photon
(d) proton
Answer:
(c) photon

Question 24.
Lymen series of Hydrogen lies in electromagnetic spectrum is
(a) X : ray
(b) Visible
(c) Infrared
(d) Ultra-violet
Answer:
(d) Ultra-violet

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 25.
When transition takes place in hydrogen atom from higher orbit to another orbit then obtain
(a) Lymen series
(b) Balmer series
(c) Paschan series
(d) Pfund series
Answer:
(b) Balmer series

Question 26.
The average binding energy per nucleon of nucleus is
(a) 8 ev
(b) 8 Mev
(c) 8 Bev
(d) 8 Joule
Answer:
(b) 8 Mev

Question 27.
The cause of emission of energy in star is
(a) chemical reaction
(b) fusion of heavy nucleus
(c) fusion of light nucleus
(d) fission of heavy nucleus
Answer:
(c) fusion of light nucleus

Question 28.
The Boolean expression of OR gate is
(a) A + B = y
(b) A.B = y
\(\text { (c) } \bar{A}=y\)
\(\text { (d) } \mathrm{C}=\overrightarrow{\mathrm{A} \cdot \mathrm{B}}\)
Answer:
(a) A + B = y

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 29.
The permeability of Ferromagnetic substance is
(a) μ > 1
(b) μ = 1
(c) μ < 1
(d) μ = 0
Answer:
(a) μ > 1

Question 30.
The light is emitted in forward biased junction diode, is called
(a) LED
(b) Photo diode
(c) Zener diode
(d) None of these
Answer:
(a) LED

Question 31.
Which expression is correct?
Bihar Board 12th Physics Model Question Paper 5 in English Medium 3
Answer:
(d)

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 32.
The majority carriers in n-type semi conductor is
(a) proton
(b) hole
(c) a-particle
(d) electron
Answer:
(d) electron

Question 33.
Which frequency range is used for TV transmission?
(a) 30 HZ-300HZ
(b) 30 KHZ-300 KHZ
(c) 30 MHZ-300MHZ
(d) 30GHZ-300GHZ
Answer:
(a) 30 HZ-300HZ

Question 34.
Ionosphere is used for radio waves in
(a) rarer mediun
(b) denser medium
(c) free space
(d) dielectric
Answer:
(a) rarer mediun

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 35.
The maximum distance upto which TV signal can be received from Antenna of height h is proportional to
(a) h1/2
(b) h
(c) h5/2
(d) h2
Answer:
(a) h1/2

Non-Objective Type Questions!
Short Answer Type Questions

Question No. 1 to 18 are short answers type question. Each question carries 2 marks. Answer any ten (10) question.

Question 1.
Electric lines of force never intersect to each other. Why?
Answer:
If two lines of force intersect at a point then it will mean that at that point electric field has two directions. As it is not possible, the lines of force here not intersect to each other.

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 2.
Write the limitations of Coulomb’s law.
Answer:
Coulomb’s law in electrostatics does not hold in all situation. It is applicable only in following situations :

  1. The electric charges must be stationary.
  2. The electric charge must be point in size. Coulomb’s law doesn’t apply to two charged bodies of finite size. It is because, the distribution of charge doesn’t remain uniform when the two bodies are brought together.

Question 3.
Prove maximum power theorem.
Answer:
It states that the output power of a source of emf is maximum, when the external resistance in the circuit is equal to the internal resistance of the source.
Bihar Board 12th Physics Model Question Paper 5 in English Medium 4
Bihar Board 12th Physics Model Question Paper 5 in English Medium 5

Bihar Board 12th Physics Model Question Paper 5 in English Medium

This is maximum power theorem.

Question 4.
Establish the relation b/w current density and drift velocity.
Answer:
Bihar Board 12th Physics Model Question Paper 5 in English Medium 6

Question 5.
What is Hysteresis loop? With its help explain the terms retentivity and co-ercivity.
Answer:
Hysteresis loop : Hysteresis loop is defined as the lagging of the magnetic induction B behind the corresponding magnetic field H.

When magnetic field H and the magnetic induction B are yarned from zero to maximum in one direction and then back through zero to a maximum in the opposite direction and finally back again through zero to the first maximum, a cycle of magnetisation is said to be completed. B(T) Y Using the value of H and B obtained above, we plot a graph by taking H along x-axis and B along y-axis. When the magnetic field H is increased from zero to the maximum value Og, the induction B follows the curve Oa. However, when the magnetic field is brought to zero, the.;induction does not follows back the curve Oa but a different curve ab. The same pattern of variations is seen for the rest of the curve. The B-H curve for decreasing H does not coincide with the B-H curve for increasing H.
Bihar Board 12th Physics Model Question Paper 5 in English Medium 7

As is clear from figure, even when the magnetising field H is zero, some magnetic induction (Ob or Oe) is still present in the speci nen. It is denoted by Br and is called remanance or retent i vity or residual magnetism.

The value of the intensity of magnetisation of a material when the magnetisi ng field is reduced to zero is called retentivity.

After the specimen has been magnetised to saturation (oh or oj) a reversed magnetising field (equal to Oc or Of) is required to reduce the magnetic induction to zero.This is called the coercivity or coercive force He. The value of the magnetising field required to reduce residual magnetism to zero is called coercivity of the material.

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 6.
Write the expression for energy stored in the
Answer:
The induced emf in coil \(e=-L \frac{d I}{d t}\)
let chasge dq passes from coil, the work done dw = e.dq
\(\begin{aligned}
&=\mathrm{L} \cdot \frac{d I}{d t} d q=\mathrm{L} \cdot d I \cdot \frac{d q}{d t} \\
d w &=\mathrm{LI} . \mathrm{dI}
\end{aligned}\)

work done after posting current O to 1
\(\omega=\mathrm{L} \int_{0}^{1} I d I=\mathrm{L}\left[\frac{I^{2}}{2}\right]_{0}^{1} ; \omega=\frac{1}{2} \mathrm{L}^{2}\)

This work done is stored in the potential energy of coil.
\(P \cdot E=\frac{1}{2} L I^{2}\)

Question 7.
What is current sensitivity and voltage sensitivity of moving coil galvanometer?
Answer:
Current sensitivity : It is defined as the deflection produced in the galvanometer, when a unit voltage is applied across its cops.
\(\frac{\alpha}{\mathrm{I}}=\frac{\mathrm{nBA}}{\mathrm{K}}\)

Voltage sensitivity : It is defined as the deflection produced in the galvanometer, when a unit voltage is applied across its COPI.
\(\frac{\alpha}{V}=\frac{n B A}{K R}\)

Question 8.
Write the expression for energy density related with electro magnetic wave and show that its ratio is equal to 1.
Answer:
Bihar Board 12th Physics Model Question Paper 5 in English Medium 9
Bihar Board 12th Physics Model Question Paper 5 in English Medium 10

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 9.
Sketch the wave front corresponding to (a)
diverging rays and (b) converging rays.
Answer:
Bihar Board 12th Physics Model Question Paper 5 in English Medium 11

Question 10.
Write the differences between magnification and magnifying power.
Answer:
Magnification :

  • It is a linear magnification which is equal to \(\frac{n_{2}}{n_{1}}\).
  • Its value increases with increase in V.
  • It may be between to -∞ to +∞.
  • Under certain conditions it is equal to magnifying power.

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Magnifying power :
(i) It is an angular magnification which is equal to \(\frac{\angle \beta}{\angle \alpha}\)
(ii) Its value decreases with increase in V.
(iii) Its value may be b/w \(\frac{d}{f}\) and \(\left(1+\frac{D}{f}\right)\)
(iv) It is a special condition of magnification when (ve = D

Question 11.
Derive the expression for de-Broglie wave length.
Answer:
A/C to quantum theory E = hv ……………………….. (i)
Bihar Board 12th Physics Model Question Paper 5 in English Medium 12
Bihar Board 12th Physics Model Question Paper 5 in English Medium 13

λ is called de-Broglie wavelength.

