Two wires A and B of same material end mass have their length is the ratio `1: 3`. On connecting them, one at a time in the same source of emf, the rate of heat dissipated in `B` is `10 W`. What is the rate of heat dissipated in A ?

Answer

`90 W`

Explanation

Let `a_(1) ,l_(1)` be the area of cross - section and length of wire A and `a_(2),l_(2)` be the area of cross - section and length of wire B .As the two wire are of same meterial and mass , so their volume is same .Hence
`a_(1) l_(1) = a_(2),l_(2) or (a_(1))/(a_(2)) = (i_(1))/(l_(2)) = (3l)/(I) = 3 or a_(1) = 3a_(2)`
If `epsilon` is the emf of the source , then the rate of dissipation of heat in wire B is
`H_(B) = (epsilon^(2))/(R_(2)) = 10 or (epsilon^(2))/(rho l_(2)//a_(2)) = 10 `
or `(epsilon^(2))/(rho 3//l a_(2)) = 10 or (epsilon^(2)a_(2))/(rho l) = 30 [," l_(2) = 3 l]`
Rate of head dissipation in wire A is
`H_(A) = (epsilon^(2))/(R_(1)) = (epsilon^(2))/(rho lla_(1)) = (epsilon^(2))/(rho l) (3a_(2))`
`= 3 xx 30= 90 W`