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 .....\(\%\)
Pick an option to check your answer.
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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