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 Maths Model Question Paper 4 in English Medium

Bihar Board 12th Maths Model Papers

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

Time : 3 Hours 15 Min
Full Marks: 100

Instructions for the candidates :

  1. Candidates are required to give their answers in their own words as far as practicable.
  2. Figure in the right-hand margin indicates full marks.
  3. While answering the questions, the candidate should adhere to the word limit as far as practicable.
  4. 15 Minutes of extra time has been allotted for the candidate 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 1-50 objective type questions which are compulsory, each carrying 1 mark. Darken the circle with blue/black ball pen against the correct option on the OMR Sheet provided to you. Do not use Whitener/Liquid/ Blade/Nail on OMR Sheet otherwise result will be invalid.
  7. In section-B, there are 25 short answer type questions (each carrying 2 marks), out of which only 15 (fifteen) questions are to be answered.
    A part from this there is 08 Long Answer Type questions (each carrying 5 marks), out of which 4 questions are to be answered.
  8. Use of any electronic device is prohibited.

Objective Type Questions

There are 1 to 50 objective type questions with 4 options, choose the correct option which is to be answered on OMR Sheet (50 × 1 = 50)

Question 1.
What type of relation on the set A = {1,2,3,4} given by R = {(1,3), (4,2), (2,4), (2,3), (3,1)} is
(a) Reflexive
(b) Transitive
(c) Symmetric
(d) None of these
Answer:
(d) None of these

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

Question 2.
If the operation * defined as (a * b) = a2 + b2 then the value of (3 * 4) * 5 is
(a) 650
(b) 125
(c) 625
(d) 3125
Answer:
(a) 650

Question 3.
If f: R → R defined as f(x) = 3x. Then the function is
(a) one-one into
(b) one-one onto
(c) Manyone into
(d) Many one onto
Answer:
(b) one-one onto

Question 4.
\(\tan ^{-1} \frac{2 x}{1-x^{2}}=\)
(a) 2sin-1x
(b) sin-12x
(c) tan-12x
(d) 2tan-1x
Answer:
(d) 2tan-1x

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

Question 5.
The value of \(\cos ^{-1}\left(-\frac{1}{2}\right)\) is
(a) \(\frac{\pi}{3}\)
(b) \(\frac{\pi}{6}\)
(c) \(\frac{2 \pi}{3}\)
(d) \(\frac{3 \pi}{4}\)
Answer:
(c) \(\frac{2 \pi}{3}\)

Question 6.
tan-1x + cot-1x = :
(a) π
(b) \(\frac{\pi}{2}\)
(c) \(\frac{2 \pi}{3}\)
(d) \(\frac{3 \pi}{4}\)
Answer:
(b) \(\frac{\pi}{2}\)

Question 7.
\(\left[\begin{array}{ccc}
b c & 1 & 1 / a \\
c a & 1 & 1 / b \\
a b & 1 & 1 / c
\end{array}\right]=?\)
(a) 1/abc
(b) 0
(c) abc
(d) None of these
Answer:
(b) 0

Question 8.
\(\left|\begin{array}{cc}
\cos 15^{\circ} & \sin 15^{\circ} \\
\sin 75^{\circ} & \cos 75^{\circ}
\end{array}\right|=?\)
(a) 1
(b) π/2
(c) 0
(d) None of these
Answer:
(c) 0

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

Question 9.
If λ ∈ R and ∆ = \(\left|\begin{array}{ll}
\mathbf{a} & \mathbf{b} \\
\mathbf{c} & \mathbf{d}
\end{array}\right|\)
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 1
Answer:
(c) \(\left|\begin{array}{ll}
\lambda a & \lambda b \\
\lambda c & \lambda d
\end{array}\right|\)

Question 10.
If A = \(\left[\begin{array}{lll}
1 & 2 & 3 \\
4 & 5 & 6
\end{array}\right]\) Then which of following is A’
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 2
Answer:
(c) \(\left[\begin{array}{ll}
1 & 4 \\
2 & 5 \\
3 & 6
\end{array}\right]\)

Question 11.
If y = sin (log x) then \(\frac{\mathrm{dy}}{\mathrm{dx}}=?\)
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 3
Answer:
\(\frac{\cos (\log x)}{x}\)

Question 12.
If y = x3 + 4x + 6 then \(\frac{d^{2} \cdot y}{d x^{2}}=?\)
(a) 3x3 + 4
(b) 3x2
(c) 6x
(d) 3x
Answer:
(c) 6x

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

Question 13.
If y = log√x then the value of \(\frac{d y}{d x}\) is
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 4
Answer:
(a) \(\frac{1}{2 x}\)

Question 14.
If y = sin2x then the value of \(\frac{d^{2} y}{d x^{2}}\) is
(a) 2 sin 4x
(b) -4 sin 2x
(c) 4 sin 2A
(d) 4 cos2 2x
Answer:
(b) -4 sin 2x

Question 15.
The ratio of rate of change x2 and log x is
(a) 2x
(b) 2x2
(c) 2
(d) 2/3
Answer:
(b) 2x2

Question 16.
If \(y=\sec ^{-1}\left[\frac{\sqrt{x}+1}{\sqrt{x}-1}\right]+\sin ^{-1}\left[\frac{\sqrt{x}-1}{\sqrt{x}+1}\right]\) then \(\frac{d y}{d x}\) = :
(a) 1
(b) π
(c) π/2
(d) 0
Answer:
(d) 0

Question 17.
The vectors a1î + a2 ĵ + a3k̂ and b1î + b2 ĵ + b3 ĵ perpendicular to each other if
(a) \(\frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\frac{a_{3}}{a_{3}}\)
(b) a1b1 + a2b2 + a3b3
(c) a1b2 + b2a1 + a3b2 = 0
(d) None of these
Answer:
(b) a1b1 + a2b2 + a3b3
Question 18.
k̂ × k̂ =
(a) 0
(b) 1
(c) |k|2
(d) None of these
Answer:
(a) 0

Question 19.
The scalar product of the vectors 5î + yĵ – 3k̂ and 3î – 4ĵ + 7k̂is
(a) 10
(b) -10
(c) 15
(d) -15
Answer:
(b) -10

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

Question 20.
The vector equation of the line which passes through the points A(3,4, -7) and B(l, -1,6) is
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 5
Answer:
(a) \(\vec{r}=(3 \hat{i}+4 \hat{j}-7 \hat{k})+\lambda(\hat{i}-\hat{j}+6 \hat{k})\)

Question 21.
For what value of x, î + ĵ + k̂ is a unit vector ?
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 6
Answer:
(b) \(\frac{1}{\sqrt{3}}\)

Question 22.
Distance between the planes r̂. n̂ = p1 and r̂. n̂ = p2 is
(a) p1 – p2
(b) |p1-p2|
(c) \(\frac{\left|p_{1}-p_{1}\right|}{2}\)
(d) None of these
Answer:
b) |p1-p2|

Question 23.
\(\int_{0}^{2 / 3} \frac{d x}{4+9 x^{2}}\) is equal to
(a) \(\frac{\pi}{6}\)
(b) \(\frac{\pi}{12}\)
(c) \(\frac{\pi}{24}\)
(d) \(\frac{\pi}{4}\)
Answer:
(c) \(\frac{\pi}{24}\)

Question 24.
\(\int_{0}^{\pi / 2} \frac{\sin x}{\sin x+\cos x} d x\) dx is equal to
(a) \(\frac{\pi}{2}\)
(b) \(\frac{\pi}{4}\)
(c) π
(d) None of these
Answer:
(b) \(\frac{\pi}{4}\)

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

Question 25.
\(\int x^{2} e^{x^{3}} d x=\)
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 7
Answer:
(a) \(\frac{1}{3} e^{x^{3}}+C\)

Question 26.
The value of \(\int_{0}^{\pi / 4} \sin 2 x d x\) is
(a) 1
(b) 1/2
(c) 1/3
(d) 1/4
Answer:
(b) 1/2

Question 27.
Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and x = 0, x = 2 is
(a) π
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{3}\)
(d) \(\frac{\pi}{4}\)
Answer:
(a) π

Question 28.
The order of differentied equation
\(\left(\frac{d^{2} y}{d x^{2}}\right)+2\left(\frac{d y}{d x}\right)^{3}+9 y=0\) is
(a) 2
(b) 3
(c) 4
(d) None of these
Answer:
(a) 2

Question 29.
The differential equation of the family of straight line which passes through the origin is
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 8
Answer:
(b) \(y \frac{d y}{d x}=x\)

Question 30.
The solution of \(\frac{d y}{d x}\) = 1 + x + y + xy is
(a) x – y = k(1 + xy)
(b) log(1 + y)=x + \(\frac{x^{2}}{2}\) + k
(c) log (1 + x) = y + \(\frac{y^{2}}{2}\) + k
(d) None of these
Answer:
(b) log(1 + y)=x + \(\frac{x^{2}}{2}\) + k

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

Question 31.
If \(\vec{a}=2 \vec{i}-5 \vec{\jmath}+\vec{k}\) and \(\vec{b}=4 \vec{i}+2 \vec{\jmath}+\vec{k}\) then \(\vec{a} \cdot \vec{b}=:\)
(a) 0
(b) -1
(c) 1
(d) 2
Answer:
(b) -1

Question 32.
If \(2 \vec{i}+\vec{j}+\vec{k}, 6 \vec{i}-\vec{j}+2 \vec{k}\) and \(14 \vec{i}-5 \vec{\jmath}+4 \vec{k}\) are the position vectors of the point A, B, C respectively, then
(a) A, B, C are collinear
(b) A, B, C are non-collinear
(c) \(\overrightarrow{A B} \perp \overrightarrow{B C}\)
(d) None of these
Answer:
(a) A, B, C are collinear

Question 33.
If \(|\vec{a} \times \vec{b}|=|\vec{a} \cdot \vec{b}|\) then angle between \(\vec{a}\) and \(\vec{b}\) is n
(a) 0
(b) \(\frac{\pi}{2}\)
(c) \(\frac{\pi}{4}\)
(d) π
Answer:
(c) \(\frac{\pi}{4}\)

Question 34.
Angle between the vectors 2î + 3ĵ + 2k̂ and î + 4ĵ + 5k̂ is
(a) 30°
(b) 90°
(c) 45°
(d) 60°
Answer:
(b) 90°

Question 35.
l, m, n and l1 m1 n1 are the direction cosines of two lines are perpendicular to each other if:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 9
Answer:
(c) ll1 + mm1 + nn1 = 0

Question 36.
A straight line passes through (2, -1,3) and its direction ratio’s are 3, -1,2. Then the equation of the line is
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 10
Answer:
(b) \(\frac{x-2}{3}=\frac{y+1}{-1}=\frac{z-3}{2}\)

Question 37.
The direction ratio’s of the normal to the plane 7x + 4y – 2z + 5 = 0 is
(a) (7,4,5)
(b) (7,4,-2)
(c) (7,4,2)
(d) (0,0,0)
Answer:
(b) (7,4,-2)

