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CBSE

Subject

Physics

Class

JEE Class 12

JEE Physics 2017 Exam Questions

Multiple Choice Questions

1.

A slender uniform rod of mass M and length

  • fraction numerator 3 straight g over denominator 2 calligraphic l end fraction space cos space straight theta
  • fraction numerator 2 space straight g over denominator 3 space calligraphic l end fraction space cos space straight theta
  • fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sin space straight theta
  • fraction numerator 2 space straight g over denominator 3 space calligraphic l end fraction space sin space straight theta

C.

fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sin space straight theta


Taking torque about pivot τ = Iα
mg space sin space straight theta space calligraphic l over 2 space equals space fraction numerator straight m calligraphic l squared over denominator 3 end fraction space straight alpha
straight alpha space equals space fraction numerator 3 straight g over denominator 2 space calligraphic l end fraction space sinθ
346 Views

2.

The variation of acceleration due to gravity g with distance d from centre of the earth is best represented by (R = Earth's radius)


B.

straight g space equals space GMx over straight R cubed space inside space the space Earth space
straight g equals space GM over straight r squared space outside space the space Earth
Where M is mass of earth



571 Views

3.

The moment of inertia of a uniform cylinder of length

  • 1

  • fraction numerator 3 over denominator square root of 2 end fraction
  • square root of 3 over 2 end root
  • fraction numerator square root of 3 over denominator 2 end fraction

C.

square root of 3 over 2 end root
straight I space equals space fraction numerator straight m calligraphic l squared over denominator straight I 2 end fraction space plus mR squared over 4
or space straight I space equals space straight m over 4 open parentheses calligraphic l squared over 3 plus straight R squared close parentheses space space.... space left parenthesis 1 right parenthesis
Also comma space straight m space equals space πR squared calligraphic l rho
rightwards double arrow space R squared space equals space fraction numerator m over denominator pi </div>		</div>
		<div class= 1287 Views

4.

A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?


A.

Velocity at any time t is given by


v = u + at

v = v0 + (–g)t
v = v0 – gt

Straight line negative slope.

595 Views

5.

A time-dependent force F = 6t acts on a particle of mass 1 kg. If the particle starts from rest, the work done by the force during the first 1 sec. will be

  • 9J

  • 18 J

  • 4.5 J

  • 22 J


C.

4.5 J

F = 6t = ma
⇒ a = 6t
rightwards double arrow space dv over dt space equals space 6 straight t
integral subscript 0 superscript straight v dv space equals space integral subscript 0 superscript 1 6 straight t space dt
straight v space equals space left parenthesis 3 straight t squared right parenthesis subscript 0 superscript 1 space equals space 3 space straight m divided by straight s
from space work space energy space theorem
straight W subscript straight F space equals space increment straight K. straight E space equals space 1 half straight m space left parenthesis straight v squared minus straight u squared right parenthesis
space equals space 1 half left parenthesis 1 right parenthesis left parenthesis 9 minus 0 right parenthesis space equals space 4.5 straight J

1105 Views

6.

A particle A of mass m and initial velocity v collides with a particle B of mass m/2 which is at rest. The collision is head on, and elastic. The ratio of the de-Broglie wavelengths λA to λB after the collision is

  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 2 over 3
  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 1 half
  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 1 third
  • straight lambda subscript straight A over straight lambda subscript straight B space equals space 2

D.

straight lambda subscript straight A over straight lambda subscript straight B space equals space 2


By conservation of linear momentum
mv space equals space mv subscript 1 space plus straight m over 2 straight v subscript 2
2 straight v space equals space 2 straight v subscript 1 space plus straight v subscript 2 space... space left parenthesis 1 right parenthesis
By space law space of space collision
straight e space equals fraction numerator straight v subscript 2 minus straight v subscript 1 over denominator straight u subscript 1 minus straight u subscript 2 end fraction
straight u space equals space straight v subscript 2 minus straight v subscript 1 space space... space left parenthesis 2 right parenthesis
by space equ. space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis
straight v subscript 1 space equals space straight v over 3 semicolon space straight v subscript 2 space equals space fraction numerator 4 straight v over denominator 3 end fraction
straight lambda subscript 1 space equals space straight h over straight p subscript 1 semicolon space straight lambda subscript 2 space equals space straight h over straight o subscript 2
straight lambda subscript 1 over straight lambda subscript 2 space equals space 2 over 1
1144 Views

7.

