A solid block of wood of density $0.6 gcm^{-3}$ weighs 3.06 N in air. Determine (a) volume of the block (b) the volume of the block immersed when placed freely in a liquid of density 0.9 $gcm^{-3}$ ? $(510 cm_{3} , 340 cm_{3})$
Difficulty: Easy
Solution:
(a) Density of wood = D = $0.6 gcm^{-3}$
Weight of the wooden block = w= 3.06 N
Since w = mg or m $=\frac{w}{g}=\frac{3.06}{10} = 0.306 kg = 306$ g
Density of liquid D = $0.9 gcm^{-3}$
(i) The volume of the block V =?
(ii) The volume of the block immersed in a liquid V =?
Density $=\frac{Mass}{Volume}$
Volume $=\frac{Mass}{Density}$
V =$\frac{ 306}{(0.6)} = 510 cm_{3}$
(b) Volume =$\frac{Mass}{Density}$
V =$\frac{306}{(0.9)} = 340 cm_{3}$
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