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A wire of a certain material is stretched slowly by ten percent. Its new resistance and specific resistance become respectively -
A
both remain the same
B
1.1 times, 1.1 times
C
1.2 times, 1.1 times
D
1.21 times, same
Detailed Solution
Resistance of a wire: $R = \rho\dfrac{l}{A}$
New length: $l' = l + \dfrac{10}{100}\,l = \dfrac{11}{10}\,l$
The volume of the wire stays constant during stretching: $A\,l = A'\,l'$
$\Rightarrow A' = \dfrac{A\,l}{l'} = \dfrac{10}{11}\,A$
New resistance: $R' = \rho\dfrac{l'}{A'} = \rho\,\dfrac{(11/10)\,l}{(10/11)\,A}$
$R' = \left(\dfrac{11}{10}\right)^2 \rho\dfrac{l}{A} = 1.21\,R$
So the resistance becomes 1.21 times the initial value.
Specific resistance (resistivity) is an intrinsic property of the material; it does not depend on the dimensions, so it remains the same.
New length: $l' = l + \dfrac{10}{100}\,l = \dfrac{11}{10}\,l$
The volume of the wire stays constant during stretching: $A\,l = A'\,l'$
$\Rightarrow A' = \dfrac{A\,l}{l'} = \dfrac{10}{11}\,A$
New resistance: $R' = \rho\dfrac{l'}{A'} = \rho\,\dfrac{(11/10)\,l}{(10/11)\,A}$
$R' = \left(\dfrac{11}{10}\right)^2 \rho\dfrac{l}{A} = 1.21\,R$
So the resistance becomes 1.21 times the initial value.
Specific resistance (resistivity) is an intrinsic property of the material; it does not depend on the dimensions, so it remains the same.