| Consider the following figures 1 and 2. In figure 1, the mass of the body A in air is 0.2 kg. The mass of the water container (with water in it) is 0.5 kg. In figure 2, when A is completely immersed in the water of the container B, the readings of the respective balances change as shown. |
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| What is the value of X? |
A) 0.55 kg
B) 0.15 kg
C) 0.3 kg
D) 0.45 kg
Correct Answer: A
Solution :
Initially, weight of A = 0.2 kg-wt Weight of B = 0.5 kg-wt After complete immersion of A m water, Weight of A = 0.15 kg-wt Reduction in weight of A = 0.2 - 0.15 = 0.05 kg-wt So, buoyant force applied on A by water = 0.05 kg-wt Required weight = weight of B + buoyant force on A = 0.5 + 0.05 = 0.55 kg-wt \[\therefore \] The reading of X = 0.55 kg
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