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The resistance of a wire is 10Ί Its length is increased10% by stretching. The new resistance will now be[CPMT 2000: Pb PET 2004: J& K CFT 0

Answer»

Let resistance of a conductor =ρL/A

Where ρ is the resistivity of the material, L is the length and A is the cross sectional area.

Let's assume the wire is cylindrical.

Volume of a cylinder = V = πr^2 x h

Where r is the radius and h is the height.

Here, A = πr^2 and h = L

So, V = AL

When you are stretching the volume will remain the same.

So V = A'L' = A'x(1.1)L (stretching by 10%)

Thus A' = V/(1.1L) = 0.909A

So new resistance = ρL'/A' = ρx1.1L/(0.909xA) = 1.21R = 12.1Ω


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