Thermal Physics - NEET Physics Questions
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Thermal Physics

Question 81: moderate

The two ends of a metal rod are maintained at temperatures $100^circ text{C}$ and $110^circ text{C}$. The rate of heat flow in the rod is found to be $4.0 text{ J/s}$. If the ends are maintained at temperatures $200^circ text{C}$ and $210^circ text{C}$, the rate of heat flow will be:
(2015)

1. $16.8 text{ J/s}$
2. $8.0 text{ J/s}$
3. $4.0 text{ J/s}$
4. $44.0 text{ J/s}$
View Answer

Rate of heat flow $frac{dQ}{dt} = frac{K A Delta T}{L}$
Since $Delta T$ is same ($10^circ text{C}$) in both cases, the rate of heat flow will remain same i.e., $4.0 text{ J/s}$.

Question 82: moderate

A slab of stone of area $0.36 text{ m}^2$ and thickness $0.1 text{ m}$ is exposed on the lower surface to steam at $100^circ text{C}$. A block of ice at $0^circ text{C}$ rests on the upper surface of the slab. In one hour $4.8 text{ kg}$ of ice is melted. The thermal conductivity of slab is: (Given latent heat of fusion of ice $= 3.36 times 10^5 text{ J/kg}$)
(2012 Mains)

1. $1.24 text{ J/m/s/}^circ text{C}$
2. $1.29 text{ J/m/s/}^circ text{C}$
3. $2.05 text{ J/m/s/}^circ text{C}$
4. $1.02 text{ J/m/s/}^circ text{C}$
View Answer

Heat transferred $frac{Q}{t} = frac{K A Delta T}{x}$
$frac{m L}{t} = frac{K A (100 - 0)}{x}$
$K = frac{m L x}{t A Delta T} = frac{4.8 times 3.36 times 10^5 times 0.1}{3600 times 0.36 times 100} = 1.24 text{ J/m/s/}^circ text{C}$

Question 83: moderate

A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat $Q$ in time $t$. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod, when placed in thermal contact with the two reservoirs in time $t$?
(2010 Pre)

1. $Q/2$
2. $Q/4$
3. $Q/16$
4. $2Q$
View Answer

$Q = frac{K A Delta T}{l} t = frac{K (pi r^2) Delta T}{l} t$
Volume is constant $Rightarrow pi r^2 l = pi (r/2)^2 l' Rightarrow l' = 4l$
$Q' = frac{K (pi (r/2)^2) Delta T}{4l} t = frac{1}{16} frac{K (pi r^2) Delta T}{l} t = frac{Q}{16}$

Question 84: moderate

The two ends of a rod of length $L$ and a uniform cross-sectional area $A$ are kept at two temperatures $T_1$ and $T_2$ ($T_1 > T_2$). The rate of heat transfer, $frac{dQ}{dt}$ through the rod in steady state is given by:
(2009)

1. $frac{dQ}{dt} = frac{k (T_1 - T_2)}{LA}$
2. $frac{dQ}{dt} = k L A (T_1 - T_2)$
3. $frac{dQ}{dt} = frac{k A (T_1 - T_2)}{L}$
4. $frac{dQ}{dt} = frac{k L (T_1 - T_2)}{A}$
View Answer

Rate of heat transfer $frac{dQ}{dt} = frac{k A (T_1 - T_2)}{L}$

Question 85: moderate

Consider a compound slab consisting of two different materials having equal thicknesses and thermal conductivities $K$ and $2K$, respectively. The equivalent thermal conductivity of the slab is:
(2003)

1. $sqrt{2} K$
2. $3 K$
3. $frac{4}{3} K$
4. $- K$
View Answer

For compound slab in series, $K_{eq} = frac{l_1 + l_2}{frac{l_1}{K_1} + frac{l_2}{K_2}}$
$K_{eq} = frac{l + l}{frac{l}{K} + frac{l}{2K}} = frac{2l}{frac{3l}{2K}} = frac{4}{3} K$

