Question

A wire of resistance 10 ohm has a length l. Half the portion of the wire is stretched so that total length becomes 2l. The resistance of the wire now will be? answer is 50 ohm , ie 5 times the original resistance. please explain how .
​

Answer 1

~$\small \blue {\sf{Answer }}$✨❤

The resistance of a wire can be expressed as R= AρL

where,

ρ is the resistivity of the wire

L is the length of the wire

A is the area of cross section of the wire

In this case, the wire is stretched so as to increase its length by 20% i.e. factor of 1.2.

Here, the wire's cross-sectional area(CSA) reduces by factor of 1.2 times of its original value as volume of wire is constant.

=1.44×10=14.4Ω

Hence, the new resistance is 14.4 ohms.

Answer 2

ACTUAL ANSWER IS 40 ohm

Explanation:

let resistance of wire = R₁ = 10Ω

length of wire = l

area of cross section of wire = A

resistivity = ρ

WHEN WIRE IS STRETCHED,

resistance of wire = R₂

length of wire = 2l

area of cross section of wire = A/2   [since wire is streched]

resistivity = ρ    [remains same always]

NOW,

R₁ = ρ*l/A --------(1)    [by formula of resistivity]

R₂ = ρ * 2l/A/2      [by formula of resistivity]

R₂ = ρ * 4l/A ------(2)    

dividing (1) by (2),

R₁/R₂ = ρ*l/A/ρ * 4l/A

R₁/R₂ = 1/4

ie. R₁:R₂ = 1:4

ie. if R₁ = 10Ω,

R₂ = 10*4 = 40Ω

mark as brainliest if this helps...