The specific resistance of a wire is p, its volume is 3m3 and its resistance is 3 Ω , then its length will be

Question

The specific resistance of a wire is p, its volume is 3m3 and its resistance is 3 Ω , then its length will be (A)  √(1/ρ )  (B)  (3/√ρ )  (C)  (√3/ρ )  (D)  ρ √(1/3)

Tardigrade Admin · Accepted Answer

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

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

$\frac{1}{ρ}$

$\frac{3}{ρ}$

$\frac{3}{ρ}$

$ρ \frac{1}{3}$

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

Q. The specific resistance of a wire is p, its volume is 3m3 and its resistance is 3 Ω , then its length will be

ρ1​​

ρ​3​

ρ3​​

ρ31​​

Volume =Al=3 A=l3​ Now, R=ρAl​ 3=3/lρ×l​ l2=ρ9​ ⇒ l=ρ​3​

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

$\frac{1}{ρ}$

$\frac{3}{ρ}$

$\frac{3}{ρ}$

$ρ \frac{1}{3}$

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