A mass of 10 kg is suspended from a spring balance. It is pulled aside by a horizontal string so that it makes an angle of `60^(@)` with the vertical. The new rading of the balance is

Options

A. (a)20 kg, vt
B. (b)10 kg, vt
C. (c)`10 sqrt(3)` kg - vt
D. (d)`20 sqrt(3)` kg, vt

Explanation

To solve the problem, we need to determine the new reading of the spring balance when a mass of 10 kg is pulled aside at an angle of 60 degrees with the vertical.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass (m) = 10 kg
- Acceleration due to gravity (g) = 10 m/s²
- Angle (θ) = 60°

2. **Calculate the Weight of the Mass:**
- The weight (W) of the mass is given by the formula:
\[ W = m \cdot g \]
- Substituting the values:
\[ W = 10 \, \text{kg} \cdot 10 \, \text{m/s}^2 = 100 \, \text{N} \]

3. **Analyze the Forces Acting on the Mass:**
- When the mass is pulled aside, the tension (T) in the string has two components:
- Vertical component (T_y) = T * cos(θ)
- Horizontal component (T_x) = T * sin(θ)
- The vertical component of the tension must balance the weight of the mass:
\[ T \cdot \cos(θ) = W \]

4. **Substituting the Known Values:**
- We know that:
\[ W = 100 \, \text{N} \quad \text{and} \quad θ = 60° \]
- Therefore:
\[ T \cdot \cos(60°) = 100 \, \text{N} \]
- Since \(\cos(60°) = \frac{1}{2}\):
\[ T \cdot \frac{1}{2} = 100 \, \text{N} \]

5. **Solving for Tension (T):**
- Rearranging the equation gives:
\[ T = 100 \, \text{N} \cdot 2 = 200 \, \text{N} \]

6. **Calculate the New Reading of the Spring Balance:**
- The reading of the spring balance (R) can be calculated using:
\[ R = \frac{T}{g} \]
- Substituting the values:
\[ R = \frac{200 \, \text{N}}{10 \, \text{m/s}^2} = 20 \, \text{kg} \]

7. **Conclusion:**
- The new reading of the spring balance is 20 kg.

### Final Answer:
The new reading of the balance is **20 kg**.

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