Question

A constantan wire of length ‘l’ and area of cross-section ‘a’, has a resistance of 8.50 Ω. If another constantan wire is of length 4l and area of cross-section `a/4`, its resistance will be:

पर्याय

• 136 Ω

• 124 Ω

• 112 Ω

• 148 Ω

Answer 1

136 Ω

Explanation:

For the same material,

$$R=\rho\frac{l}{a} = 8.50\Omega$$

$$R'=\rho\frac{4l}{a/4}$$

$$=16\rho \frac{l}{a}$$

= 16 R

= 16 × 8.50

= 136 Ω