A block of wood of density 0.6 gcm\(^{-3}\), weighing 3.06N in air, floats freely in a liquid of density 0.9 gcm\(^{-3}\). Calculate volume of the portion immersed (g = 10ms\(^{-2}\))
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Correct Answer: Option B
Explanation:
V = \(\frac{M}{D}\)
= \(\frac{3.06}{10} \times \frac{1000}{0.9} = 340\)
V = \(\frac{M}{D}\)
= \(\frac{3.06}{10} \times \frac{1000}{0.9} = 340\)