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 $$4.0$$ 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:
1. $$44.0$$ J/s
2. $$16.8$$ J/s
3. $$8.0$$ J/s
4. $$4.0$$ J/s

Subtopic:  Conduction |
83%
From NCERT
NEET - 2015
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Two metal rods $$1$$ and $$2$$ of same lengths have the same temperature difference between their ends. Their thermal conductivities are $${K}_1$$ and $${K}_2$$ and cross-sectional areas $${A}_{1}$$ and $${A}_{2},$$ respectively. If the rate of heat conduction in $$1$$ is four times that in $$2,$$ then:
1. $$K_1 A_1=4K_2 {A}_2$$
2. $$K_1 {A}_1=2 {K}_2 {A}_2$$
3. $$4 {K}_1{A}_1={K}_2 {A}_2$$
4. $${K}_1 {A}_1={K}_2 {A}_2$$
Subtopic:  Conduction |
74%
From NCERT
NEET - 2013
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A slab of stone with an area $$0.36~\text{m}^{2}$$${}^{}$ 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 $$= 3.36\times10^{5}~\text{JKg}^{-1}$$${}^{}$)
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. $$1.24~\text{J/m/s/}^{\circ}\text{C}$$

Subtopic:  Conduction |
58%
From NCERT
AIPMT - 2012
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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 at the same time?

1. Q /4

2. Q/16

3. 2Q

4. Q/2

Subtopic:  Conduction |
60%
From NCERT
AIPMT - 2010
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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. $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KL}\left({\mathrm{T}}_{1}-{\mathrm{T}}_{2}\right)}{\mathrm{A}}$

2. $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{K}\left({\mathrm{T}}_{1}-{\mathrm{T}}_{2}\right)}{\mathrm{LA}}$

3. $\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{KLA}\left({\mathrm{T}}_{1}-{\mathrm{T}}_{2}\right)$

4. $\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{KA}\left({\mathrm{T}}_{1}-{\mathrm{T}}_{2}\right)}{\mathrm{L}}$

Subtopic:  Conduction |
89%
From NCERT
AIPMT - 2009
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