Two rods P and Q of same length and same diameter having thermal conductivity ratio 2 : 3 joined end to end.If temperature at one end of Pis 100°C and at one and of Q 0°C, then the temperature of the interface is

40°C
50°C
60°C
70°C

Solution

In a surrounding medium of temperature 10°C, a body takes 7 min for a fall of temperature from 60°C to 40°C. In what time the temperature of the body will fall from 40°C to 28°C?

7 min
11 min
14 min
21 min

Solution

In a thermocouple,the temperature of the cold junction and the neutral temperature are –40°C and 275° Crespectively. If the cold junction temperature is increased by60°C, the neutral temperature and temperature of in version respectively become

275°C, 530°C
355°C, 530°C
375°C, 590°C
355°C, 590°C

Solution

When the temperature of cold junction is increased,the neutral temperature remains constant for a given thermocouple and it is independent of the temperature of the cold junction.

∴ θ_n= 275°C, θ_i– θ_n= θ_n– θ_c

∴ θ_i= 2θ_n– θ_c= 530°C.

Calculate the surface temperature of the planet, if the energy radiated by unit area in unit time is 5.67 × 10⁴ watt.

1273°C
1000°C
727°C
727K

Solution

The tempearture of an isolated black body falls from T₁ to T₂ in time t, then t is (Let c be a constant)

\(t=c\left ( \frac{1}{T_{2}}-\frac{1}{T_{1}} \right )\)
\(t=c\left ( \frac{1}{T_{2}^{2}}-\frac{1}{T_{1}^{2}} \right )\)
\(t=c\left ( \frac{1}{T_{2}^{3}}-\frac{1}{T_{1}^{3}} \right )\)
\(t=c\left ( \frac{1}{T_{2}^{4}}-\frac{1}{T_{1}^{4}} \right )\)

A metal cube of length 10.0 mm at 0°C (273K) is heated to 200°C (473K). Given :its coefficient of linear expansion is 2 × 10^-5 K^-1. The percent change of its volume is

Solution

Percentage change in volume = γt × 100

= 3α × 100 = 300 α t = 300 × 2 × 10^-5 × 200 = 1.2

A black body rediates energy at the rate of E watt permetre2at a high temperature T K. when the temperature is reduced to (T/2) K,the radiant energy will be

E/16
E/4
E/2
2E

The temperature of a furnace is 2324ºC and the intensity is maximum in its radiation spectrum nearly at 12000 Å. If the intensity in the spectrum of a star is maximum nearly at 4800Å, then the surface temperature of the star is

8400ºC
7200ºC
6492.5ºK
5900ºC

The maximum energy in the thermal radiation from a hot source occurs at a wavelength of 11 × 10^-5 cm. According to Wien’s law, the temperature of this source(on Kelvin scale)will be n times the temperature of another source(on Kelvin scale) for which the wavelength at maximum energy is 5.5 × 10^-5 cm. The value n is

Solution

11.0 × 10^-5 T₁ = 5.5 × 10^-5 T₂

^T₁⁄_T₂ = ¹⁄₂

A bucket full of hot water is kept in a room and it cools from 75ºC to 70ºC in T1minutes, from 70ºC to 65ºC in T₂ minutes and from 65ºC to 60ºC in T₃ minutes. Then

T₁ = T₂ = T₃
T₁ < T₂ < T₃
T₁ > T₂ > T₃
T₁ < T₂ > T

Solution

The time of cooling increases as the difference between the temperature of body & surrounding is reduced. So T₁ < T₂ < T₃(according to Newton’s Law of cooling).

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