Calculate the new resistance of the wire.

ICSE Class 10 Physics 2022 Paper

If a wire of resistance $2 Ω$ gets stretched to thrice its original length:

Question : 44 of 45

Marks: +1, -0

Solution:

Given :

R_1=2\,\Omega

;

Let

l_1=l

and

a_1=a'

\because

The wire is stretched three time, so the new length

(l_2)=3l

Let new area of crossection be

a_2

\because \;\; l_1 \times a_1 = l_2 \times a_2

(Volume of wire is constant)

\therefore \; l \times a = 3 l \times a_2

\Rightarrow \; a_2 = \frac{l \times a}{3 l} = \frac{a}{3}

We know,

R = \frac{\rho l}{a}

\therefore \;\; \frac{R_2}{R_1} = \frac{l_2}{l_1} \times \frac{a_1}{a_2} (\because \rho = \text{ constant })

\Rightarrow \; \frac{R_2}{2} = \frac{3l}{l} \times \frac{a}{\frac{a}{3}}

\Rightarrow \; \frac{R_2}{2} = 3 \times 3

\Rightarrow R_2 = 18 \Omega

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