A uniform box of height `2 m` and having a square base of side `1 m`, weight `150 kg`, is kept on one end on the floor of a truck. The maximum speed with which the truck can round a curve of radius `20 m` without causing the block to tip over is (assume that friction is sufficient is no sliding).

Options

A. `15m//s`
B. `10m//s`
C. `8m//s`
D. depends on the value of coefficient of friction

Explanation

To solve the problem, we need to determine the maximum speed at which the truck can round a curve without causing the box to tip over. We will analyze the forces and torques acting on the box.

### Step-by-Step Solution:

1. **Identify the Dimensions and Weight of the Box:**
- Height (H) = 2 m
- Side of the base (a) = 1 m
- Weight (W) = 150 kg
- The gravitational force (mg) = 150 kg × 9.81 m/s² = 1471.5 N (approximately 1500 N for simplicity).

2. **Determine the Center of Mass:**
- The center of mass of the box is located at a height of H/2 = 2 m / 2 = 1 m from the base.

3. **Identify the Forces Acting on the Box:**
- The weight of the box (W = mg) acts downward through the center of mass.
- The normal force (N) acts upward at the base of the box.
- When the truck rounds a curve, a centripetal force is required to keep the box moving in a circular path.

4. **Torque Calculation:**
- For the box to remain stable and not tip over, the torque due to the weight must be balanced by the torque due to the normal force.
- The torque due to the weight (W) about the tipping edge is given by:
\[ \text{Torque due to W} = W \times \left(\frac{a}{2}\right) = mg \times \left(\frac{1}{2}\right) = 0.5 \times mg \]
- The maximum torque due to the normal force (N) is:
\[ \text{Torque due to N} = N \times H = N \times 2 \]

5. **Setting Up the Torque Balance Equation:**
- For equilibrium (no tipping), we set the torques equal:
\[ N \times 2 = 0.5 \times mg \]
- Solving for N gives:
\[ N = \frac{0.5 \times mg}{2} = \frac{mg}{4} \]

6. **Centripetal Force Requirement:**
- The centripetal force required to keep the box moving in a circular path is:
\[ F_c = \frac{mv^2}{r} \]
- This centripetal force is provided by the normal force:
\[ N = \frac{mv^2}{r} \]

7. **Equating the Forces:**
- From the previous equations, we have:
\[ \frac{mg}{4} = \frac{mv^2}{r} \]
- Canceling mass (m) from both sides gives:
\[ \frac{g}{4} = \frac{v^2}{r} \]

8. **Solving for Maximum Speed (v):**
- Rearranging gives:
\[ v^2 = \frac{g \cdot r}{4} \]
- Substituting g = 10 m/s² and r = 20 m:
\[ v^2 = \frac{10 \cdot 20}{4} = 50 \]
- Taking the square root:
\[ v = \sqrt{50} \approx 7.07 \text{ m/s} \]

### Final Answer:
The maximum speed with which the truck can round a curve of radius 20 m without causing the box to tip over is approximately **7.07 m/s**.