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A metal piece of mass 160 gm lies in equilibrium inside a glass of water. The piece touches the glass at small number of points. If the density of the metal is $8000 k g / m^{3}$ , then the normal force exerted by the bottom of the glass on the metal piece is _______ N $[T a k e g = 10 m / s^{2}]$

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answer is 1.4.

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Detailed Solution

A piece touches glass at small number of points, water would be there on bottom of piece also and hence buoyancy force acts on it.

From equilibrium condition, $m g = F_{B} + N$
$F_{B} = V ρ_{w a t e r} g = \frac{m}{ρ_{m e t a l}} \times ρ_{w a t e r} \times g = \frac{0.16}{g} \times 10 = 0.2 N$
So $N = m g - F_{B} = \frac{160}{1000} \times 10 - 0.2 = 1.4 N$

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