A wire of $1 Ω$ has a length of $1 m$ . It is stretched till its length increases by $25$ %. The percentage change in the resistance to the nearest integer is -

25 %

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12.5 %

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76 %

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56 %

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Solution

The correct option is D $56 %$
We know that,
$R = ρ \frac{l}{A}$
So,
$\frac{R_{1}}{R_{2}} = \frac{l_{1} A_{2}}{A_{1} l_{2}} . . . (1)$

Also, volume remain constant. So,
$A_{1} l_{1} = A_{2} l_{2}$
$\Rightarrow \frac{A_{2}}{A_{1}} = \frac{l_{1}}{l_{2}} . . . (2)$

From $(1)$ and $(2)$ ,

$\frac{R_{1}}{R_{2}} = \frac{l_{1}^{2}}{l_{2}^{2}}$

$\Rightarrow \frac{1}{R_{2}} = \frac{(l)^{2}}{(1.25 l)^{2}}$

$\Rightarrow R_{2} = 1.5625 Ω$

Hence, percentage change in the resistance,
$= \frac{1.5625 - 1}{1} \times 100 = 56.25 % \approx 56 %$

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The correct option is D 56% We know that, R=ρlA So, R1R2=l1A2A1l2 ...(1) Also, volume remain constant. So, A1l1=A2l2 ⇒A2A1=l1l2 ...(2) From (1) and (2), R1R2=l21l22 ⇒1R2=(l)2(1.25l)2 ⇒R2=1.5625 Ω Hence, percentage change in the resistance, =1.5625−11×100=56.25%≈56%

A wire of $1 Ω$ has a length of $1 m$ . It is stretched till its length increases by $25$ %. The percentage change in the resistance to the nearest integer is -