A force F is acting on area A. If the force is double and the area is halved, the resultant pressure will be of the initial pressure :

Q: A force F is acting on area A. If the force is double and the area is halved, the resultant pressure will be of the initial pressure :

When a force $ F $ is acting on an area $ A $, the pressure $ P $ is given by the formula: \[ P = \frac{F}{A} \] Now, according to the problem, the force is doubled, and the area is halved. Let's denote the initial force as $ F $ and the initial area as $ A $. The initial pressure $ P_1 $ is: \[ P_1 = \frac{F}{A} \] After doubling the force and halving the area, the new force becomes $ 2F $ and the new area becomes $ \frac{A}{2} $. The new pressure $ P_2 $ is: \[ P_2 = \frac{2F}{\frac{A}{2}} \] \[ P_2 = \frac{2F \times 2}{A} \] \[ P_2 = \frac{4F}{A} \] Comparing the new pressure $ P_2 $ with the initial pressure $ P_1 $: \[ P_2 = 4 \times P_1 \] This shows that the resultant pressure is four times the initial pressure. Therefore, Option 4 is correct. **Explanation for Incorrect Options:** 1. **Option 1 (half):** If the pressure were half of the initial pressure, it would imply that the changes in force and area resulted in a decrease in pressure. However, doubling the force and halving the area actually increases the pressure significantly, not decreases it. 2. **Option 2 (double):** If the pressure were double the initial pressure, it would mean that the changes resulted in a pressure that is twice the original. However, as calculated, the pressure becomes four times the original, not just double. 3. **Option 3 (one-fourth):** If the pressure were one-fourth of the initial pressure, it would suggest a significant decrease in pressure. This is contrary to the actual effect of doubling the force and halving the area, which increases the pressure. Thus, the only correct conclusion is that the resultant pressure is four times the initial pressure, supporting Option 4 as the correct answer.

Question

Accepted Answer

When a force $ F $ is acting on an area $ A $, the pressure $ P $ is given by the formula:

$$ P = \frac{F}{A} $$

Now, according to the problem, the force is doubled, and the area is halved. Let's denote the initial force as $ F $ and the initial area as $ A $. The initial pressure $ P_1 $ is:

$$ P_1 = \frac{F}{A} $$

After doubling the force and halving the area, the new force becomes $ 2F $ and the new area becomes $ \frac{A}{2} $. The new pressure $ P_2 $ is:

$$ P_2 = \frac{2F}{\frac{A}{2}} $$

$$ P_2 = \frac{2F \times 2}{A} $$

$$ P_2 = \frac{4F}{A} $$

Comparing the new pressure $ P_2 $ with the initial pressure $ P_1 $:

$$ P_2 = 4 \times P_1 $$

This shows that the resultant pressure is four times the initial pressure. Therefore, Option 4 is correct.

**Explanation for Incorrect Options:**

1. **Option 1 (half):** If the pressure were half of the initial pressure, it would imply that the changes in force and area resulted in a decrease in pressure. However, doubling the force and halving the area actually increases the pressure significantly, not decreases it.

2. **Option 2 (double):** If the pressure were double the initial pressure, it would mean that the changes resulted in a pressure that is twice the original. However, as calculated, the pressure becomes four times the original, not just double.

3. **Option 3 (one-fourth):** If the pressure were one-fourth of the initial pressure, it would suggest a significant decrease in pressure. This is contrary to the actual effect of doubling the force and halving the area, which increases the pressure.

Thus, the only correct conclusion is that the resultant pressure is four times the initial pressure, supporting Option 4 as the correct answer.