The specific resistance of a wire is ρ. If its volume is 3 m 3 and its resistance is 3 ohms, then its length will be

Question

The specific resistance of a wire is ρ. If its volume is 3 m 3 and its resistance is 3 ohms, then its length will be (A) √(1/ρ) (B) (3/√ρ) (C) (1/ρ) √3 (D) ρ √(1/3)

Tardigrade Admin · Accepted Answer

Volume =A l=3 ⇒ A=(3/l) Now R=ρ (l/A) ⇒ 3=(ρ × l/3 / l)=(ρ l2/3) ⇒ l2=(9/ρ)=(3/√ρ)

Q. The specific resistance of a wire is $ρ$ . If its volume is $3 m^{3}$ and its resistance is 3 ohms, then its length will be

$\frac{1}{ρ}$

$\frac{3}{ρ}$

$\frac{1}{ρ} 3$

$ρ \frac{1}{3}$

Volume $= A l = 3 \Rightarrow A = \frac{3}{l}$
Now $R = ρ \frac{l}{A} \Rightarrow 3 = \frac{ρ \times l}{3/ l} = \frac{ρ l ^{2}}{3}$
$\Rightarrow l^{2} = \frac{9}{ρ} = \frac{3}{ρ}$

Q. The specific resistance of a wire is ρ. If its volume is 3m3 and its resistance is 3 ohms, then its length will be

ρ1​​

ρ​3​

ρ1​3​

ρ31​​