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Owais Bhai Chadha
Asked: 2022-11-08T22:48:47+05:30 In:

Put a brick on sponge as show in Figure . Now, put the brick vertical on sponge. The applied force is equal on sponge in both situations.

However, we see that sponge squeezes more in vertical position of brick. In which situation, the contract area between sponge and brick is less ?

If the area is reduced and value of force is kept constant then pressure increases. This is the reason why the balloon is punctured by sharp pin head and sponge squeezes more in vertical position of brick.

We may say that the value of pressure depends on the contact area of object. When contact area is reduced pressure increases.

1. In which situation, contact area of the brick and the sponge is less?

2. Why sponge gets more depressed when brick is made to stand on it?

Put a brick on sponge as show in Figure . Now, put the brick vertical on sponge. The applied force is equal on sponge in both situations.

However, we see that sponge squeezes more in vertical position of brick. In which situation, the contract area between sponge and brick is less ?

If the area is reduced and value of force is kept constant then pressure increases. This is the reason why the balloon is punctured by sharp pin head and sponge squeezes more in vertical position of brick.

We may say that the value of pressure depends on the contact area of object. When contact area is reduced pressure increases.

1. In which situation, contact area of the brick and the sponge is less?

2. Why sponge gets more depressed when brick is made to stand on it?

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1 Answer

  1. 1. When brick input vertical on the sponge, contact area of both is less.

    2. When a brick is made to stand on the sponge, contact area gets decreased, hence more pressure is exerted on the sponge. Therefore it gets more depressed.

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