Question

A cylindrical wire of resistance ‘R’ is pulled to thrice its length keeping the volume constant.It’s new resistance will be xR. What is the value of x ?

Answer 1

Given:

A cylindrical wire of resistance ‘R’ is pulled to thrice its length keeping the volume constant. Its new resistance is xR.

To find:

Value of "x" ?

Calculation:

Let initial resistance be

$\boxed{ \bold{R = \rho \times \dfrac{l}{ (area)_{1}} }}$

Let new radius of wire be r_(2):

Since volume is constant:

$\therefore \: \pi {r}^{2} l = \pi {(r_{2})}^{2} (3l)$

$\implies \:\pi{r}^{2} =3 \times \bigg \{ \pi {(r_{2})}^{2} \bigg \}$

$\implies \: \pi {(r_{2})}^{2} = \dfrac{1}{3} \bigg(\pi {r}^{2} \bigg)$

$\boxed{ \implies \: (area)_{2} = \dfrac{1}{3} \bigg \{ (area)_{1} \bigg \}}$

Now, new resistance will be:

$\therefore \: xR = \rho \times \bigg \{\dfrac{3l}{ (area)_{2} } \bigg \}$

$\implies\: xR = \rho \times \bigg \{\dfrac{3l}{ \frac{1}{3} (area)_{1} } \bigg \}$

$\implies\: xR = 9 \bigg [\rho \times \bigg \{\dfrac{l}{ (area)_{1} } \bigg \} \bigg]$

$\implies\: xR = 9R$

$\implies \: x = 9$

So, value of x is 9.

Answer 2

Answer:

ur answer hope helps you