Subject

Physics

Class

JEE Class 12

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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.

387 Views

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.

314 Views

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:

  • straight A over 3 square root of 41
  • 3A

  • straight A square root of 3
  • 7 over 3 straight A

D.

7 over 3 straight A

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

straight v space equals space straight omega square root of straight A squared minus straight x squared end root
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. 
straight v squared space equals space straight omega squared space open square brackets straight A squared minus open parentheses fraction numerator 2 straight A over denominator 3 end fraction close parentheses squared close square brackets
Where A, is initial amplitude and ω is angular frequency.
Final velocity,
left parenthesis 3 straight v squared right parenthesis space equals space straight omega squared open square brackets straight A squared minus open parentheses fraction numerator 2 straight A over denominator 3 end fraction close parentheses squared close square brackets
From space Eqs. space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get
1 over 9 space equals space fraction numerator straight A squared minus begin display style fraction numerator 4 straight A squared over denominator 9 end fraction end style over denominator straight A apostrophe squared space minus begin display style fraction numerator 4 straight A squared over denominator 9 end fraction end style end fraction
straight A apostrophe space equals space fraction numerator 7 straight A over denominator 3 end fraction




1229 Views

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:

  • fraction numerator straight Q over denominator 2 πa squared end fraction
  • fraction numerator straight Q over denominator 2 straight pi left parenthesis straight b to the power of 2 minus end exponent straight a squared right parenthesis end fraction
  • fraction numerator 2 straight Q over denominator straight pi left parenthesis straight a squared minus straight b squared right parenthesis end fraction
  • fraction numerator 2 straight Q over denominator πa squared end fraction

A.

fraction numerator straight Q over denominator 2 πa squared end fraction

A Gaussian surface at distance r from centre.




fraction numerator straight Q space plus integral subscript straight a superscript straight r begin display style straight A over straight r end style 4 straight pi squared dr over denominator straight epsilon subscript straight o end fraction space equals space straight E 4 πr squared
straight E 4 straight epsilon subscript straight o straight r squared space equals space straight Q space plus space straight A fraction numerator 4 straight pi over denominator straight r squared end fraction open parentheses fraction numerator straight r squared minus straight a squared over denominator 2 end fraction close parentheses
straight E space equals fraction numerator 1 over denominator 4 πε subscript straight o end fraction space open square brackets straight Q over straight r squared plus straight A space 2 straight pi open parentheses fraction numerator straight r squared minus straight a squared over denominator straight r squared end fraction close parentheses close square brackets
straight E space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction open square brackets straight Q over straight r squared plus straight A 2 straight pi minus fraction numerator straight A 2 πa squared over denominator straight r squared end fraction close square brackets
straight E space equals space fraction numerator 1 over denominator 4 πε subscript straight o end fraction space straight x space straight A space straight x space 2 straight pi
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

1179 Views

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:

  • fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator 4 space nR end fraction
  • fraction numerator 3 straight p subscript straight o straight V subscript straight o over denominator 2 nR end fraction
  • 9 over 2 fraction numerator straight p subscript straight o straight V subscript straight o over denominator nR end fraction
  • fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator nR end fraction

A.

fraction numerator 9 straight p subscript straight o straight V subscript straight o over denominator 4 space nR end fraction

