A load of 5N gives an extension of 0.65cm in a wire which obeys the hookes law. what is the extension caused by the load of 20N

2.6 cm

Explanation:

Hookes Law said

F = c ∆x

F = Force

c = elasticity constant of the spring / wire

∆x = change in the length

5 = c * 0.65

c = 5/0.65

if we change the Force to 20 then

20 = c ∆x

20 = (5/0.65) * ∆x

∆x = 20 * 0.65/5

∆x = 2.6 cm