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Question

A block of wood 40 cm × 20 cm × 10 cm) is kept on a tabletop in three different positions: (a) with its side of dimensions 20 cm × 10 cm; (b) with its side of dimensions 10 cm × 40 cm; and (c) with its side of dimensions 40 cm × 20 cm. The pressure exerted by the wooden block on the tabletop in these positions is represented by PA, PB and Pc, respectively. The pressure follows the trend

This question was previously asked in
NDA I 2023 GAT Previous Year Paper (16-Apr-2023)
The correct answer is

PA > PB > Pc

Understanding Pressure Exerted by a Wooden Block on a Tabletop

Pressure is defined as the force applied perpendicularly to a surface per unit area. The formula for pressure is:

\( \text{Pressure (P)} = \frac{\text{Force (F)}}{\text{Area (A)}} \)

In this problem, the force exerted by the wooden block on the tabletop is its weight. The weight of the block is constant, regardless of which face is in contact with the tabletop. Let \( W \) represent the weight of the block.

The dimensions of the wooden block are given as 40 cm \(\times\) 20 cm \(\times\) 10 cm. The area of contact between the block and the tabletop changes depending on the block's orientation. We need to calculate the area of contact for each of the three given positions.

Calculating Contact Area for Each Position

  • Position (a): The block is placed with its side of dimensions 20 cm \(\times\) 10 cm on the tabletop.
  • The area of contact \( A_A \) is the product of these dimensions:
  • \( A_A = 20 \text{ cm} \times 10 \text{ cm} = 200 \text{ cm}^2 \)
  • Position (b): The block is placed with its side of dimensions 10 cm \(\times\) 40 cm on the tabletop.
  • The area of contact \( A_B \) is the product of these dimensions:
  • \( A_B = 10 \text{ cm} \times 40 \text{ cm} = 400 \text{ cm}^2 \)
  • Position (c): The block is placed with its side of dimensions 40 cm \(\times\) 20 cm on the tabletop.
  • The area of contact \( A_C \) is the product of these dimensions:
  • \( A_C = 40 \text{ cm} \times 20 \text{ cm} = 800 \text{ cm}^2 \)

Relating Area and Pressure Exerted

We can now express the pressure exerted in each position using the formula \( P = \frac{W}{A} \):

  • Pressure in position (a): \( P_A = \frac{W}{A_A} = \frac{W}{200} \)
  • Pressure in position (b): \( P_B = \frac{W}{A_B} = \frac{W}{400} \)
  • Pressure in position (c): \( P_C = \frac{W}{A_C} = \frac{W}{800} \)

Since the weight \( W \) is the same in all three cases, the pressure exerted is inversely proportional to the area of contact. This means that a larger contact area results in lower pressure, and a smaller contact area results in higher pressure.

Let's compare the calculated areas:

\( A_A = 200 \text{ cm}^2 \)

\( A_B = 400 \text{ cm}^2 \)

\( A_C = 800 \text{ cm}^2 \)

Comparing these values, we see that \( A_C > A_B > A_A \).

Because pressure is inversely proportional to area (\( P \propto \frac{1}{A} \) when \( W \) is constant), the relationship between the pressures will be the opposite of the relationship between the areas.

Therefore, the pressure trend is \( P_A > P_B > P_C \).

\( \frac{W}{200} > \frac{W}{400} > \frac{W}{800} \)

Conclusion on Pressure Trend

The pressure exerted by the wooden block on the tabletop in positions (a), (b), and (c) follows the trend \( P_A > P_B > P_C \).

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Important Questions from Pressure

  1. The pressure inside the cabin of the aircraft flying at an altitude is

    A. The same as that outside.

    B. Less than that outside.

    C. More than that outside.

    D. Normal atmospheric pressure at sea level.
  2. How many Pascals are equivalent to $0.25$ bar?
  3. A camel can walk/run in deserts very easily as compared to horse, donkey etc, because is-

  4. What is the thrust on unit area called?

  5. Calculate the pressure (in Pa) if a thrust of 1000 N is applied to an area of 5 m 2.

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