SSC JE Electrical Engineering 2010 Paper

Question : 44 of 50

Marks: +1, -0

A wire has a resistance 10Ω. It is stretched by one-tenth of its original length. Then its resistance will be

Solution:

According to question

R_1= ρl_1/A_1

...(i)

R_2=ρl_2/A_2

...(ii)

l_2=l_1+l_1/10={11l_1}/10

l_2/l_1=11/10

...(iii)
∵ On stretching, Volume will be constant.

l_1A_1=l_2A_2

A_2/A_1=l_1/l_2

...(iv)
From (i) & (ii)

R_1/R_2=l_1/l_2×A_2/A_1

=10/11×10/11

From (iii) & (iv)

R_2=R_1 121/100

R_2=1.21R_1

R_2=1.21×10=12.1Ω

So the value of resistance is 12.1Ω

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