Two uniform wires A and B of same metal and have equal masses, the radius of wire A is twice that of wire B. The total resistance of A and B when connected in parallel is

Options

A. `4 Omega`, when resistance of wire A is `4.25 Omega`
B. `5 Omega`, when the resistance of wire A is `4 Omega`
C. `4Omega`, when the resistance of wire B is `4.25 Omega`
D. `5Omega`, when the resistance of wire B is `4 Omega`

Answer

A

Explanation

`r_(A)=2r_(B)`
`rho_(A)=rho_(B)`
In parallel, `(1)/(R )=(1)/(R_(A))+(1)R_(B))`
`A_(A)l_(A)=A_(B)l_(B) implies pi r_(A)^(2)l_(A)=pir_(B)^(2)l_(B)`
`(l_(A))/(l_(B))=(1)/(4) " and " (A_(A))/(A_(B))=4 implies (R_(A))/(R_(B))=(rho(l_(A))/(A_(A)))/(rho(l_(B))/(A_(B)))`
`(R_(A))/(R_(B))=(l_(A))/(l_(B))xx(A_(B))/(A_(A))=(1)/(4)xx(1)/(4)=(1)/(16)`
`16R_(A)=R_(B)`
`(1)/(R )=(1)/(R_(A))+(1)/(R_(B))`
`(1)/(R_(A))+(1)/(16R_(A))=(16+1)/(16R_(A))=(17)/(16xxR_(A))implies R=-(16R_(A))/(17)`
If `" " R_(A)=4.25 Omega implies R=(16xx4.25)/(17)=(68)/(17)`
` therefore R=4 Omega`