The same mass of copper is drawn into two wires 2 mm and 3mm thick. The two wires are connected in series and current is passed through them. The ratio of heat produced in the two wires is

Options

A. `(9)/(4)`
B. `(3)/(2)`
C. `(2)/(3)`
D. `(81)/(16)`

Answer

D

Explanation

`m=A. lx =pi (d^(2))/(4) l. x, x`= Density
` implies " " d_(1)^(2)l_(1)=d_(2)^(2)l_(2)` (as mass is same)
In series, current will be same.
So, `(H_(1))/(H_(2))=(i^(2)R_(1))/(i^(2)R_(2))=((rho(l_(1))/(A_(1))))/(rho((l_(2))/(A_(2))))=(l_(1))/(l_(2))(A_(2))/(A_(1))=((d_(2)^(2))/(d_(1)^(2)))(pi(d_(2)^(2))/(4))/(pi(d_(1)^(2))/(4))`
`(d_(2)^(4))/(d_(1)^(4))=((3)/(2))^(4)=(81)/(16)`