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If the volume of $1.32 \;kg$ of an iron ball is $0.17\;m^3$ find the density.

$\begin{array}{1 1}(A)\;0.0153 \; kg\;m^{-3} \\(B)\;2.3 \;kg m^{-3} \\(C)\;7.8 \;kgm^{-3} \\(D)\;2.5\;kgm^{-3} \end{array} $

1 Answer

$d= \large\frac{M}{V}$
$\qquad= \large\frac{1.32kg}{0.17m^3}$
$\qquad= 7.764 70 kgm^{-3}$
Since the factor with lowest significant figures is 0.17 and it has two significant figures, rounding the result to two significant figures, we have the density of iron ball = $7.8 kg m^{-3}$
Hence C is the correct answer.
answered Jul 11, 2014 by meena.p

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