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What is the pressure inside the drop of mercury of radius 3mm at room temperature?S.T of mercury at that temperature $(20^{\large\circ}C)$ is $4.65\times 10^{-1}Nm^{-1}$.The $P_{atm}=1.01\times 10^4pa$.Also give excess pressure inside the drop

$\begin{array}{1 1}(A)\;1.01\times 10^5pa;310pa\\(B)\;1\times 10^5pa;220pa\\(C)\;3\times 10^5pa;330pa\\(D)\;1\times 10^5pa;300pa\end{array} $

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1 Answer

Pressure inside the mercury drop=Excess pressure inside mercury +$P_{atm}$
$\Rightarrow \large\frac{2S}{r}$$+P_0$
$\Rightarrow \large\frac{2\times 4.65\times 10^{-1}}{3\times 10^{-3}}$$+1.01\times 10^{5}$
$\Rightarrow 1.01\times 10^5$pa
Excess pressure =$\large\frac{2S}{r}$
$\Rightarrow 310pa$
Hence (A) is the correct answer.
answered May 9, 2014 by sreemathi.v
 

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