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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

D.

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

B.

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

In steady state flow of heat

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

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

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