A block of volume 3 x 105m3 and density 2.5 x 103 kg m-3 is suspended from a spring balance with \(\frac{2}{3}\) of its volume immersed in a liquid of density 900 kg m-3. Determine the reading of the spring balance [g = 10ms-2]
- 0.18N
-
0.57N
- 0.75n
- 0.93N
The correct answer is: B
Explanation
Volume immersed solid = 3 x 105 x \(\frac{2}{3}\)= 2 x 105m3
Mass of liquid displaced = ev
= 2 x 105 x 900 = 18 x 107 kg
Upthrust = 18 x 107 x 10
= 7.8 x 109N
Mass of solid = ev
= 2.5 x 103 x 105
= 7.5 x 108
Weight of solid = 7.5 x 108 x10 = 7.8 x 109
Reading = 0.75 - 0.18 = 0.57N