Question

The resistance (R) of wire of length is halved and area of cross section (A) is doubled,its new resistance (R) will be-​

Answer 1

Explanation:

The resistance (R) of wire of length is halved and area of cross section (A) is doubled,its new resistance (R) will be-R=Pl/A whre P is the specific resistance of the conductor

If it is 3m q u can answer like this,

Given l'= l/2

A'= 2A

Hence ,

R'=P(l/2)/2A

= P(l/A)*1/4

R'=1/4(R)

Hence the new resistance will be quarter the value of original resistance

If it is 2m....u can answer like this

Resistance of wire depends upon resistivity, length and area of cross section,in simple words resistance is directly proportional to lenght and inversely proportional to cross sectional area. In this case since lenght is halfed and area is doubled resistance decreases by 4 times.

R~l/a(R is proportional to l/a). So numerator becomes 1/2 and denominator gets multiplied by 2 final answer will be 1/4 times the previous value.

Answer 2

Answer:

The new resistance of the wire will be one fourth of the initial resistance of the wire.

Explanation:

We know that,

The resistance of the wire R $= \rho \frac{L}{A}$

where ρ is the resistivity of the wire,

L is the length of the wire,

A is the cross section area of the given wire.

Now, given that length of the wire will be halved. So, the length becomes $L'=\frac{L}{2}$ and area of cross section of wire will be doubled, $A' =2A$.

Consider that R' is the new resistance of the wire.

$R'=\rho \frac{L'}{A'}$

$R'=\rho( \frac{L}{2}) (\frac{1}{2A} )$

$R'=\frac{1}{4} (\rho \frac{L}{A} )$

$R'=\frac{R}{4}$

Therefore, the new resistance (R') of the wire will be one fourth of initial resistance (R).