﻿ ‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be: from Physics Class 12 JEE Year 2016 Free Solved Previous Year Papers

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# JEE Physics 2016 Exam Questions

#### Multiple Choice Questions

11.

Hysteresis loops for two magnetic materials A and B are given below:

These materials are used to make magnets for electric generators, transformer core and electromagnet core. Then it is proper to use:

• A for electric generators and transformers.

• A for electromagnets and B for electric generators

• A for transformers and B for electric generators.

• B for electromagnets and transformers.

D.

B for electromagnets and transformers.

Area of the hysteresis loop is proportional to the net energy absorbed per unit volume by the material, as it is taken over a complete cycle of magnetisation.

For electromagnets and transformers, energy loss should be low.

i.e thin hysteresis curves.

Also |B|→0 When H = 0 and |H| should be small when B →0.

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12.

The temperature dependence of resistances of Cu and undoped Si in the temperature range 300-400 K is best described by:

• Linear increase for Cu, linear increase for Si.

• Linear increase for Cu, exponential increase for Si.

• Linear increase for Cu, exponential decrease for Si.

• Linear decrease for Cu, linear decrease for Si.

C.

Linear increase for Cu, exponential decrease for Si.

As we know Cu is a conductor, so increase in temperature, resistance will increase.  Then, Si is a semiconductor, so with the increase in temperature, resistance will decrease.

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13.

A particle performs simple harmonic motion with amplitude A. It's speed is trebled at the instant that it is at a distance 2A/3 from the equilibrium position. The new amplitude of the motion is:

• 3A

D.

The velocity of a particle executing SHM at any instant is defined as the time rate of change of its displacement at that instant.

Where ω is angular frequency, A is amplitude and x are displacements of a particle.

Suppose that the new amplitude of the modules be A'.

The initial velocity of a particle performs SHM.

Where A, is initial amplitude and ω is angular frequency.
Final velocity,

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14.

The region between two concentric spheres of radii ‘a’ and ‘b’, respectively (see figure), has volume charge density ρ = A/r , where A is a constant and r is the distance from the centre. At the centre of the spheres is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is:

A.

A Gaussian surface at distance r from centre.

At the centre of the sphere is a point charge Q. The value of A such that the electric field in the region between the spheres will be constant is
As, Q = 2πAa2
i.e A = Q/2πa2

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15.

‘n’ moles of an ideal gas undergoes a process A→B as shown in the figure. The maximum temperature of the gas during the process will be:

A.

As, T will be maximum temperature where product of pV is maximum

Equation of line AB, we have

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16.

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge Q (having a charge equal to the sum of the charges on the 4 µF and 9 µF capacitors), at a point distant 30 m from it, would equal:

• 240N/C

• 360N/C

• 420N/C

• 480N/C

C.

420N/C

Resultant circuit,

As, charge on 3μF = 3μF x 8V = 24μC
Charge on 3μF = 3μF x 2V = 18 μC
charge on 4μF +Charge on 9μF
= (24 + 18)μC = 42μC
therefore,
Electric field at a point distant 30 m

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17.

Arrange the following electromagnetic radiations per quantum in the order of increasing energy:

A: Blue light
B: Yellow light
C: X-ray

• D, B, A, C

• A, B, D, C

• C, A, B, D

• B, A, D, C

A.

D, B, A, C

As, we know energy liberated, E = hc/λ

i.e
So, lesser the wavelength greater will be energy liberated by electromagnetic radiations per quantum.

As, order of wavelength is given by

therefore, the order of electromagnetic radiations per quantum.

D<B<A<C

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18.

Two identical wires A and B, each of length ‘l’, carry the same current I. Wire A is bent into a circle of radius R and wire B is bent to form a square of side ‘a’. If BA and BB are the values of the magnetic field at the centres of the circle and square respectively, then the ratio BA /BB is:

C.

Magnetic field in case of circle of radius R, we have

Magnetic field in case of square of side we get

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19.

An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to:

• 80 H

• 0.08 H

• 0.044 H

• 0.065 H

D.

0.065 H

I = 10 A, V = 80 V
R = V/I = 80/10 = 8Ω and ω = 50 Hz

For AC circuit, we have

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20.

A galvanometer having a coil resistance of 100 Ω gives a full-scale deflection, when a current of 1 mA is passed through it. The value of the resistance, which can convert this galvanometer into ammeter giving a full-scale deflection for a current of 10 A, is:

• 0.01 Ω

• 2 Ω

• 0.1 Ω

• 3 Ω

A.

0.01 Ω

Maximum voltage that can be applied across the galvanometer coil = 100 Ω x 10-3 A = 0.1

If Rs is the shunt resistance, then
Rs x 10 A = 0.1 V
Rs = 0.01 Ω

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