QJEEM21Phcf07: JEE Main 2021 Physics - Class 12 JEEM21Phcf07 Solution: A wire of 1 Ω has a length of 1 m. It is stretched till its…

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A wire of 1

Ω

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

Accepted Answer

R₀ = 1

Ω

R₁ = ?

l₀ = 1m

l₁ = 1.25 m

A₀ = A

As volume of wire remains constant so

A₀l₀ = A₁l₁

\Rightarrow

A₁ =

\frac{l_{0} A_{0}}{l_{1}}

Now

Resistance (R) =

\frac{p l}{A}

\frac{R_{0}}{R_{1}} = \frac{l_{0}}{A_{0}} (\frac{l_{0} A_{0}}{l_{1} \times l_{1}})

\Rightarrow

R_{1} = \frac{l_{1}^{2}}{l_{0}^{2}} = 1.5625 Ω

So % change in resistance

= \frac{R_{1} - R_{0}}{R_{0}} \times 100 %

= \frac{1.5625 - 1}{1} \times 100 %

= 56.25 %

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