Question

The length of an elastic string is $x$ metre when the tension is $8N$. Its length is $y$ metre when the tension is $10N$. What will be its length, when the tension is $18N$?

• A. $2x+y$
• B. $5y-4x$
• C. $7y-5x$
• D. $7y+5x$

Answer 1

The correct option is B $5y-4x$

Let, original length of the spring is L metre and, Y = $\frac{F.L}{A.l}$

Now, when F = 8N, and l = (x - l)m then, $Y=\frac{8.L}{A.(x-L)}⟶\left(I\right)$

and when F=10N, and l = (y - l)m then, $y=\frac{10.L}{A.(y-L)}m⟶\left(II\right)$

From equation (I) and (II) we get,

$\frac{8L}{A(x-L)}=\frac{10L}{A(y-L)}$

or, $8(y-L)=10(x-L)$

or, $4y-4L=5x-5L$

or, $L=5x-4y$

When, F=18N,

Let, length of the wire will be Z metre.

$\therefore Y=\frac{18.L}{A.(Z-L)}⟶\left(III\right)$

From equation (I) and (III) we get,

$\frac{8L}{A(x-L)}=\frac{18L}{A(Z-L)}$

or, $9(x-L)=4(Z-L)$

or, $4Z=9x-9L+4L$

$=9x-5L$

$=9x-25x+20y$ [putting value of L]

or, $Z=5y-4x$