Assume that the distance that a car runs on one litre of petrol varies inversely as the square of the speed at which it is driven. It gives a run of 9 km per litre at speed of 30 km/h. At what speed should it be driven to get a run of 100 km/L?

Options

A. 150 km/h
B. 225 km/h
C. 36 km/h
D. 9 km/h

Answer

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

### Step 1: Understand the relationship
The distance a car runs on one litre of petrol varies inversely as the square of the speed. This can be expressed mathematically as:
\[ D = \frac{K}{V^2} \]
where \( D \) is the distance per litre, \( V \) is the speed, and \( K \) is a constant.

### Step 2: Find the constant \( K \)
We are given that the car runs 9 km per litre at a speed of 30 km/h. We can substitute these values into the equation to find \( K \):
\[ 9 = \frac{K}{30^2} \]
Calculating \( 30^2 \):
\[ 30^2 = 900 \]
Now substituting back:
\[ 9 = \frac{K}{900} \]
To find \( K \), multiply both sides by 900:
\[ K = 9 \times 900 = 8100 \]

### Step 3: Set up the equation for the new distance
Now we need to find the speed \( V_1 \) at which the car runs 100 km per litre. Using the same relationship:
\[ 100 = \frac{K}{V_1^2} \]
Substituting \( K = 8100 \):
\[ 100 = \frac{8100}{V_1^2} \]

### Step 4: Solve for \( V_1^2 \)
Rearranging the equation to solve for \( V_1^2 \):
\[ V_1^2 = \frac{8100}{100} \]
Calculating the right side:
\[ V_1^2 = 81 \]

### Step 5: Find \( V_1 \)
Taking the square root of both sides gives us:
\[ V_1 = \sqrt{81} = 9 \]
Since speed cannot be negative, we take the positive value.

### Conclusion
The speed at which the car should be driven to get a run of 100 km/L is:
\[ \boxed{9 \text{ km/h}} \]

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