When `50Omega` resistance is connected to one voltmeter, its range is obtained upto V volt. Now, instead of `50Omega`, when `500Omega` resistance is connected to this voltmeter, its range is obtained upto 2 V volt. Find resistance of this voltmeter.

Options

A. `100Omega`
B. `200Omega`
C. `300Omega`
D. `400Omega`

Answer

D

Explanation

We have `R=G(n-1)" "...(1)`
(i) `R_(1)=G(n_(1)-1)`
`therefore50=G(n_(1)-1)=Gn_(1)-G" "...(2)`
(ii) `R_(2)=G(n_(2)-1)`
`therefore500=G(2n_(1)-1)=2Gn_(1)-G" "...(3)`
Multiplying equation (2) by 2, we get,
`100=2Gn_(1)-2G" "...(4)`
Subtracting equation (4) from equation (3) we get,
400 = G
`thereforeG=400Omega`