A book exerts a pressure of 2 Pa on a table due to its weight. If Rahul places an identical book above it, then what would be the pressure (in Pa) on the table?

4

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Solution

The correct option is A 4

Initial pressure, $P_{1}$ = $\frac{F o r c e (F)}{A r e a (A)}$

When Rahul puts an identical book on the first book, the force on the table doubles, but the area remains the same.

Therefore, Final pressure, $P_{2}$ = $\frac{2 F}{A}$ = 2 $P_{1}$ = 2 x 2 = 4 Pa

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