A spring balance has a scale that reads from 0 to 50 kg. the length of the scale is 20 cm. a block of mass m is suspended from this balance, when displaced and released, it oscillates with a period 0.5 s. the value of m is (Take `g=10ms^(-2)`)

Options

A. 8 kg
B. 12 kg
C. 16 kg
D. 20 kg

Explanation

To solve the problem step by step, we will follow these instructions:

### Step 1: Calculate the maximum force exerted by the spring balance
The maximum mass that the spring balance can read is 50 kg. To find the maximum force (F_max) exerted by this mass, we use the formula:
\[ F_{\text{max}} = m \cdot g \]
Where:
- \( m = 50 \, \text{kg} \)
- \( g = 10 \, \text{m/s}^2 \)

Calculating:
\[ F_{\text{max}} = 50 \, \text{kg} \times 10 \, \text{m/s}^2 = 500 \, \text{N} \]

### Step 2: Calculate the spring constant (k)
The spring constant (k) can be calculated using the formula:
\[ k = \frac{F_{\text{max}}}{L} \]
Where:
- \( F_{\text{max}} = 500 \, \text{N} \)
- \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)

Calculating:
\[ k = \frac{500 \, \text{N}}{0.2 \, \text{m}} = 2500 \, \text{N/m} \]

### Step 3: Use the formula for the period of oscillation
The period (T) of oscillation for a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

### Step 4: Rearranging the formula to find mass (m)
Squaring both sides gives:
\[ T^2 = 4\pi^2 \frac{m}{k} \]
Rearranging for mass (m):
\[ m = \frac{T^2 \cdot k}{4\pi^2} \]

### Step 5: Substitute the known values
Given that \( T = 0.5 \, \text{s} \) and \( k = 2500 \, \text{N/m} \):
\[ m = \frac{(0.5)^2 \cdot 2500}{4\pi^2} \]

Calculating:
\[ m = \frac{0.25 \cdot 2500}{4\pi^2} \]
\[ m = \frac{625}{4\pi^2} \]

### Step 6: Calculate the numerical value of m
Using \( \pi \approx 3.14 \):
\[ m = \frac{625}{4 \cdot (3.14)^2} \]
\[ m = \frac{625}{4 \cdot 9.8596} \]
\[ m = \frac{625}{39.4384} \approx 15.8 \, \text{kg} \]

### Conclusion
The value of \( m \) is approximately 16 kg.