Mathematics in Physics

JEE Main level, single correct, medium

94 of 121

Question 94 of 121

A wire of 1Ω1\Omega has a length of 1m1m. It is stretched till its length increases by 2525%. The percentage change in resistance to the nearest integer is :

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Answer: (D) 5656%

R0=1ΩR1={R}_{0}=1\Omega {R}_{1}=
ℓ0=1mℓ1=1.25m{\ell }_{0}=1m{\ell }_{1}=1.25m
A0=A{A}_{0}=A
As volume of wire remains constant so
A0ℓ0=A1ℓ1⇒A1=ℓ0A0ℓ1{A}_{0}{\ell }_{0}={A}_{1}{\ell }_{1}\Rightarrow {A}_{1}=\dfrac{{\ell }_{0}{A}_{0}}{{\ell }_{1}}
Now, Resistance (R)=ρℓA\left(R\right)=\dfrac{\rho \ell }{A}
R0R1=ℓ0/A0ρℓ1/A1\dfrac{{R}_{0}}{{R}_{1}}=\dfrac{{\ell }_{0}/{A}_{0}}{\rho {\ell }_{1}/{A}_{1}}
1R1=ℓ0A0(ℓ0A0ℓ1×ℓ1)R1=ℓ12ℓ02=1.5625Ω\dfrac{1}{{R}_{1}}=\dfrac{{\ell }_{0}}{{A}_{0}}\left(\dfrac{{\ell }_{0}{A}_{0}}{{\ell }_{1}\times {\ell }_{1}}\right){R}_{1}=\dfrac{{\ell }_{1}^{2}}{{\ell }_{0}^{2}}=1.5625\Omega
So % change in resistance
=R1−R0R0×100=\dfrac{{R}_{1}-{R}_{0}}{{R}_{0}}\times 100%
=1.5625−11×100=\dfrac{1.5625-1}{1}\times 100%
=56.25=56.25%

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