A car of weight W is on an inclined road that rises by 100 m over a distance of 1 km and applies a constant frictional force `(W)/(20)` on the car.While moving uphill on the road at a speed of 10 `ms^(-1)` ,the car needs power P.If it needs power `(P)/(2)` while moving downhill at speed v then value of v is:

Options

A. 20`ms^(-1)`
B. `5 ms^(-1)`
C. `15 ms^(-1)`
D. `10 ms^(-1)`

Explanation

To solve the problem, we need to analyze the forces acting on the car while it is moving uphill and downhill, and then calculate the required power in both scenarios.

### Step-by-Step Solution:

1. **Determine the angle of inclination (θ):**
The car rises by 100 m over a distance of 1 km (1000 m).
\[ \sin \theta = \frac{\text{rise}}{\text{hypotenuse}} = \frac{100}{1000} = \frac{1}{10} \]

2. **Calculate the forces acting on the car while moving uphill:**
- The gravitational force component acting down the slope is \( W \sin \theta \).
- The frictional force acting against the motion is \( \frac{W}{20} \).
- The total force required to move uphill is:
\[ F_{\text{uphill}} = W \sin \theta + \frac{W}{20} = W \left( \frac{1}{10} + \frac{1}{20} \right) \]
To combine these fractions:
\[ \frac{1}{10} = \frac{2}{20} \quad \Rightarrow \quad F_{\text{uphill}} = W \left( \frac{2}{20} + \frac{1}{20} \right) = W \left( \frac{3}{20} \right) \]

3. **Calculate the power required to move uphill:**
The power \( P \) required to move uphill at a speed of 10 m/s is given by:
\[ P = F_{\text{uphill}} \times v = \left( \frac{3W}{20} \right) \times 10 = \frac{3W}{2} \]

4. **Determine the forces acting on the car while moving downhill:**
- The gravitational force component acting down the slope is still \( W \sin \theta \).
- The frictional force acting against the motion is \( \frac{W}{20} \).
- The net force while moving downhill is:
\[ F_{\text{downhill}} = W \sin \theta - \frac{W}{20} = W \left( \frac{1}{10} - \frac{1}{20} \right) \]
Combining these fractions:
\[ F_{\text{downhill}} = W \left( \frac{2}{20} - \frac{1}{20} \right) = W \left( \frac{1}{20} \right) \]

5. **Calculate the power required to move downhill:**
The power \( \frac{P}{2} \) required to move downhill at speed \( v \) is given by:
\[ \frac{P}{2} = F_{\text{downhill}} \times v = \left( \frac{W}{20} \right) \times v \]
We know from the uphill calculation that \( P = \frac{3W}{2} \), so:
\[ \frac{P}{2} = \frac{3W}{4} \]
Setting the two expressions for power equal gives:
\[ \frac{3W}{4} = \left( \frac{W}{20} \right) \times v \]

6. **Solve for \( v \):**
Cancel \( W \) from both sides (assuming \( W \neq 0 \)):
\[ \frac{3}{4} = \frac{v}{20} \quad \Rightarrow \quad v = 20 \times \frac{3}{4} = 15 \text{ m/s} \]

### Final Answer:
The value of \( v \) is \( 15 \text{ m/s} \).