A 2 cm cubical block of mass 2 kg is lying on the ground. What is the pressure exerted by the block on the ground?

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Solution

The correct option is C

$Weight of the block = m g = 2 k g \times$ 9.8m $s^{- 2} = 19.6 N$

Area of the surface in contact = $(2 c m)^{2} = 4 {c m}^{2} = 4 \times (10^{- 4}) m^{2}$

Pressure exerted = $\frac{F}{A} = (\frac{19.6}{4}) \times 10^{4} N m^{- 2} = 4.9 \times 10^{4} N m^{- 2}$

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A 2 cm cubical block of mass 2 kg is lying on the ground. What is the pressure exerted by the block on the ground?

The correct option is C Weight of the block=mg=2kg× 9.8m s−2=19.6N Area of the surface in contact = (2cm)2=4cm2=4×(10−4)m2 Pressure exerted = FA=(19.64)×104Nm−2=4.9×104Nm−2

The correct option is C

$Weight of the block = m g = 2 k g \times$ 9.8m $s^{- 2} = 19.6 N$

Area of the surface in contact = $(2 c m)^{2} = 4 {c m}^{2} = 4 \times (10^{- 4}) m^{2}$

Pressure exerted = $\frac{F}{A} = (\frac{19.6}{4}) \times 10^{4} N m^{- 2} = 4.9 \times 10^{4} N m^{- 2}$