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 Multiple Choice QuestionsMultiple Choice Questions

1.

A radiation of energy E falls normally on a perfectly reflecting surface. The momentum transferred to the surface is

  • E/c

  • 2E/c

  • Ec
  • Ec

B.

2E/c

Initial momentum of surface
pi = E/c
where c = velocity of light (constant)
Since, the surface is perfectly so, the same momentum will be reflected completely
Final momentum
pf= E/c
therefore
Change in momentum
∆p = pf − pi
= -E/c - E/c = -2E/c
Thus, momentum transferred to the surface is
∆p' = |∆p| = 2E/c

171 Views

2.

Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of nitrogen at temperature T0, while Box B contains one mole of helium at temperature (7/3) T0. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature. (Ignore the heat capacity of boxes). Then, the final temperature of the gases, Tf, in terms of T0 is

  • straight T subscript straight f space equals space 5 over 2 straight T subscript 0
  • straight T subscript straight f space equals space 3 over 7 straight T subscript 0
  • straight T subscript straight f space equals space 7 over 3 straight T subscript 0
  • straight T subscript straight f space equals space 7 over 3 straight T subscript 0

D.

straight T subscript straight f space equals space 7 over 3 straight T subscript 0
increment straight U space equals space 0
rightwards double arrow 3 over 2 straight R left parenthesis straight T subscript straight f minus straight T subscript 0 right parenthesis space plus space 1 space straight x space 5 over 2 space straight R space left parenthesis straight T subscript straight f minus 7 over 3 straight T subscript 0 right parenthesis equals 0
straight T subscript straight f space equals space 3 over 2 straight T subscript 0
343 Views

3.

The temperature of two outer surfaces of a composite slab, consisting of two materials having coefficients of thermal conductivity K and 2K and thickness x and 4x, respectively are T2 and T1 (T2 > T1). The rate of heat transfer through the slab, in a steady state isopen parentheses fraction numerator straight A left parenthesis straight T subscript 2 minus straight T subscript 1 right parenthesis straight K over denominator straight x end fraction close parentheses straight f with f, equal to

  • 1

  • 1/2

  • 2/3

  • 2/3


D.

2/3



increment straight q space equals kA over straight x space open square brackets straight T subscript 2 space minus fraction numerator 2 straight T subscript 2 minus straight T subscript 1 over denominator 3 end fraction close square brackets
space equals space fraction numerator kA over denominator 3 straight x end fraction space left square bracket straight T subscript 2 minus straight T subscript 1 right square bracket
571 Views

4.

Assuming the sun to be a spherical body of radius R at a temperature of T K, evaluate the total radiant power, incident on Earth, at a distance r from the Sun.

where r0 is the radius of the Earth and σ is Stefan’s constant

  • fraction numerator straight R squared σT to the power of 4 over denominator straight r squared end fraction
  • fraction numerator 4 πr subscript 0 superscript 2 straight R squared σT to the power of 4 over denominator straight r squared end fraction
  • fraction numerator πr subscript 0 superscript 2 straight R squared σT to the power of 4 over denominator straight r squared end fraction
  • fraction numerator πr subscript 0 superscript 2 straight R squared σT to the power of 4 over denominator straight r squared end fraction

C.

fraction numerator πr subscript 0 superscript 2 straight R squared σT to the power of 4 over denominator straight r squared end fraction
627 Views

5.

A light ray is incident perpendicular to one face of a 90° prism and is totally internally reflected at the glass-air interface. If the angle of reflection is 45°, we conclude that the refractive index n

  • n<2

  • straight n greater than square root of 2
  • straight n space greater than space fraction numerator 1 over denominator square root of 2 end fraction
  • straight n space greater than space fraction numerator 1 over denominator square root of 2 end fraction

B.

straight n greater than square root of 2

Angle of incidence i > C for total internal reflection.
Here i = 45° inside the medium. ∴ 45° > sin−1 (1/n)
⇒ n > √2.

657 Views

6.

If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio of the radiant energy received on earth to what it was previously will be

  • 4

  • 16

  • 32

  • 32


D.

