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A plant virus is found to consist of uniform cylindrical particles of 150A in diameter and 5000A long.The specific volume of the virus is $0.75cm^3/g$.If the virus is considered to be a single particle and its molecular weight

$\begin{array}{1 1}(a)\;6\times 10^7&(b)\;7.095\times 10^7\\(c)\;5.095\times 10^7&(d)\;9\end{array}$

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Volume of virus =$\pi r^2l$
$\qquad\qquad\qquad=\large\frac{22}{7}\times \frac{150}{2}\times \frac{150}{2}$$\times 10^{-6}\times 5000\times 10^{-8}$
$\qquad\qquad\qquad=0.884\times 10^{-16}cm^3$
Weight of one virus $=\large\frac{0.884\times 10^{-16}}{0.75}$g
$\Rightarrow 1.178\times 10^{-16}g$
Mol.wt.of.virus=$1.178\times 10^{-16}\times 6.023\times 10^{23}$
$\Rightarrow 7.095\times 10^7$
Hence (b) is the correct answer.
answered Oct 30, 2013 by sreemathi.v

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