A cylindrical metallic rod in thermal contact with two reservoirs 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 reservoirs in time t?

Options

A. `(Q)/(4)`
B. `(Q)/(16)`
C. 2Q
D. `(Q)/(20`

Answer

B

Explanation

The amount of heat flows in time t through a cylindrical metallic rod of length L and uniform area of cross-section `A(= pi R^(2))` with its ends maintained at temperature `T_(1) and T_(2) (T_(1) gt T_(2))` is given by `Q = (KA (T_(1) - T_(2))t)/(L)` ..(i)
where K is the thermal conductivity of the material of the rod. Area of cross-section of new rod `A.= pi ((R )/(2))^(2)`
`= (pi R^(2))/(4) = (A)/(4)`..(ii)
As the volume of the rod remains unchanged
`:. AL= A.L.`
where L. is the length the new rod
or `L. = L (A)/(A.)` ...(iii)
4L (Using(ii))
Now the amount of heat flows in same time t in the new rod with its ends maintained at the same temperature `T_(1) and T_(2)` is givne by `Q. = (KA. (T_(1) - T_(2))t)/(L.)` ..(iv)
Substituting the value of A. and L. from equations (ii) and (iii) in the above equation, we get
`Q.= (K(A//4) (T_(1) - T_(2))t)/(4L) = (1)/(16) (KA (T_(1) - T_(2))t)/(L)`
` = (1)/(16)Q` (Using (i))