The Young's modulus of steel is twice that of brass. Two wires of same length and of same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of

Question

A 2 $\text{kg}$ mass is attached to a spring with spring constant $ k = 200, \text{N/m} $. If the mass is displaced by $ 0.1, \text{m} $, what is the potential energy stored in the spring?

Answer 1

To solve this problem, we need to understand the relationship between Young's modulus, stress, and strain.

The Young's modulus (Y) is a measure of the ability of a material to withstand changes in length when under lengthwise tension or compression. It is defined as the ratio of tensile stress to tensile strain:

Y = \frac{\text{Stress}}{\text{Strain}}

The formula for extension ($\Delta L$) of a wire under a load is given by:

\Delta L = \frac{F \cdot L}{A \cdot Y}

where:

F is the force applied (weight added).
L is the original length of the wire.
A is the area of cross section of the wire.
Y is the Young's modulus of the wire material.

Given that the Young's modulus of steel is twice that of brass, we have:

Y_{\text{steel}} = 2Y_{\text{brass}}

Since the extensions for the wires need to be the same (so their lower ends remain at the same level), we set up the equation:

\frac{F_{\text{steel}} \cdot L}{A \cdot 2Y_{\text{brass}}} = \frac{F_{\text{brass}} \cdot L}{A \cdot Y_{\text{brass}}}

On simplifying, we find:

\frac{F_{\text{steel}}}{2} = F_{\text{brass}}

This implies:

F_{\text{steel}} = 2F_{\text{brass}}

Thus, the weights added to the steel and brass wires must be in the ratio 2:1.

The correct answer is 2:1.

Answer 2