CGPET 2019 Solved Paper

Resistance of wire, R=10Ω.
Let the initial length of wire =l
Resistance of wire, R=

ρl

A

......(i)
According to the question,
New length of wire, l′=l+

l×10

100

l′=l+

l

10

or l′=

11l

10

The volume of wire remains constant.
∴ Volume of wire before stretching = Volume of wire after stretching Al=A′l′
Al=A′(

11l

10

)⇒A′=

10

11

A
New resistance of wire,
R′=

ρl′

A′

⇒R′=

ρ( 11 10 l)

10 11 A

R′=

ρl

A

×

11

10

×

11

10

From Eq. (i)
R′=R×(

11

10

)2=10×(

11

10

)2 =12.1Ω≈12Ω