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

Question 38.
Angle between the lines \(\vec{r}=(4 \hat{i}-\hat{j})+s(2 \hat{i}+\hat{j}-3 k)\) and
\(\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+t(\hat{i}-3 \hat{j}+2 \hat{k})[latex] is
3n n 2n n
(a) [latex]\frac{3 \pi}{2}\)
(b) \(\frac{\pi}{3}\)
(c) \(\frac{2 \pi}{3}\)
(d) \(\frac{\pi}{6}\)
Answer:
(b) \(\frac{\pi}{3}\)

Question 39.
The vectors î + ĵ and ĵ + k̂ are perpendicular then unit vectors is equal to
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 11
Answer:
(c) \(\begin{aligned}
&\wedge \wedge \quad \wedge\\
&\frac{i+j+k}{3}
\end{aligned}\)

Question 40.
Which of the following is value of constant c of Rolle’s theorem when fix) = 2x3 – 5x2 – 4x + 3, x ∈\(\left[\frac{1}{3}, 3\right]\)
(a) \(-\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) -2
(d) 2
Answer:
(d) 2

Question 41.
The modulus of the vector \(2 \vec{i}-7 \vec{j}-3 \vec{k}\) is
(a) √61
(b) √62
(c) √64
(d) √32
Answer:
(b) √62

Question 42.
The projection of the vector î – 2ĵ + k̂ on 2î – ĵ + k̂ is
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 12
Answer:
(b) \(\frac{5}{\sqrt{6}}\)

Question 43.
If \(|\vec{a}|=\sqrt{26}|\vec{b}|=7\) and \(|\vec{a} \times \vec{b}|=35\) then \(\vec{a} \cdot \vec{b}=\)
(a) 8
(b) 7
(c) 9
(d) 12
Answer:
(b) 7

Question 44.
The function Z = ax + by in Linear programming problem is called
(a) objective Function
(b) constraints
(c) Feasible solution
(d) corner point
Answer:
(a) objective Function

Question 45.
If P(A) = \(\frac{1}{2}\) ,P(B) = 0 then P\(\frac{A}{B}\) is
(a) 0
(b) 1/2
(c) 1
(d) Not defined
Answer:
(d) Not defined

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

Question 46.
If P( A) = 0.8, P(B) = 0.5 and P(B/A) = 0.4 then P(A∪B) equal to :
(a) 0.78
(b) 0.88
(c) 0.98
(d) 1.08
Answer:
(c) 0.98

Question 47.
Two events A and B are mutually independent if
(a) P(A’ B’) = [1 – P(A)][1 – P(B)]
(b) P(A) = P(B)
(c) P(A) + P(B) = 1
(d) A and B are muttually exclusive
Answer:
(a) P(A’ B’) = [1 – P(A)][1 – P(B)]

Question 48.
The probability of getting doublet with one die is
(a) \(\frac{2}{3}\)
(b) \(\frac{1}{6}\)
(c) \(\frac{5}{6}\)
(d) \(\frac{5}{36}\)
Answer:
(b) \(\frac{1}{6}\)

Question 49.
If A and B are two events such that if P(A∪B) = P(a) Then
(a) P(A/B)= 1
(b) P(B/A)= 1
(c) P(A/B) = 0
(d) P(A/B) = 0
Answer:
(d) P(A/B) = 0

Question 50.
If A and B are imitualy exclusive events, then
(a) P (A/B) = 0
(b) P (B/A) = 0
(c) P(A∪B) = 0
(d) P (A ∩ B) = 0
Answer:
(d) P (A ∩ B) = 0

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

Non-Objective Type Questions

Short Answer Type Questions

Q. No. 1 to 25 are of Short Answer type carry of 2 marks each. Answer any 15 questions. (15 × 2 = 30)

Question 1.
Let a function f: R→R defined as \(f(x)=\frac{x^{2}}{x^{2}+1}\) ∀ x ∈ R . Then find f-1.
Answer:
Let f: R→R \( defined as f(x)=\frac{x^{2}}{x^{2}+1}\)
Taking 1, 1, ∈ R, we get, Clearly f(1) = f(-1) = 1/2
So, f is manyone function. Thus f-1 is not exists.

Question 2.
If \(\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x\) , then find x.
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 13

Question 3.
Applying properties of determinant, prove that:
\(\left|\begin{array}{lll}
1 & 1 & 1 \\
a & b & c \\
b c & c a & a b
\end{array}\right|=(a-b)(b-c)(c-a)\)
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 14
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 15

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

Question 4.
If \(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+a
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\) then find a,b,c and d
Answer:
\(\left[\begin{array}{cc}
a-b & 2 a+c \\
2 a-b & 3 c+a
\end{array}\right]=\left[\begin{array}{cc}
-1 & 5 \\
0 & 13
\end{array}\right]\)
a – b = -1 …………(1)
2a – b = 0 …………(2)
2a + c = 5…….(3)
3c + d = 13……(4)
Subtract (1) from (2), we get
2a – b – a + b = 0 + 1 ⇒ a = 1
Frorn (2), 2a – b = 0 ⇒ 2.1 – b ⇒ b = 2
Putting the value ofa in(3), we get 2a + c = 5 ⇒ 2.1 +c = 5
⇒ c = 5 – 2=3 , Lastly putting the value of c in (4), we get
3c + d = 13 ⇒ 3 x 3 + d = 13 ⇒ d= 13 – 9 = 4
Thus a= 1,b=2,c=3and d=4

Question 5.
If y = ex(sin x + cos x) then prove that
\(\frac{d^{2} y}{d x^{2}}-2 \frac{d y}{d x}+2 y=0\) we have
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 16

Question 6.
Verify the continuity at x = 0 of the function \(f(x)=\left\{\begin{aligned}
\frac{|\sin x|}{x}, & x \neq 0 \\
1, & x=0
\end{aligned}\right.\)
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 17

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

Question 7.
Differentiate: sec {tan ( √x)} with respect to x.
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 18

Question 8.
Find the projection of the vector î + ĵ + k̂) in the direction ĵ
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 19

Question 9.
Show that the tangents to the curve y = 7x3 +11 at the points where x = 2 and x= 2 are parallel.
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 20

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

Question 10.
Integrate: \(\int \frac{(\log x)^{2}}{x} d x\)
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 21

Question 11.
Find the value of \(\int_{0}^{\pi / 4} \tan ^{2} x d x\)
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 22

Question 12.
Solve the differential equation \(\frac{d y}{d x}\) = 1 + x + y + xy
Answer:
We have, \(\frac{d y}{d x}\) = (1 + x) + y(1 + x) = (1 + x)(1 + y)
\(\int \frac{d y}{1+y}\) = (1 + x)dx = ∫dx + ∫ xdx
log(1 + y) = x + \(\frac{x^{2}}{2}+c\)

Question 13.
Form the differential equation of the family of circles toucingthey-axis at Origin
Answer:
The equation of circle of radius a and centre (a, 0) and which passes through the origin with y-axis is
(x – a)2 + y2 = a2
or x2 + y2 = 2ax …(1)
Differentiating w.r.t. x, we get
2x + 2yy’= 2a
or, x + yy’ = a ……(2)
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 23
Putting the value of a from (2) in (1), we get
x2 + y2 = 2x(x + yy’) = 2x2 + 2 xy y’
or \(2 x y \frac{d y}{d x}+x^{2}-y^{2}=0\)
Which is the required differential equation

Question 14.
Show that the lines \(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(3 \hat{i}-\hat{j})\) and
\(\vec{r}=(4 \hat{i}-\hat{k})+\mu(2 \hat{i}+3 \hat{k})\) intersect each other. Find their point of intersection.
Answer:
Since the lines intersects for the any value of λ and μ the refore the value of \(\vec{r}\) is equal for them
Then (î + ĵ – k̂) + λ(3î – ĵ) =(4î – k̂) + µ(2î+3k)
⇒ (î + 3λ)î+(1 – λ)ĵ – k̂= (4 + 2µ)î + (3µ – 1)k̂)
⇒ 1 + 3λ = 4 + 2µ, 1 – λ = 0 and 3µ – 1 = -1
⇒ 3λ – 2µ = 3 …………(1)
λ = 1 ………….. (2)
and µ = 0 …(3)
clearly, λ = 1 and µ = O are sastis fled
Putting λ = 1. we get r̂ = (î + ĵ – k̂) + λ(3î – ĵ)
r̂ = (4î + 0ĵ – k̂)
Thus point of intersection of line is (4, 0, -1)

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

Question 15.
Find the equation of the through the point (5, -2, 4) and which is parallell to (2î – ĵ + 3k̂)
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 24
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 25
Question 16.
What is the maximum value of the function sin x + cos x
Answer:
Let the interval be [0, 2π] is given
Then f(x) = sin A’ + cos x
f'(x) = cos x – sin x
For maximum and minimum value,f'(A) = 0
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 26

Question 17.
Prove that : \(\overrightarrow{(\vec{a}-\vec{b})} \times(\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})\)
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 27

Question 18.
Solve : \(\frac{d y}{d x}+2 y \tan x=\sin x, y=0\) when x = π/3
Answer:
We have \(\frac{d y}{d x}\) + 2ytan x = sin x
This is linear differential equation, where P = 2tanx and Q = sin.x
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 28

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

Question 19.
If A and B are independent events, then prove that A and B’ are independent events.
Answer:
Given, A and B are independent then
P(A∩B) = P(A).P(B)
Now, A = (A∩B) ∪ (A∩B’)
⇒ P(A) = P(A∩B) + P(A∩B’)
P(A) = P(A).P(B) +P(A∩B’)
⇒ P(A∩B’) = P(A)-P(A) P(B)
⇒ P(A∩B’) = P(A) [ 1 -P(B) ]
⇒ P(A∩B’)= P(A).P(B’)
Thus A and B’ are independent.

Question 20.
If \(|\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}|=|\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}|\) the prove that \(\overrightarrow{\mathbf{a}} \perp \overrightarrow{\mathbf{b}}\)
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 29

Question 21.
Solve the differential equation \(\frac{d y}{d x}\)= 1 – x + y – xy
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 30

Question 22.
If P(A) = 0.8, P(B) = 0.5 and P(B/A) =0.4, then find
P(A ∩ B)
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 31

Question 23.
Show that the relation R in the set Z of integers given by R = {(a, b): 3 divides a – b} is an equivalence relation.
Answer:
Let a,b,c ∈ Z
(i) (a, a) ∈ R ⇒ aRa ⇒ a – a = O ∈ Z, divisible by 3 since 0∈Z, which is divisible by 3, so R is reflexive relation.
(ii) (a, b) ∈ R ⇒ (b, a) ∈ R ⇒ (a – b) is divisible by ⇒ (b – a) is divisible by 3. Since (a – b) is divisible by 3 so (b- a) is also divisible by 3, ∀ a, b ∈ Z. so R is symmetric relation.
(iii) (a, b) ∈ R, (b, c) ∈ R ⇒ (a, c) ∈ R ⇒ (a – b) is divisible by 3, (b – c) is divisible by 3
⇒ a – b + b – c = (a – c) is divisible by 3.
Since (a – b) is divisible by 3, (b – c) is divisible by 3
⇒ (a – c) is divisible by 3, ∀ a, b, c ∈ Z. SO R is transitive relation.
Since R is reflexsive, symmetric and transitive relations. Therefore R is an equivalence relation.

