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 18Let 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 l−x. 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(l−x)A+ρ(l2+x)A′ On substituting the value of A′ ⇒Rf=ρ(l−x)A+ρ(l2+x)2xA Given, Rf=4Rin ⇒ρ(l−x)A+ρ(l2+x)2xA=4×ρlA ⇒xl−x2+l24+lx+x2=4lx ⇒l24=2lx ⇒x=l8 ∴ value of fraction xl=18 Hence, option (B) is the correct answer.

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