when the metal sphere is suspended at the end of a meal wire its extension is 0.4mm. If another metal sphere of the same material with its radius half as the previous is suspended then extension would be

ans:
1)0.05mm

0.05mm

0.05 To find the answer, you need to understand the concept of Hooke's law, which states that the extension of an object is directly proportional to the force applied to it, as long as the elastic limit of the material is not exceeded.

In this case, since both metal spheres are made of the same material and suspended at the end of a wire, we can assume they have the same spring constant (k).

Let's denote the extension of the first metal sphere as x1 and the radius as R1. We are given that x1 = 0.4 mm.

Now, let's denote the extension of the second metal sphere as x2 and the radius as R2. As per the question, R2 = (1/2) * R1.

According to Hooke's law, the extension is directly proportional to the radius. So, we can say:

x1/R1 = x2/R2

Substituting the values we have:

0.4/R1 = x2/[(1/2) * R1]

Simplifying the equation:

0.4 = 2x2

Dividing both sides by 2:

x2 = 0.4/2

x2 = 0.2 mm

Therefore, the extension of the second metal sphere with half the radius is 0.2 mm.