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A cylindrical conductor of length ‘l’ and uniform area of cross-section ‘A’ has resistance ‘R’. Another conductor of length 2·5 l and resistance 0·5 R of the same material has area of cross-section:
(A) 5 A (B) 2·5 A (C) 0·5 A (D) 1 A
(A) 5 A
Explanation
Given:
For first conductor
Length of the cylinder, $L_1$ = $l$
Resistance of the cylinder, $R_1$ = $R$
Cross-sectional area of the cylinder, $A_1$ = $A$
For seccond conductor
Length of the cylinder, $L_2$ = $2.5l$
Resistance of the cylinder, $R_2$ = $0.5R$
To find: Area of cross-section of the second conductor, $A_2$.
Solution:
We know that the resistance of a wire is expressed as-
$R=ρ\frac{l}{A}$
where,
$R-$ Resistance of the conductor.
$ρ(rho)-$ Resistivity (constant).
$l-$ Length of the conductor.
$A-$ Area of a cross-section of the conductor.
Putting the values of the first conductor in the formula, its resistivity will be-
$ρ_1=\frac{R\times A}{l}$ -------------------- (i)
Now, putting the values of the second conductor in the formula, its resistivity will be-
$ρ_2=\frac{0.5\times A_2}{2.5}$ -------------------- (ii)
The resistivity $(ρ)$ of the given conductor depends only on the material of the conductor, and here it is given that both conductors have the same material. Hence, the resistivity of both materials will be the same.
$\therefore ρ_1=ρ_2 $
Equating (i) and (ii) we get-
$\frac{R_1\times A_1}{L_1}=\frac{R_2\times A_2}{L_2}$
$\frac{R\times A}{l}=\frac{0.5R\times A_2}{2.5l}$
$\frac{R\times A}{l}=\frac{5R\times A_2}{25l}$
$\frac{R\times A}{l}=\frac{R\times A_2}{5l}$
${A_2}=\frac{R\times {A}\times 5l}{R\times l}$
${A_2}=5A$
Thus, the area of cross-section of the second conductor will be 5A.