Question

Q.175 A wire extends by 1 mm when a force is applied. Double the force is applied to another wire of same material and length but half the radius of cross-section. The elongation of the wire in mm
will be:
​

Answer 1

Answer:

8 mm

Explanation:

From Young's modulus formula

Y = FL / (Ae)

Y = FL / (πr²e)

It's given that both are of same material (same value of "Y") and and of same length (L)

Y/L = F/(πr² e)

Yπ/L = F/(r² e)

From above

F/(r² e) is same for both the wires

So,

F1/(r1² e1) = F2/(r2² e2)

e2 = F2/(r2)² × (r1² e1)/F1

= (2F1)/(r1/2)² × (r1² × 1 mm)/(F1)

= 8 mm

Answer 2

Answer: 8 mm

Explanation: