Rate of heat flow through cylindrical rod is $Q_{1}$ . Temperatures of ends of rod are $T_{1}$ and $T_{2}$ . If all the linear dimensions of the rod become double and temperature difference remains same, its rate of heat flow is $Q_{2}$ , then

Q_{1} = 2 Q_{2}

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Q_{2} = 2 Q_{1}

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Q_{2} = 4 Q_{1}

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Q_{1} = 4 Q_{2}

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Solution

The correct option is B $Q_{2} = 2 Q_{1}$
$H = k A \frac{(Q_{1} - Q_{2})}{L} T$
Rate of neat flow
$Q = \frac{H}{T} = \frac{K A (Q_{1} - Q_{2})}{t}$
$Q_{1} \times \frac{A}{L}$ Dimension of $A r e a = A = (L^{2})$ Dimension of distance $= (L)$
$Q Q L$
$\Rightarrow \frac{Q_{1}}{Q_{2}} = \frac{L}{L_{2}} = 2$
$\Rightarrow Q = 2 Q_{2}$ .

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Rate of heat flow through cylindrical rod is Q1. Temperatures of ends of rod are T1 and T2. If all the linear dimensions of the rod become double and temperature difference remains same, its rate of heat flow is Q2, then

The correct option is B Q2=2Q1H=kA(Q1−Q2)LTRate of neat flowQ=HT=KA(Q1−Q2)tQ1×AL Dimension of Area=A=(L2) Dimension of distance =(L)QQL⇒Q1Q2=LL2=2⇒Q=2Q2.

Rate of heat flow through cylindrical rod is $Q_{1}$ . Temperatures of ends of rod are $T_{1}$ and $T_{2}$ . If all the linear dimensions of the rod become double and temperature difference remains same, its rate of heat flow is $Q_{2}$ , then