The increase in length \( l \) of a wire when subjected to a force \( F \) is given by:
\(l = \frac{FL}{AY}\)
where:
- \( L \) is the original length of the wire,
- \( A = \pi r^2 \) is the cross-sectional area of the wire,
- \( Y \) is Young’s modulus of the material of the wire.
If both the force \( F \) and the radius \( r \) are reduced to half, let’s see how \( l \) changes.
Step 1: New Force:
\(F' = \frac{F}{2}\)
Step 2: New Radius and Area:
Since the radius is reduced to half:
\(r' = \frac{r}{2}\)
The new cross-sectional area \( A' \) is:
\(A' = \pi (r')^2 = \pi \left(\frac{r}{2}\right)^2 = \frac{\pi r^2}{4} = \frac{A}{4}\)
Step 3: New Increase in Length \( l' \):
Substituting the new values of \( F' \) and \( A' \):
\(l' = \frac{F'L}{A'Y} = \frac{\frac{F}{2} \cdot L}{\frac{A}{4} \cdot Y} = \frac{FL}{AY} \cdot 2 = 2l\)
Thus, the increase in length will become 2 times the original increase in length.
The Correct Answer is: 2 Times
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?
Match List-I with List-II: List-I