There are two springs P and Q . Spring constant for P is K and that for Q is 3K . Both the springs are elongated by applying force of same magnitude . If U is energy stored in spring P then energy stores in spring Q is

Options

A. 3U
B. U/3
C. 9U
D. U/9

Explanation

To solve the problem, we need to find the energy stored in spring Q when both springs P and Q are elongated by the same force.

### Step-by-step Solution:

1. **Understanding the Energy Stored in a Spring**:
The energy (U) stored in a spring when it is elongated or compressed is given by the formula:
\[ U = \frac{1}{2} k x^2 \]
where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position.

2. **Finding the Displacement (x)**:
Since both springs are elongated by the same force \( F \), we can express the displacement \( x \) in terms of the force and the spring constant. According to Hooke's Law:
\[ F = k x \implies x = \frac{F}{k} \]

3. **Calculating Energy Stored in Spring P**:
For spring P, with spring constant \( K \):
\[ U_P = \frac{1}{2} K x^2 \]
Substituting \( x = \frac{F}{K} \):
\[ U_P = \frac{1}{2} K \left(\frac{F}{K}\right)^2 = \frac{1}{2} K \cdot \frac{F^2}{K^2} = \frac{F^2}{2K} \]

4. **Calculating Energy Stored in Spring Q**:
For spring Q, with spring constant \( 3K \):
\[ U_Q = \frac{1}{2} (3K) x^2 \]
Again substituting \( x = \frac{F}{3K} \):
\[ U_Q = \frac{1}{2} (3K) \left(\frac{F}{3K}\right)^2 = \frac{1}{2} (3K) \cdot \frac{F^2}{(3K)^2} \]
Simplifying this:
\[ U_Q = \frac{1}{2} (3K) \cdot \frac{F^2}{9K^2} = \frac{3F^2}{18K} = \frac{F^2}{6K} \]

5. **Relating Energy Stored in Spring Q to U**:
We already found that:
\[ U_P = \frac{F^2}{2K} \]
Now, we can express \( U_Q \) in terms of \( U_P \):
\[ U_Q = \frac{F^2}{6K} = \frac{1}{3} \cdot \frac{F^2}{2K} = \frac{1}{3} U_P \]

### Final Result:
Thus, the energy stored in spring Q is:
\[ U_Q = \frac{1}{3} U_P \]