If the work done is stretching a wire by 1 mm is 2J, the work necessary for stretching another wire of samme material by 1 mm but with double the radius and half the length is joule is
Answers
Answer:
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 dlW=
2
1
F⋅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}
W
2
W
1
=
2
1
F
2
⋅dl
2
1
F
1
⋅dl
\dfrac{2 J}{W_{2}}=\dfrac{F_{1}}{F_{2}}
W
2
2J
=
F
2
F
1
W_{2}=2\times\dfrac{F_{2}}{F_{1}}W
2
=2×
F
1
F
2
.....(I)
Now, the force is
F = YA\dfrac{dl}{l}F=YA
l
dl
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}}}
F
2
F
1
=
YA
l
2
dl
YA
l
1
dl
\dfrac{F_{1}}{F_{2}}=\dfrac{r_{1}^2}{r_{2}^2}\dfrac{l_{2}}{l_{1}}
F
2
F
1
=
r
2
2
r
1
2
l
1
l
2
.....(II)
We know that,
r_{2}= 2r_{1}r
2
=2r
1
l_{2}=\dfrac{l_{1}}{2}l
2
=
2
l
1
Put the value of r_{2}r
2
and l_{2}l
2
in equation (II)
\dfrac{F_{1}}{F_{2}}=\dfrac{r_{1}^2}{4r_{1}^2}\dfrac{l_{1}}{2l_{1}}
F
2
F
1
=
4r
1
2
r
1
2
2l
1
l
1
\dfrac{F_{1}}{F_{2}}=\dfrac{1}{8}
F
2