JEE MAIN - Physics (2021 - 26th February Evening Shift - No. 8)
A wire of 1$$\Omega$$ 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 :
76%
12.5%
25%
56%
Explanation
R0 = 1$$\Omega$$
R1 = ?
l0 = 1m
l1 = 1.25 m
A0 = A
As volume of wire remains constant so
A0l0 = A1l1 $$ \Rightarrow $$ A1 = $${{{l_0}{A_0}} \over {{l_1}}}$$
Now
Resistance (R) = $${{pl} \over A}$$
$${{{R_0}} \over {{R_1}}} = {{{l_0}} \over {{A_0}}}\left( {{{{l_0}{A_0}} \over {{l_1} \times {l_1}}}} \right)$$
$$ \Rightarrow $$ $${R_1} = {{l_1^2} \over {l_0^2}} = 1.5625 \,\Omega $$
So % change in resistance
$$ = {{{R_1} - {R_0}} \over {{R_0}}} \times 100\% $$
$$ = {{1.5625 - 1} \over 1} \times 100\% $$
$$ = 56.25\% $$
R1 = ?
l0 = 1m
l1 = 1.25 m
A0 = A
As volume of wire remains constant so
A0l0 = A1l1 $$ \Rightarrow $$ A1 = $${{{l_0}{A_0}} \over {{l_1}}}$$
Now
Resistance (R) = $${{pl} \over A}$$
$${{{R_0}} \over {{R_1}}} = {{{l_0}} \over {{A_0}}}\left( {{{{l_0}{A_0}} \over {{l_1} \times {l_1}}}} \right)$$
$$ \Rightarrow $$ $${R_1} = {{l_1^2} \over {l_0^2}} = 1.5625 \,\Omega $$
So % change in resistance
$$ = {{{R_1} - {R_0}} \over {{R_0}}} \times 100\% $$
$$ = {{1.5625 - 1} \over 1} \times 100\% $$
$$ = 56.25\% $$
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