A cylindrical metallic rod in thermal contact with two reservation of heat at its two ends conducts an amount of heat `Q` in time `t`. The metallic rod is melted and the material is formed into a rod of half the radius of the original rod. What is the amount of heat conducted by the new rod when placed in thermal contact with the two reservation in time t?

Options

A. `Q//4`
B. `Q//16`
C. `2Q`
D. `Q//2`

Answer

B

Explanation

In 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 coefficent of thermal conductivity of material of rod
`rArr (Q)/(t) prop (A)/(l) prop (r^2)/(l)`..(i)
As the metallic rod is melted and the material is formed into a rod of half the radius.
`V_(1)=V_(2)`
`pir_(1)^(2)l_(1) = pir_(2)^(2)l_(2)`
`rArr l_(1) = (l_2)/(4) `...(ii)
Now from equs. (i) and (ii)
`(Q_1)/(Q_2)=(r_(1)^(2))/(l_1) xx (l_2)/(r_(2)^(2)) = (r_(1)^(2))/(l_1) xx (4l_1)/((r_(1)//2)^(2))`
`rArr Q_(1) = 16Q_(2)`.