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Conduction Law

A cylindrical...

Question

A cylindrical rod has a hot end at $70^{\circ} C$ and cold end at $30^{\circ} C$ . The rate of heat flow is $9 W$ . If the dimensions of the cylindrical rod are tripled keeping the temperature difference constant, then find the value of new rate of heat flow.

9 W

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81 W

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18 W

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27 W

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Solution

The correct option is D $27 W$
Let the length of the cylindrical rod be $l_{1}$ ,
Area of cross-section of cylindrical rod be $A_{1}$ ,
Temperature difference between the ends of rod:
$(T_{1} - T_{2}) = 40^{\circ} C$
Given, Rate of heat flow:
$(Q_{1}) = 9 W$
If the dimensions tripled, then:
$l_{2} = 3 l_{1}; A_{2} = π r_{2}^{2} = π (3 r_{1})^{2} = 9 (π r_{1}^{2}) = 9 A_{1}$
$∴$ Applying the equation of thermal conductivity:
$\frac{Q_{1}}{Q_{2}} = (\frac{A_{1}}{A_{2}}) \times (\frac{l_{2}}{l_{1}}) = \frac{1}{9} \times 3 = \frac{1}{3}$
$\Rightarrow Q_{2} = 3 Q_{1} = 3 \times 9 = 27 W$

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Conduction Law

Standard XIII Physics

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A cylindrical rod has a hot end at 70∘C and cold end at 30∘C. The rate of heat flow is 9 W. If the dimensions of the cylindrical rod are tripled keeping the temperature difference constant, then find the value of new rate of heat flow.

A cylindrical rod has a hot end at $70^{\circ} C$ and cold end at $30^{\circ} C$ . The rate of heat flow is $9 W$ . If the dimensions of the cylindrical rod are tripled keeping the temperature difference constant, then find the value of new rate of heat flow.