The density of water at \(20^\circ \text{C}\) is \(998\) kg/m³ and at \(40^\circ \text{C}\) is \(992\) kg/m³. The coefficient of volume expansion of water is: 
1. \(3 \times 10^{-4} / ^\circ\text C\)
2. \(2 \times 10^{-4} / ^\circ\text C\)
3. \(6 \times 10^{-4} / ^\circ\text C\)
4. \(10^{-4} / ^\circ\text C\)

A slab of stone with an area \(0.36~\text{m}^{2}\)${\mathrm{}}^{}$ and thickness of \(0.1~\text{m}\) is exposed on the lower surface to steam at \(100​​^{\circ}\mathrm{C}\)$$. A block of ice at \(0^{\circ}\mathrm{C}\) rests on the upper surface of the slab. In one hour \(4.8~\text{kg}\) of ice is melted. The thermal conductivity of the slab will be: (Given latent heat of fusion of ice \(= 3.36\times10^{5}~\text{JKg}^{-1}\)${\mathrm{}}^{}$)
1. \(1.29~\text{J/m/s/}^{\circ}\text{C}\)
2. \(2.05~\text{J/m/s/}^{\circ}\text{C}\)
3. \(1.02~\text{J/m/s/}^{\circ}\text{C}\)
4. \(1.24~\text{J/m/s/}^{\circ}\text{C}\)

A black body at 1227 °C emits radiations with maximum intensity at a wavelength of 5000 Å. If the temperature of the body is increased by 1000 °C, the maximum intensity will be observed at:

1. 4000 Å

2. 5000 Å

3. 6000 Å

4. 3000 Å

A black body is at 727 °C. It emits energy at a rate that is proportional to:

1. $(727{)}^2$

2. $(1000{)}^4$

3. $(1000{)}^2$

4. $(727{)}^4$

Assuming the sun to have a spherical outer surface of radius r, radiating like a black body at temperature t °C, the power received by a unit surface of the earth (normal to the incident rays) at a distance R from the centre of the sun is: 
(where σ is Stefan’s constant.)

1. $\frac{4{\mathrm{\pi r}}^2{\mathrm{\sigma t}}^4}{R^2}$

2. $\frac{r^2\sigma (t+273{)}^4}{4\mathrm{\pi }{R}^2}$

3. $\frac{16{\pi }^2{r}^2{\mathrm{\sigma t}}^4}{R^2}$

4. $\frac{r^2\sigma (t+273{)}^4}{R^2}$

If the cold junction of a thermocouple is kept at 0 °C and the hot junction is kept at T °C, then the relation between neutral temperature (${\mathrm{T}}_{\mathrm{n}}$) and temperature of inversion (${\mathrm{T}}_{\mathrm{i}}$) is:

1. ${\mathrm{T}}_{\mathrm{n}}=\frac{{\mathrm{T}}_{\mathrm{i}}}{2}$

2. ${\mathrm{T}}_{\mathrm{n}}$= 2${\mathrm{T}}_{\mathrm{i}}$

3. ${\mathrm{T}}_{\mathrm{n}}$= ${\mathrm{T}}_{\mathrm{i}}$ - T

4. ${\mathrm{T}}_{\mathrm{n}}$= ${\mathrm{T}}_{\mathrm{i}}$ + T

On a new scale of temperature, which is linear and called the \(\mathrm{W}\) scale, the freezing and boiling points of water are \(39^\circ ~\mathrm{W}\)$$and \(239^\circ ~\mathrm{W}\) respectively. What will be the temperature on the new scale corresponding to a temperature of \(39^\circ ~\mathrm{C}\) on the Celsius scale?
1. \(78^\circ ~\mathrm{C}\)
2. \(117^\circ ~\mathrm{W}\)
3. \(200^\circ ~\mathrm{W}\)
4. \(139^\circ ~\mathrm{W}\)

The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T₁ and T₂ (T₁> T₂). The rate of heat transfer $\frac{\mathrm{dQ}}{\mathrm{dt}}$ through the rod in a steady state is given by:

1. $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KL}({\mathrm{T}}_1-{\mathrm{T}}_2)}{\mathrm{A}}$

2. $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{K}({\mathrm{T}}_1-{\mathrm{T}}_2)}{\mathrm{LA}}$

3. $\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{KLA}({\mathrm{T}}_1-{\mathrm{T}}_2)$

4. $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KA}({\mathrm{T}}_1-{\mathrm{T}}_2)}{\mathrm{L}}$

A black body at \(227^{\circ}~\mathrm{C}\) radiates heat at the rate of \(7~ \mathrm{cal-cm^{-2}s^{-1}}\).  At a temperature of \(727^{\circ}~\mathrm{C}\), the rate of heat radiated in the same units will be:
1. \(60\)
2. \(50\)
3. \(112\)
4. \(80\)