CBSE DelhiSet12016
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Q1
A point charge +Q is placed at point O as shown in the figure. Is the potential difference V_{A} – V_{B} positive, negative or zero?
Marks:1Answer:

Q2
How does the electric flux due to a point charge enclosed by a spherical Gaussian surface get affected when its radius is increased?
Marks:1Answer:
According to Gauss’s law, flux through a closed surface is given by
Where q is the charge enclosed by the Gaussian surface.
Since, on increasing the radius of the Gaussian surface, charge, q remains unchanged, so the flux through the spherical Gaussian surface will not be affected when its radius is increased.

Q3
Write the underlying principle of a moving coil galvanometer.
Marks:1Answer:
When a currentcarrying coil is placed in a magnetic field, it experiences a torque. This is the underlying principle of a moving coil galvanometer.

Q4
Why are microwaves considered suitable for radar systems used in aircraft navigation?
Marks:1Answer:
Microwaves are considered suitable for radar system used in aircraft navigation because they have a short wavelength range from 10^{3} m to 0.3m, which makes them suitable for long distance communication.
Another reason for using microwaves is that they can penetrate through clouds also.

Q5
Define’ quality factor’ of resonance in series LCR circuit. What is its SI unit?
Marks:1Answer:
The Q factor of series resonance circuit is defined as the ratio of the voltage developed across the inductor or capacitor at resonance to the impressed voltage, which is the voltage across R
It is dimensionless hence, it has no units. 
Q6
Explain the terms (i) Attenuation and (ii) Demodulation used in Communication System.
Marks:2Answer:
(i) Attenuation: The loss of strength of a signal while propagating through a medium.
(ii) Demodulation: The process of retrieval of information from the carrier wave at the receiver end. This is the reverse process of modulation.

Q7
Plot a graph showing variation of deBroglie wavelength , where V is accelerating potential for two particles A and B carrying same charge but of masses m_{1}, m_{2} (m_{1}> m_{2}). Which one of the two represents a particle of smaller mass and why?
Marks:2Answer:
The slope of the smaller mass is larger, therefore, plot A in the above graph is for mass m_{2}. 
Q8
A nucleus with mass number A = 240 and BE/A = 7.6 MeV breaks into tow fragments each of A = 120 with BE/A = 8.5 MeV. Calculate the released energy.
Marks:2Answer:
The binding energy of the nucleus of mass number 240 is, B_{1} = 7.6 x 240 = 1824 MeV
The binding energy of each product nucleus is, B_{2} = 8.5 x 120 = 1020 MeV
The energy released as the nucleus breaks is
E = 2B_{2} – B_{1} = 2 x 1020 – 1824 = 216 MeV
Alternatively: Gain in binding energy for nucleon is about 0.9 MeV.
Hence, the total gain in binding energy
DE = 240 × 0.9
DE = 216 MeV.

Q9
Calculate the energy in fusion reaction.
Marks:2Answer:

Q10
Two cells of emfs 1.5 V and 2.0 V having internal resistances 0.2 Ω and 0.3 Ω respectively are connected in parallel. Calculate the emf and internal resistance of the equivalent cell.
Marks:2Answer:
Given,
E_{1} = 1.5 V
E_{2} = 2 V
r_{1} = 0.2 Ω
r_{2} = 0.3 Ω
Effective EMF of two cells connected in parallel can be calculated as: