In the above question, if equal forces are applied on two springs, then

Options

A. (a) More work is done on Q
B. (b) More work is done on P
C. (c) Heir force constants will become equal
D. (d) Equal work is done on both the springs

Explanation

To solve the problem step by step, we will analyze the situation with two springs having different spring constants when equal forces are applied to them.

### Step 1: Understand the Given Information
We have two springs with spring constants \( k_1 \) and \( k_2 \), where \( k_1 > k_2 \). When equal forces \( F \) are applied to both springs, we need to determine the elongation in each spring.

**Hint:** Remember that the elongation in a spring is related to the force applied and the spring constant.

### Step 2: Apply Hooke's Law
According to Hooke's Law, the force exerted by a spring is given by:
\[ F = k \cdot x \]
where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the elongation (displacement) of the spring.

For spring 1:
\[ F = k_1 \cdot x_1 \]
For spring 2:
\[ F = k_2 \cdot x_2 \]

**Hint:** Since the forces are equal, we can set the two equations equal to each other.

### Step 3: Set the Forces Equal
Since \( F \) is the same for both springs:
\[ k_1 \cdot x_1 = k_2 \cdot x_2 \]

**Hint:** Rearranging this equation will help us find the relationship between \( x_1 \) and \( x_2 \).

### Step 4: Solve for Elongations
Rearranging the equation gives us:
\[ x_1 = \frac{k_2}{k_1} \cdot x_2 \]

Since \( k_1 > k_2 \), we can conclude that:
\[ \frac{k_2}{k_1} < 1 \]
This means:
\[ x_1 < x_2 \]

**Hint:** This indicates that the elongation in the spring with the larger spring constant (spring 1) is less than that in the spring with the smaller spring constant (spring 2).

### Step 5: Calculate Work Done on Each Spring
The work done on a spring is given by:
\[ W = F \cdot x \]
For spring 1:
\[ W_1 = F \cdot x_1 \]
For spring 2:
\[ W_2 = F \cdot x_2 \]

**Hint:** Since \( x_1 < x_2 \), the work done on spring 1 will be less than the work done on spring 2.

### Step 6: Compare Work Done
Since \( x_1 < x_2 \) and the force \( F \) is the same for both springs:
\[ W_1 < W_2 \]

Thus, we conclude:
\[ W_1 < W_2 \]

**Final Conclusion:** More work is done on spring 2 (the one with the smaller spring constant).

### Summary of the Answer
The work done on spring 2 (with spring constant \( k_2 \)) is greater than the work done on spring 1 (with spring constant \( k_1 \)).

**Final Answer:** More work is done on spring 2.