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# JEE Physics 2014 Exam Questions

#### Multiple Choice Questions

11.

In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be

• 8 A

• 10 A

• 12 A

• 14 A

C.

12 A

Total power (P) consumed
= (15 x 40) + ( 5x 100) + ( 5x 80) + (1 x 1000)  = 2500
Power, P = VI

Minimum capacity should be 12 A.

233 Views

12.

A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.

• 12

• 8

• 6

• 4

C.

6

i.e,

therefore, possible frequencies = 100 Hz, 300 Hz, 500 Hz, 700Hz, 900Hz, 1100Hz, below 1250 Hz.
356 Views

13.

The current voltage relation of the diode is given by I = e(1000V/T-1) mA, where the applied voltage V is in volt and the temperature T is in kelvin. If a student makes an error measuring ± 0.01V while measuring the current of 5mA at 300K, what will be the error in the value of current in mA?

• 0.2 mA

• 0.02 mA

• 05 mA

• 0.05 mA

A.

0.2 mA

Given, I =(e1000V/T-1)mA
dv =±0.01 V
T = 300 K
I = 5 mA
I = e1000V/T-1
I +1 = e1000V/T
Taking log on both sides, we get
log(l+T) = 1000V/T
On differentiating,

So, error in the value of current is 0.2 mA

283 Views

14.

One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in the figure. The process BC is adiabatic. The temperatures at A, B and C are 400 K, 800 K and 600 K respectively. Choose the correct statement:

• The change in internal energy in the process AB is -350 R.

• The change in internal energy in the process BC is -500 R.

• The change in internal energy in the whole cyclic process is 250 R.

• The change in internal energy in the process CA is 700 R.

D.

The change in internal energy in the process CA is 700 R.

According to first law of thermodynamics,
(i) change in internal energy from A to B i.e,

573 Views

15.

Match List-I (Electromagnetic wave type) with List-II (Its association/application) and select the correct option from the choices given below the lists:

 List I List II A. Infared waves 1. To treat muscular strain B. Radio waves 2. For broadcasting C. X -rays 3. To detect fracture of bones D. Ultraviolet 4. Absorbed by the ozone layer of the atmosphere
•  A B C D 4 3 2 1
•  A B C D 1 2 4 3
•  A B C D 3 2 1 4
•  A B C D 1 2 3 4

D.

 A B C D 1 2 3 4

Infrared rays are used to treat muscular strain.
X- rays are used to detect fracture of bones.
Ultraviolet rays are absorbed by ozone.

536 Views

16.

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is 3 × 104 V/m, the charge density of the positive plate will be close to

• 6 x 10-7 C/m2

• 3  x 10-7 C/m2

• 3 x 104 C/m2

• 6x 104 C/m2

A.

6 x 10-7 C/m2

When free space between parallel plate capacitor,
When dielectric is introduced between parallel plates of capacitor,

Electric field inside dielectric

where, K = dielectric constant of medium = -2.2
εo = permitivity of free space = 8.85 x 10-12
σ = 2.2 x 8.85 x 10-12 x 3 x 104
= 6.6 x 8.85 x 10-8 = 5.841 x10-7
= 6 x 10-7 C/m2

185 Views

17.

An open glass tube is immersed in mercury in such a way that a length of 8 cm extends above the mercury level. The open end of the tube is then closed and sealed and the tube is raised vertically up by additional 46 cm. What will be the length of the air column above mercury in the tube now? (Atmospheric pressure = 76 cm of Hg)

• 16 cm

• 22 cm

• 38 cm

• 6 cm

A.

16 cm

Since the system is accelerating horizontally such that no component of acceleration in the vertical direction. Hence, the pressure in the vertical direction will remain unaffected.

.
For air trapped in tube, p1V1 = p2V2
p1 = patm = pg76
V1 = A.8 [ A = area of cross section]
p2 = patm - ρg(54-x) =  ρg(22+x)
V2 = A.x
ρg76 x 8A = ρg (22+x) Ax
x2+22x-78x 8
by solving, x = 16

668 Views

18.

Three rods of Copper, Brass and Steel are welded together to from a Y –shaped structure. Area of cross – section of each rod = 4 cm2. End of the copper rod is maintained at 100oC where as ends of brass and steel are kept at 0oC. Lengths of the copper, brass and steel rods are 46, 13 and 12 cms respectively. The rods are thermally insulated from surroundings except at ends. Thermal conductivities of copper, brass and steel are 0.92, 0.26 and 0.12 CGS units respectively. Rate of heat flow through copper rod is:

• 1.2 cal/s

• 2.4 cal/s

• 4.8 cal/s

• 6.0 cal/s

C.

4.8 cal/s

590 Views

19.

Assume that an electric field  exists in space. Then the potential difference VA – VO, where VO is the potential at the origin and VA the potential at x = 2 m is:

• 120 J

• -120 J

• -80 J

• 80 J

C.

-80 J

As we know, potential difference VA -Vo is
dV = - Edx

172 Views

20.

A particle moves with simple harmonic motion in a straight line. In first τ s, after starting from rest it travels a distance a, and in next τ s it travels 2a, in the same direction, then

• amplitude of motion is 3a

• time period of oscillations is 8τ

• amplitude of motion is 4a

• time period of oscillations is 6τ

D.

time period of oscillations is 6τ

711 Views