Question

[Physics]

Q) A wire of resistance 3 ohm and length 10 cm is stretched to length 30 cm. Assuming that it has a uniform cross section, what will be it's new resistance?
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Answer 1

Answer:

Explanation:

We will consider this question case by case. That means first the original value as Case 1 and required value in Case 2.

Case 1 :- When things depict originality.

Given resistance as 3 ohm.

Wire length is 10 cm.

We know that:-

$\sf{ R = \dfrac{ \rho \times l}{a}}$

Where Rho is a resistivity constant

Where l is length

Where a is cross section or area, whichever considered.

Now, "stretching" the wire means there will be or ought to be a change in the wire.

So, what will happen is :-

Length (original) = l (let)

So, length (final) = 3 times the original one ,

So, length = 3l.

Case 2 :- Stretching of wire which affects its area also.

$\sf{ R' = \dfrac{ \rho \times 3l}{ \dfrac{a}{3}}}$

Remember, the ratio in which length increases, so the area will decrease. Only only and only in the case of stretching of a wire, not cutting.

Solving further :-

$\sf{ R' = \dfrac{ \rho \times 3l \times 3 }{ a}}$

[By 1st-4th and 2nd-3rd division method]

$\sf{ R' = \dfrac{ 9 \times \rho \times l \times }{ a}}$

Treasure hunted- Can you notice the formula which was the originality of all the cases?

( $\sf{ R = \dfrac{ \rho \times l}{a}}$ ) This one.

It's there. So, we can replace it by R, right?

So, it will be 3R.

That's the answer? NO.

We need to multiply it with 3 [See above in the question, the wire is of 3 ohm]

So, 9R = 9 × 3 = $\sf{ 27 \ \Omega}$

Hence, 27 ohm the required answer.

Answer 2

See attachment

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