Question

a cylindrical wire of resistance r is pulled to thrice its length keeping the volume constant its new volume would be xr . what is the volume of x​
plz answer

Answer 1

Given:

a cylindrical wire of resistance r is pulled to thrice its length keeping the volume constant. The new value of resistance is xr.

To find:

Value of x.

Calculation:

Volume is constant , so let new Radius be R_(2)

$\therefore \: \pi {R}^{2} l = \pi {(R_{2})}^{2} (3l)$

$= > \: \pi {(R_{2})}^{2} = \dfrac{1}{3} ( \pi {R}^{2} ) \: \: \: ......(1)$

Now , initial resistance is r ;

$\therefore \: r = \rho( \dfrac{l}{a} ) = \rho( \dfrac{l}{\pi {r}^{2} } )$

Let new resistance be

$\therefore \: r_{2} = \rho( \dfrac{3l}{a_{2}}) \\ = >r_{2} = \rho( \dfrac{3l}{\pi {(R_{2})}^{2} } )$

$= >r_{2} = \rho( \dfrac{3l}{ \frac{1}{3} \pi {(R)}^{2} } )$

$= >r_{2} = 9 \times \rho( \dfrac{l}{ \pi {(R)}^{2} } )$

$= >r_{2} = 9 \times r$

$= >x \times r= 9 \times r$

$= >x = 9$

So, value of x = 9.

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Answer 2

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