A block of mass 20 kg is hung with the help of ideal string, pulleys and spring (Spring constant k = 1000 N/m) as shown in figure. If block is in equilibrium position then extension in the spring will be `(g = 10 ms^-2)`

Options

A. 10 cm
B. 2.5 cm
C. 5 cm
D. 20 cm

Answer

To solve the problem of finding the extension in the spring when a block of mass 20 kg is in equilibrium, we can follow these steps:

### Step 1: Identify the forces acting on the block
When the block is in equilibrium, the forces acting on it are:
- The gravitational force (weight) acting downwards, which is given by \( mg \).
- The spring force acting upwards, which is given by Hooke's law as \( kx \), where \( x \) is the extension in the spring.

### Step 2: Write the equilibrium condition
In equilibrium, the net force acting on the block is zero. Therefore, we can write the equation:
\[ mg - kx = 0 \]
This implies that:
\[ mg = kx \]

### Step 3: Substitute the known values
We know:
- Mass \( m = 20 \, \text{kg} \)
- Gravitational acceleration \( g = 10 \, \text{m/s}^2 \)
- Spring constant \( k = 1000 \, \text{N/m} \)

Now substituting these values into the equation:
\[ 20 \, \text{kg} \times 10 \, \text{m/s}^2 = 1000 \, \text{N/m} \times x \]
This simplifies to:
\[ 200 \, \text{N} = 1000 \, \text{N/m} \times x \]

### Step 4: Solve for the extension \( x \)
To find \( x \), we rearrange the equation:
\[ x = \frac{200 \, \text{N}}{1000 \, \text{N/m}} = 0.2 \, \text{m} \]

### Step 5: Convert to centimeters
Since the question asks for the extension in centimeters:
\[ x = 0.2 \, \text{m} = 20 \, \text{cm} \]

### Final Answer
The extension in the spring when the block is in equilibrium is **20 cm**.

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