Question 24.
Solve tan-12x + tan -13x = π/4
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 32
or, 5x = 1 – 6x2
or, 6x2+ 5x – 1 = 0
or, 6x2 + 6x – x – 1 = 0
or, 6x (x + 1) – 1(x+ 1) = 0 – or,
(6x- 1)(x+ 1) = 0
∴ 6x – 1 = 0
or x + 1 = 0
i.e, x = 1/6 or, x = -1
Since x ≠ -1 so x = 1/6

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

Question 25.
Find X and Y if X + Y = \(\)
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 33

Long Answer Type Questions

Questions No. 26 to 33 are Long Answer Type Questions. Each question carry 5 marks. Answer any 4 question. (4 × 5 = 20)

Question 26.
Show that semi-vertical angle of right circular cone of given surface are anf maximum value is sin -1(1/3)
Answer:
Let r be the radius, Z be sland height and h be height of the cone. Then s = πr2 + πrl
or \(l=\frac{s-\pi r^{2}}{\pi r}\)
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 34
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 35
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 36

Question 27.
Prove that the function sinx(1 + cosx),x ∈ [0,π] is maximum at x = \(\frac{\pi}{3}\)
Answer:
Let f(x) = sinx(1 + cosx)
⇒ f'(x) = -sin x + cosx(1 + cosx)
⇒ f'(x) = 2cos2x + cosx – 1 = (2cosx – 1)(cosx + 1)
and f”(x) = -4cosx sin x – sin x
= -sinx(1 + 4cosx)
Now f'(x) = 0 ⇒ (2cosx – 1)(cosx + 1) = 0
⇒ 2cos x = 1 or, cos x = – 1
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 37

Question 28.
Find the area of the parabola’s enclosed by the region y2 = 4ax and x2 = 4ay, where a
Answer:
Given parabola’s y2 = 4ax …(1)
and x2 = 4ay …(2)
On solving (1) and (2), we get point of intersection So, substituting the value of x from (1) in (2), we get
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 38
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 39

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

Question 29.
Find the value at \(\int_{0}^{\pi / 4} \log (1+\tan x) d x\)
Answer:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 40

Question 30.
Find the vector equation of plane which pases through the points A(3, -5, -1); B(-1, 5,7) and parallel to the
vector 3î – ĵ + 7k̂
Answer:
Let the position vector of given points A and B are (3î – 5ĵ – k̂) and -î + 5ĵ + 7k̂
Let P be any point on the plane with position vector x
xî + yĵ + zk̂
AP=(x — 3)î + (y + 5)ĵ + (z +1)k̂
and AB = -4î + 10ĵ+ 8k̂
Clearly AP, AB and 3î – ĵ + 7k̂ are coplanar . So the equation of plane is
\(\left|\begin{array}{ccc}
x-3 & y+5 & z+1 \\
-4 & 10 & 8 \\
3 & -1 & 7
\end{array}\right|=0\)
⇒(x – 3) (70 + 8) – (y + 5)(-28 – 24) + (z +1)(4 – 30)= 0
⇒ 78 (x-3) + 52(y + 5)-26(z + l) = 0
⇒ 78x – 234 + 52y + 260 – 26z – 26 = 0
⇒ 78x + 52y-26z = 0
⇒ 3x + 2y-z = 0
The vector equation of plane is r. (3î – 5ĵ – k̂) = 0

Question 31.
Prove that the lines \(\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}\) are \(\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z-5}{5}\) coplanar.
Answer:
Here x1 = -3, y1 = 1, z1 = 5, a1 = -3, b1 = 1, c1 =5
and x2 = -1 y2= 2, z2 = 5,a2 = -1 b2 = 2, c2 = 5
Given lines are copalanar if
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 41
=2(5 – 10) – 1 (-15 + 5) + 0(-6 + 1)
= 2 x (-5) + 10 + 0 = 0
Hence the given lines are coplanar.

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

Question 32.
Using LPP method, maximize Z = 10x + 6y subject to
constraints 3x +y ≤ 12,2x + 5y ≤ 34, x ≥ 0,y ≥ 0.
Answer:
Maximize objeftive function Z = 10x + 6y
Subject to constraints
3x + y≤ 12 ⇒ 3x + y = 12 -..(1)
2x + 5y ≤ 34 ⇒ 2x + 5y = 34 …(2)
x ≥ 0, y ≥ 0 ⇒ x = 0,y = 0 …(3)
First of all draw graph of equations (1) to (3) corTes ponds to the inequations, we see from figure, the feasible, region OAPD be form, which is bounded also.
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 42
The co-ordinate of the comer point of the feasible region, are 0(0, 0), A(4, 0), P(2, 6) and D(0, 34/5). Lastly, by using comer point method to find the maximum value of Z are as follows :
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 43
It is obvious from the above table that maximum value of Z is 56 at the point (2,6). Thus the maximum value of Z is 56.

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

Question 33.
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs. 17.50 per packages on nuts and Rs. 7.00 per package on bolts. How many package of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours a day ?
Answer:
Let x nuts and y bolts are produced. We have the given data, as follows:
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 44
Now, the time of machine A is used. = (x + 3y) hours Avilable time = 12 hours,
So, x + 3y ≤ 12 ⇒ x + 3y = 12 …(1)
The time of machine B is used = (3x + y) hours
Available time = 12 hours, So 3x + y ≤ 12
⇒ 3x + y = 12 …. (2)
Profit function Z = 17.5x + 7y is objective function
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 45
Also, non-negative constraints
x ≥ 0, y ≥ 0 ⇒ x = 0, y = 0 …(3)
Now, linear equation (1) corresponds to inequation passes through the points A( 12, 0) and B(0, 4)
Putting x = 0, y = 0 in x + 3y ≤ 12,
we get 0 ≤ 12, which is true
So x + 3y ≤ 12 lies on and below AB The linear equation (2) corresponds to inequation passes through C(4, 0) and D(0. 12)
Putting x = 0, v = 0 in 3x + y ≤ 12,
we get 0 ≤ 12 which is true
So, 3x + y ≤ 12 lies on and below CD.
x ≥ 0 is the region which lies on and to the right on y-axis
y ≥ 0 is the region which lies on and above the x-axis
The line AB and CD intersects at the point P(3,3) Clearly it is obvious from figure the region OCPB is feasible region and is bounded. Lastly, calculate the value of Z= 17.5x + 7y as follows.
Bihar Board 12th Maths Model Question Paper 4 in English Medium - 46
From above table, it is clear that the maximum profit is Rs. 73.50 when 3 nuts and 3 bolts packages are produced.

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Bihar Board 12th Chemistry Objective Questions and Answers

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Question 1.
Homogeneous mixture of two or more than two components is called
(a) solute
(b) solvent
(c) both (a) and (b)
(d) solution
Answer:
(d) solution

Question 2.
Which of the following is an example of gaseous solution ?
(a) Camphor is nitrogen gas
(b) Solution of hydrogen in palladium
(c) Chloroform mixed with nitrogen gas
(d) Both (a) and (c)
Answer:
(d) Both (a) and (c)

Question 3.
In amalgam of mercury with sodium, solvent is
(a) mercury
(b) sodium
(c) amalgam
(d) none of these
Answer:
(b) sodium

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Question 4.
What is the mole fraction of glucose in 10% w/W ‘ glucose solution ?
(a) 01
(b) 0.02
(c) 0.03
(d) 0.04
Answer:
(a) 01
No. of moles of glucose = \(\frac{10}{180}\) = 0.0555 mol
No. of moles of water = \(\frac{90}{18}\) = 5 mol
Number of moles of solution = 5.0555 mol
Mole fraction of glucos
\(=\frac{\text { No. of moles of glucose }}{\text { No. of moles of solution }}=\frac{0.0555}{5.0555}=0.01\)

Question 5.
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%
Solute in 200 g of 30% solution = 60 g
Solute in 300 g of 20% solution = 60 g
Total grams of solute = 120 g
Total grams of solution = 200 + 300 = 500 g
% of solute in the final solution = \(=\frac{120}{500} \times 100=24 \%\)

Question 6.
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) 0.0104 ppm
(d) 104 ppm
Answer:
(b) 10.4 ppm

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Question 7.
What will be the mole fraction of ethanol in a sample of spirit containing 85% ethanol by mass ?
(a) 0. 69
(b) 0.82
(c) 0.85
(d) 0.60
Answer:
(a) 69
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 1

Question 8.
What is the molarity of a solution containing 10 g of NaOH in 500 mL of solution ?
(a) 25 mol L-1
(b) 0.75 mol L-1
(c) 0.5 mol L-1
(d) 1.25 mol L-1
Answer:
(c) 0.5 mol L-1
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 2

Question 9.
What will be the molarity of 30 mL of 0.5 M H2SO4 solution diluted to 500 mL ?
(a) 3 M
(b) 0.03 M
(c) 3 M
(d) 0.103 M
Answer:
(b) 0.03 M
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 3
NaCl → Na+ + Cl
No. of moles of Na+ ions = 0.025
No. of Na+ ions = 03325 x 6.023 x 1023
= 1.505 x 1022

Question 10.
How many Na+ ions are present in 100 mL of 0.25 M of NaCl solution ?
(a) 025 x 1023
(b) 1.505 x 1022
(c) 15 x 1022
(d) 2.5 x 1023
Answer:
(b) 1.505 x 1022

Question 11.
How many grams of NaOH are present in 250 mL of 0.5 M NaOH solution ?
(a) 32 g
(b) 3.8 g
(c) 5 g
(d) 0.5 g
Answer:
(c) 5 g

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Question 12.
What will be the molality of a solution of glucose in water which is 10% w/W ?
(a) 01 m
(b) 0.617 m
(c) 0.668 m
(d) 1.623 m
Answer:
(b) 0.617 m
(b) Mass of solution = 100 g
Mass of glucose = 10 g, Mass of water = 90 g
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 4

Question 13.
The molality of 648 g of pure water is …………….
(a) 36 m
(b) 5 m
(c) 3.6 m
(d) 5.55 m
Answer:
(d) 5 m

Question 14.
What is the mass of urea required for making 2.5 kg of 0.25 molal aqueous solution ?
(a) 37 g
(b) 25 g
(c) 125 g
(d) 27.5 g
Answer:
(a) 37 g

Question 15.
The law which indicates the relationship between solubility of a gas in liquid and pressure is_____________
(a) Raoult’s law
(b) Henry’s law
(c) Lowering of vapour pressure
(d) van’t Hoff law
Answer:
(b) Henry’s law