Question 12.
What is mass defect?
Answer:
Mass defect : The difference b/w sum of the masses of neutrons and protons forming a nucleus and actual mass of the nucleus is called mass defect.
Δm = [mpz + mn (A – Z) – M]
The unit of mass defect is amu.

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 13.
Why does thermionic emission takes place from metal surface only?
Answer:
Thermionic emission takes place from metal surface only because metal has free electrons on its surface.This free electrons requires less energy to come out from one surface. Also sometimes, to reduce work function of metal by coating oxide layer on metal.

Question 14.
Why is semi conductor damaged by a strong current?
Answer:
A strong current, when passed through a semi conductor, heat up the semi conductor and covalent bond break up. It results in a large number of free electrons. The material behaves as a conductor. As now the semi conductor no longer possesses the property of conduction, it is said to be damaged.

Question 15.
Write the differences between Intrinsic and extrinsic semi conductor.
Answer:
Intrinsic semi conductor :

  • They are the crystals of pure elements like Germanium and Silicon.
  • ne = nh
  • The electrical conductivity of Intrinsic semi conductor is low.
  • Resistivity is higher.

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Extrinsic semi conductor:

  • When some impurity is added in Intrinsic semi conductor, we get an extrinsic semi conductor.
  • ne ≠ nh
  • Its electrical conductivity is high.
  • Resistivity is lower.

Question 16.
Define mean value and root square value A.C. Write its expression.
Answer:
Mean value of Average value : It is that steady current when passes through the circuit for half time period, sends the same amount of charge of the alternating current, sends the charge in same circuit and in same time period. It is denoted by Im.
\(\begin{aligned}
&I_{m}=\frac{2 I_{0}}{\pi}\\
&I_{m}=0.636 I_{0}
\end{aligned}\)

R.M.S. value or virtual value: It is that steady current when passes through the circuit for half time period, produces the same amount of heat as the alternating current produces the heat in same time and in same circuit. It is denoted by Irms.
\(\begin{aligned}
&I_{r m s}=\frac{I_{0}}{\sqrt{2}}\\
&I_{r m s}=0.7071
\end{aligned}\)

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 17.
What is polaroids? Write its important uses.
Answer:
It is a device which produces plane polarised light. Use of polaroids :

  • Sun glasses fitted with polaroid sheets protect the eye from glass.
  • Wind shields of automobiles are also made of Polaroid sheets.
  • They are useful in three dimensional motion pictures.
  • It is used in L.C.D., calculator, watches, T.V. etc.

Question 18.
Write the condition for sustained interference.
Answer:
Condition for sustained interference :

  • The two source should emf the light wave continuously.
  • The light wave should be of same wave length.
  • The two light source should be narrow.
  • The light waves emitted should be of the same amplitude.
  • The two source of light must lie very close to each other.

Long Answer Type Questions

Question No. 19 to 24 are long answers type question. Each question of this category carries 5 marks. Answer any three (3) question. (3 x 5 = 15)

Question 19.
Explain any two of the following terms :
(i) Ground Waves
(ii) Sky Waves
(iii) Space Waves.
Answer:
(i) Ground Waves : The ground waves which progress along the surface of the earth are called ground waves or surface waves. The ground propagation is suitable for low and medium frequency, i.e., up to 2 MHz only, hence it is also called medium wave propagation: The maximum range of ground or surface wave propagation depends on :

Bihar Board 12th Physics Model Question Paper 5 in English Medium

  • The frequency of the radio waves and
  • Power of the transmitter.

(ii) The Sky Waves : The sky waves are the radio waves of frequency between 2 MHz to 30 MHz. The ionosphere reflects those radio waves so that they can propagate through atmosphere. The sky waves propagation is also known as ionosphere propagation.

The sky waves are used for very long distance radio communication at medium high frequencies (i.e. at medium waves and short waves). The radio waves can cover a distance approximately 400 km in a single reflection from the ionosphere.

(iii) Space Waves : The space waves are the radio waves of very high frequency (i.e. between 30 MHz to 300 MHz or more). The space waves can travel through atmosphere from transmitter antenna to receive antenna either directly or after reflection from ground in the earth’s troposphere region. It is also called Tropospherical propagation or line of sight propagation. This propagation is limited (a) to the line of sight distance, (b) by the curvature of the earth.

It is utilized in T.V. communication, Radar communication, etc.

Question 20.
Find an expression for the electric field strength at a distant point situated (i) on the axis and (ii) along the equatorial line of an electric dipole.
Answer:
Consider an electric dipole AB. The charges -q and +q of dipole are situated at A and B respectively. The separation between the charges is 21.

Electric dipole moment, p = Q 21
The direction of dipole moment is from -q to + Q

Bihar Board 12th Physics Model Question Paper 5 in English Medium

(i) At axial or end-on position : Consider a point P on the axis of dipole at a distance r from mid-point O of electric dipole.
Bihar Board 12th Physics Model Question Paper 5 in English Medium 14

The distance of point P from charge + q is BP = r – l and distance of point P from charge – q is, AP = r + l. Let E1 and E2 be the electric field strengths at point P due to charges +<7 and -q respectively. We know that the direction of electric field due to a point charge is away from positive charge and towards the negative charge. Therefore
Bihar Board 12th Physics Model Question Paper 5 in English Medium 15

Clearly the directions of electric field strengths \(\overrightarrow{\mathrm{E}}_{1}\) and \(\overrightarrow{\mathrm{E}}_{2}\) are along the same line but opposite to each other and E1 > E2 because positive charge is nearer.

∴ The resultant electric field due to electric dipole has magnitude equal to the difference of E1 and E2 and direction from 13 to P i.e.
Bihar Board 12th Physics Model Question Paper 5 in English Medium 16
Bihar Board 12th Physics Model Question Paper 5 in English Medium 17

Bihar Board 12th Physics Model Question Paper 5 in English Medium

If the dipole is infinitely small and point P is far away from the dipole, then r >> l, therefore equation (i) may be expressed as
\(\mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 p r}{r^{4}} \text { or } \mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{2 p}{r^{3}}\) …………………………… (ii)

This is the expression for the electric field strength at axial position due to a short electric dipole.

Question 21.
Explain the principle and working of a Cyclotron with the help of a neat diagram. Write the expression for Cyclotron frequency.
Or,With the help of a labelled diagram, state the underlying principle of a Cyclotron. Explain clearly how it works to accelerate the charged particles.
Bihar Board 12th Physics Model Question Paper 5 in English Medium 18

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Show that Cyclotron frequency is independent of energy of the particle. Is there an upper limit on the energy acquired by the particle? Give reason.
Answer:
(a) Cyclotron : The cyclotron, devised by Lawrence and Livingston, is a device for accelerating ions to high speed by the repeated application of accelerating potentials.

Construction:
The cyclotron consists of two flat semi-circular metal boxes called ‘dees’ and are arranged with a small gap between them. A source of ions is located near the mid-point of the gap between the dees (fig). The dees are connected to the terminals of a radio frequency oscillator, so that a high frequency alternating potential of several million cycles per second exists between the dees. Thus dees act as electrodes. The dees are enclosed in an insulated metal box containing gas at low pressure. The whole apparatus is placed between the poles of a strong electro¬magnet which provides a magnetic field perpendicular to the plane of the dees.

