
The length of the wire increases by 8 mm when a weight of 5 kg is hung. If all the conditions are the same but the radius of the wire is doubled, what will be the increase in its length?
(A) 2 mm
(B) 1 mm
(C) $0.5$ mm
(D) $1.5$ mm
Answer
577.5k+ views
Hint:
For solid matter operating in the elastic region, the force on a component depends on its Young’s modulus. A change in the dimensions of the component is bound to change other parameters, too.
Formula used: $F = YA\dfrac{{\Delta l}}{l}$, where F is the force on the component, A is the area of the cross section, l is the original length of the wire, Y is the Young’s modulus, and $\Delta l$ is the change in the length of the wire due to the force applied.
Complete step by step answer:
In this question, we are provided with a wire on which force F is exerted in the form of a weight hung from it. We are required to find the change in length of the wire, when all the conditions remain the same except the cross section of the wire. The data given to us includes:
Change in length before $\Delta l = 8$mm
Force on the wire $F = 5$kg
We know that the Young’s modulus and the force on a wire depend as:
$\Rightarrow F = YA\dfrac{{\Delta l}}{l}$
Bringing the unchanging parameters on the LHS according to the question:
$\Rightarrow \dfrac{{Fl}}{Y} = A\Delta l$ [Eq. 1]
We are told that the radius R of the wire is doubled. Thus, the area will change as:
$\Rightarrow A' = \pi R{'^2} = \pi {(2R)^2}$
$\Rightarrow A' = 4\pi {R^2} = 4A$
As the LHS of Eq. 1 will remain constant, we can have the relation between the new and old dimensions as:
$\Rightarrow A\Delta l = A'\Delta l'$
To find the value of the new change in length:
$\Rightarrow \Delta l' = \dfrac{{A\Delta l}}{{A'}}$
We put the known values in this equation to get:
$\Rightarrow \Delta l' = \dfrac{{A \times 8}}{{4A}} = \dfrac{8}{4} = 2$mm
Hence, the new change in length is given by option (A); 2 mm.
Note:
Young’s modulus of a material is defined as its ability to withstand the changes in length or cross section when under tension or compression. Since it is a material property, it is not constant. It also has different values in different directions.
For solid matter operating in the elastic region, the force on a component depends on its Young’s modulus. A change in the dimensions of the component is bound to change other parameters, too.
Formula used: $F = YA\dfrac{{\Delta l}}{l}$, where F is the force on the component, A is the area of the cross section, l is the original length of the wire, Y is the Young’s modulus, and $\Delta l$ is the change in the length of the wire due to the force applied.
Complete step by step answer:
In this question, we are provided with a wire on which force F is exerted in the form of a weight hung from it. We are required to find the change in length of the wire, when all the conditions remain the same except the cross section of the wire. The data given to us includes:
Change in length before $\Delta l = 8$mm
Force on the wire $F = 5$kg
We know that the Young’s modulus and the force on a wire depend as:
$\Rightarrow F = YA\dfrac{{\Delta l}}{l}$
Bringing the unchanging parameters on the LHS according to the question:
$\Rightarrow \dfrac{{Fl}}{Y} = A\Delta l$ [Eq. 1]
We are told that the radius R of the wire is doubled. Thus, the area will change as:
$\Rightarrow A' = \pi R{'^2} = \pi {(2R)^2}$
$\Rightarrow A' = 4\pi {R^2} = 4A$
As the LHS of Eq. 1 will remain constant, we can have the relation between the new and old dimensions as:
$\Rightarrow A\Delta l = A'\Delta l'$
To find the value of the new change in length:
$\Rightarrow \Delta l' = \dfrac{{A\Delta l}}{{A'}}$
We put the known values in this equation to get:
$\Rightarrow \Delta l' = \dfrac{{A \times 8}}{{4A}} = \dfrac{8}{4} = 2$mm
Hence, the new change in length is given by option (A); 2 mm.
Note:
Young’s modulus of a material is defined as its ability to withstand the changes in length or cross section when under tension or compression. Since it is a material property, it is not constant. It also has different values in different directions.
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