Question

wire of resistance 5 ohm is stretched to twice its original length. What is the reaistance of wire now ?
10 ohm
15 ohm
25 ohm
20 ohm
can not be determined

Answer 1

\[R=\frac{\rho l}{A}\]Multiplying numerator and denominator by l (length)\[R=\frac{\rho l^2}{V}\]where V is volume
Now, let \(R_1\) and \(R_2\) be the resistances
so we have \[\frac{R_1}{R_2}=\frac{l^2}{4l^2}\]Here \(R_1\)=5\(\Omega\)
So, \(R_2\)=20\(\Omega\)

Answer 2

Resistance=pl/a ( p is resistivity, l=lenght of wire, a= area of crossection)
if L=2l
then new resistance =pL/A ( pie*r*r*l=pie*R*R*(2l)
new radius of wire, R=r/root2
new area after streching, A=pie*r*r/2 )
new resitance = p(2l)/A
substituting value of A, we get new resistance= 4times the old resistance
new resistance =4*5ohms
= 20 ohms

Answer 3

Nicely done ujjwal

Answer 4

thanks @JamesJ