A RECTANGULAR IRON BLOCK OF MASS 10 KG IS PLACED ON THE TOP OF THE TABLE. THE DIMENSIONS OF THE BLOCK ARE 20 CM X 10 CM X 5 CM. FIND THE PRESSURE EXERTED BY THE BLOCK ON THE TABLE . IF THE BLOCK LIES ON THE TABLE WITH ITS SIDES OF DIMENSIONS 10 CM X 5 CM.

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Solution

$Pressure = \frac{Force}{Area} = \frac{mg}{A} = \frac{10 \times 9.8}{0.1 \times 0.05} = 19600 Pa$

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A RECTANGULAR IRON BLOCK OF MASS 10 KG IS PLACED ON THE TOP OF THE TABLE. THE DIMENSIONS OF THE BLOCK ARE 20 CM X 10 CM X 5 CM. FIND THE PRESSURE EXERTED BY THE BLOCK ON THE TABLE . IF THE BLOCK LIES ON THE TABLE WITH ITS SIDES OF DIMENSIONS 10 CM X 5 CM.

Pressure=Force Area =mgA =10×9.80.1×0.05 =19600 Pa

$Pressure = \frac{Force}{Area} = \frac{mg}{A} = \frac{10 \times 9.8}{0.1 \times 0.05} = 19600 Pa$