If same tensile force is applied on two wires, there will an extension of `1xx10^(-3)` m in them. The Young's moduli and radii of these wires are `10xx10^(10)N//m^(2)` and `20xx10^(10)N//m^(2)` and `R_(1)` and `R_(2)` respectively then `R_(2)` is equal to

Options

A. `sqrt(2)R_(1)`
B. `R_(1)//sqrt(2)`
C. `4R_(1)`
D. `R_(1)//4`

Answer

B

Explanation

`(y_(1))/(y_(2))=((R_(2))/(R_(1)))^(2)`
`=(10xx10^(10))/(20xx10^(10))=((R_(2))/(R_(1)))^(2)`
`=(1)/(sqrt(2)) = (R_(2))/(R_(1)) " " therefore R_(2)=(R_(1))/(sqrt(2))`