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


Equation of line AB, we have

straight y minus straight y subscript 1 space equals space fraction numerator straight y subscript 2 minus straight y subscript 1 over denominator straight x subscript 2 minus straight x subscript 1 end fraction space left parenthesis straight x minus straight x subscript 1 right parenthesis
rightwards double arrow space straight p minus straight p subscript straight o space equals fraction numerator 2 straight p subscript straight o minus straight p subscript straight o over denominator straight V subscript straight o minus 2 straight V subscript straight o end fraction space left parenthesis straight V minus 2 straight V subscript straight o right parenthesis
straight p minus straight p subscript straight o space equals space fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction left parenthesis straight V minus 2 straight V subscript straight o right parenthesis
straight p space equals space fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction straight V squared space plus 3 straight p subscript straight o straight V
nRT space equals space fraction numerator negative straight p subscript straight o over denominator straight v subscript straight o end fraction straight V squared space plus space 3 straight p subscript straight o straight V
straight T space equals space 1 over nR open parentheses fraction numerator negative straight p subscript straight o straight V squared over denominator straight v subscript straight o end fraction space plus 3 straight p subscript straight o straight V close parentheses
For space maximum space temperature
fraction numerator partial differential straight T over denominator partial differential straight V end fraction space equals space 0
fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction left parenthesis 2 straight V right parenthesis space equals space minus space 3 straight p subscript straight o
straight V space equals space 3 divided by 2 straight V subscript straight o
left parenthesis condition space for space maximum space temperature right parenthesis
Thus comma space the space maximum space temperature space of space the space gas space during space the space process space will space be
straight T subscript max space equals space 1 over nR open parentheses fraction numerator negative straight p subscript straight o over denominator straight V subscript straight o end fraction space straight x 9 over 4 straight V subscript 0 superscript 2 space plus space 3 straight p subscript straight o space straight x space 3 over 2 straight V subscript straight o close parentheses
equals space 1 over nR open parentheses negative 9 over 4 straight p subscript straight o straight V subscript straight o space plus space 9 over 2 straight p subscript straight o straight V subscript straight o close parentheses space equals space 9 over 4 fraction numerator straight p subscript straight o straight V subscript straight o over denominator nR end fraction

1416 Views

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

fraction numerator 9 space straight x space 10 cubed straight x space 42 space straight x space 10 to the power of negative 6 end exponent over denominator 30 space straight x space 30 end fraction space equals space 420 space straight N divided by straight C

260 Views

17.

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

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

  • 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 straight E space proportional to space 1 over straight lambda
So, lesser the wavelength greater will be energy liberated by electromagnetic radiations per quantum.

As, order of wavelength is given by

X- rays, VIBGYOR, Radio waves
therefore, the order of electromagnetic radiations per quantum.

D<B<A<C

294 Views

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:

  • straight pi squared over 8
  • fraction numerator straight pi squared over denominator 16 square root of 2 end fraction
  • fraction numerator straight pi squared over denominator 8 square root of 2 end fraction

C.

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



straight B subscript straight A space equals space fraction numerator straight mu subscript straight o straight I over denominator 2 straight R end fraction
As comma space 2 πR space equals space straight l space left parenthesis straight l space is space length space of space straight a space wire right parenthesis
straight R space equals fraction numerator straight l over denominator 2 straight pi end fraction
straight B subscript straight A space equals space fraction numerator straight mu subscript straight o straight I over denominator 2 space straight x space begin display style fraction numerator straight l over denominator 2 straight pi end fraction end style end fraction space equals space fraction numerator straight mu subscript straight o Iπ over denominator straight l end fraction space... space left parenthesis straight i right parenthesis

Magnetic field in case of square of side we get



281 Views

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 




straight I space equals space fraction numerator straight V over denominator square root of 8 squared plus straight X subscript straight L superscript 2 end root end fraction
rightwards double arrow space 10 space equals space fraction numerator 220 over denominator square root of 64 plus straight X subscript straight L superscript 2 end root end fraction
rightwards double arrow space square root of 64 space plus straight X subscript straight L superscript 2 end root space equals space 22
squaring space on space both space sides comma space we space get
64 space plus space straight X subscript straight L superscript 2 space equals space 484
straight X subscript straight L superscript 2 space equals space 484 minus 64 space equals space 420
straight X subscript straight L space equals space square root of 420
2 straight pi space straight x space ωL space equals space square root of 420
Series space induction space on space an space arc space lamp.
straight L space equals space fraction numerator square root of 420 over denominator left parenthesis 2 straight pi space straight x space 50 right parenthesis end fraction space equals space 0.065 space straight H

225 Views

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

192 Views

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