A cylindrical rod with one end in steam and other end in ice results in melting of `0.1` g of ice per second. If the rod is replaced by another with half the length and double the radius and `1//4th` the conductivity than that of the first, the rate at which the ice melts in g/sec will be :

Options

A. 3.2
B. 1.6
C. 0.2
D. 0.1

Answer

C

Explanation

`(Q)/(t)=(kA(T_(1)-T_(2)))/(d)`
This heat is used in melting ice `:.Q=mL`.
`rArr" "(mL)/(t)=(kA(T_(1)-T_(2)))/(d)` or `m/t=(kA(T_(1)-T_(2)))/(Ld)`
`:." "(m_(1))/(t_(1))=(k_(1)pir_(1)^(2)(T_(1)-T_(2)))/(Ld_(1))`
`(m_(2))/(t_(2))=(k_(2)pir_(2)^(2)(T_(1)-T_(2)))/(Ld_(2))`
`:.(m_(1))/(t_(1))xx(t_(2))/(m_(2))=(k_(1))/(k_(2))((r_(1))/(r_(2)))^(2)*(d_(2))/(d_(1))=0.1xx(t_(2))/(m_(2))=(1)/(1//4)xx(1/2)^(2)xx(1//2)/(1)=1/2`
`(m_(2))/(t_(2))=0.2" g/sec"`
Thus correct choice is (c ).