Book Online Demo

Check Your IQ

Try Test

Courses

A wire of resistance $4 Ω$ is redrawn by pulling it doubled. Its new resistance is [[1]] $Ω$ .

see full answer

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

A wire of resistance

4 Ω

is redrawn by pulling it doubled. Its new resistance is

16 Ω

.
Let

L and L'

be the length of wire before and after stretching respectively.

A and A'

be the area of cross section before and after stretching and

R and R'

be the resistance of the wire before and after stretching respectively.
As volume of wire does not change,

AL = A' L'

Now, it is given that the wire is pulled to double, so,

L^{'} = 2 L

So,

AL = A^{'} \times 2 L

\Rightarrow A = 2 A'

Now, resistance,

R = ρ \frac{L}{A}

And final resistance,

R' = ρ \frac{L'}{A'}

So,

\frac{R'}{R} = \frac{ρL'}{A'} \times \frac{A}{ρL}

\Rightarrow \frac{R'}{R} = \frac{ρ \times 2 L \times A}{\frac{A}{2} ρL}

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

{\Rightarrow R}^{'} = 4 R

\Rightarrow R^{'} = 4 \times 4

\Rightarrow R^{'} = 16 Ω

Watch 3-min video & get full concept clarity