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Question 16.
H2S is a toxic gas used in qualitative analysis. If solubility of H2S in water at STP is 0.195 m, what is the value of Kh ?
(a) 0263 bar
(b) 69.16 bar
(c) 192 bar
(d) 282 bar
Answer:
(d) 282 bar
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 6

Question 17.
Henry’s law constant for molality of methane in benzene at 298 K is 4.27 x 105 mm Hg. The mole fraction of methane in benzene at 298 K under 760 mm Hg is
(a) 1.78 x 10-3
(b) 17.43
(c) 0.114
(d) 2.814
Answer:
(a) 1.78 x 10-3
(a) Accorading to Henry’s law P = Kx
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 7

Question 18.
Partial pressure of a solution component is directly proportional to its mole fraction. This is known as
(a) Henry’s law
(b) Raoult’s law
(c) Distribution law
(d) Ostwald’s dilution law
Answer:
(b) Raoult’s law

Question 19.
Among the following substances the lowest vapour pressure is exerted by
(a) water
(b) alcohol
(c) ether
(d) mercury
Answer:
(d) mercury

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Question 20.
Which of the following solutions shows positive deviation from Raoult’s law ?
(a) Acetone + Aniline
(b) Acetone + Ethanol
(c) Water + Nitric acid
(d) Chloroform + Benzene
Answer:
(b) Acetone + Ethanol

Question 21.
Which of the following solutions is an example of negative deviation from Raoult’s law ?
(a) Acetone + Ethanol
(b) Carbon tetrachloride + Chloroform
(c) Acetone + Chloroform
(d) Water + Ethanol
Answer:
(c) Acetone + Chloroform

Question 22.
The system that forms maximum boiling azeotrope is
(a) acetone – chloroform
(b) ethanol – acetone
(c) n-hexane – n-heptane
(d) carbon disulphide – acetone.
Answer:
(a) acetone – chloroform

Question 23.
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molecular mass of the solute ?
(a) 23.4 g mol-1
(b) 41.35 g mol-1
(c) 10 g mol-1
(d) 20.8 g mol-1
Answer:
(b) 41.35 g mol-1
Vapour pressure of pure water at boiling point = 1 atm = 1.013 bar
Vapour pressure of solution (ps) = 1.004 bar
Let mass of solution = 100 g
Mass of solute = 2 g
Mass of solvent = 100 – 2 = 98 g
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 8

Question 24.
Sprinkling of salt helps in clearing the snow covered roads in hills. The phenomenon involved in the process is ………..
(a) lowering in vapour pressure of snow
(b) depression in freezing point of snow
(c) increase in freezing point of snow
(d) melting of ice due to increase in temperature by putting salt.
Answer:
(b) depression in freezing point of snow

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Question 25.
What weight of glycerol should be added to 600 g of water in order to lower its freezing point by 10°C ?
(Kf = 1.86° Cm-1)
(a) 496 g
(b) 297 g
(c) 310 g
(d) 426 g
Answer:
(b) 297 g
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 9

Question 26.
The osmotic pressure of a solution can be increased by
(a) increasing the volume
(b) increasing the number of solute molecules
(c) decreasing the temperature
(d) removing semipermeable membrane.
Answer:
(b) increasing the number of solute molecules

Question 27.
10% solution of urea is isotonic with 6% solution of a non-volatile solute X. What is the molecular mass of solute X ?
(a) 6 g mol-1
(b) 60 g mol-1
(c) 36 g mol-1
(d) 32 g mol-1
Answer:
(c) 36 g mol-1
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 10

Question 28.
A solution containing 10.2 g glycerine per litre is isotonic , with a 2% solution of glucose. What is the molecular mass of glycerine ?
(a) 91.8 g
(b) 1198 g
(c) 83.9 g
(d) 890.3 g
Answer:
(a) 91.8 g
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 11

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Question 29.
A plant cell shrinks when it is kept in a …………..
(a) hypotonic solution
(b) hypertonic solution
(c) isotonic solution
(d) pure water.
Answer:
(b) hypertonic solution

Question 30.
Why is the molecular mass determined by measuring colligative property in case of some solutes is abnormal ?
(a) Due to association or dissociation of solute molecules.
(b) Due to insolubility of solute molecules.
(c) Due to decomposition of solute molecules.
(d) Due to large size of solute molecules.
Answer:
(a) Due to association or dissociation of solute molecules.

Question 31.
Which of the following representations of i (van’t Hoff factor) is not correct ?
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 14
Answer:
(d)

Question 32.
Which of the following will have same value of van’t Hoff factor as that of K4[Fe(CN)6] ?
(a) Al2(SO4)
(b) AlCl
(c) Al(NO3)3
(d) Al(OH)3
Answer:
(a) Al2(SO4)

Question 33.
Which of the following has the highest freezing point ?
(a) 1m NaCl solution
(b) 1m KCl solution
(c) 1m AlCl3 solution
(d) 1m C6HpO6 solution
Answer:
(d) 1m C6HpO6 solution

Question 34.
What will be the degree of dissociation of 0.1 M Mg(NO3)2 solution if van’t Hoff factor is 2.74 ?
(a) 75%
(b) 87%
(c) 100%
(d) 92%
Answer:
(b) 87%
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 12

Question 35.
What will be the freezing point of a 0.5 m KCl solution ? The molal freezing point constant of water is 1.86 °C m-1.
(a) -1.86 °C
(b) -0.372 °C
(c) -3.2 °C
(d) 0°C
Answer:
(a) -1.86 °C

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Question 36.
What amount of CaCl2 (i = 2.47) is dissolved in 2 litres of water so that its osmotic pressure is 0.5 atm at 27 °C ?
(a) 3.42 g
(b) 9.24 g
(c) 2.834 g
(d) 1.820 g
Answer:
(d) 1.820 g
Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions 13
Question 37.
The van’t Hoff factor of 0.005 M aqueous solution of KCl is 1.95. The degree of ionisation of KCI is …………..
(a) 0.95
(b) 0.97
(c) 0.94
(d) 0.96
Answer:
(a) 0.95

Question 38.
At equilibrium the rate of dissolution of a solid solute in a volatile liquid solvent is
(a) less than the rate of crystallisation
(b) greater than the rate of crystallisation
(c) equal to the rate of crystallisation
(d) zero
Answer:
(c) equal to the rate of crystallisation

Question 39.
Maximum amount of a solid solute that can be dissolved in a specified amount of a given liquid solvent does not depend upon
(a) temperature
(b) nature of solute
(c) pressure
(d) nature of solvent
Answer:
(c) pressure

Question 40.
Low concentration of oxygen in the blood and tissues of people living at high altitude is due to_______________
(a) low temperature
(b) low atmospheric pressure
(c) high atmospheric pressure
(d) both low temperature and high atmospheric pressure
Answer:
(b) low atmospheric pressure

Bihar Board 12th Chemistry Objective Answers Chapter 2 Solutions

Question 41.
Which of the following aqueous solutions should have the highest boiling point ?
(a) 1.0 MNaOH
(b) 1.0 MNa2SO4
(c) 1.0 MNH4NO3
(d) 1.0 MKNO3
Answer:
(a) 1.0 MNaOH

Question 42.
In comparison to a 0.01 M solution of glucose, the depression in freezing point of a 0.01 M MgCl2 solution is
(a) the same
(b) about twice
(c) about three times
(d) about six times.
Answer:
(c) about three times

Question 43.
The values of van’t Hoff factors for KCI, NaCl and K2SO4, respectively, are_______________ .
(a) 2, 2 and 2
(b) 2, 2 and 3
(c) 1,1 and 2
(d) 1, 1 and 1
Answer:
(b) 2, 2 and 3

Question 44.
Value of Henry’s constant KH _____________
(a) increases with increase in temperature
(b) decreases with increase in temperature
(c) remains constant
(d) first increases then decreases
Answer:
(a) increases with increase in temperature

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Bihar Board 12th Biology Objective Questions and Answers

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 1.
Temperature of the scrotum which is necessary for the functioning of testis is always __________ around below body temperature.
(a) 2°C
(b) 4°C
(c) 6°C
(d) 8°C
Answer:
(a) 2°C

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 2.
Which of the following is correct about mammalian testes ?
(a) Graafian follicles, Sertoli cells, Leydig’s cells
(b) Graafian follicles, Sertoli cells, Seminiferous tubules
(c) Sertoli cells, Seminiferous tubules, Leyding’s cells
(d) Graafian follicle, leyding’s cells, Seminiferous tubule
Answer:
(c) Sertoli cells, Seminiferous tubules, Leyding’s cells

Question 3.
The nutritive cells found in seminiferous tubules are
(a) Leydig’s cells
(b) atretic follicular cells
(c) Sertoli cells
(d) chromaffin cells.
Answer:
(c) Sertoli cells

Question 4.
Sertoli cells are regulated by the pituitary hormone known as
(a) LH
(b) FSH
(c) GH
(d) prolactin.
Answer:
(b) FSH

Question 5.
The head of the epididymis at the head of the testis is called
(a) cauda epididymis
(b) vas deferens
(c) caput epididymis
(d) gubernaculum.
Answer:
(c) caput epididymis

Question 6.
Seminal plasma in humans is rich in
(a) fructose and calcium but has no enzymes
(b) glucose and certain enzymes but has no calcium
(c) fructose and certain enzymes but poor in calcium
(d) fructose, calcium and certain enzymes.
Answer:
(d) fructose, calcium and certain enzymes.

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 7.
Prostate glands are located below
(a) gubernaculum
(b) seminal vesicles
(c) epididymis
(d) bulbourethral glands
Answer:
(b) seminal vesicles

Question 8.
The function of the secretion of prostate gland is to
(a) inhibit sperm activity
(b) attract sperms
(c) stimulate sperm activity
(d) none of these.
Answer:
(c) stimulate sperm activity

Question 9.
Lower narrow end of uterus is called
(a) urethra
(b) cervix
(c) clitoris
(d) vulva.
Answer:
(b) cervix

Question 10.
Bartholin’s glands are situated
(a) on the either side of vagina in humans
(b) on either side of vas deferens in humans
(c) on either side of penis in humans
(d) on either side of Fallopian tube in humans.
Answer:
(a) on the either side of vagina in humans

Question 11.
In humans, at the end of the first meiotic division, the male germ cells differentiate into the
(a) spermatids
(b) spermatogonia
(c) primary spermatocytes
(d) secondary spermatocytes.
Answer:
(d) secondary spermatocytes.

