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. `5:9:5`
B. `25:9:1`
C. `125:15:1`
D. `25:15:9`

Answer

C

Explanation

Mass of wire, m = Volume `xx` Density = `Alxxd`
Resistance of wire, `R=(rhol)/(A)`
or `A=(rhol)/(R)therefore m=(rhol)/(R)xxldorR=(rhol^(2)d)/(m)i.e.,Rprop(l^(2))/(m)`
When wires are in series, there will be same current in each wire. The heat produced `HpropR`.
`therefore H_(1):H_(2):H_(3)=(l_(1)^(2))/(m_(1)):(l_(2)^(2))/(m_(2)):(l_(3)^(2))/(m_(3))=(5^(2))/(1):(3^(2))/(3):(1^(2))/(5)`
`=25:3:(1)/(5)=125:15:1`