Question

If a wire of resistance R is melted and recast into half of its length, the new resistance of
wire will be

a) R/4
b) R/2
c) R
d) 2 R​

Answer 1

We know that,

Resistance of a wire is directly proportional to it's length i.e. R ∝ l.

And R = p l/A

Here; R is Resistance, p is rho, l is length and A is Area of cross-section.

Also, Volume = Area × Length

So,

R = p l/A × l/l

R = p l²/V ...............(1st equation)

Therefore, R is directly proportional to square of it's length.

As per given condition, a wire of resistance R is melted and recast into half of its length.

l' = l/2

New Resistance (R') = p (l')²/V

Substitute l' as l/2

R' = p (l/2)²/V

R' = p (l/4)/V

R' = 1/4 p l/V

From (1st equation)

R' = 1/4 R

R' = R/4

Therefore, new resistance is R/4.

Option a) R/4

Answer 2

Answer

If a wire of resistance R and resistivity ρ, having length l and area of cross section A is ,

• $\bold{R=\frac{\rho l}{A} }$

We know that volume = Area * length

⇒ V = Al

Now multiply the first equation with l/l , we get ,

• $\bold{R=\frac{\rho l}{A}*\frac{l}{l} }$

$\implies \bold{R=\frac{\rho l^2}{v} ...(1)}$

If a wire of resistance R is melted and recast into half of its length, the new resistance of  wire will be ,

then the new length , l = 1/2 ( length )

$\implies \bold{R^{'}=\frac{\rho (\frac{l}{2}l )^2}{v} }$

$\implies \bold{R^{'}=\frac{1}{4}(\frac{\rho l^2}{v} )}$

$\implies \bold{R^{'}= \frac{1}{4}\;R }\;\;\bold{[From\;(1)]}\\\\\implies \bold{R^{'}=\frac{R}{4} }$

So option a) is correct .