JEE MAIN - Physics (2017 - 9th April Morning Slot - No. 15)
A uniform wire of length 1 and radius r has a resistance of 100 $$\Omega $$. It is recast into a wire of radius $${r \over 2}.$$ The resistance of new wire will be :
1600 $$\Omega $$
400 $$\Omega $$
200 $$\Omega $$
100 $$\Omega $$
Explanation
Resistance of a wire of length l and radius r is given
by
R = $${{\rho l} \over A}$$ = $${{\rho l} \over A} \times {A \over A} = {{\rho V} \over {{A^2}}} = {{\rho V} \over {{\pi ^2}{r^4}}}$$
$$ \Rightarrow $$ R $$ \propto $$ $${1 \over {{r^4}}}$$
$$ \therefore $$ $${{{R_1}} \over {{R_2}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^4}$$
Given, R1 = 100 $$\Omega $$, r1 = r, r2 = $${r \over 2}$$ , R2 = ?
$$ \therefore $$ R2 = R1$${\left( {{{{r_1}} \over {{r_2}}}} \right)^4}$$ = 16R1 = 1600 $$\Omega $$
R = $${{\rho l} \over A}$$ = $${{\rho l} \over A} \times {A \over A} = {{\rho V} \over {{A^2}}} = {{\rho V} \over {{\pi ^2}{r^4}}}$$
$$ \Rightarrow $$ R $$ \propto $$ $${1 \over {{r^4}}}$$
$$ \therefore $$ $${{{R_1}} \over {{R_2}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^4}$$
Given, R1 = 100 $$\Omega $$, r1 = r, r2 = $${r \over 2}$$ , R2 = ?
$$ \therefore $$ R2 = R1$${\left( {{{{r_1}} \over {{r_2}}}} \right)^4}$$ = 16R1 = 1600 $$\Omega $$
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