Question

The resistance of a 10m long wire is 10ohm.its length is increased by 25% by stretching the wire uniformly.the resistance of wire will change to

Answer 1

Final Answer : 15.625 ohm

Steps and Understanding:
1) On stretching the wire, it's length and Area of Cross section changes but Volume remains constant.
We know.
R = pl/A
R = pl * l /(A*l ) = pl^2/V
R α l^2
where
l. = length of wire
A = Area of cross section of wire.
V = Volume of wire.

2) After increasing the length of wire by 25%,then
Final length of wire, l(f) = 10 + 10 * (25/100)
= 10 + 2.5 = 12.5 m
We got that ,
R α l^2. -(1)
And
R(i) = 10ohm
R(f) = R
l(i) = 10m
l(f) = 12.5 ohm.

3)
$\frac{r(i)}{r(f)} = \frac{ {l(i)}^{2} }{ {l(f)}^{2} } \\ = > \frac{10}{r(f)} = \frac{ {10}^{2} }{ {12.5}^{2} } \\ = > r(f) = \frac{ {12.5}^{2} }{10} = 15.625 \: ohm$

Therefore, final resistance of wire is 15.625ohm.

Answer 2

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