Question 12.
How many sperms are formed from a secondary spermatocyte ?
(a) 4
(b) 8
(c) 2
(d) 1
Answer:
(c) 2

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 13.
How many sperms are formed from 4 primary spermatocytes ?
(a) 4
(b) 1
(c) 16
(d) 32
Answer:
(c) 16

Question 14.
In spermatogenesis, reduction division of chromosome occurs during conversion of
(a) spermatogonia to primary spermatocytes
(b) primary spermatocytes to secondary spermatocytes
(c) secondary spermatocytes to spermatids
(d) spermatids to sperms.
Answer:
(b) primary spermatocytes to secondary spermatocytes

Question 15.
Which of the following groups of cells in the male gonad, represent haploid cells ?
(a) Spermatogonial cells
(b) Germinal epithelial cells
(c) Secondary spermatocytes
(d) Primary spermatocytes
Answer:
(c) Secondary spermatocytes

Question 16.
The process of release of spermatozoa from Sertoli cells into cavity of the seminiferous tubules is called
(a) spermiogenesis
(b) spermatogenesis
(c) spermatocytogenesis
(d) spermiation.
Answer:
(d) spermiation.

Question 17.
The correct sequence of spermatogenetic stages leading to the formation of sperms in a mature human testis is
(a) spermatogonia → spermatocyte → spermatid → sperms
(b) spermatid → spermatocyte → spermatogonia → sperms
(c) spermatogonia → spermatid → spermatocyte → sperms
(d) spermatocyte → spermatogonia → spermatid → sperms
Answer:
(a) spermatogonia → spermatocyte → spermatid → sperms

Question 18.
Spermatogenesis is induced by
(a) FSH
(b) ICSH
(c) STH
(d) ATH
Answer:
(a) FSH

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 19.
The middle piece of the sperm contains
(a) proteins
(b) mitochondria
(c) centriole
(d) nucleus.
Answer:
(b) mitochondria

Question 20.
The principal tail piece of human sperm shows the microtubular arrangement of
(a) 7 + 2
(b) 9 + 2
(c) 11 + 2
(d) 13 + 2
Answer:
(c) 11 + 2

Question 21.
Acrosome is a type of
(a) lysosome
(b) flagellum
(c) ribosome
(d) basal body.
Answer:
(a) lysosome

Question 22.
Which of the following contains the actual genetic part of a sperm ?
(a) Whole of it
(b) Tail
(c) Middle piece
(d) Head
Answer:
(d) Head

Question 23.
The sperms undergo physiological maturation, acquiring increased motility and fertilising capacity in
(a) seminiferous tubules
(b) vasa efferentia
(c) epididymis
(d) vagina.
Answer:
(c) epididymis

Question 24.
At what stage of life is oogenesis initiated in a human female ?
(a) At puberty
(b) During menarch
(c) During menopause
(d) During embryonic development
Answer:
(d) During embryonic development

Question 25.
1st polar body is formed at which stage of oogenesis ?
(a) 1st meiosis
(b) 2nd mitosis
(c) 1st mitosis
(d) Differentiation
Answer:
(a) 1st meiosis

Question 26.
Which one is released from the ovary ?
(a) Primary oocyte
(b) Secondary oocyte
(c) Graafian follicle
(d) Oogonium
Answer:
(b) Secondary oocyte

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 27.
During oogenesis, each diploid cell produces
(a) four functional eggs
(b) two functional eggs and two polar bodies
(c) one functional egg and three polar bodies
(d) four functional polar bodies.
Answer:
(c) one functional egg and three polar bodies

Question 28.
In oogenesis haploid egg is fertilised by sperm at which stage ?
(a) Primary oocyte
(b) Secondary oocyte
(c) Oogonium
(d) Ovum
Answer:
(b) Secondary oocyte

Question 29.
Layers of an ovum from outside to inside is
(a) corona radiata, zona pellucida and vitelline membrane
(b) zona pellucida, corona rodiata and vitelline membrane
(c) vitelline membrane, zona pellucida and corona radiata
(d) zona pellucida, vitelline membrane and corona radiata.
Answer:
(a) corona radiata, zona pellucida and vitelline membrane

Question 30.
Which part of ovary in mammals acts as an endocrine gland after ovulation ?
(a) Stroma
(b) Germinal epithelium
(c) Vitelline membrane
(d) Graafian follicle
Answer:
(d) Graafian follicle

Question 31.
The phase of menstrual cycle in humans that last for 7 – 8 days, is
(a) follicular phase
(b) ovulatory phase
(c) luteal phase
(d) menstruation.
Answer:
(a) follicular phase

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 32.
During proliferative phase uterine wall undergoes certain changes, these are
(a) myometrium wall is sloughed off
(b) endometrium wall is sloughed off
(c) blood vessels in endometrium become long and coiled
(d) proliferation of myometrial epithelial lining.
Answer:
(c) blood vessels in endometrium become long and coiled

Question 33.
Repair of endometrium is undertaken by
(a) LH
(b) FSH
(c) estrogen
(d) prolactin.
Answer:
(c) estrogen

Question 34.
In the 28 days human ovarian cycle, the ovulation takes place typically on
(a) day 1 of the cycle
(b) day 14 of the cycle
(c) day 5 of the cycle
(d) day 28 of the cycle
Answer:
(b) day 14 of the cycle

Question 35.
The time for optimum chances of conception in a woman is _________ starting from the day of menstruation.
(a) 1st day
(b) 4th day
(c) 14th day
(d) 26th day
Answer:
(c) 14th day

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 36.
Ovulation in the human female normally takes place during the menstrual cycle
(a) at the mind secretory phase
(b) just before the end of the secretory phase
(c) at the beginning of the proliferative phase
(d) at the end of the proliferative phase.
Answer:
(d) at the end of the proliferative phase.

Question 37.
After ovulation Graafian follicle regresses into
(a) corpus atresia
(b) corpus callosum
(c) corpus luteum
(d) corpus albicans
Answer:
(c) corpus luteum

Question 38.
Immediately after ovulation, the mammalian egg is covered by a membrane known as
(a) chorion
(b) zona pellucida
(c) corona radiata
(d) vitelline membrane.
Answer:
(c) corona radiata

Question 39.
Which one of the following events is correctly matched with the time period in a normal menstrual cycle ?
(a) Release of egg : 5 th day
(b) Endometrium regenerates : 5 – 10 days
(c) Endometrium secretes nutrients for implantation: 11 – 18 days
(d) Rise in progesterone level : 1 – 15 days
Answer:
(b) Endometrium regenerates : 5 – 10 days

Question 40.
If mammalian ovum fails to get fertilised, which one of the following is unlikely ?
(a) Corpus luteum will distintegrate.
(b) Progesterone secretion rapidly declines.
(c) Estrogen secretion increases.
(d) Primary follicle starts developing.
Answer:
(c) Estrogen secretion increases.

Question 41.
A human female reaches menopause aroung the age of
(a) 50 years
(b) 15 years
(c) 70 years
(d) 25 years.
Answer:
(a) 50 years

Question 42.
Cessation of menstrual cycle in a woman is called
(a) lactation
(b) ovulation
(c) menopause
(d) parturition
Answer:
(c) menopause

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 43.
Name the hormone that has no role in menstruation,
(a) LH
(b) FSH
(c) Estradiol
(d) TSH
Answer:
(d) TSH

Question 44.
Withdrawal of which of the following hormones is the immediate cause of menstruation ?
(a) Progesterone
(b) Estrogen
(c) FSH
(d) FSH-RH
Answer:
(a) Progesterone

Question 45.
At menopause there is rise in urinary excretion of
(a) FSH
(b) STH
(c) MSH
(d) none of these.
Answer:
(a) FSH

Question 46.
A reaction of granules content which harden the zona pellucida and ensures sure block to polyspermy is
(a) acrosomal reaction
(b) cortical reaction
(c) acrosin reaction
(d) bindin reaction.
Answer:
(b) cortical reaction

Question 47.
Which part of the sperm plays an important role in penetrating the egg membrane ?
(a) Allosome
(b) Tail
(c) Autosome
(d) Acrosome
Answer:
(d) Acrosome

Question 48.
In oocyte secondary maturation occurs in
(a) ovary
(b) abdominal cavity
(c) Fallopian tube
(d) uterus.
Answer:
(c) Fallopian tube

Question 49.
Besides activating the egg another role of a sperm is to carry to egg
(a) RNA
(b) mitochondria
(c) DNA
(d) ribosomes.
Answer:
(c) DNA

Question 50.
Preparation of sperm before penetration of ovum is
(a) spermiation
(b) cortical reaction
(c) spermiogenesis
(d) capacitation.
Answer:
(a) spermiation

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 51.
The sex of the fetus will be decided at
(a) fertilisation by male gamete
(b) implantation
(c) fertilisation by female gamete
(d) the start of cleavage.
Answer:
(a) fertilisation by male gamete

Question 52.
What is true about cleavage in the fertilised egg in humans ?
(a) It starts while the egg is in Fallopian tube.
(b) It starts when the egg reaches uterus.
(c) It is meroblastic
(d) It is identical to the normal mitosis.
Answer:
(a) It starts while the egg is in Fallopian tube.

Question 53.
Cleavage differs from mitosis in lacking
(a) synthetic phase
(b) growth phase
(c) both (a) and (b)
(d) none of these.
Answer:
(b) growth phase

Question 54.
The solid mass of 8-16 cells formed from zygote after successive mitotic divisions is called
(a) blastula
(b) gastrula
(c) morula
(d) none of these.
Answer:
(c) morula

Question 55.
Implantation takes place after __________ of fertilisation.
(a) 5 days
(b) 6 days
(c) 7 days
(d) 8 days
Answer:
(c) 7 days

Question 56.
Structure connecting the fetus to placenta is
(a) umbilical cord
(b) amnion
(c) yolk sac
(d) chorion.
Answer:
(a) umbilical cord

Question 57.
The main function of trophoectoderm in mammalian embryo is
(a) formation of future endoderm
(b) formation of the body of developing embryo
(c) formation of future ectoderm
(d) formation of placenta.
Answer:
(d) formation of placenta.

Question 58.
Fetus gets nourishment and oxygen through
(a) allantois
(b) placenta
(c) yolk sac
(d) chorion.
Answer:
(b) placenta

Question 59.
hCG, hPL and relaxin are produced in women
(a) at the time of puberty
(b) only during pregnancy
(c) at the time of menopause
(d) during menstruation.
Answer:
(b) only during pregnancy

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 60.
Which of the following hormones is not a secretory product of human placenta ?
(a) Human chorionic gonadotropin
(b) Prolactin
(c) Estrogen
(d) Progesterone
Answer:
(b) Prolactin

Question 61.
Urine test during pregnancy determines the presence of
(a) human chorionic gonadotropin hormone
(b) estrogen
(c) progesterone
(d) luteinising hormone.
Answer:
(a) human chorionic gonadotropin hormone

Question 62.
In the event of pregnancy, the corpus luteum persists under the influence of
(a) LH
(b) FSH
(c) chorionic gonadotropin
(d) progesterone.
Answer:
(c) chorionic gonadotropin

Question 63.
During the development of embryo, which of the following occurs first?
(a) Differentiation of organ
(b) Differentiation of tissue
(c) Differentiation of organ system
(d) Differentiation of cells
Answer:
(d) Differentiation of cells

Question 64.
The structures derived from ectoderm are
(i) pituitary gland
(ii) cornea
(iii) kidneys
(iv) notochord
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (ii)
(d) (ii) and (iv).
Answer:
(c) (i) and (ii)

Question 65.
Gastrula is the embryonic stage in which
(a) cleavage occurs
(b) blastocoel form
(c) germinal layers form
(d) villi form.
Answer:
(c) germinal layers form

Question 66.
In the development of the human body, the ectoderm is responsible for the formation of
(a) lens of the eye
(b) nervous system
(c) sweat glands
(d) all of these.
Answer:
(d) all of these.