The temperature of an open room of volume 30 m3 increases from 17°C to 27°C due to sunshine. The atmospheric pressure in the room remains 1 × 105 Pa. If ni and nf are the numbers of molecules in the room before and after heating, then nf – ni will be

  • 2.5 x 1025

  • -2.5 x 1025

  • -1.61 x 1023

  • 1.38 x 1023


B.

-2.5 x 1025

Using ideal gas equation
PV = nRT
(N is number of moles)
P0V0 = niR × 290 ...... (1)
[Ti = 273 + 17 = 290 K]
After heating
P0V0 = NfR × 300 ...... (2)
[Tf = 273 + 27 = 300 K]
from equation (1) and (2)
straight N subscript straight f minus straight N subscript straight i space equals space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 300 end fraction space minus space fraction numerator straight P subscript straight o straight V subscript 0 over denominator straight R space straight x 290 end fraction
difference space in space number space of space moles space minus fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 300 end fraction space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R space straight x space 290 end fraction
difference space in space number space of space moles space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R end fraction space open square brackets fraction numerator 10 over denominator 290 space straight x space 300 end fraction close square brackets
Hence comma space straight n subscript straight f space minus straight n subscript straight i space is
equals space minus space fraction numerator straight P subscript straight o straight V subscript straight o over denominator straight R end fraction space open square brackets fraction numerator 10 over denominator 290 space straight x space 300 end fraction close square brackets straight x space 6.023 space straight x space 10 to the power of 23

putting P0 = 105 PA and V0 = 30 m3
Number of molecules nf – ni = – 2.5 × 1025

824 Views

8.

The following observations were taken for determining surface tension T of water by the capillary method :
Diameter of capillary, D = 1.25 × 10–2 m rise of water, h = 1.45 × 10–2 m
Using g = 9.80 m/s2 and the simplified relation T = (rhg/2) x103 n/m, the possible error in surface tension is closest to:

  • 2.4%

  • 10%

  • 0.15%

  • 1.5%


D.

1.5%

straight T space equals space rhg over 2 space straight x space 10 cubed
fraction numerator increment straight T over denominator straight T end fraction space equals space fraction numerator increment straight r over denominator straight r end fraction space plus space fraction numerator increment straight h over denominator straight h end fraction space plus 0
100 space straight x space fraction numerator increment straight T over denominator straight T end fraction space equals space open parentheses fraction numerator 10 to the power of negative 2 end exponent space straight x space.01 over denominator 1.25 space straight x space 10 to the power of negative 2 end exponent end fraction space plus fraction numerator 10 to the power of negative 2 end exponent space straight x space 0.1 over denominator 1.45 space straight x space 10 to the power of negative 2 end exponent end fraction close parentheses 100
space equals space 0.8 space plus 0.689
equals space 1.489
space 100 space straight x space fraction numerator increment straight T over denominator straight T end fraction space equals space 1.489 percent sign
space approximately equal to space 1.5 space percent sign
1509 Views

9.

A particle is executing simple harmonic motion with a time period T. AT time t = 0, it is at its position of equilibrium. The kinetic energy-time graph of the particle will look like 


B.

582 Views

10.

A body of mass m = 10–2 kg is moving in a medium and experiences a frictional force F = –kv2. Its initial speed is v0 = 10 ms–1. If, after 10 s, its energy is 1/8 mv02,the value of k will be

  • 10-4 kg m-1

  • 10–1 kg m–1 s–1

  • 10-3 kg m-1

  • 10-3 kg s-1


A.

10-4 kg m-1

1 half mv subscript straight f superscript 2 space equals space 1 over 8 mv subscript 0 superscript 2
straight v subscript straight f space equals space straight v subscript 0 over 2 space equals space 5 space straight m divided by straight s
left parenthesis 10 to the power of negative 2 end exponent right parenthesis space dv over dt space equals space minus space kv squared
integral subscript 10 superscript 5 space dv over straight v squared space equals negative 100 space straight k integral subscript 0 superscript 10 space dt
1 fifth space minus space 1 over 10 space equals space 100 space straight k space left parenthesis 10 right parenthesis
straight k space equals space 10 to the power of negative 4 end exponent
829 Views

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