Subject

Physics

Class

JEE Class 12

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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

straight I space equals space 2500 over 220 space straight A space equals space 125 over 11 space equals space 11.3 space straight A
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, 
straight lambda over 4 space equals space 0.85
rightwards double arrow straight lambda space 4 space straight x space 0.85
As space we space know comma space straight v space equals space straight c over straight lambda
rightwards double arrow space fraction numerator 340 over denominator 4 space straight x space 0.85 end fraction space equals space 100 space Hz
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,
fraction numerator dl over denominator 1 plus 1 end fraction space equals space 1000 over straight T space dV
dl space equals space fraction numerator 1000 over denominator straight l plus straight T end fraction space equals space 1000 over straight T space dV
dl space equals space 1000 over straight T space straight x space left parenthesis straight l plus 1 right parenthesis space dV
rightwards double arrow space dl space equals space 1000 over 300 space straight x space left parenthesis 5 plus 1 right parenthesis space straight x 0.01 space equals space 0.2 space mA
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, 

increment straight U subscript AB space equals space nC subscript straight V left parenthesis straight T subscript straight B minus straight T subscript straight A right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 800 minus 400 right parenthesis space equals space 1000 straight R
increment straight U subscript BC space equals space nC subscript straight V left parenthesis straight T subscript straight C minus straight T subscript straight B right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 600 minus 400 right parenthesis space equals space minus 500 straight R
increment straight U subscript Total space equals space 0
increment straight U subscript CA space equals space nC subscript straight V left parenthesis straight T subscript straight A minus straight T subscript straight C right parenthesis space equals space 1 space straight x space fraction numerator 5 straight R over denominator 2 end fraction left parenthesis 400 minus 600 right parenthesis space equals space minus 500 straight R space

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.
Radio waves are used for broadcasting
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, straight E space equals straight sigma over straight epsilon subscript 0
When dielectric is introduced between parallel plates of capacitor, 

straight E to the power of apostrophe space equals space straight sigma over Kε subscript 0
Electric field inside dielectric 

straight sigma over Kε subscript 0 space equals space 3 space straight x space 10 to the power of 4
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.

straight p subscript 1 space equals straight p subscript 0 space plus ρgh

.
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


dQ subscript 1 over dt space equals space dQ subscript 2 over dt space equals space dQ subscript 3 over dt
rightwards double arrow fraction numerator 0.92 left parenthesis 100 minus straight T right parenthesis over denominator 46 end fraction space equals space fraction numerator 0.26 space left parenthesis straight T minus 0 right parenthesis over denominator 13 end fraction plus fraction numerator 0.12 space left parenthesis straight T minus 0 right parenthesis over denominator 12 end fraction
rightwards double arrow straight T space equals space 40 to the power of straight o straight C
dQ subscript 1 over dt space equals space fraction numerator 0.92 space straight x space 4 space left parenthesis 100 minus 40 right parenthesis over denominator 40 end fraction space equals space 4.8 space cal divided by straight s
590 Views

19.

Assume that an electric field straight E equals space 30 space straight x squared space straight i with hat on top 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
integral subscript straight V subscript straight o end subscript superscript straight V subscript straight A end superscript dV space equals negative integral subscript 0 superscript 2 30 space straight x squared dx
straight V subscript straight A space minus straight V subscript straight O space equals space minus space 30 space straight x space open square brackets straight x cubed over 3 close square brackets squared space equals negative 10 space straight x left square bracket 2 cubed minus 0 cubed right square bracket
space equals space minus 10 space straight x 8 space equals space minus 80 space straight J

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

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