Masses fo three are in the ratio `1:3:5` their lengths are in the ratio `5:3:1` when they are connected in series to an external source, the amounts of heats produced in them are in the ratio

Options

A. `1:3:5`
B. `5:3:1`
C. `1:15:125`
D. `125:15:1`

Answer

d

Explanation

`H = I^(2)R t = I^(2)(rho//t)/(A) = (I^(2)rho//t)/((m//ld)) = (I^(2)rhol^(2)dt)/(m)( :' m = A l d)`
So `H prop l^(2)//m`
Hence `H_(1): H_(2) : H_(3) = (5^(2))/(1) : (3^(2))/(3) : (1^(2))/(5) = 125: 15:1`