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A metal piece of mass 160g lies in equlibrium inside a glass of $H_2O$.The piece touches the bottom of the glass at a small number of points.If the density of the metal is $8000Kg/m^3$.Find the normal force exerted by the bottom of the glass on the metal piece

$\begin{array}{1 1}(A)\;1.5N\\(B)\;1.4N\\(C)\;1N\\(D)\;1.8N\end{array} $

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$mg=U+R$
Where U-upthrust force
$\rightarrow mg-U=R$
$\Rightarrow R=mg-V\rho_{\omega}g$
$\Rightarrow R=mg-\large\frac{m}{\rho}$$\rho_{\omega}g$
$\Rightarrow R=1.4N$
Hence (B) is the correct answer.
answered May 9, 2014 by sreemathi.v
 

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