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A rod of radius ' r ' & length ' l ' is beaten such that its length doubles . If it took $\;t_{0}\;$ seconds to transfer heat Q , how much time will it take now ?


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Answer : $\;4t_{0}$
Explanation :
Since , the volume remains same
Since , $\;\rho = \large\frac{KA}{l}\; \bigtriangleup T$
$t \propto \large\frac{l}{A}$
$\large\frac{t_{1}}{t_{2}}= \large\frac{l_{1}}{A_{1}}\; . \large\frac{A_{2}}{l_{2}}$
$t_{2} = t_{0} \large\frac{A_{1} l_{2}}{A_{2} l_{1}}=t_{0} (2.2)=4t_{0}$
answered Mar 13, 2014 by yamini.v

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