Ice starts forming in a lake with water at 0ºC when the atomspheric temperature is –10ºC. If the time taken for the first 1 cm of ice to be formed is 7 hours, then the time taken for the thickness ofice to change from 1 cm to 2 cm is

7 hours
14 hours
21 hours
3.5 hours

Two solid spheres, of radii R1 and R2are made of the same material and have similar surfaces.The spheres are raised to the same temperature and then allowed to cool underidentical conditions. Assuming spheres to be perfect conductors of heat, their initial rates of loss of heat are

\(R_{1}^{2}/R_{2}^{2}\)
R₁/R₂
R₂/R₁
\(R_{2}^{2}/R_{1}^{2}\)

Solution

Initial rate of loss of heat = \(\frac{\sigma T^{4}\times A_{1}\times C}{\sigma T^{4}\times A_{2}\times C}=\frac{R_{1}^{2}}{R_{2}^{2}}\)

A body of mass 5kg falls from a height of 20 metres on the ground and it rebounds to a height of 0.2 m. If the loss in potential energy is used up by the body,then what will be the temperature rise?(specific heat of material = 0.09 cal gm^-1ºC^-1)

0ºC
4ºC
8ºC
None of these

Solution

Steam at 100ºC is passed into 1.1 kg of water contained in acalorimeter of water equivalent 0.02 kg at 15ºC till the temperature of the calorimeter and its contents rises to80ºC.The mass of the steam condensed in kg is

0.130
0.065
0.260
0.135

Solution

mL + m(100 - 80)

= 1.1 × 1 × (80 –15) + 0.02 × (80 - 15)

m × 540 + 20 m = 71.5 + 1.30

560m = 72.80 ∴ m = 0.130

A beaker contains 200 gm of water. The heat capacity of the beaker is equal to that of 20 gm of water. The initial temperatue of water in the beaker is 20ºC. If 440 gm of hot water at 92ºC is poured in it, the final temperature, neglect in gradiation loss, will be nearest to

58ºC
68ºC
73ºC
78ºC

Solution

Let the final temperature be T.

Then 200 × 1 ×(T – 20)+ 20 ×(T– 20)

= 440 (92 –T)

Solving it, we get T = 68ºC.

The coefficient of apparent expansion of mercury in a glass vessel is 153 × 10^–6/ºC and in a steel vessel is 144 × 10^–6/ºC.If α for steel is 12 × 10^–6/ºC, then, that of glass is

9 × 10^–6/ºC
6 × 10^–6/ºC
36 × 10^–6/ºC
27 × 10^–6/ºC

Solution

A rectangular block is heated from 0ºC to 100ºC. The percentage increase in its length is 0.10%. What will be the percentage increase in its volume?

0.03%
0.10%
0.30%
None of these

Solution

A glass flask of volume 1 litre is fully filled with mercury at0ºC. Both the flask and mercury are now heated to 100ºC. If the coefficient of volume expansion of mercury is 1.82 × 10^–4/ºC, volume coefficient of linear expansion of glass is 10 × 10^–6/ºC, the amount of mercury which is spilted out is

15.2 ml
17.2 ml
19.2ml
21.2 ml

Solution

ΔV = V₀(γ_m-γ_g)ΔT

=1[1.82 × 10^-4 - 3 ×(10 × 10^-6)] = 100

=1[1.82 × 10^-4 - 0.3 × 10^-4] = 15.2ml

A crystal has a coefficient of expansion 13×10^-7 in one direction and 231 × 10^-7 in every direction at right angles to it. Then the cubical coefficient of expansion is

462 × 10^-7
244 × 10^-7
475 × 10^-7
257 × 10^-7

Solution

γ = α₁ + α₂ + α₃

=13 × 10^-7 + 231 × 10^-7 + 231 × 10^-7

= 475 × 10^-7

Two straight metallic strips each of thickness t and length l are rivetted together. Their coefficients of linear expansions are α₁ and α₂. If they are heated through temperature ΔT,the bimetallic strip will bend to form an arc of radius

t/{(₁+α₂)ΔT}
t/{(α₂-α₁)ΔT}
t(α₁+α₂)ΔT
t(α₂-α₁)ΔT

Solution

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