Working :
The principle of action of the apparatus is shown in fig.The positive ions produced from a source S at the centre are accelerated by a dees which is at negative potential at that moment. Due to the presence of perpendicular magnetic field the ion will move in a circular path inside the dees. The magnetic field and the frequency of the applied voltages are so chosen that as the ion comes out of a dees, the dees change their polarity (positive becoming negative and vice-versa) and the ion is further accelerated and moves with high velocity along a circular path of greater radius. The phenomenon is continued till the ion reaches at the periphery of the deeS where an auxiliary negative electrode (deflecting plate) deflects the accelerated ion on the target to be bombarded.

Bihar Board 12th Physics Model Question Paper 5 in English Medium

The function of electric field is to accelerae the charged particle and so to impart energy to the charged particle.

The function of magnetic field is to provide circular path to charged particle and so to provide the location where charged particle is capable of gaining energy from electric field.

Expression for Period of Revolution and Frequency :

Suppose the positive ion with charge q moves in a dees with a velocity v, then,
\(q v \mathrm{B}=\frac{m v^{2}}{r} \text { or }=\frac{m v}{q \mathrm{B}}\) …………………………… (i)

where m is the mass and r is the radius of the path of ion in the dees and B is the strength of the magnetic field. The angular velocity to of the ion is given by.
\(\omega=\frac{v}{r}=\frac{q \mathrm{B}}{m}( \text { from eq. }\) ………………… (ii)

The time taken by the ion in describing a semi-circle, i.e., in turning through an angle π is.
\(t=\frac{\pi}{\omega}=\frac{\pi m}{\mathrm{B} q}\)

Thus the time is independent of the speed of the ion i.e., although the speed of the ion goes on increasing with increase in the radius (from eQuestion i) when it moves from one dees to the other, yet it take the same time in each dees.

From eQuestion (iii) it is clear that for a particular ion, \(\frac{m}{q}\)

being known, B can be calculated for producing resonance with the high frequency alternating potential.

(b) Resonance Condition : The condition of working of cyclotron is that the frequency of radio frequency alternating potential must be equal to the frequency of revolution of charged particles within the dees. This is called resonance condition.

Now for the cyclotron to work, the applied alternating potential should also have the same semiperiodic time (T/2) as that taken by the ion to cross either dees, i.e.,

This is the expression for period of revolution. Obviously, period of revolution is independent of speed of charged particle and radius of circular path.

Bihar Board 12th Physics Model Question Paper 5 in English Medium

∴ Frequency of revolution of particles.
\(f=\frac{1}{T}=\frac{q B}{2 \pi m}\)

This frequency is called the cyclotron frequency. Clearly the cyclotron frequency is independent of speed of particle.

(c) Expression for K.E. attained : If R be the radius of the path and Vmax, the velocity of the ion when it leaves the periphery, then in accordance with eq (ii)
\(v_{\max }=\frac{q \mathrm{BR}}{m}\) …………………………….. (vi)

The kinetic energy of the ion when it leaves the apparatus is,
\(\text { K.E. }=\frac{1}{2} m v^{2} \max =\frac{q^{2} B^{2} R^{2}}{2 m}\) ………………….. (vii)

When charged particle crosses the gap between dees it gains KE = qV

In one revolution, it crosses the gap twice, therefore if it completes ^-revolutions before emerging the dees, the kinetic energy gained = 2nqV ………………. (viii)
\(\text { Thus, K.E. }=\frac{g^{2} \mathrm{B}^{2} \mathrm{R}^{2}}{2 m}=2 n q \mathrm{V}\)

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 22.
Using Ampere’s circuital law, derive an expression for the magnetic field along the axis of a toroidal solenoid.
Answer:
Magnetic field due to a toroidal solenoid : A long solenoid shaped in the form of closed ring is called toroidal solenoid (or endless solenoid). Let n be the number of turns per unit of toroid and I the current flowing through it. The current causes the magnetic field inside the turns of the solenoid. The magnetic lines of force inside the toroid are in the form of concentric circle. By symmetry of the magnetic field it has the same magnitude at each point of circle and is along the tangent at every point on the circle.
Bihar Board 12th Physics Model Question Paper 5 in English Medium 19

For points inside the core of toroid: Consider a circle of radius r in the region enclosed by turns of toroid. Now we apply Ampere’s circuital law to this circular path, i.e.,
\(\vec{f} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=\mu_{0} I\) ………………. (i)

\(\vec{f} \overrightarrow{\mathrm{B}} \cdot \overrightarrow{d l}=f B d l \cos 0=B \cdot 2 \pi r\) Length of toroid = 2πr Number of turns in toroid = n (2πr) and current in one- turn = I.

∴ Current enclosed by circular path = (n 2πr) • 1
∴ Equation (i) gives
\(\mathrm{B} 2 \pi r=\mathrm{m}_{0}(n 2 \pi r l) \Rightarrow \mathrm{B}=\mu_{0} n \mathrm{l}\)

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Derivation of formula for magnetic field due to a current carrying wire using Biot-Savart law : Consider a wire EF carrying current I in upward direction. The point of observation is P at a finite distance R from the wire. If PM is perpendicular dropped from P on wire; then PM = R. The wire may be supposed to be formed of a large number of small current elements. Consider a small element CD of length δf at a distance f from M.
Bihar Board 12th Physics Model Question Paper 5 in English Medium 20

\(\text { Let } \angle \mathrm{CPM}=\phi \text { and } \angle \mathrm{CPD}=\delta \phi, \angle \mathrm{PDM}=\theta\) The length δf is very small, so that ∠PCM may also be taken as equal to θ.

The perpendicular dropped from C on PD is CN.

The angle formed between element I \(\overrightarrow{\delta l}\) and \(\overrightarrow{r}\) (= cp is (π – θ), Therefore according to Biot-Savart law, the magnetic field due to current element I \(\overrightarrow{\delta l}\) at P is
Bihar Board 12th Physics Model Question Paper 5 in English Medium 21
Bihar Board 12th Physics Model Question Paper 5 in English Medium 22

Bihar Board 12th Physics Model Question Paper 5 in English Medium

If the wire is of finite length and its ends make angles a and p with line MP, then net magnetic field (B) at P is obtained by summing over magnetic fields due to all current elements, i.e.,
Bihar Board 12th Physics Model Question Paper 5 in English Medium 23

This is expression for magnetic field due to current carrying wire of finite length (or very long), then α = β ⇒ π/2.
Bihar Board 12th Physics Model Question Paper 5 in English Medium 24

Question 23.
Draw the labelled ray diagram for the formation of image by a compound microscope.
Derive an expression for its total magnification (or magnifying power), when the find image is formed at the near point.
Why both objective and eye piece of a compound microscope must have their short focal lengths?
Answer:
Compound Microscope :
It consists of a long cylindrical tube, containing at one end of a convex lens of small aperture and small focal length. This is called the objective lens (O). At the other end of the tube another co¬axial smaller and wide tube is fitted which carries a convex lens (E) at its outer end. This lens is towards the eye and is called the eye-piece. The focal length and aperture of eye piece are somewhat large than those of objective lens. Cross¬wires are mounted at a definite distance before the eye piece. The entire tube can be moved forward and backward by the rack and pinion arrangement.

