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

Question

The specific resistance of a wire is ρ , its volume is 3m3 and its resistance is 3Ω , then its length will be (A)  (3/√ρ )  (B)  (√3/ρ )  (C)  (5/√ρ )  (D)  (√5/ρ )

Tardigrade Admin · Accepted Answer

Volume = Area × length 3= Al A =(3/1) Resistance R =ρ (1/ A ) 3=(ρ l/((3)1))=(ρ l2/3) 12=(9/ρ) ⇒ 1=(3/√ρ)

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

$\frac{3}{ρ}$

$\frac{3}{ρ}$

$\frac{5}{ρ}$

$\frac{5}{ρ}$

Volume $=$ Area $\times$ length
$3 = A l$
$A = \frac{3}{1}$
Resistance $R = ρ \frac{1}{A}$
$3 = \frac{ρl}{( \frac{3}{1} )} = \frac{ρ l ^{2}}{3}$
$1^{2} = \frac{9}{ρ}$
$\Rightarrow 1 = \frac{3}{ρ}$

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

ρ​3​

ρ3​​

ρ​5​

ρ5​​

Volume = Area × length 3=Al A=13​ Resistance R=ρA1​ 3=(13​)ρl​=3ρl2​ 12=ρ9​ ⇒1=ρ​3​

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

$\frac{3}{ρ}$

$\frac{3}{ρ}$

$\frac{5}{ρ}$

$\frac{5}{ρ}$

Volume $=$ Area $\times$ length
$3 = A l$
$A = \frac{3}{1}$
Resistance $R = ρ \frac{1}{A}$
$3 = \frac{ρl}{( \frac{3}{1} )} = \frac{ρ l ^{2}}{3}$
$1^{2} = \frac{9}{ρ}$
$\Rightarrow 1 = \frac{3}{ρ}$