The length of an elastic string is x metre when the tension is 8N, and Ym when the tension is 10N . What is the length in metres when the tension is 18N ?

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Let ‘L’ be the original length of the string. ‘A’ be its area of cros-section, ‘Y’ be the Young’s modulus of the material of the string.

When 8 N force is applied, the length becomes x m. So, the strain is = (x – L)/L.

Thus we can write,

Y = (8/A)/[ (x – L)/L]

Similarly, when force is 10 N we have,

Y = (10/A)/[(y – L)/L]

Thus we can equate,

(8/A)/[ (x – L)/L] = (10/A)/[(y – L)/L]

=> L = 5x – 4y

Let the length be ‘z’ when the force applied is 18 N.

So,

Y = (18/A)/[(z – L)/L]

Lets equate it as,

(10/A)/[(y – L)/L] = (18/A)/[(z – L)/L]

=> z = 5y – 4x