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NEET 2010
Exam
NEET
Year
2010
Subject
Physics
Chapter
Thermal Properties of Matter

NEET 2010 · Physics · Thermal Properties of Matter

A cylindrical metallic rod in thermal contact with two reservoirs...

A cylindrical metallic rod in thermal contact with two reservoirs of heat at its two ends conducts an amount of heat QQ 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\mathrm{t} ?

Options

  1. AQ/4Q / 4
  2. BQ/16Q / 16
  3. C2Q2 \mathrm{Q}
  4. DQ/2Q / 2

Correct answer

B. Q/16Q / 16

Step-by-step solution

In steady state the amount of heat flowing from one face to the other face in time tt is given by Q=KA(θ1−θ2)tlQ=\frac{K A\left(\theta_1-\theta_2\right) t}{l} where KK is coefficient of thermal conductivity of material of rod ⇒Qt∝Al∝r2l\Rightarrow \quad \frac{\mathrm{Q}}{\mathrm{t}} \propto \frac{\mathrm{A}}{\mathrm{l}} \propto \frac{\mathrm{r}^2}{\mathrm{l}} As the metallic rod is melted and the material is formed into a rod of half the radius \begin{aligned}\mathrm{V}_{1} & = \mathrm{V}_{2} \\ \Rightarrow \quad \pi \mathrm{r}_{1}^2 \mathrm{l}_{1} & =\pi \mathrm{r}_{2}^2 \mathrm{l}_{2} \\ \quad \quad \quad \mathrm{l} }_{1} & =\frac{\mathrm{l}_2}{4} \end{aligned}Now,fromfromEqs. (i) and (ii) ⇒Q1Q2=r12l1×l2r22=r12l1×4l1(r1/2)2⇒Q1=16Q2\begin{aligned} \Rightarrow \quad & \frac{Q_1}{Q_2}=\frac{r_1^2}{l_1} \times \frac{l_2}{r_2^2}=\frac{r_1^2}{l_1} \times \frac{4 l_1}{\left(r_1 / 2\right)^2} \\ \Rightarrow \quad Q_1 & =16 Q_2 \end{aligned}$
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