The resistance of a wire of iron is `10 ohm` and temperature coefficient of resistivity is `5 xx 10^-3//.^@C`, At `20^@C` it carries `30 mA` of current. Keeping constant potential difference between its ends. The temperature of the wire is raised to `120^@C`. The current in `mA` that flows in the wire now is.

Options

A. 20
B. 15
C. 10
D. 40

Answer

A

Explanation

`R_(120) = R_(20) [1 + alpha(120 - 20)]`
=`10[1 + 5 xx 10^-3 xx 100] = 15 Omega`
`V = i R_(20) = i_2 R_(120)`
`30 xx 10 = i_2 xx 15 rArr i_2 = 20 mA`.