Current Electricity Ques 106
106. A wire of a certain material is stretched slowly by ten per cent. Its new resistance and specific resistance become respectively:
[2008]
(a) 1.2 times, 1.3 times
(b) 1.21 times, same
(c) both remain the same
(d) 1.1 times, 1.1 times
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Answer:
Correct Answer: 106.(b)
Solution:
- (b) Resistance of a wire is given by $R=r \frac{l}{a}$ If the length is increased by $10 \%$ then new
length $l^{\prime}=l+\frac{1}{10}=\frac{11}{10} l$
In that case, area of cross-section of wire would decrease by $10 \%$
$\therefore$ New area of cross-section
$A^{\prime}=A-\frac{A}{10}=\frac{9}{10} A$
$\therefore R^{\prime}=r \frac{\ell^{\prime}}{A^{\prime}}=r \frac{\frac{11}{10} l}{\frac{9}{10} A}$
$R^{\prime}=\frac{11}{9} r \frac{l}{R} \quad R^{\prime}=1.21 R$
The new resistance increases by 1.21 times. The specific resistance (resistivity) remains unchanged as it depends on the nature of the material of the wire.
After stretching if length of a conductor increases by $x \%$ then resistance will increase by $2 x \%$ (Valid only if $x<10 \%$ )
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