Resistance of a resistor wire is 5 `Omega " at " 50" "^(@)C and 6 Omega " at " 100 " "^(@)C` , then its resistance at `0" "^(@) C ` will be .... .

Options

A. `2 Omega`
B. `1 Omega`
C. `3 Omega`
D. `4 Omega`

Answer

D

Explanation

`4 Omega`
`R_(t) = R_(0) (1 + alpha (t - t_(0))`
`therefore 5 = R_(0) (1 + alpha (50 - 0)) `
`therefore 5 = R_(0) (1 + 50 alpha) " " `.... (1)
and 6 = `R_(0) (1 + alpha (100- 0)) `
` therefore 6 = R_(0) ( 1+ 100 alpha)` .... (2)
`therefore (5)/(6) = (1 + 50 alpha)/(1 + 100 alpha) `
`therefore 5 + 500 alpha = 6 + 300 alpha`
`therefore 200 alpha = 1 `
`therefore alpha = (1)/(200 )`
Now , by using `alpha = (1)/(200)` in equation (1),
`5 = R_(0) [ 1 + 50 xx (1)/(200) ]`
`therefore 5 = R_(0) [ 1 + (1)/(4) ]`
`therefore 5 = R_(0) xx (5)/(4)`
`therefore R_(0) = 4 Omega`