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JEE Mains · Physics · STD 12 - Current electricity
  • JEE Mains 2021
  • Hard
  • Numerical

A wire of \(1 \,\Omega\) has a length of \(1\, m\). It is stetched till its length increases by \(25\, \%\). The percentage change in resistance to the neartest integer is .....\(\%\)

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Answer & Solution

Correct Answer

(A) \(56\)

Step-by-step Solution

Detailed explanation

\(R _{0}=1\, \Omega\)

\(\ell_{0}=1\, m\)

\(A _{0}= A\)

\(R _{1}=?\)

\(\ell_{1}=1.25\, m\)

As volume of wire remains constant so

\(A _{0} \ell_{0}= A _{1} \ell_{1} \Rightarrow A _{1}=\frac{\ell_{0} A _{0}}{\ell_{1}}\)

Now

Resistance \(( R )=\frac{\rho \ell}{ A }\)

\(\frac{ R _{0}}{ R _{1}}=\frac{\ell_{0} / A _{0}}{\rho \ell_{1} / A _{1}}\)

\(\frac{1}{ R _{1}}=\frac{\ell_{0}}{ A _{0}}\left(\frac{\ell_{0} A _{0}}{\ell_{1} \times \ell_{1}}\right) R _{1}=\frac{\ell_{1}^{2}}{\ell_{0}^{2}}=1.5625\, \Omega\)

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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