Question 86: moderate

Consider two rods of same length and different specific heats ($S_1$, $S_2$), conductivities ($K_1$, $K_2$) and area of cross-sections ($A_1$, $A_2$) and both having temperature $T_1$ and $T_2$ at their ends. If rate of loss of heat due to conduction is equal, then
(2002)

1. $K_1 A_1 = K_2 A_2$
2. $frac{K_1 A_1}{S_1} = frac{K_2 A_2}{S_2}$
3. $K_2 A_1 = K_1 A_2$
4. $frac{K_2 A_1}{S_2} = frac{K_1 A_2}{S_1}$
View Answer

Rate of heat loss $frac{dQ}{dt} = frac{K A (T_1 - T_2)}{l}$
Given $(frac{dQ}{dt})_1 = (frac{dQ}{dt})_2 Rightarrow frac{K_1 A_1 (T_1 - T_2)}{l} = frac{K_2 A_2 (T_1 - T_2)}{l}$
$Rightarrow K_1 A_1 = K_2 A_2$

Question 87: moderate

A cylindrical rod having temperature $T_1$ and $T_2$ at its ends. The rate of flow of heat $Q_1 text{ cal/sec}$. If all the linear dimensions are doubled keeping temperature constant, then rate of flow of heat $Q_2$ will be:
(2001)

1. $4 Q_1$
2. $2 Q_1$
3. $frac{Q_1}{4}$
4. $frac{Q_1}{2}$
View Answer

$Q_1 = frac{K A Delta T}{l} = frac{K (pi r^2) Delta T}{l}$
When dimensions are doubled, $r' = 2r$, $l' = 2l$
$Q_2 = frac{K (pi (2r)^2) Delta T}{2l} = frac{4 K (pi r^2) Delta T}{2l} = 2 Q_1$

Question 88: moderate

Certain quantity of water cools from $70^\circ\text{C}$ to $60^\circ\text{C}$ in the first $5$ minutes and to $54^\circ\text{C}$ in the next $5$ minutes. The temperature of the surroundings is: (2014)

1. $45^\circ\text{C}$
2. $20^\circ\text{C}$
3. $42^\circ\text{C}$
4. $10^\circ\text{C}$
View Answer

Using Newton's law of cooling: $\frac{70-60}{5} = K(65-T_s) \Rightarrow 2 = K(65-T_s)$ and $\frac{60-54}{5} = K(57-T_s) \Rightarrow 1.2 = K(57-T_s)$. Dividing gives $\frac{2}{1.2} = \frac{65-T_s}{57-T_s} \Rightarrow 5(57-T_s) = 3(65-T_s)$. Solving for $T_s$, we get $T_s = 45^\circ\text{C}$.

Question 89: moderate

A beaker full of hot water is kept in a room. If it cools from $80^\circ\text{C}$ to $75^\circ\text{C}$ in $t_1$ minutes, from $75^\circ\text{C}$ to $70^\circ\text{C}$ in $t_2$ minutes and from $70^\circ\text{C}$ to $65^\circ\text{C}$ in $t_3$ minutes, then: (1995)

1. $t_1 < t_2 < t_3$
2. $t_1 > t_2 > t_3$
3. $t_1 = t_2 = t_3$
4. $t_1 < t_2 = t_3$
View Answer

According to Newton's law of cooling, the rate of cooling is directly proportional to the temperature difference between the body and surroundings. As the water cools, the temperature difference decreases, slowing the rate of cooling. Thus, successive intervals take more time, giving $t_1 < t_2 < t_3$.

Question 90: moderate

Three stars A, B, C have surface temperatures $T_A$, $T_B$, $T_C$ respectively. Star A appears bluish, star B appears reddish and star C yellowish. Hence, (2020-Covid)

1. $T_B > T_C > T_A$
2. $T_C > T_B > T_A$
3. $T_A > T_C > T_B$
4. $T_A > T_B > T_C$
View Answer

According to Wien's displacement law, $\lambda_{max} \propto \frac{1}{T}$. The wavelength of red is greater than yellow, which is greater than blue ($\lambda_B > \lambda_C > \lambda_A$). Therefore, the temperatures will be in the reverse order: $T_A > T_C > T_B$.