Question

If work done in stretching a wire by 1mm is 2j, the work necessary for stretching another wire of same material, but with double the radius and half the length by 1mm in joule is

Answer 1

Answer: The work done will be 16 J.

Explanation:

Given that,

Work done W = 2 J

We know that,

We have a two wires of same material and same stretch length.

The work done in stretching  a wire is defined as:

$W = \dfrac{1}{2}F\cdot dl$

Where,

F = required force

dl = increases length of wire

We can have,

$\dfrac{W_{1}}{W_{2}}=\dfrac{\dfrac{1}{2}F_{1}\cdot dl}{\dfrac{1}{2}F_{2}\cdot dl}$

$\dfrac{2 J}{W_{2}}=\dfrac{F_{1}}{F_{2}}$

$W_{2}=2\times\dfrac{F_{2}}{F_{1}}$.....(I)

Now, the force is

$F = YA\dfrac{dl}{l}$

Here, A = Area of cross section of the wire

Y = Young's modulus

l = length of the wire

The force for both case

$\dfrac{F_{1}}{F_{2}}=\dfrac{YA\dfrac{dl}{l_{1}}}{YA\dfrac{dl}{l_{2}}}$

$\dfrac{F_{1}}{F_{2}}=\dfrac{r_{1}^2}{r_{2}^2}\dfrac{l_{2}}{l_{1}}$.....(II)

We know that,

$r_{2}= 2r_{1}$

$l_{2}=\dfrac{l_{1}}{2}$

Put the value of $r_{2}$ and $l_{2}$ in equation (II)

$\dfrac{F_{1}}{F_{2}}=\dfrac{r_{1}^2}{4r_{1}^2}\dfrac{l_{1}}{2l_{1}}$

$\dfrac{F_{1}}{F_{2}}=\dfrac{1}{8}$

Put the value of $\dfrac{F_{1}}{F_{2}}$ in equation(I)

$W_{2}=2\times8$

$W_{2}=16 J$

Hence, The work done will be 16 J.

Answer 2

this may help........ for the question ..........

the answrr is 16 joule