32

straight E space proportional to space AT to the power of 4
straight E proportional to space straight r squared straight T to the power of 4
straight E subscript 1 over straight E subscript 2 space equals space fraction numerator straight r subscript 1 superscript 2 space straight T subscript 1 superscript 4 over denominator straight t subscript 2 superscript 2 straight T subscript 2 superscript 4 end fraction
straight E subscript 1 over straight E subscript 2 space space equals space fraction numerator straight R squared straight T to the power of 4 over denominator left parenthesis 2 straight R right parenthesis squared space straight x space left parenthesis 2 straight T right parenthesis to the power of 4 end fraction

straight E subscript 1 over straight E subscript 2 space space equals space fraction numerator straight R squared straight T to the power of 4 over denominator 4 straight R squared space straight x space 16 straight T to the power of 4 end fraction
straight E subscript 1 over straight E subscript 2 space space equals 64
493 Views

7.

The figure shows a system of two concentric spheres of radii r1 and r2 and kept at temperatures T1 and T2 respectively. The radial rate of flow of heat in a substance between the two concentric sphere is proportional to

  • fraction numerator straight r subscript 2 minus straight r subscript 1 over denominator straight r subscript 1 straight r subscript 2 end fraction
  • ln space open parentheses straight r subscript 2 over straight r subscript 1 close parentheses
  • fraction numerator straight r subscript 1 straight r subscript 2 over denominator straight r subscript 2 minus straight r subscript 1 end fraction
  • fraction numerator straight r subscript 1 straight r subscript 2 over denominator straight r subscript 2 minus straight r subscript 1 end fraction

C.

fraction numerator straight r subscript 1 straight r subscript 2 over denominator straight r subscript 2 minus straight r subscript 1 end fraction
open parentheses dQ over dt close parentheses space equals space open parentheses straight T subscript 1 minus straight T subscript 2 close parentheses fraction numerator 4 πr subscript 1 straight r subscript 2 straight K over denominator left parenthesis straight r subscript 2 minus straight r subscript 1 right parenthesis end fraction
511 Views

8.

One end of a thermally insulated rod is kept at a temperature T1 and the other at T2. The rod is composed of two sections of lengths

  • (k2

  • (k2

  • (k1

  • (k1


C.

(k1

fraction numerator left parenthesis straight T subscript 1 minus straight T right parenthesis straight k subscript 1 over denominator calligraphic l subscript 1 end fraction space equals space fraction numerator left parenthesis straight T minus straight T subscript 2 right parenthesis straight k subscript 2 over denominator calligraphic l subscript 2 end fraction
straight T space equals space fraction numerator straight T subscript 1 straight k subscript 1 calligraphic l subscript 2 space plus space straight T subscript 2 straight k subscript 2 calligraphic l subscript 1 over denominator straight k subscript 1 calligraphic l subscript 2 space plus straight k subscript 2 calligraphic l subscript 1 end fraction
203 Views

9.

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

  • doubled

  • four times

  • one fourth

  • one fourth


A.

doubled

straight H space equals space fraction numerator straight V squared increment straight t over denominator straight R end fraction
straight H apostrophe space equals space space fraction numerator straight V squared over denominator straight R apostrophe end fraction. increment straight t
Given space straight R apostrophe space equals space straight R divided by 2
311 Views

10.

Time taken by a 836 W heater to heat one litre of water from 10°C to 40°C is

  • 50 s

  • 100 s

  • 150 s

  • 150 s


C.

150 s

Let time taken in boiling the water by the heater is t sec. Then
Q = ms ∆ T

fraction numerator 836 space over denominator 4.2 end fraction straight t space equals space 1 space straight x space 1000 space left parenthesis 40 to the power of 0 minus 10 to the power of 0 right parenthesis
fraction numerator 836 over denominator 4.2 end fraction space straight t space equals space 1000 space straight x space 30
space straight t space equals space fraction numerator 1000 space space straight x space 30 space straight x space 4.2 over denominator 836 end fraction space equals space 150 space straight s

309 Views