A cylindrical rod with one end in steam chamber and the other 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 second rod is 1/4of that of first, the rate at which ice melts in g/s will bea) 0.4b) 0.05c) 0.2d) 0.01

Since the rate of melting of ice is directly proportional to the amount of heat transfered per unit time,

Then calculating the mag. of heat currents set up in the two rod would give u a fair estimate how fast is the rate.

dQ/dt=(KA/L)T where t is the temperature difference b/w the two.

For First rod , dQ₁/dt=(K₁A₁/L₁)T

For the second rod, dQ₂/dt=(K₂A₂/L₂)T

 Since for the second rod , L₂=L₁/2 , R₂= 2R₁ === > A₂=4A₁ , K₂= (1/4)K₁

Then dQ₂/dt=(K₁8A₁/4L₁)T = 2(K₁A₁/L₁)T

Since the rate of heat tranfer is 2 times the first. then the rate of melting would be 2 times as the first = 0.2g/s