A) \[g(1-K)\]
B) \[g(1+K)\]
C) \[g\left( 1-\frac{1}{K} \right)\]
D) \[g\left( 1+\frac{1}{K} \right)\]
Correct Answer: C
Solution :
The net force acting on the stone in a downward direction would be equal to \[F=V\sigma g-V\rho g,\]where m is the mass of the stone, the force acting downwards would be equal to \[V\sigma g\]due to its weight and the force acting upwards would be equal to \[V\rho g\]due to the buoyant force. Hence, we get \[F=V\sigma g-V\rho g=V\sigma g\left( 1-\frac{\rho }{\sigma } \right)\] \[=mg\left( 1-\frac{\rho }{\sigma } \right)=mg\left( 1-\frac{1}{K} \right)\] Thus, \[a=g\left( 1-\frac{1}{K} \right)\]You need to login to perform this action.
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