Bihar Board 12th Physics Model Question Paper 5 in English Medium
Bihar Board 12th Physics Model Question Paper 5 in English Medium 25

Magnifying power of a microscope:
is defined as the ratio of angle ((3) subtended by final image on the eye to the angle (a) subtended by the object on eye, when the object is placed at the least distance of distinct vision, i.e.,
Bihar Board 12th Physics Model Question Paper 5 in English Medium 26

Magnifying power \(M=\frac{\beta}{\alpha}\)

As object is very small, angles α and β are very small and so tan α = α and tan β = β. By  efinition the object AB is placed at the least distance of distinct vision
Bihar Board 12th Physics Model Question Paper 5 in English Medium 27
Bihar Board 12th Physics Model Question Paper 5 in English Medium 28

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Question 24.
State the principle of working of p-n diode as a rectifier. Explain with the help of a circuit diagram, the use of p-n diode as full wave rectifier. Draw a sketch of the input and output wave forms. Or, Draw a circuit diagram of a full wave rectifier. Explain the working principle. Draw the input/output wave forms indicating clearly the functions of the two diodes used.
Answer:
Rectification :
Rectification means conversion of a.c. into d.c. A p-n diode acts as a rectifier because an ac changes polarity periodically and a p-n diode conducts only when it is forward biased; it does not conduct when reverse biased.
Bihar Board 12th Physics Model Question Paper 5 in English Medium 29

In a full wave rectifier, if input frequency is / hertz, then output frequency will be 2/ hertz because for each cycle of input, two positive half cycles of output are obtained.

Working:
The AC input voltage across secondary s1 and s2 cli anges polarity after each half cycle. Suppose during the first half cycle of input AC signal, the terminal s1 is positive relative to centre tap O and s2 is negative relative to O. Then diode D1 is forward biased and diode D2 is reverse biased. Therefore diode D1 conducts while diode D2 does not. The direction of current (i1) due to diode D1 in load rsistance RL is directed from A to 13. In next half cycle, the terminal s is negative and s2 is positive relative to centre tap O.

Bihar Board 12th Physics Model Question Paper 5 in English Medium

The diode D is reverse biased and diode D2 is forward biased. Therefore, diode D2 conducts while D1 does not. The direction of current (i2) due to diode D2 in load resistance RL is still from A to B. Thus the current in load resistance RL is in the same direction for both half cycles of input AC voltage. Thus for input AC signal the output current is a continuous series of unidirectional pulses.
Bihar Board 12th Physics Model Question Paper 5 in English Medium 30

In a full wave rectifier, if input frequency is f hertz, then output frequency will be 2f hertz because for each cycle of input, two positive half cycles of output are obtained.

Bihar Board 12th Physics Model Question Paper 5 in English Medium

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Bihar Board 12th Biology Objective Questions and Answers

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 1.
Single-celled animals are said to be immortal because
(a) they grow indefinitely in size
(b) they can tolerate any degree of change in temperature
(c) they can reproduce throughout their life span
(d) they continue to live as their daughter cells.
Answer:
(d) they continue to live as their daughter cells.

Question 2.
Which of the following has the longest life span?
(a) Banyan tree
(b) tortoise
(c) parrot
(d) Elephant
Answer:
(a) Banyan tree

Question 3.
Select the option which arranges the given organisms in ascending order of their life span.
(a) Parrot < Crow < Butterfly < Banyan tree
(b) Butterfly < Crow < Parrot < Crocodile
(c) Fruit fly < Crocodile < Parrot < Banyan tree
(d) Parrot < Tortoise < Dog < Crow
Answer:
(c) Fruit fly < Crocodile < Parrot < Banyan tree

Question 4.
________ is a life process that is not essential for an individual’s survival but for survival of the species.
(a) Growth
(b) Reproduction
(c) Respiration
(d) Nutrition
Answer:
(b) Reproduction

Question 5.
‘Clones’ are individuals that have exactly the same
(a) Lifespan
(b) physiology
(c) growth rate
(d) genetic makeup.
Answer:
(d) genetic makeup.

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 6.
Which one of the following processes results in the formation of clone of bacteria?
(a) Regeneration
(b) Budding
(c) Binary fission
(d) Fragmentation
Answer:
(c) Binary fission

Question 7.
Asexual reproduction is seen in members of Kingdom
(a) Monera
(b) Plantae
(c) Animalia
(d) All of these.
Answer:
(d) All of these.

Question 8.
During binary fission in Amoeba which of the following organelles is duplicated?
(a) Plasma membrane
(b) Nucleus
(c) Contractile
(d) All of these
Answer:
(b) Nucleus

Question 9.
Vegetative propagation is the term used for
(a) sexual reproduction in animals
(b) sexual reproduction in plants
(c) asexual reproduction in animals
(d) asexual reproduction in plants.
Answer:
(d) asexual reproduction in plants.

Question 10.
Which of the following is not used for vegetative propagation?
(a) Bud
(b) Bulbil
(c) Turion
(d) Antherozoid
Answer:
(d) Antherozoid

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 11.
Identify the given organism and find its maximum life span.
Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms 1
(a) Sparrow, 25 years
(b) Crow, 30 years
(c) Crow, 15 years
(d) Eagle, 40 years
Answer:
(c) Crow, 15 years

Question 12.
Which of the following options shows two plants in which new plantlets arise from the same organ?
(a) Dahlia and ginger
(b) Potato and sweet potato
(c) Dahlia and rose
(d) Potato and sugarcane
Answer:
(d) Potato and sugarcane

Question 13.
Refer to the given figure and identify X in it.
Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms 2
(a) Offset
(b) Eyes
(c) Runner
(d) Bulb
Answer:
(b) Eyes

Question 14.
Fleshy buds produced in the axil of leaves, which grow to form new plants when shed and fall on ground, are called
(a) bulbs
(b) bulbils
(c) tubers
(d) offsets.
Answer:
(b) bulbils

Question 15.
In which one pair, both the plants can be vegetatively propagated by leaf pieces?
(a) Bryophyllum and Kalanchoe
(b) Chrysanthemum and Agave
(c) Agave and Dioscorea
(d) Bryophyllium and Asparagus
Answer:
(a) Bryophyllum and Kalanchoe

Question 16.
Identify the given vegetative propagule.
Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms 3
(a) Bulb
(b) Runner
(c) Rhizome
(d) Bulbil
Answer:
(d) Bulbil

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 17.
If a leaf cell of Agave has x chromosomes then what will
be the number of chromosomes in a cell of its bulbil?
(a) 2 x
(b) x/2
(c) x/4
(d) x
Answer:
(d) x

Question 18.
Which of the following cannot serve as a vegetative propagule?
(a) A piece of potato tuber with eyes
(b) A middle piece of sugarcane internode
(c) A piece of ginger rhizome
(d) A marginal piece of Bryophyllum leaf
Answer:
(b) A middle piece of sugarcane internode

Question 19.
Which of the following options correctly identifies artificial and natural methods of vegetative propagation?
Artificial methods – Natural methods
(a) Grafting – Cutting
(b) Layering – Bulbils
(c) Offset – Tissue culture
(d) Tubers – Rhizomes
Answer:
(b) Layering – Bulbils

Question 20.
Sexual reproduction is considered more beneficial than asexual reproduction because
(a) it is not affected by adverse environmental conditions
(b) fertilization is a chance factor
(c) it rapidly multiplies the population
(d) it assists in evolution by producing variations.
Answer:
(d) it assists in evolution by producing variations.

