A uniform rectangular block of wood $20 cm \times 7.5 cm \times 7.5 cm$ and of mass 1000g stands on a horizontal surface with its longest edge vertical.
Find:
(i) The pressure exerted by the block on the surface
(ii) Density of the wood.
$1778 Nm^{-2}, 889 kgm^{-3}$
Difficulty: Hard
Solution:
Length of the smallest side of the block = 7.5 cm
Mass of the block m = 1000g = 1kg
The pressure exerted by the block P =?
The density of wood $ \rho $ =?
Calculations:
(i) since the smallest edge of the block is rested on the horizontal surface. Therefore, area of the block will be:
Area = A = $7.5 cm \times 7.5 cm$ = $56.25 cm^{2}$
= $56.25 \times \frac {1}{100} \times \frac{1}{100} m^{2}$ =$56.25 \times 10^{-4} m ^{2}$
Pressure under the thumb = P =?
P = $\frac{F}{A} =\frac{mg}{A}$
P = $\frac{1 \times 10}{56.25 \times 10^{-4} }$ = $0.1778\times 10^{4} = 1778 Nm^{-2}$
(ii) Volume = V = $ 20 cm \times 7.5 cm \times 7.5$ cm $ = 1125 cm^{3} $
= $1125 \times \frac{1}{100} m \times \frac{1}{100}$ m $ \times\frac{1}{100} m $
= $1125 \times 10^{6} m^{3}$
Or V = $1.125 \times 10 ^{-3} m^{3}$
Density = $\frac{Mass}{Volume}$
Density $ =\frac{1}{(1.125 \times 10(^{-3})} (1 ) = 0.8888 \times 10^{3} = 888.8 kgm^{-3} $
Density = $889 kgm^{-3}$
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