Question

If length of a metallic wire becomes n times,
its resistance becomes
(a) {n}^{2} times
(b) \sqrt{n} times
(c) \frac{1}{ \sqrt{n} } times
(d) \frac{1}{ {n}^{4} } times
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Answer 1

Question -

If length of a metallic wire becomes n times,

its resistance becomes

(a) ${n}^{2}$ times

(b) $\sqrt{n}$ times

(c) $\frac{1}{ \sqrt{n} }$ times

(d) $\frac{1}{ {n}^{4} }$ times

​

Answer -

Option A is the correct answer .

Solution -

Suppose that the initial length of the metallic wire is x units.

Now, the length of the metallic wire is stretched by n times.

Hence the new length of the wire becomes nx units .

Now,  Resistance is directly proportional to the length of the wire and is inversely proportional to the cross sectional area.

So, from the above statement , we can state that  length is inversely proportional to the cross sectional area.

So, the product of the length and the cross sectional area of any wire is a constant .

Using this,

x.A = nx NA

So, NA = n times less than original area .

So. nX / NA = n^2.

This is the required answer .

Hence, option A is correct .

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