First slide

Conduction

Question

The ratio of radii of two cylindrical rods of same material is 2 :1 and ratio of their lengths is 2 : 3. Their ends are maintained at same temperature difference. If rate of flow of heat in the longer rod is 2 Cal s^-1, then that in the shorter rod will be

Easy

A

4 Cal s^-1
B

12 Cal s^-1
C

8 Cal s^-1
D

1 Cal s^-1

Solution

$\large \frac {r_1}{r_2}=\frac 21,\frac {l_1}{l_2}=\frac 23,\left ( \frac Q2 \right )_t=2cal/sec$
$\large \frac Qt\propto \frac {r^2}{l}, i.e \frac {\left ( \frac Qt \right )_1}{\left ( \frac Qt \right )_2}=\left ( \frac {r_1}{r_2} \right )^2\frac {l_2}{l_1}$
$\large \frac {\left ( \frac Qt \right )_1}{2}=\left ( \frac 21 \right )^2\left ( \frac 32 \right )\Rightarrow \left ( \frac Qt \right )_1=12cal/sec$

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