First slide

Ohm's law

Question

A wire has a resistance 10 $Ω$ . It is stretched by one-tenth of its original length. Then its resistance will be

Moderate

A

10 $Ω$
B

12 $Ω$
C

9 $Ω$
D

11 $Ω$

Solution

Here volume remains constant. Thus
${πr}_{1}^{2} l = {πr}_{2}^{2} (11 l / 10)$
or ${πr}_{2}^{2} = \frac{10}{11} ({πr}_{1}^{2})$
{ $∴$ When wire is stretched by 1/10 of its original length, the new length of wire becomes (11 / / 10)}
Let the new resistance be R_2. Then
$\begin{array}{l} R_{2} = \frac{ρ \cdot (\frac{11}{10} l)}{{πr}_{2}^{2}} = \frac{(11 / 10) ρl}{(10 / 11) {πr}_{1}^{2}} \\ = \frac{(11 / 10)}{10 / 11)} [\frac{ρl}{{πr}_{1}^{2}}] \\ = \frac{(11 / 10)}{(10 / 11)} \times 10 = \frac{121}{10} = 12 \cdot 1 Ω \end{array}$

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