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A cubical block of wood of side 5cm is placed on a table. Find the thrust by the block of wood on the table: (density of wood is $5 g / {cm}^{3}$ and g is $10 m / s^{2}$ )

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3.75 N

4.25 N

5.50 N

6.25 N

answer is D.

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Detailed Solution

$A = 5 \times 5 = 25 \times 10^{- 4} m^{2} m = PV = 5000 \times 125 \times 10^{- 6} kg$
Thrust $F = mg = 5000 \times 125 \times 10^{- 6} \times 10$
$= 625 \times 10^{- 2} N = 6 .25 N$

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