Correct Answer - Option 4 : 40 Ω
Concept:
Resistance (R): The resistance offered to the flow of current is known as the resistance.
The SI unit of resistance is the ohm (Ω).
Mathematically resistance can be written as:
\(R = \frac{{\rho l}}{A}\)
Where,
R = resistance,
l = length,
A = area of cross-section and
ρ = resistivity
Calculation:
Given: R1 = 10Ω and l2 = 2l1
As we know that the resistance of the wire is,
\(R = \frac{{\rho l}}{A}\)
When the wire is stretched then, then its area will decrease automatically. But the volume of the wire will be the same.
∴ The volume of original wire = volume of new wire
⇒ A1l1 = A2l2
⇒ A1l1 = A2 2l1
⇒ A2 = A1/2
The resistance of the wire in 1st case
\(\Rightarrow {R_1} = R = \frac{{\rho {l_1}}}{{{A_1}}}\) ----(1)
The resistance of the wire in 2nd case
\(\Rightarrow {R_2} = \frac{{\rho {l_2}}}{{{A_2}}} = \frac{{\rho 2{l_1}}}{{\frac{{{A_1}}}{2}}} = \frac{{4\rho {l_1}}}{{{A_1}}}\) -----(2)
Divide equation 1 and 2, we get
\(\frac{R_1}{{{R_2}}} = \frac{{\frac{{\rho {l_1}}}{{{A_1}}}}}{{\frac{{4\rho {l_1}}}{{{A_1}}}}} = \frac{1}{4}\)
R2 = 4R1 = 4 x 10 = 40Ω