Question 21.
The growth phase of an organism before attaining sexual maturity is referred to as
(a) juvenile phase
(b) vegetative phase
(c) both (a) and (b)
(d) none of these.
Answer:
(c) both (a) and (b)

Question 22.
Select the monocarpic plant out of the following.
(a) Bamboo
(b) Lite hi
(c) Mango
(d) All of these
Answer:
(a) Bamboo

Question 23.
Clear cut vegetative, reproductive and senescent phases cannot be observed in
(a) annual plants
(b) perennial plants
(c) biennial plants
(d) ephemeral plants.
Answer:
(b) perennial plants

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 24.
Strobilanthus kunthiana flowers once in
(a) 5 years
(b) 12 years
(c) 20 years
(d) 50 years.
Answer:
(b) 12 years

Question 25.
Strobilanthus kunthiana differs from bamboo in
(a) being monocarpic
(b) length of juvenile phase
(c) being polycarpic
(d) none of these.
Answer:
(b) length of juvenile phase

Question 26.
Oestrous cycle is reported in
(a) cows and sheep
(b) humans and monkeys
(c) chimpanzees and gorillas
(d) none of these.
Answer:
(a) cows and sheep

Question 27.
Which of the following animals show menstrual cycle ?
(a) Gorillas and chimpanzees
(b) Monkeys and humans
(c) Orangutans and monkeys
(d) All of these
Answer:
(d) All of these

Question 28.
Senescent phase of an organism’s life span can be recognised by
(a) slow metabolism
(b) cessation of reproduction
(c) decreased immunity
(d) all of these
Answer:
(d) all of these

Question 29.
If a fungal thallus has both male and female reproductive structures, it will be called
(a) heterothallic
(b) homothallic
(c) dioecious
(d) monoecious
Answer:
(b) homothallic

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 30.
Staminate flowers produce
(a) eggs
(b) antherozoids
(c) fruits
(d) all of these
Answer:
(b) antherozoids

Question 31.
Which of the following is a unisexual organisam?
Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms 4
Answer:
Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms 5

Question 32.
Which of the following groups is formed only of the hermaphrodite organisms?
(a) Earthworm, tapeworm, housefly, frog
(b) Earthworm, tapeworm, sea horse, housefly
(c) Earthworm, leech, sponge, roundworm
(d) Earthworm, tapeworm, leech, sponge
Answer:
(d) Earthworm, tapeworm, leech, sponge

Question 33.
Which of the following options shows bisexual animals only?
(a) Amoeba, sponge, leech
(b) Sponge, cockroach, Amoeba
(c) Earthworm, sponge, leech
(d) Tapeworm, earthworm, honeybee
Answer:
(c) Earthworm, sponge, leech

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 34.
Read the following statements and select the incorrect one.
(a) Cucurbits and coconuts are monoecious plants.
(b) Papayas and date palms are dioecious plants.
(c) Leeches and tapeworms are bisexual animals.
(d) Sponges and coelenterates are unisexual animals.
Answer:
(d) Sponges and coelenterates are unisexual animals.

Question 35.
Meiosis does not occur in
(a) asexually reproducing diploid individuals
(b) sexually reproducing haploid individuals
(c) sexually reproducing diploid individuals
(d) all of these.
Answer:
(a) asexually reproducing diploid individuals

Question 36.
A diploid parent plant body produces ________ gametes and a haploid parent plant body produces ________ gametes.
(a) diploid, haploid
(b) haploid, diploid
(c) diploid, diploid
(d) haploid, haploid
Answer:
(d) haploid, haploid

Question 37.
Which of the following organisms has the highest number of chromosomes?
(a) Housefly
(b) Butterfly
(c) Ophioglossum
(d) Onion
Answer:
(c) Ophioglossum

Question 38.
In maize, a meiocyte has 20 chromosomes. What will be the number of chromosomes in its somatic cell?
(a) 40
(b) 30
(c) 20
(d) 10
Answer:
(c) 20

Question 39.
If a butterfly has chromosome number 360 in its meiocyte (2n). What will be the chromosome number in its gametes?
(a) 380
(b) 190
(c) 95
(d) 760
Answer:
(b) 190

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 40.
In flowering plants, both male and female gametes are non-motile. The method to bring them together for fertilisation is
(a) water
(b) air
(c) pollination
(d) apomixis
Answer:
(c) pollination

Question 41.
Development of new individual from female gamete without fertilisation is termed as
(a) syngamy
(b) embryogenesis
(c) oogamy
(d) parthenogenesis.
Answer:
(d) parthenogenesis.

Question 42.
Fertilisation cannot occur in absence of surface water in
(a) Fucus
(b) Funaria
(c) Marsilea
(d) all of these.
Answer:
(d) all of these.

Question 43.
Spirogyra is a sexually reproducing alga in which vegetative thallus is haploid. In Spirogyra, meiosis
(a) never occurs
(b) occurs at time of gamete production
(c) occurs after fertilisation
(d) occurs during vegetative growth.
Answer:
(c) occurs after fertilisation

Question 44.
Life begin in all sexually reproducing organisms as a
(a) single-celled zygote
(b) double-celled zygote
(c) haploid zygote
(d) haploid gametes.
Answer:
(a) single-celled zygote

Question 45.
Which of the following is not correct regarding sexual reproduction ?
(a) It is usually biparental.
(b) Gametes are always formed.
(c) It is a slow process
(d) It involves only mitosis.
Answer:
(a) It is usually biparental.

Question 46.
Offsprings of oviparous animals are at greater risk of survival as compared to those of viviparous animals because
(a) proper embryonic care and protection is absent
(b) embryo does not develop completely
(c) progenies are of smaller size
(d) genetic variations do not occur.
Answer:
(a) proper embryonic care and protection is absent

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 47.
Deposition of calcareous shell around zygote occurs in
(a) birds and reptiles
(b) birds and mammals
(c) mammals and reptiles
(d) all of these.
Answer:
(a) birds and reptiles

Question 48.
Select the option which shows viviparous animals only,
(a) Lizard, Turtile
(b) Platypus, Crocodile
(c) Cow, Crocodile
(d) Whale, Mouse
Answer:
(d) Whale, Mouse

Question 49.
Which of the following animals give birth to young ones?
(a) Ornithorhynchus and Echidna
(b) Macropus and Pteropus
(c) Balaenoptera and Homo sapiens
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)

Question 50.
Viviparity is found in
(a) Sharks
(b) lizards
(c) frogs
(d) birds
Answer:
(a) Sharks

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 51.
In which of the following plants, sepals do not fall off after fertilisation and remain attached to the fruit?
(a) Brinjal
(b) Cucumber
(c) Papaya
(d) Bitter gourd
Answer:
(a) Brinjal

Question 52.
Which of the labelled parts in the transverse section of tomato fruit, is/are diploid?
Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms 6
(a) X
(b) Y
(c) Both X and Y
(d) None of these
Answer:
(c) Both X and Y

Question 53.
The wall of the ovary forms
(a) pericarp
(b) fruit wall
(c) fruit
(d) both (a) and (b).
Answer:
(d) both (a) and (b).

Question 54.
The term ‘clone’ cannot be applied to offspring formed by sexual reproduction because
(a) offspring do not possess exact copies of parental DNA
(b) DNA of only one parent is copied and passed on to the offspring
(c) offspring are formed at different times
(d) DNA of parent and offspring are completely different
Answer:
(a) offspring do not possess exact copies of parental DNA

Question 55.
The male gametes of rice plant have 12 chromosomes in their nucleus. The chromosome number in the female gamete, zygote and the cells of the seedling will be, respectively
(a) 12,24,12
(b) 24,12,12
(c) 12, 24, 24
(d) 24, 12, 24.
Answer:
(c) 12, 24, 24

Question 56.
Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because
(a) nodes are shorter than intemodes
(b) nodes have meristematic cells
(c) nodes are located near the soil
(d) nodes have non-photosynthetic cells.
Answer:
(b) nodes have meristematic cells

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Question 57.
There is no natural death in single celled organisms like Amoeba and bacteria because
(a) they cannot reproduce sexually
(b) they reproduce be binary fission
(c) parental body is distributed among the offspring
(d) they are microscopic.
Answer:
(c) parental body is distributed among the offspring

Question 58.
There are various types of reproduction. The type of reproduction adopted by an organism depends on
(a) the habitat and morphology of the organism
(b) morphology of the organism
(c) morphology and physiology of the organism
(d) the organisms habitat, physiology and genetic makeup.
Answer:
(d) the organisms habitat, physiology and genetic makeup.

