# 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 ?

$(a)\;t_{0}\qquad(b)\;2t_{0}\qquad(c)\;4t_{0}\qquad(d)\;\large\frac{t_{0}}{2}$

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