Subject

Physics

Class

JEE Class 12

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JEE Physics 2009 Exam Questions

Multiple Choice Questions

11.

A transparent solid cylindrical rod has a refractive index of 2√3. It is surrounded by air. A light ray is incident at the midpoint of one end of the rod as shown in the figure.θ The incident angle θ for which the light ray grazes along the wall of the rod is 

  • sin space to the power of minus space open parentheses 1 half close parentheses
  • sin space to the power of minus space open parentheses fraction numerator square root of 3 over denominator 2 end fraction close parentheses
  • sin space to the power of minus space open parentheses fraction numerator 2 over denominator square root of 3 end fraction close parentheses
  • sin space to the power of minus space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses

D.

sin space to the power of minus space open parentheses fraction numerator 1 over denominator square root of 3 end fraction close parentheses
422 Views

12.

A current loop ABCD is held fixed on the plane of the paper as shown in the figure. The arcs BC (radius = b) and DA (radius = a) of the loop are joined by two straight wires AB andCD. A steady current I is flowing in the loop. Angle made by AB and CD at the origin O is 30º. Another straight thin wire with steady current I1 flowing out of the plane of the paper is kept at the origin.



The magnitude of the magnetic field (2) due to the loop ABCD at the origin (O) is

  • zero

  • fraction numerator straight mu space left parenthesis straight b minus straight a right parenthesis over denominator 24 ab end fraction
  • fraction numerator straight mu subscript straight o straight I over denominator 4 space straight pi end fraction space open square brackets fraction numerator straight b minus straight a over denominator ab end fraction close square brackets
  • fraction numerator straight mu subscript 0 straight I over denominator 4 straight pi end fraction space open square brackets 2 left parenthesis straight b minus straight a right parenthesis space plus straight pi over 3 left parenthesis straight a plus straight b right parenthesis close square brackets

B.

fraction numerator straight mu space left parenthesis straight b minus straight a right parenthesis over denominator 24 ab end fraction
straight B space equals space 1 over 12 space of space fraction numerator straight mu subscript 0 straight I over denominator 2 end fraction space open parentheses 1 over straight a minus 1 over straight b close parentheses
space equals space fraction numerator straight mu subscript 0 straight I space left parenthesis straight b minus straight a right parenthesis over denominator 24 space ab end fraction
125 Views

13.

A long metallic bar is carrying heat from one of its ends to the other end under steady-state. The variation of temperature θ along the length x of the bar from its hot end is best described by which of the following figure. 


B.

We know that dQ over dt space equals space KA dθ over dx

In steady state flow of heat

dθ space equals space dQ over dt.1 over KA. dx
rightwards double arrow straight theta subscript straight H space equals straight theta space equals space straight k apostrophe straight x
rightwards double arrow space straight theta space equals straight theta subscript straight H space minus space straight k apostrophe straight x
Equation space straight theta space equals space straight theta subscript straight H space minus space straight k apostrophe space straight x space represents space straight a space straight space line

357 Views

14.

This question contains Statement-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.

Statement – 1: For a charged particle moving from point P to point Q, the net work done by an electrostatic field on the particle is independent of the path connecting point P to point Q.
Statement-2: The net work done by a conservative force on an object moving along a closed loop is zero.

  • Statement-1 is true, Statement-2 is false

  • Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

  • Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

  • Statement-1 is false, Statement-2 is true.


B.

Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

Work done by conservative force does not depend on the path. The electrostatic force is a conservative force.

138 Views

15.

Two moles of helium gas are taken over the cycle ABCDA, as shown in the P–T diagram.

The work done on the gas in taking it from D to A is

  • – 414 R

  • 414 R

  • -690 R

  • 690 R


A.

– 414 R

straight W subscript DA space equals space nRT space In space straight P subscript straight i over straight P subscript straight f space equals space 2 space straight x space straight R space straight x space 300 space In 1 half
261 Views

16.

Two moles of helium gas are taken over the cycle ABCDA, as shown in the P–T diagram. 



Assuming the gas to be ideal the work done on the gas in taking it from A to B is

  • 200 R

  • 300 R

  • 400 R

  • 500 R


C.

400 R

WAB = nR
(Tf–Ti) = × − 2 R(500 300)

 = 400 R
208 Views

17.

Two wires are made of the same material and have the same volume. However wire 1 has cross-section area A and wire 2 has cross–sectional area 3A. If the length of wire 1 increases by ∆x on applying force F, how much force is needed to stretch wire 2 by the same amount?

  • F

  • 4F

  • 9F

  • 6F


C.

9F

calligraphic l subscript 1 space equals space 3 calligraphic l subscript 2
straight Y space equals space straight F over straight A space straight x fraction numerator calligraphic l subscript 1 over denominator increment straight x end fraction space..... space left parenthesis straight i right parenthesis
straight Y space equals fraction numerator straight F apostrophe over denominator straight A end fraction space straight x space fraction numerator calligraphic l subscript 1 divided by 3 over denominator increment straight x end fraction space..... space left parenthesis ii right parenthesis
from space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space we space get
straight F over straight A space straight x fraction numerator calligraphic l subscript 1 over denominator increment straight x end fraction space equals space fraction numerator straight F apostrophe over denominator 3 straight A end fraction space straight x fraction numerator calligraphic l subscript 1 over denominator 3 increment straight x end fraction
straight F apostrophe space equals space 9 straight F
198 Views

18.

This question contains Statement-1 and Statement-2. Of the four choices given after the statements,choose the one that best describes the two statements.

Statement-1: The temperature dependence of resistance is usually given as R = Ro(1 + α∆t). The resistance of a wire changes from 100 Ω to 150 Ω when its temperature is increased from 27o C to 227o C. This implies that 3o 2.5 10 / C − α= ×.
Statement 2: R = Ri(1 + α∆T) is valid only when the change in the temperature ∆T is small and ∆R =(R - Ro) << Ro.

  • Statement-1 is true, Statement-2 is false.

  • Statement-1 is true, Statement-2 is true; Statement-2 is the correct explanation of Statement-1.

  • Statement-1 is true, Statement-2 is true; Statement-2 is not the correct explanation of Statement-1.

  • Statement-1 is false, Statement-2 is true.


A.

Statement-1 is true, Statement-2 is false.

R0 is the resistance at 0ºC.

207 Views

19.

Three sound waves of equal amplitudes have frequencies (ν – 1), ν, (ν + 1). They superpose to give beats. The number of beats produced per second will be.

  • 4

  • 3

  • 2

  • 1


C.

2

p1 =po sin 2π(x- 1)t
p2 =po sin 2π(x)t
p3 =po sin 2π(x+ 1)t
p=p1+p3+p2
=posin2π(x-1)t +po sin 2π(x+1)t +po sin 2π(x)t
= 2posin2πxtcos2πt +posin2πxt
= 2posin2πxt[2cos πt +1]
⇒ fbeat = 2

898 Views

20.

Two moles of helium gas are taken over the cycle ABCDA, as shown in the P–T diagram.

The net work done on the gas in the cycle ABCDA is

  • zero

  • 276 R 

  • 1076 R

  • 1904 R


B.

276 R 

Net work done in a cycle = WAB + WBC + WCB + WBA
= 400 R + 2 × 2.303 × 500 R ln 2 – 400R – 414 R
= 1000R x ln 2 – 600R x ln 2 = 400R x ln 2 = 276R

497 Views

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