Question 59.
Which of the following is a post-fertilisation event in flowering plants?
(a) Transfer of pollen grains
(b) Embryo development
(c) Formation of flower
(d) Formation of pollen grains
Answer:
(b) Embryo development

Question 60.
The number of chromosomes in the shoot tip cells of a maize plant is 20. The number of chromosomes in the microspore mother cells of the same plant shall be
(a) 20
(b) 10
(c) 40
(d) 15
Answer:
(a) 20

Bihar Board 12th Biology Objective Answers Chapter 1 Reproduction in Organisms

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Bihar Board 12th Biology Model Papers

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Time : 3 Hours 15 Min
Total Marks : 70

Instructions for the candidates:

  1. Candidates are required to give answers in their own words as far as practicable.
  2. Figures in the right-hand margin indicate full marks.
  3. While answering the questions, the candidate should adhere to the word limit as for as practicable.
  4. 15 Minutes of extra time has been allotted for the candidates to read the questions carefully.
  5. This question paper is divided into two sections: Section – A and Section – B
  6. In Section – A, there are 35 objective type questions which are compulsory, each carrying 1 mark. Darken the circle with blue/black ball pen against the correct option on OMR Sheet provided to you. Do not use Whitener/Liquid/ Blade/Nail on OMR Sheet otherwise result will be treated as invalid.
  7. In Section – B, there are Non-objective type questions. There are 18 Short answer type questions, out of which any 10 questions are to be answered. Each question carries 2 marks. Apart from this, there are 6 long answer type questions, out of which any 3 of them are to be answered. Each question carries 5 marks
  8. Use of any electronic device is prohibited.

Objective Type Questions

There are 1 to 35 objective type questions with 4 options. Choose the correct option which is to be answered on OMR Sheet. (35 x 1 = 35)

Question 1.
Pollination by birds is called
(a) Entomophily
(b) Cheiropterophily
(c) Ornithophily
(d) All of the above
Answer:
(c) Ornithophily

Question 2.
Which one is the hotspot of biodiversity?
(a) Eastern Ghat
(b) Western Ghat
(c) Aravalli Hills
(d) None of the above
Answer:
(b) Western Ghat

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Question 3.
Who is considered as Father of Genetics?
(a) Morgan
(b) Darwin
(e) Hugo-de-Vries
(d) Mendel
Answer:
(d) Mendel

Question 4.
Noise pollution is measured in
(a) Heartz
(b) Fathoms
(c) Decibels
(d) Nanometer
Answer:
(c) Decibels

Question 5.
Indicator of water pollution is
(a) Entamoeba histolytica
(b) E. Coli
(c) Vibrio cholera
(d) All of the above
Answer:
(b) E. Coli

Question 6.
mRNA formation takes place in
(a) Ribosome
(b) Nucleus
(c) Cytoplasm
(d) Mitochondria
Answer:
(b) Nucleus

Question 7.
Cancer affecting connective tissue is called
(a) Lymphoma
(b) Sarcoma
(c) Carcinoma
(d) None of the above
Answer:
(b) Sarcoma

Question 8.
IgM antibody is found in
(a) Serum
(b) Saliva
(c) Blood
(d) All of the above
Answer:
(c) Blood

Question 9.
How many Peptide chains are found in each antibody molecule?
(a) 4
(b) 3
(c) 2
(d) None of the above
Answer:
(a) 4

Question 10.
Green plants are known as producers, which convert
(a) Chemical energy into light energy.
(b) Light energy into chemical energy
(c) All of the above
(d) None of the above
Answer:
(b) Light energy into chemical energy

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Question 11.
Which of the following cry gene prevents crop from borer
(a) Cry AC
(b) Cry / Ab
(c) Cry // Ab
(d) All of the above
Answer:
(d) All of the above

Question 12.
Human blood groups are
(a) A, B, AB, O
(b) A, B, C, O
(c) B, C, O
(d) None of the above
Answer:
(a) A, B, AB, O

Question 13.
Khorana got Nobel prize for
(a) Discovery of RNA
(b) Discovery of DNA
(c) Chemical synthesis of gene
(d) None ‘
Answer:
(c) Chemical synthesis of gene

Question 14.
Secondary pollutant is
(a) CO
(b) C02
(c) PAN
(d) None of these
Answer:
(c) PAN

Question 15.
Chipko movement is launched for protection of
(a) Grass land
(b) Forest
(c) Animal Community
(d) All of the above
Answer:
(b) Forest

Question 16.
TTxttis
(a) Test cross
(b) Back cross
(c) Hybridization
(d) None of the above
Answer:
(a) Test cross

Question 17.
How many types of DNA polymerases are associated with eukaryotic cell ?
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(d) 5

Question 18.
Example of In Situ conservation is
(a) Zoological garden
(b) Botanical garden
(c) Biological safari park
(d) All of the above
Answer:
(d) All of the above

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Question 19.
Which of the following is diploid ?
(a) Male gamete
(b) Ovum
(c) Zygote
(d) All of the above
Answer:
(c) Zygote

Question 20.
Pattern of any population depends on
(a) Density
(b) Distribution
(c) Specific structure
(d) All of the above
Answer:
(d) All of the above

Question 21.
Montreal Protocol is related to control the deleterious effect of
(a) Acid rain
(b) Ozone depletion
(c) Deforestation
(d) Global warming
Answer:
(b) Ozone depletion

Question 22.
Fertilization takes place in
(a) Ovary
(b) Uterus
(c) Fallopian tube
(d) All of the above
Answer:
(c) Fallopian tube

Question 23.
Hugo de-Vries worked on which plant ?
(a) Pea
(b) Evening Primrose
(c) Both
(d) None
Answer:
(b) Evening Primrose

Question 24.
Salmonella is associated with
(a) Polio
(b) T.B,
(c) Typhoid
(d) All of the above
Answer:
(c) Typhoid

Question 25.
AIDS is identified first of all in
(a) USA
(b) France
(c) Russia
(d) All of the above
Answer:
(a) USA

Question 26.
Bones of fore limbs of Bat, and humans are
(a) Homologous
(b) Analogous
(c) Both
(d) None
Answer:
(a) Homologous

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Question 27.
Which enzyme is used to join cut ends of DNA?
(a) Ligase
(b) Cellulose
(c) Pectinase
(d) None of the above
Answer:
(a) Ligase

Question 28.
Which of the following serve as biofertilizer in paddy fields?
(a) Yeast
(b) Bacteria
(c) Cyano bacteria
(d) Fungi
Answer:
(c) Cyano bacteria

Question 29.
Which of the following is used in preparing Cheese?
(a) Viruses
(b) Algae
(c) Fungi
(d) Micorobes
Answer:
(d) Micorobes

Question 30.
Statins used for lowering blood cholesterol level
are extracted from
(a) Yeast
(b) Algae
(c) Virus
(d) All of the above
Answer:
(a) Yeast

Question 31.
Microbes are found in
(a) air
(b) water
(c) soil
(d) All of the above
Answer:
(d) All of the above

Question 32.
Ranikhet disease is connected with
(a) Pigs
(b) Cow
(c) Fish
(d) Hens
Answer:
(d) Hens

Question 33.
Induced breeding is done in case of
(a) Apiculture
(b) Pisci culture
(c) Lac culture
(d) Sericulture
Answer:
(b) Pisci culture

Question 34.
Taichung Native-1 is a variety of
(a) Rice
(b) Wheat
(c) Maize
(d) None
Answer:
(a) Rice

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Question 35.
Zeba cattle is
(a) Goat
(b) Cat
(c) Cow
(d) Sheep
Answer:
(c) Cow

Non-Objective Type Questions

Short Answer Type Questions

In this section, there are 18 short answer type question (each carryies 2 marks) out of which answers any ten (10) questions. (10 x 2 = 20)

Question 1.
What is co-dominance? Explain with example.
Answer:
In co-dominance both dominant alleles of a gene express themselves equally in F, hybrids. The phenotype ratio matches with genotypic ratio i.e, 1: 2: 1 in F2 generation of offspring. Example of co-dominance can be seen in goat colour in
Bihar Board 12th Biology Model Question Paper 5 in English Medium 1

Question 2.
What is significance of double fertilization?
Answer:
Significance of double fertilization:
(i) Double fertilization is very important in angiospermic plants. If only syngamy occurs in plants and triple fusion does not occur, only 3 zygote will be formed. Endosperm will not be formed and this will result seed with undeveloped embryo or seed without embryo.

