Question

A cylindrical rod with one end in a steam chamber and the other end in ice results in melting of 0.1 g of ice per second. If the rod is replaced by another rod with half the length and double the radius of the first and if the thermal conductivity of material of the second rod is 0.25 times that of first, the rate at which ice melts in g s^–1 will be –

• A. 0.1
• B. 0.2
• C. 1.6
• D. 3.2

Answer 1

Answer: (B)

0.2

Answer 2

For first rod $\frac{\Delta Q}{\Delta t} = \frac{\Delta m}{\Delta t}L = \frac{KA(T_{1} - T_{2})}{L}$

0.1 × L = $\frac{KA(T_{1} - T_{2})}{L}$ ….(i)

For second rod

$\frac{\Delta m}{\Delta t}$L = $\frac{0.25K \times 4A(T_{1} - T_{2})}{L/2}$….(ii)

\ $\frac{\Delta m}{\Delta t}$ = 0.2 g/s