Question

The length of an elastic string is x meter when the tension is 8N and y meter when the tension is 10N. The length in the meter when the tension is 18N is
$A.\,4x-5y$
$B.\,5y-4x$
$C.\,9x-4y$
$D.\,4y-9y$

Answer 1

This question is based on the concept of Young’s modulus of a material. We are given with two cases, so, we will first find the expressions for the same in terms of Young’s modulus and then will equate. Then, we will consider the third case, and use the equation obtained in the first 2 cases in this case equation to find the value of the length.
Formula used:
$Y=\dfrac{FL}{A\Delta L}$

Complete step by step answer:
Young's modulus (modulus of elasticity) of a material is given as follows.
$Y=\dfrac{FL}{A\Delta L}$
Where F is the force exerted on an object under tension, A is the original cross-sectional area through which the force is applied, L is the original length of the object and $\Delta L$is the change in the length of the material.
We are given the two cases. So, we will continue the calculation, considering these 2 cases.
Case I: The length of an elastic string is x meter when the tension is 8N
Thus, the data is, the length of the string after elongation is x meters and the force exerted on the string under tension is 8 N.
Substitute these values in the formula for calculating the value of Young’s modulus of the elastic string.
$Y=\dfrac{8L}{A(x-L)}$…… (1)
Case II: The length of an elastic string is y meter when the tension is 10N
Thus, the data is, the length of the string after elongation is y meters and the force exerted on the string under tension is 10 N.
Substitute these values in the formula for calculating the value of Young’s modulus of the elastic string.
$Y=\dfrac{10L}{A(y-L)}$…… (2)
We can equate the equations obtained from the cases I and II, as the LHS part of the equations are the same. So, we get,
$\dfrac{8L}{A(x-L)}=\dfrac{10L}{A(y-L)}$
Continue the further calculation.

& 8(y-L)=10(x-L) \\\ & \Rightarrow 4y-4L=5x-5L \\\ & \Rightarrow L=5x-4y \\\ \end{aligned}$$ Now, compute the equation for Young's modulus when the value of the force equals 18 N. Let the length of the string after elongation be z meters $$Y=\dfrac{18L}{A(z-L)}$$…… (3) We can equate the equations (1) and (3), as the LHS part of the equations are the same. So, we get, $$\begin{aligned} & \dfrac{8L}{A(x-L)}=\dfrac{18L}{A(z-L)} \\\ & \Rightarrow 9(x-L)=4(z-L) \\\ & \Rightarrow 4z=9x-5L \\\ \end{aligned}$$ Substitute the value of L obtained in the above equation. $$\begin{aligned} & 4z=9x-5(5x-4y) \\\ & \Rightarrow 4z=20y-16x \\\ & \Rightarrow z=5y-4x \\\ \end{aligned}$$ The length in meters when the tension is 18N is $$5y-4x$$. At the length in the meter when the tension is 18N is $$5y-4x$$. **Thus, option (B) is correct.** **Note:** The change in the lengths of the elastic string are considered to be x, y and z. The value of the length z is the value that we need to obtain. We will consider Young’s modulus equation, as it becomes easy to equate the equations, when one of the parts of the equation, that is, LHS or RHS, will be the same.