logo

Ask Questions, Get Answers

X
 

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} $

1 Answer

$mg=U+R$
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
 

Related questions

...