Correct Answer - B
In a steady state the amount of heat flowing from one face to the other face in time t is given by `Q = (KA(theta_(1)-theta_(2))t/l)` where K is coefficient of thermal conductivity of material of rod
`rArr alpha/l alpha (A/l) alpha (r^(2)/l)`................(i)
As the metallic rod is melted and hte material is formed into a rod of half the radius
`V_(1) = V_(2)rArr pir_(1)^(2)l_(1)=pir_(2)^(2)l_(2) rArr l_(1) = (l_(2)/4)`............(ii)
Now, from eqs (i) and (ii), we get
`Q_(1)/Q_(2) = r_(1)^(2)/l_(1) xx (l_(2)/r_(2)^(2) = (r_(1)^(2)/l_(1) xx (4l_(1))/(r_(2)//2)^(2) rArr` `Q_(1) = 16Q_(2) rArr Q_(2) = Q/16`