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Question

In order to increase the resistance of a given wire of uniform cross-section to four times its value, a fraction of its length is stretched uniformly till the full length of the wire becomes 32 times the original length. What is the value of this fraction?

A
14
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B
18
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C
116
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D
16
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Solution

The correct option is B 18
Let us choose the length of the resistance wire initially as l and cross sectional area as A.Rin=ρlA


Let us choose a part whose length is x.

So, the length of remaining portion is lx.

Final length of the wire is 32l.

After we stretch the chosen part whose initial length was x.

As volume of this chosen part remains same.


A=xAl2+x

We finally have two resistances in series as same amount of current will flow through them. So, the final resistance of the wire,

Rf=R1+R2

Rf=ρ(lx)A+ρ(l2+x)A

On substituting the value of A

Rf=ρ(lx)A+ρ(l2+x)2xA


Given, Rf=4Rin

ρ(lx)A+ρ(l2+x)2xA=4×ρlA

xlx2+l24+lx+x2=4lx

l24=2lx

x=l8

value of fraction xl=18

Hence, option (B) is the correct answer.

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