200 `Omega` resistor is connected in one of the gaps of the meter bridge. Series combination of X `Omega` and 50 `Omega` resistors is connected in the second gap. Here unknown resistance X `Omega` is kept in a heat bath at a certain temperature. The unknown resistance and its temperature is....... and....... respectively if the balance point is obtained at 50 cm. The total length of the wire of the meter bridge is equal to 1 meter. The resistance of the unknown resistance at `0" "^(@)C ` temperature is equal to 100 `Omega . alpha = 0.5 xx 10^(-3) ""^(@) C^(-1)` for the material of the X `Omega` resistors.

Explanation

`R_(1) = 200 Omega, " " R_(2) = (X + 50) Omega`
`l_(1) = 50 cm , " " l_(2) = 100 - 50 = 50 cm `
Here , we have `(R_(1))/(R_(2)) = (l_(1))/(l_(2))`
`therefore (200)/(X + 50) = (50)/(50) `
`therefore X +50 = 200 `
`therefore X = 150 Omega`
Now, X = `X_(0) [ 1 + alpha (theta - 0 ) ] `
`therefore 150 = 100 [ 1 + 5 xx 10^(-3) theta ]`
`therefore 1.5 = 1+ 5 xx 10^(-3) theta`
`therefore theta = 100^(@) `C