Two cubes of masses 3 g and 6 g and sides of length $2 cm$ and $3 cm$ respectively are placed on a smooth table as shown in the figure. Calculate the pressure experienced by the table due to the system of boxes. [Take $g = 10 m / s^{2}]$

100 Pa

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60 Pa

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120 Pa

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80 Pa

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Solution

The correct option is A 100 Pa
Mass of the two boxes = 6 g + 3 g
= 9 g
$= 0.009 kg$
$∴$ Weight of the two boxes exerted on the table $= 0.009 \times g$
$W = 0.09 N$
Pressure $= \frac{Force}{Area}$
$= \frac{0.09}{(3 \times 10^{- 2})^{2}}$
$= \frac{0.09 \times 10^{4}}{9}$
$= 100 P a$

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Two cubes of masses 3 g and 6 g and sides of length 2 cm and 3 cm respectively are placed on a smooth table as shown in the figure. Calculate the pressure experienced by the table due to the system of boxes. [Take g=10m/s2]

The correct option is A 100 PaMass of the two boxes = 6 g + 3 g = 9 g =0.009 kg ∴ Weight of the two boxes exerted on the table =0.009×g W=0.09N Pressure =ForceArea =0.09(3×10−2)2 =0.09×1049 =100 Pa

Two cubes of masses 3 g and 6 g and sides of length $2 cm$ and $3 cm$ respectively are placed on a smooth table as shown in the figure. Calculate the pressure experienced by the table due to the system of boxes. [Take $g = 10 m / s^{2}]$

The correct option is A 100 Pa
Mass of the two boxes = 6 g + 3 g
= 9 g
$= 0.009 kg$
$∴$ Weight of the two boxes exerted on the table $= 0.009 \times g$
$W = 0.09 N$
Pressure $= \frac{Force}{Area}$
$= \frac{0.09}{(3 \times 10^{- 2})^{2}}$
$= \frac{0.09 \times 10^{4}}{9}$
$= 100 P a$