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A beaker of water kept in the left pan of a common balance is counter poised with weights in the right pan. Now a body of mass 12 g and density 3 g/cc is suspended inside the beaker from an independent support and the body is completely immersed in the water without touching the sides of the beaker. To restore equilibrium the weight to be added in the right pan will be:

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zero

12 g

4 g

8 g

answer is D.

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Detailed Solution

For equilibrium $T_{h} = mg \Rightarrow \frac{12}{3} \times 1 \times g = mg \Rightarrow m = 4 g$

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