The length of a wire is doubled and the radius is doubled. By what factor does the resistance change

4 times

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2 times

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Unchanged

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0.5 times

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Solution

The correct option is D 0.5 times
$L^{'} = 2 L$

$r^{'} = 2 r ⟹ A^{'} = 4 A$

We know,

R $= \frac{ρ \times L}{A}$

Hence,

$\frac{R^{'}}{R} = \frac{L^{'}}{L} \times \frac{A}{A^{'}} = \frac{2 L}{L} \times \frac{A}{4 A} = \frac{1}{2}$

Hence,
$R^{'} = \frac{R}{2}$

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The length of a wire is doubled and the radius is doubled. By what factor does the resistance change

The correct option is D 0.5 timesL′=2L r′=2r⟹A′=4AWe know,R =ρ×LAHence,R′R=L′L×AA′=2LL×A4A=12Hence,R′=R2

The correct option is D 0.5 times
$L^{'} = 2 L$

$r^{'} = 2 r ⟹ A^{'} = 4 A$

We know,

R $= \frac{ρ \times L}{A}$

Hence,

$\frac{R^{'}}{R} = \frac{L^{'}}{L} \times \frac{A}{A^{'}} = \frac{2 L}{L} \times \frac{A}{4 A} = \frac{1}{2}$

Hence,
$R^{'} = \frac{R}{2}$