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 67.
The first movements of the fetus and appearance of hair on its head are usually observed during which month of pregnancy ?
(a) Fourth month
(b) Fifth month
(c) Sixth month
(d) Third month
Answer:
(b) Fifth month

Question 68.
The early stage human embryo distinctly possesses
(a) gills
(b) gill slits
(c) external ear (pinna)
(d) eyebrows.
Answer:
(b) gill slits

Question 69.
Delivery of developed fetus is scientifically called
(a) parturition
(b) oviposition
(c) abortion
(d) ovulation.
Answer:
(a) parturition

Question 70.
In human adult females oxytocin
(a) stimulates pituitary to secrete vasopressin
(b) causes strong uterine contractions during parturition
(c) is secreted by anterior pituitary
(d) stimulates growth of mammary glands.
Answer:
(b) causes strong uterine contractions during parturition

Question 71.
The third stage of parturition is called “after-birth”. In this stage
(a) excessive bleeding occurs
(b) fetus is bom and cervix and vagina contraction to normal condition happens
(c) fetus is bom and contraction of uterine wall prevents excessive bleeding
(d) placenta is expelled out.
Answer:
(d) placenta is expelled out.

Question 72.
After birth, colostrum is released from mammary glands which is rich in
(a) fat and low in proteins
(b) proteins and low in fat
(c) proteins, antibodies and low in fat
(d) proteins, fat and low in antibodies.
Answer:
(c) proteins, antibodies and low in fat

Question 73.
Spot the odd one out from the following structures with reference to the male reproductive system.
(a) Rate testis
(b) Epididymis
(c) Vasa efferentia
(d) Isthmus
Answer:
(d) Isthmus

Question 74.
Seminal plasma, the fluid part of semen, is contributed by
(i) seminal vesicle
(ii) prostate
(iii) urethra
(iv) bulbourethral gland
(a) (i) and (ii)
(b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)
(d) (i) and (iv)
Answer:
(b) (i), (ii) and (iv)

Question 75.
Spermiation is the process of the release of sperms from
(a) seminiferous
(b) vas deferens
(c) epididymis
(d) prostate gland
Answer:
(a) seminiferous

Question 76.
Mature Graafian follicle is generally present in the ovary of a healthy human female around
(a) 5-8 day of menstrual cycle
(b) 11-17 day of menstrual cycle
(c) 18-23 day of menstrual cycle
(d) 24-28 day of menstrual cycle.
Answer:
(b) 11-17 day of menstrual cycle

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Question 77.
Acrosomal reaction of the sperm occurs due to
(a) its contact with zona pellucida of the ova
(b) reactions within the uterine environment of the female
(c) reactions within the epididymal environment of the male
(d) androgens produced in the uterus.
Answer:
(a) its contact with zona pellucida of the ova

Question 78.
Which one of the following is not a male accessory gland ?
(a) Seminal vesicle
(b) Ampulla
(c) Prostate
(d) Bulbourethral gland
Answer:
(b) Ampulla

Question 79.
Which among the following has 23 chromosomes ?
(a) Spermatogonia
(b) Zygote
(c) Secondary oocyte
(d) Oogonia
Answer:
(c) Secondary oocyte

Question 80.
Which of the following hormones is not secreted by human placenta ?
(a) hCG
(b) Estrogens
(c) Progesterone
(d) LH
Answer:
(d) LH

Question 81.
The vas deferens receives duct from the seminal vesicle and opens into urethra as
(a) epididymis
(b) ejaculatory duct
(c) efferent ductule
(d) ureter
Answer:
(b) ejaculatory duct

Question 82.
Urethral meatus refers to the
(a) urinogenital duct
(b) opening of vas deferens into urethra
(c) external opening of the urinogenital duct
(d) muscles surrounding the urinogenial duct.
Answer:
(c) external opening of the urinogenital duct

Question 83.
Morula is a developmental stage
(a) between the zygote and blastocyst
(b) between the blastocyst and gastrula
(c) after the implantation
(d) between implantation and parturition.
Answer:
(a) between the zygote and blastocyst

Question 84.
The membranous cover of the ovum at ovulation is
(a) corona radiata
(b) zona radiata
(c) zona pellucida
(d) chorion.
Answer:
(a) corona radiata

Question 85.
Identify the odd one from the following
(a) Labia minora
(b) Fimbriae
(c) Infundibulum
(d) Isthmus
Answer:
(a) Labia minora

Bihar Board 12th Biology Objective Answers Chapter 3 Human Reproduction

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Bihar Board 12th Biology Objective Questions and Answers

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Question 1.
Science of cultivation, breeding, marketing and arrangement of flowers is called
(a) arboriculture
(b) floriculture
(c) horticulture
(d) anthology
Answer:
(b) floriculture

Question 2.
Nonessential floral organs in a flower are
(a) sepals and petals
(b) anther and ovary
(c) stigma and filament
(d) petals only.
Answer:
(a) sepals and petals

Question 3.
The stamens represent
(a) microsporangia
(b) male gametophyte
(c) male gametes
(d) microsporophylls.
Answer:
(d) microsporophylls

Question 4.
Anther is generally
(a) monosporangiate
(b) bisporangiate
(c) letrasporangiate
(d) trisporangiate.
Answer:
(c) letrasporangiate

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Question 5.
The anther wall consists of four wall layers where
(a) tapetum lies just inner to endothecium
(b) middle layers lie between endothecium and tapetum
(c) endothecium lies inner to middle layers
(d) tapetum lies next to epidermis.
Answer:
(b) middle layers lie between endothecium and tapetum

Question 6.
The innermost layer of anther is tapetum whose function is
(a) dehiscence
(b) mechanical
(c) nutrition
(d) protection.
Answer:
(c) nutrition

Question 7.
Callase enzyme which dissolves callose of pollen tetrads to separate four pollens is provided by
(a) pollens
(b) tapetum
(c) middle layers
(d) endothecium.
Answer:
(b) tapetum

Question 8.
In angiosperms various stages of reductional division can best be studied in
(a) young anthers
(b) mature anthers
(c) young ovules
(d) endosperm cells.
Answer:
(a) young anthers

Question 9.
Study of pollen grains is called
(a) micrology
(b) anthology
(c) palynology
(d) pomology
Answer:
(c) palynology

Question 10.
Several pollen grains form a unit designated as pollinium in Family
(a) Asteraceae
(c) Asclepiadaceae Pollen
(b) Cucurbitaceae
(d) Brassicaceae
Answer:
(c) Asclepiadaceae Pollen

Question 11.
Pollen grain is a
(a) megaspore
(b) microspore
(b) microspore
(d) microsporangium.
Answer:
(b) microspore

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Question 12.
How many pollen mother cells should undergo meiotic division to produce 64 pollen grains ?
(a) 64
(b) 32
(c) 16
(d) 8
Answer:
(c) 16

Question 13.
How many meiotic divisions are required for the formation of 100 pollen grains ?
(a) 100
(b) 50
(c) 25
(d) 26
Answer:
(c) 25

Question 14.
One of the most resistant biological material present in the exine of pollen grain is
(a) pectocellulose
(b) sporopollenin
(c) suberin
(d) cellulose.
Answer:
(b) sporopollenin

Question 15.
What is the function of germ pore ?
(a) Emergence of radicle
(b) Absorption of water for seed germination
(c) Initiation of pollen tube
(d) All of these .
Answer:
(c) Initiation of pollen tube

Question 16.
_______of the pollen grain divides to form two male gametes.
(a) Vegetative cell
(b) Generative cell
(c) Microspore mother cell
(d) None of these
Answer:
(b) Generative cell

Question 17.
The three cells found in a pollen grain when it is shed at 3-celled stage are
(a) 1 vegetative cell, 1 generative cell, 1 male gamete
(b) 1 vegetative cell, 2 male gametes
(c) 1 generative cell, 2 male gametes
(d) either (a) or (b).
Answer:
(b) 1 vegetative cell, 2 male gametes

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Question 18.
Megasporangium along with its protective integuments is called
(a) ovary
(b) ovule
(c) funicle
(d) chalaza
Answer:
(b) ovule

Question 19.
Mature ovules are classified on the basis of funiculus. If micropyle comes to lie close to the funiculus the ovule is termed as
(a) orthotropous
(b) anatropous
(c) hemitropous
(d) campylotropous
Answer:
(b) anatropous

Question 20.
When micropyle, chalaza and hilum lie in a straight line, the ovule is said to be
(a) anatropous
(b) orthotropous
(c) amphitropous
(d) campylotropous.
Answer:
(b) orthotropous

Question 21.
What is the function of filiform apparatus in an angiospermic embryo sac?
(a) Brings about opening of the pollen tube
(b) Guides the pollen tube into a synergid
(c) Prevents entry of more than one pollen tube into a synergid
(d) None of these
Answer:
(b) Guides the pollen tube into a synergid

Question 22.
The female gametophyte of a typical dicot at the time of fertilisation is
(a) 8 – celled
(b) 7 – celled
(c) 6 – celled
(d) 5 – celled
Answer:
(b) 7 – celled

Question 23.
Polygonum type of embryo sac is
(a) 8 – nucleate, 7 – celled
(b) 8 – nucleate, 8 – celled
(c) 7 – nucleate, 7 – celled
(d) 4 – nucleate, 3 – celled
Answer:
(a) 8 – nucleate, 7 – celled

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Question 24.
Both chasmogamous and cleistogamous flowers are present in
(a) Helianthus
(b) Commelina
(c) Rosa
(d) Gossypium
Answer:
(b) Commelina

Question 25.
Even in absence of pollinating agents seed-setting is assured in
(a) Commelina
(b) Zostera
(c) Salvia
(d) Fig
Answer:
(a) Commelina

Question 26.
Male and female flowers are present on different plants (dioecious) to ensure xenogamy, in
(a) papaya
(b) bottle gourd
(c) maize
(d) all of these.
Answer:
(a) papaya

Question 27.
Feathery stigma occurs in
(a) pea
(b) wheat
(c) Datura
(d) Caesalpinia
Answer:
(b) wheat

Question 28.
Plants with ovaries having only one or a few ovules are generally pollinated by
(a) bees
(b) butterflies
(c) birds
(d) wind
Answer:
(d) wind

Question 29.
Which of the following is not a water pollinated plant ?
(a) Zostera
(b) Vallisneria
(c) Hydrilla
(d) Cannabis
Answer:
(d) Cannabis

