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\(^{-2}\))
- 510cm3
-
340cm3
- 254cm3
- 170cm3
The correct answer is: B
Explanation
V = \(\frac{M}{D}\)
= \(\frac{3.06}{10} \times \frac{1000}{0.9} = 340\)