(ii) Endosperm is formed due to double fertilization. This provides nutrition to embryo and it contains maternal and paternal chromosomes. A body shows physiological aggressiveness due to hybrid vigour in inner endospermic cells.

Question 3.
What is radio-active pollution? What are the sources of radio-active pollution?
Answer:
Radio-active Pollution: Certain elements such as Radium, Thorium, Uranium emit protons (Alpha particles), electrons (Beta particles) and Gamma particles by disintegration of atomic nuclei. This phenomenon is called radio-activity and these elements are known as radio-active. When the radio-active radiations contaminate to water, air, soil and food materials it is called radio-active pollution.

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Question 4.
Define Commensalism and Parasitism.
Answer:
Commensalism: An interaction between two species in which one is benefitted and the other one is neither harmed nor benefitted.
e.g. Orchids growing on trunk/branch of a tree.

Parasitism: An interaction between two species where parasite not only obtains food from its host but also completes its life cycle on the host. e.g. Bacteria, Fungi etc.

Question 5.
What are the ways for controlling amoebiosis?
Answer:
Amoebiosis is caused by a protozoan parasite Entamoeba histolytica, in which pain in stomach and stool with mucus and blood are seen as specific symptom.
Following are the control major of amoebiosis:

  1. Patient should take rest and after some interval electrolytic fluid must be taken.
  2. For control of disease, emetin, vioform, chinoform and diodoquin should be taken as drug and antibiotics like Terramycine, Erythromycine, Metronidazole and Aureomycin may be helpful in treatment and control of this disease.

Question 6.
What is Jumping Gene (Transposons)?
Answer:
These are capable of changing their locations from one place to another, so it is called jumping genes. It is a mobile fragment of DNA that can change position in a genome, a cause of sequence insertions and deletions in some organisms.

Question 7.
Name four living fossils.
Answer:
A living fossil is a living animal of ancient origin with many primitive characters. A living fossil has been living as such from the time of origin without many changes.
Examples are:

  • Peripatus of phylum Arthropoda.
  • Limulus of phylum Arthropoda.
  • Latimeria-a bony fish.
  • Sphenodon-a reptile.

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Question 8.
Write any four benefits of micro-propagation.
Answer:
Benefits of micro-propagation:

  • By such method crop production increases.
  • By such method storage of germplasm can be taken place.
  • By Meristem tip culture method disease resistant plants are produced.
  • Micropropagation is also beneficial in conservation of biodiversity.

Question 9.
What is fishery? Name few common freshwater and marine edible fishes.
Answer:
Fishery: It is an industry concerned with catching, processing or selling of fish, shell fish or other aquatic animals.
Freshwater edible fishes:

  • Catla
  • Rohu
  • Cirrhina mrigala.

Marine water edible fishes: (a) Hilsa (b) Sardines (c) Mackerel (d) Pomfrets.

Question 10.
Write short notes on gene cloning.
Answer:
The production of living structures genetically identical to their parent structure is known as gene cloning. It is therefore an exact copy of a single living parent. Gene cloning is the act of making copies of single gene. Once a gene is identified clones can be used.

Question 11.
Why is the Human Genome Project called a megaproject?
Answer:
Human Genome Project is called a mega project because it is aimed to sequence every base in human genome. It has developed the ways of mapping the human genome at increasing fine level of precision and to store this information in database and develops tools for data analysis. The magnitude and the requirements for this project are more for opening up new areas and avenues.

Question 12.
What are the functions of t-RNA in the protein synthesis?
Answer:

  1. It carries specific amino acids from the cytoplasm to the ribosomal sites for the formation of polypeptide chain according with the sequence specified by m-RNA.
  2. The t-RNA charged with the amino acid serves as a adaptor molecule to decode the information on the m- RNA.
  3. It comes to elongate the polypeptide chain by the addition of several newly synthesised amino acids.

Question 13.
Define homologous orgAnswer: Give examples.
Answer:
These are the organs of similar structure and origin but dissimilar in functions and form.This is due to common ancestary.
Examples:
(i) The forelimbs of a frog, the wings of a bird, legs of a horse, the hands of a man and the flipper of a whale are homologous organs because all of them have similar pattern of basic plan (pentadacty) i.e. same number of bones, muscles, nerves and blood vessels etc. but they do the different functions such as hopping (frog), flying (bird), running (horse), graping (man) and swimming (whale).

(ii) Phylloclade of Opuntia and Cladode of Ruscus are homologous organs as both are modified stems. Similarly, a thorn of Bougainvillea and a tendril of Cucurbita are homologous as both arise in axillary position.

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Question 14.
What is green revolution? List two factors that have led green revolution in India.
Answer: Green revolution is the enormous increase in the food production due to intensive cultivation by use of improved varieties of seeds and fertilizers.
The factors that contributed to the green revolution in India are:

  • The development and introduction of high yielding varieties of crop and extension of the high yielding varieties over large area in the country.
  • Transfer of technology of scientific farming from research farm to village farmers.

Question 15.
What do you understand by restriction enzyme?
Answer:
Restriction enzymes: These enzymes are nucleases which cut DNA into short pieces containing identifiable genes at specific sites. These pieces are then introduced into plastids, yeasts or plant cells.
Ex. ECORI, Hind-III.

Question 16.
Write short notes on plasmids.
Answer:
A plasmid is small circular double stranded DNA molecule that is distinct from a cell, chromosomal DNA.
Plasmids naturally exist in bacterial cells and they also occur in some eukaryotes. Often, the genes carried in plasmid provide bacteria with genetic advantage, such as antibiotic resistance.

Question 17.
Write short notes on organic evolution.
Answer:
The process by which changes in the genetic composition of populations of organism occur in response to environmental changes is known as organic evolution. Examples of organic evolution include organs that are different in structure but perform similar function. A Dolphin’s flipper’ and human arms are example of organic evolution.

Question 18.
Discuss the concept of IPM.
Answer:
Integrated Pest Management (IPM) aims at minimum use of pesticides to prevent agro-chemical pollution and to adopt natural methods of pest control as far as possible. The natural methods of pest control are the part of a larger agricultural strategy and these include:

  1. Use of resistant varieties of crop to local pests.
  2. To carry the practice of crop rotation and improved sanitation method.
  3. To adopt biological control method and starvation method.
  4. To grow planting or a preferred target crop to lure away insects.
  5. Use of sterilization strategy, mechanical control, repellents, chemical attractants and natural insecticides etc.

