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A cylindrical conductor of length ‘l’ and uniform area of cross-section ‘A’ has resistance ‘R’. The area of cross-section of another conductor of same material and same resistance but of length ‘2l’ is(a) $\frac {A}{2}$ (b) $\frac {3A}{2}$(c) 2A (d) 3A
(c) 2A
Explanation
Given:
Length of the cylinder = $l$
Resistance of the cylinder = $R$
Cross-sectional area of the cylinder = $A$
To find: Area of cross-section, when the length is $2l$.
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.
From this relation, we conclude that the length is directly proportional to the resistance $(l\propto R)$, and the area of cross-section is inversely proportional to the resistance $(A\propto \frac{1}{R})$.
Thus, in this case, the length of the conductor is doubled $2l$, so the resistance will be $2R$. For the resistance to remain the same as $R$, the area of cross-section will also get doubled as $2A$.
Hence, the area of cross-section is $2A$, when the length is $2l$.
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