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Question 30.
Spiny or sticky pollen grains and large, attractively coloured flowers are associated with
(a) hydrophily
(b) entomophily
(c) ornithophily
(d) anemophily
Answer:
(b) entomophily

Question 31.
Fragrant flowers with well developed nectaries are an adaptation for
(a) hydrophily
(b) anemophily
(c) entomophily
(d) none of these
Answer:
(c) entomophily

Question 32.
Pollen kitt is generally found in
(a) anemophilous flowers
(b) entomophilous flowers
(c) ornithophilous flowers
(d) malacophilous flowers
Answer:
(b) entomophilous flowers

Question 33.
Which of these is a condition that makes flowers invariably autogamous ?
(a) Dioecy
(b) Self incompatibility
(c) Cleistogamy
(d) Xenogamy
Answer:
(c) Cleistogamy

Question 34.
Heterostyly as a contrivance for cross-pollination is found in
(a) Pennisetum
(b) Impatiens
(c) Primula vulgaris
(d) Oenothera
Answer:
(c) Primula vulgaris

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Question 35.
The part of gynoecium that determines the compatible nature of pollen is
(a) stigma
(b) style
(c) ovary
(d) synergids
Answer:
(a) stigma

Question 36.
Part of the gynoecium which receives the pollen is called
(a) style
(b) stigma
(c) ovule
(d) ovary
Answer:
(b) stigma

Question 37.
Growth of pollen tube towards embryo sac is
(a) chemotropic
(b) thigmotaxis
(c) geotropic
(d) none of these
Answer:
(a) chemotropic

Question 38.
During the process of fertilisation the pollen tube of the pollen grain usually enters the embryo sac through
(a) integument
(b) nucellus
(c) chalaza
(d) micropyle
Answer:
(d) micropyle

Question 39.
Fusion of one of the male gametes with egg nucleus is referred to as
(a) generative fertilisation
(b) syngamy
(c) vegetative fertilisation
(d) both (a) and (b)
Answer:
(d) both (a) and (b)

Question 40.
The total number of nuclei involved in double fertilisation in angiospersm are
(a) two
(b) three
(c) four
(d) five
Answer:
(d) five

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Question 41.
Triple fusion in Capsella bursa pastoris is fusion of male gamete with
(a) egg
(b) synergid
(c) secondary nucleus
(d) antipodal.
Answer:
(c) secondary nucleus

Question 42.
Double fertilisation was first discovered in 1898 by _______ in Fritillaria and Lilium.
(a) Nawaschin
(b) Strasburger
(c) Amici
(d) Focke
Answer:
(a) Nawaschin

Question 43.
If an endosperm cell of an angiosperm contains 24 chromosomes, the number of chromosomes in each cell of the root will be
(a) 8
(b) 4
(c) 16
(d) 24
Answer:
(c) 16

Question 44.
The cells of endosperm have 24 chromosomes. What will be the number of chromosomes in the gametes ?
(a) 8
(b) 16
(c) 23
(d) 32
Answer:
(a) 8

Question 45.
The true embryo develops as a result to fusion of
(a) two polar nuclei of embryo sac
(b) egg cell and male gamete
(c) synergid and male gamete
(d) male gamete and antipodals.
Answer:
(b) egg cell and male gamete

Question 46.
Father of Indian embryology is
(a) P. Maheshwari
(b) Swaminathan
(c) R. Misra
(d) Butler
Answer:
(a) P. Maheshwari

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Question 47.
The portion of embryonal axis between plumule (future shoot) and cotyledons is called
(a) hypocotyl
(b) epicotyl
(c) coleorhiza
(d) coleoptile.
Answer:
(b) epicotyl

Question 48.
Coleoptile and coleorhiza are the protective sheaths _______ covering _______ and _______ respectively.
(a) plumule, epicotyl
(b) radicle, plumule
(c) plumule, radicle
(d) radicle, hypocotyl
Answer:
(c) plumule, radicle

Question 49.
_______ is not an endospermic seed.
(a) Pea
(b) Castor
(c) Maize
(d) Wheat
Answer:
(a) Pea

Question 50.
Endosperm is completely consumed by the developing embryo in
(a) pea and groundnut
(b) maize and castor
(c) castor and groundnut
(d) maize and pea.
Answer:
(a) pea and groundnut

Question 51.
Endospermic seeds are found in
(a) castor
(b) barley
(c) coconut
(d) all of these
Answer:
(d) all of these

Question 52.
In albuminous seeds, food is stored in _______ and in non albuminous seeds, it is stored in _______.
(a) endosperm, cotyledons
(b) cotyledons, endosperm
(c) nucellus, cotyledons
(d) endosperm, radicle
Answer:
(a) endosperm, cotyledons

Question 53.
Persistent nucellus is called as _______ and is found in _______.
(a) perisperm, black pepper
(b) perisperm, groundnut ‘
(c) endosperm, black pepper
(d) endosperm groundnut
Answer:
(a) perisperm, black pepper

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

Question 54.
Indentify the wrong statement regarding post-fertilisation development.
(a) The ovary wall develops into pericarp.
(b) The outer integument of ovule develops into tegmen.
(c) The fusion nucleus (triple nucleus) develops into endosperm.
(d) The ovule develops into seed.
Answer:
(b) The outer integument of ovule develops into tegmen.

Question 55.
Polyembryony commonly occurs in
(a) banana
(b) tomato
(c) potato
(d) citrus.
Answer:
(d) citrus.

Question 56.
An embryo may sometimes develop from any cell of embryo sac other than egg. It is termed as
(a) apospory
(b) apogamy
(c) parthenogenesis
(d) parthenocarpy
Answer:
(b) apogamy

Question 57.
Embryo sac is to ovule as _______ is to an anther.
(a) Stamen
(b) filament
(c) pollen grain
(d) androecium
Answer:
(c) pollen grain

Question 58.
The outermost and innermost wall layers of microsporangium in an anther are respectively
(a) endothecium and tapetum
(b) epidermis and endodermis
(c) epidermis and middle layer
(d) epidermis and tapetum.
Answer:
(d) epidermis and tapetum.

Question 59.
During microsporogenesis, meiosis occurs in
(a) endothecium
(b) microspore mother cells
(c) microspore tetrads
(d) pollen grains
Answer:
(b) microspore mother cells

Question 60.
From among the sets of terms given below, identify those that are associated with the gynoecium.
(a) Stigma, ovule, embryo sac, placenta
(b) Thalamus, pistil, style, ovule
(c) Ovule, ovary, embryo sac, tapetum
(d) Ovule, stamen, ovary, embryo sac
Answer:
(a) Stigma, ovule, embryo sac, placenta

Question 61.
Starting from the innermost part, the correct sequence of parts in an ovule are
(a) egg, nucellus, embryo sac, integument
(b) egg, embryo sac, nucellus, integument
(c) embryo sac, nucellus, integument, egg
(d) egg. integument, embryo sac, nucellus.
Answer:
(b) egg, embryo sac, nucellus, integument

Question 62.
The phenomenon wherein, the ovary develops into a fruit without fertilisation is called
(a) parthenocarpy
(b) apomixis
(c) asexual reproduction
(d) sexual reproduction
Answer:
(a) parthenocarpy

Bihar Board 12th Biology Objective Answers Chapter 2 Sexual Reproduction in Flowering Plants

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 Chemistry Objective Answers Chapter 1 The Solid State

Bihar Board 12th Chemistry Objective Questions and Answers

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 1.
Which among the following will show anisotropy ?
(a) Glass
(b) NaBr
(c) Plastic
(d) Rubber
Answer:
(b) NaBr

Question 2.
Which of the following forms a molecular solid when solidified ?
(a) Calcium fluoride
(b) Silicon dioxide
(c) Carbon dioxide
(d) Sodium chloride
Answer:
(c) Carbon dioxide

Question 3.
The major binding force in diamond, silicon and quartz is ……………
(a) electrostatic force
(b) electrical attraction
(c) covalent bond force
(d) van der Waals force
Answer:
(c) covalent bond force

Question 4.
Solid X is a very hard solid which is electrical insulator in solid as well as in molten state and has extremely high melting point. What type of solid is it ?
(a) Ionic solid
(b) Covalent solid
(c) Metallic solid
(d) Molecular solid
Answer:
(b) Covalent solid

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 5.
Examples of few solids are given below. Find out the example which is not correctly matched.
(a) Ionic solids – NaCl, ZnS
(b) Covalent solids – H2,l2
(c) Molecular solids – H2O(s)
(d) Metallic solids – Cu, Sn
Answer:
(b) Covalent solids – H2,l2

Question 6.
Which of the following solids is the structure of CsCl crystal ?
(a) Body centred cubic
(b) Simple cubic
(c) Face centred cubic
(d) Edge centred cubic
Answer:
(a) Body centred cubic

Question 7.
Which of the following does not represent a type of crystal system ?
(a) Triclinic
(b) Monoclinic
(c) Rhombohedral
(d) Isotropical
Answer:
(d) Isotropical

Question 8.
Which of the following primitive cells show the given parameters ?
a # b # c, α = β = Y = 90°
(a) Cubic
(b) Tetragonal
(c) Orthorhombic
(d) Hexagonal
Answer:
(c) Orthorhombic

Question 9.
Which type of crystals contains more than one Bravais lattice ?
(a) Hexagonal
(b) Triclinic
(c) Rhombohedral
(d) Monoclinic
Answer:
(d) Monoclinic

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 10.
For the structure given below the site marked as S is a ……………
(a) tetrahedral void
(b) cubic void
(c) octahedral void
(d) none of these

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State
Answer:
(c) octahedral void

Question 11.
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 12.
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 13.
The coordination number of metal crystallising in a hexagonal close packing is ………
(a) 12
(b) 4
(c) 8
(d) 6
Answer:
(a) 12

Question 14.
A crystalline structure has radius ratio (r, / r) in the range of 0.225 – 0.414. The coordination number and arrangement of anions around the cations are ………..
(a) 3, plane triangular
(b) 6, octahedral
(c) 4, tetrahedral
(d) 8, cubic
Answer:
(c) 4, tetrahedral

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 15.
A crystal lattice with alternative +ve and -ve ions has radius ratio 0.524. The coordination number of lattice is …………
(a) 4
(b) 6
(c) 8
(d) 12
Answer:
(b) 6

Question 16.
A solid AB has a rock salt structure. If radius of cation A+ is 120 pm, what is the minimum value of radius of B+anion (a) 120 pm
(b) 240 pm
(c) 290 pm
(d) 360 pm
Answer:
(c) 290 pm
Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State 2
Question 17.
How many chloride ions are surrounding sodium ion in sodium chloride crystal ?
(a) 4
(b) 8
(c) 6
(d) 12
Answer:
(c) 6

Question 18.
In NaCl structure, ……………
(a) all octahedral and tetrahedral sites are occupied
(b) only octahedral sites are occupied
(c) only tetrahedral sites are occupied
(d) neither octahedral nor tetrahedral sites are occupied.
Answer:
(b) only octahedral sites are occupied