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Long Answer Type Questions

There are 6 long answers type question (each carrying 05 marks) out of which answers any three questions. (5 x 3 = 15)

Question 19.
Discuss the characters chosen by Mendel in garden pea. Mention the advantages of selecting this plant for experiment.
Answer:
Gregor Mendel, conducted hybridisation experiments on garden peas for seven years (1856-1863) and proposed the laws of inheritance. During Mendel’s investigations into inheritance patterns it was for the first time that statistical analysis and mathematical logic were applied to problems in biology. His experiments had a large sampling size, which have greater credibility to the data

that he collected. Also, the confirmation of his inferences from experiments on successive generations of his test plants, proved that his results pointed to general rules of inheritance rather than being unsubstantiated ideas. Mendel selected only pea plants for his experiment due to presence of many types of contrasting characters like:

  • Round and wrinkled seeds
  • Yellow and green seeds
  • Violet and white flowers
  • Inflated and constricted pods
  • Green and yellow pods
  • Axillary and terminal flowers
  • Tall and dwarf plants

Mendel investigated characters in the garden pea plants that were manifested as two opposite traits which allowed him to set up a basic frame work of rules governing inheritance. Mendel conducted such artificial pollination/ cross pollination experiments using several true breeding pea lines.

Advantages of selecting pea plants for experiment by Mendel:

  1. Due to short height of pea plants, it was easy to do experiment on that.
  2. Pea plants had a short life cycle so that results could be had within a year.
  3. Pea plants produced many seeds in one generation which helped in drawing correct conclusions.
  4. Artificial cross-pollination could be easily achieved because androecium and gynoecium were completely enclosed.
  5. Pea plants having each of the seven characters he selected were readily available.
  6. Mendel took only one character at one time for his experiment.
  7. Pea plants are self pollinated and pure.

Question 20.
Give an accout of Neo-Darwinism/Modern concept of evolution.
Answer:
Neo-Darwinism is the improvement of Darwin’s orginal theory of Natural Selection in order to remove its defects.
Important mechanisms for Neo-Darwinism are:
(a) Variations (b) Natural selection (c) Isolation

(a) Variations: Variations which influence evolution are the ones that develop in the genetic machinery of the germ cells. They are formed due to Mutations, Recombinations, Migrations, Hybridisation and Genetic Drift.

(b) Natural Selection: It operates through differential reproduction. Individuals with advantageous variations survives, reach maturity and leave there genes in the gene pool through differential reproduction. Repeated reproductive selection of individuals with specific traits increases the frequency of their genes in the gene pool. Individuals with harmful neutral variations are eliminated at various stages.

(c) Isolation: It is the separation of a segment of population from the rest due to apearance of a barrier. Gene pool of the isolated population changes due to genetic drift and accumulation of different types of reproduction.

Question 21.
Give a labelled diagram of the electron microscopic view of a human sperm and describe it briefly.
Answer:
The human sperm is a microscopic, long, flagellated and haploid motile cell and is formed of head, neck, middle piece and tail.

(i) Head: It constitutes the anterior region of sperm containing nucleus and acrosome. The nucleus is a narrow, flat and oval structure consisting of densely packed nuclear chromatin material. The latter is formed of DNA and nucleo protein. A small pointed sheathy acrosome is found at the anterior region of head and is derived from the golgi bodies. It contains hydrolytic enzyme-hyalourinidase which helps the sperm to penetrate into ovum during fertilization by dispersing the cells of corona radiata.

Bihar Board 12th Biology Model Question Paper 5 in English Medium 2

(ii) Neck: It is very short or indistinguishable region lying in between head and middle piece. It contains the proximal and distal centriole with 9 + 0 arrangement of micro-tubules. The proximal centriole forms the spindle in the first cleavage divison of the fertilised ovum. The distal centriole gives rise to many fine micro-tubules which run upto sperm tail by passing through middle piece.

(iii) Middle piece: It is a cylindrical region lying behind the neck. It contains number of mitochondrial spiral (nebenkern) which encircles the axial fdament arising from distal centriole. The mitochondrial spiral contains the oxidative enzymes which provide energy for movement of sperm by the process called oxidative phosphorylations.

(iv) Tail: It is the longest part of the sperm and is formed of main central axial filament and the outer protoplasmic sheath with small amount of cytoplasm. At the posterior end of tail, the axial filament is naked without any sheath and is called as end piece. The tail undulates rapidly and provides mobility to the sperm with the head forward in the fluid medium.

Bihar Board 12th Biology Model Question Paper 5 in English Medium

Question 22.
What is deforestation? Write its causes and effects.
Answer:
Deforestation is the conversion of forest areas to non-forest ones. It is a serious threat to the quality of life, national economy and future of environment. According to an estimate, almost 40>percent forests have been lost in the tropics, compared to only one percent in the temperate region. The present-scenario of deforestation is particularly grim in India.

The main cause of decrease in forest cover in India is deforestation due to expansion of agriculture, urbanization, industrialization, excessive commercial use of timber, fuel, wood etc. The causes of deforestation are population explosion, fires, pests, grazing and gnawing mammals, weather, jhum cultivation, etc.

Effects of deforestation are as follows:

  1. Soil erosion has increased;
  2. Flood and droughts have become more frequent;
  3. Pattern of rain fall is changing;
  4. Land slides and avalanches are on the increase;
  5. Climate has become warmer;
  6. Forest dwilling species are becoming extinct;
  7. Consumption of CO2 and production of O2 are getting adversely affected.

Question 23.
Give an account of Apiculture.
Answer:
The maintenance of hives of honey bees for the production of honey is called bee keeping or apiculture.
Bee keeping is a small scale industry for the production of honey which have following economic importance:

  • Bee keeping gives honey which is an important nutritive substance.
  • Bee-hives gives bee-wax which is utilized in making polishes.
  • Bee keeping helps in pollination of various crops useful for mankind

Question 24.
Write short notes on Homology and Analogy organ, vestigial organ, Atavism and Connecting links.
Answer:
(i) Homology and Analogy: The organs of similar structure and origin but dissimilar in functions and forms are called as homologous organs and this phenomenon is called homology. The presence of homologous organs implies a common evolutionary origin of amphibians, reptiles, birds and mammals from some ancient fish ancestor. The homologous structures seen in successive generations indicate actual relationship and the possessors are the diverse descendants of common ancestry (Divergent evolution).

The organs that perform the same function but differ in their origin and structure, are called as analogous organs and the phenomenon is called analogy. The wings of an insect are analogous to those of birds and bats because they perform the same function but have dissimilar structure and origin. The wings of an insect are modified outgrowth of the body wall whereas wings of birds and bats are forelimbs.

These organs have arisen in evolutionary process through adaptation of quite different organisms to a similar mode of life (Convergent evolution). On the same line, the similarities in proteins are genes performing a given function among diverse organisms give clues to common ancestry. These biochemical similarities point to the same shared ancestry as structural similiarities among diverse organisms.

(ii) Vestigial organs: These are the reduced and functionless organs which are of no use to the possessor but they still persists generation after generation in reduced form in an individual. They were complete and functional in the ancestors e.g. appendix in man is considered as the remnant of large intestine (caecum) but it is considered to be storage organ for cellulose digestion in herbivorous mammals. The vestigial organs which used to perform a normal function in the ancestor but during the course of evolution, they have been reduced to vestiges.

(iii) Atavism: It is reappearance of ancestral characters other than parents in the newly born offspring, which have either completely disappeared or reduced. The reappearance of short tail in some babies, multiple mammals in some individuals and dense hairy body etc. are the examples of atavism. The reappearance of such ancestral characters favour evolution.

(iv) Connecting links: The organisms which possess the characters of two different groups of organisms are known as connecting links e.g., duckbilled platypus and spiny ant-eater serve as a connecting link between the mammals and reptiles. Similarly lung fishes (Protopterus) is a connecting link between fishes and amphibians.

Bihar Board 12th Biology Model Question Paper 5 in English Medium