Question 19.
NaCl type crystal (with coordination no. 6 : 6) can be converted into CsCl type crystal (with coordination no. 8 : 8) by applying ……
(a) high temperature
(b) high pressure
(c) high temperature and high pressure
(d) low temperature and low pressure.
Answer:
(b) high pressure

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 20.
Coordination numbers of Cs+ and Cl in CsCl crystal are ………….
(a) 8, 8
(b) 4,4
(c) 6,6
(d) 8,4
Answer:
(a) 8, 8

Question 21.
A unit cell of BaCl2 (fluorite structure) is made up of ………..
(a) four Ba2+ ions and four Cl ions
(b) four Ba2+ ions and eight Cl ions
(c) eight Ba2+ ions and four Cl ions
(d) four Ba2+ ions and six Cl
Answer:
(c) eight Ba2+ ions and four Cl ions

Question 22.
In face centred cubic unit cell, edge length is …………..
(a) \(\frac{4}{\sqrt{3}} \mathrm{r}\)
(b) \(\frac{4}{\sqrt{2}} \mathrm{r}\)
(c) 2r
(d) \(\frac{\sqrt{3}}{2} \mathrm{r}\)
Answer:
(b) \(\frac{4}{\sqrt{2}} \mathrm{r}\)

Question 23.
The edge length of fee cell is 508 pm. If radius of cation is 110 pm, the radius of anion is …………
(a) 110 pm
(b) 220 pm
(c) 285 pm
(d) 144 pm
Answer:
(d) 144 pm
Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State 3

Question 24.
The edge length of sodium chloride unit cell is 564 pm. If the size of Cl ion is 181 pm. The size of Na+ ion will be …….
(a) 101 pm
(b) 181 pm
(c) 410 pm
(d) 202 pm
Answer:
(a) 101 pm

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 25.
The radii of Na+ and Cl ions are 95 pm and 181 pm respectively. The edge length of NaCl unit cell is ………..
(a) 276 pm
(b) 138 pm
(c) 552 pm
(d) 415 pm
Answer:
(c) 552 pm

Question 26.
Copper crystallises in fee with a unit cell length of 361 pm. What is the radius of copper atom ?
(a) 157 pm
(b) 181 pm
(c) 127 pm
(d) 108 pm
Answer:
(c) 127 pm

Question 27.
Total volume of atoms present in a fee unit cell of a metal with radius r is …………
(a) \(\frac{12}{3} \pi r^{2}\)
(b) \(\frac{16}{3} \pi r^{3}\)
(c) \(\frac{20}{3} \pi r^{3}\)
(d) \(\frac{24}{3} \pi r^{3}\)
Answer:
(b) \(\frac{16}{3} \pi r^{3}\)

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State 4

Question 28.
Relationship between atomic radius and the edge length a of a body-centred cubic unit cell is ……..
(a) r = a/2
(b) \(r=\sqrt{a / 2}\)
(c) \(r=\frac{\sqrt{3}}{4} a\)
(d) \(r=\frac{3 a}{2}\)
Answer:
(c) \(r=\frac{\sqrt{3}}{4} a\)
Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State 5

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State 6
Question 29.
Edge length of a unit cell of chromium metal is 287 pm with bcc arrangement. The atomic radius is of the order
(a) 124.27 pm
(b) 287 pm
(c) 574 pm
(d) 143.5 pm
Answer:
(a) 124.27 pm
Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State 8

Question 30.
The fraction of the total volume occupied by the atoms present in a simple cube is ………….
(a) \(\frac{\pi}{4}\)
(b) \(\frac{\pi}{6}\)
(c) \(\frac{\pi}{3 \sqrt{2}}\)
(d) \(\frac{\pi}{4 \sqrt{2}}\)
Answer:
(b) \(\frac{\pi}{6}\)

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 31.
A unit cell of sodium chloride has four formula units. The edge length of the unit cell is 0.564 nm. What is the density of sodium chloride ?
(a) 3.89 g cm-3
(b) 2.16 gem-3
(c) 3 g cm-3
(d) 1.82 g cm-3
Answer:
(b) 2.16 gem-3

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State 9

Question 32.
The distance between Na+ & Cl ions in NaCl with a density 3.165 g cm-3 is …………..
(a) 497 pm
(b) 248.5 pm
(c) 234 pm
(d) 538.5 pm
Answer:
(b) 248.5 pm
Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State 10

Question 33.
How many lithium atoms are present in a unit cell with edge length 3.5A and density 0.53 g cm-3 ?
(a) 2
(b) 1
(c) 4
(d) 6
Answer:
(a) 2
Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State 11

Question 34.
Alkali halides do not show Frenkel defect because …………….
(a) cations and anions have almost equal size
(b) there is a large difference in size of cations and anions
(c) cations and anions have low coordination number
(d) anions cannot be accommodated in voids.
Answer:
(a) cations and anions have almost equal size

Question 35.
Which of the following crystals does not exhibit Frenkel defect ?
(a) AgBr
(b) AgCl
(c) KBr
(d) ZnS
Answer:
(c) KBr

Question 36.
What type of stoichiometric defect is shown by ZnS ?
(a) Schottky defect
(b) Frenkel defect
(c) Both Frenkel and Schottky defects
(d) Non-stoichiometric defect
Answer:
(b) Frenkel defect

Question 37.
In a Schottky defect, …………..
(a) an ion moves to interstitial position between the lattice points
(b) electrons are trapped in a lattice site
(c) some lattice sites are vacant
(d) some extra cations are present in interstitial spaces.
Answer:
(c) some lattice sites are vacant

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 38.
Silver halides generally show ………….
(a) Schottky defect
(b) Frenkel defect
(c) both Frenkel and Schottky defects
(d) cation excess defect.
Answer:
(c) both Frenkel and Schottky defects

Question 39.
An electron trapped in an anion site in a crystal is called ………….
(a) F-centre
(b) Frenkel defect
(c) Schottky defect
(d) interstitial defect
Answer:
(a) F-centre

Question 40.
Which of the following will have metal deficiency defect ?
(a) NaCl
(b) FeO
(c) KCl
(d) ZnO
Answer:
(b) FeO

Question 41.
Pure silicon and germanium behave as …………
(a) conductors
(b) semiconductors
(c) insulators
(d) piezoelectric crystals
Answer:
(c) insulators

Question 42.
To get n-type of semiconductor, germanium should be doped with ………….
(a) gallium
(b) arsenic
(c) aluminium
(d) boron
Answer:
(b) arsenic

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 43.
p-type semiconductors are formed when Si or Ge are doped with …………
(a) group 14 elements
(b) group 15 elements
(c) group 13 elements
(d) group 18 elements.
Answer:
(c) group 13 elements

Question 44.
Paramagnetic substances are magnetised in a magnetic field in the same direction. Paramagnetism is due to the presence of……………….
(a) one or more unpaired electrons
(b) all paired electrons
(c) permanent spin and orbital motion
(d) parallel and anti-parallel spins in equal numbers.
Answer:
(a) one or more unpaired electrons

Question 45.
Fe3O4 is ferrimagnetic at room temperature but at 850 K, it becomes
(a) diamagnetic
(b) ferrimagnetic
(c) paramagnetic
(d) anti-ferromagnetic
Answer:
(c) paramagnetic

Question 46.
Which of the following metal oxides is antiferromagnetic in nature ?
(a) MnO2
(b) TiO2
(c) NO2
(d) CrO2
Answer:
(a) MnO2

Question 47.
Which of the following conditions favours the existance of a substance in the solid state ?
(a) High temperature
(b) Low temperature
(c) High thermal energy
(d) Weak cohesive forces
Answer:
(b) Low temperature

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 48.
Which of the following is an amorphous solid ?
(a) Graphite (c)
(b) Quartz glass (SiO2)
(c) Chrome alum
(d) Silicon carbide (SiC)
Answer:
(b) Quartz glass (SiO2)

Question 49.
Iodine molecules are held in the crystals lattice by
(a) London forces
(b) dipole-dipole interactions
(c) covalent bonds
(d) coulombic forces
Answer:
(a) London forces

Question 50.
Which of the following is a network solid ?
(a) SO2(Solid)
(b) I2
(c) Diamond
(d) H2O(Ice)
Answer:
(c) Diamond

Question 51.
Graphite is a good conductor of electricity due to the presence of_______
(a) lone pair of electrons
(b) free valence electrons
(c) cations
(d) anions
Answer:
(b) free valence electrons

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 52.
Which of the following oxides behaves as conductor or insulator depending upon temperature ?
(a) TiO
(b) SiO2
(c) TiO3
(d) MgO
Answer:
(c) TiO3

Question 53.
Which of the following oxides shows electrical properties like metals ?
(a) SiO2
(b) MgO
(c) SO2(s)
(d) CrO2
Answer:
(d) CrO2

Question 54.
The lattice site in a pure crystal cannot be occupied by
(a) molecule
(b) ion
(c) electron
(d) atom
Answer:
(c) electron

Question 55.
Graphite cannot be classified as_____________ .
(a) conducting solid
(b) network solid
(c) covalent solid
(d) ionic solid
Answer:
(d) ionic solid

Question 56.
Cations are present in the interstitial sites in___________
(a) Frenkel defect
(b) Schottky defect
(c) vacancy defect
(d) metal deficiency defect
Answer:
(a) Frenkel defect

Question 57.
Which of the following is true about the charge acquired by p-type semiconductors ?
(a) positive
(b) neutral
(c) negative
(d) depends on concentration of p impurity
Answer:
(b) neutral

Bihar Board 12th Chemistry Objective Answers Chapter 1 The Solid State

Question 58.
In which pair most efficient packing is present ?
(a) hep and bcc
(b) hep and ccp
(c) bcc and ccp
(d) bcc and simple cubic cell
Answer:
(c) bcc and ccp

Question 59.
What is the coordination number in a square close packed structure in two dimensions ?
(a) 2
(b) 3
(c) 4
(d) 6
Answer:
(c) 4

Question 60.
Which kind of defects are introduced by doping ?
(a) Dislocation defects
(b) Schottky defects
(c) Frenkel defects
(d) Electronic defects
Answer:
(d) Electronic defects

Question 61.
The edge lengths of the unit cells in terms of the radius of spheres constituting fee, bcc and simple cube unit cell are respectively.
(a) \(2 \sqrt{2} r, \frac{4 r}{\sqrt{3}}, 2 r\)
(b) \(\frac{4 r}{\sqrt{3}}, 2 \sqrt{2} r, 2 r\)
(c) \(2 \mathrm{r}, 2 \sqrt{2} \mathrm{r}, \frac{4 \mathrm{r}}{\sqrt{3}}\)
(d) \(2 \mathrm{r}, \frac{4 \mathrm{r}}{\sqrt{3}}, 2 \sqrt{2} \mathrm{r}\)
Answer:
(a) \(2 \sqrt{2} r, \frac{4 r}{\sqrt{3}}, 2 r\)