Question

A lead ball of mass 230g, when immersed into a measuring cylinder, the water level rises from 35 ml to 55 ml mark. Find (a) The volume of the ball (b) the density of ball in g/cubic cm​

Answer 1

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \bf \to \: given \: : \\ \\ \sf \hookrightarrow \: mass \: of \: the \:ball \: \: m = 230 \: g \\ \\ \sf \hookrightarrow \: initial \: volume \: of \: water \: v_ 1 = 35\: ml \\ \\ \sf \hookrightarrow \: final \: volume \: of \: water \: \: v_2 = 55 \: ml \\ \\ \\ \sf \: a) \:volume \: of \: the \: ball \: v = \: to \: find \\ \\ \sf \: b) \: density \: of \: the \: ball \: d = to \: find \\ \\ \\ \: \: \red{ \underline{ \sf \: solution}} \\ \\ \sf \: a) \\ \\ \: \sf \: volume \: of \: the \: ball\: \: v = v_2 - v_1 \\ \\ = 65 - 40 \\ \\ = 25 \: ml \\ \\ \sf \: \therefore \: v = 25 \: {cm}^{3} \: \: \: \: \: \: \{\pink{ \because \: \sf \: 1 \: ml = 1 {cm}^{3}} \} \\ \\ \boxed{\therefore \: \sf \: volume \: of \: the \: ball \: v = 20 \: cm {}^{3} } \\ \\ \\ \sf \:b ) \\ \\ \sf \hookrightarrow \: density \: of \: the \: object \: = \: d \\ \\ \\ \bf \: we \: know \: that : \\ \\ \blue{ \boxed{ \sf \: density(d) = \frac{mass(m)}{volume(v)}}} \\ \\ \sf \implies \:d = \frac{m}{v} \\ \\ \sf \implies \:d= \frac{230}{20} \\ \\ \sf \implies \:d = 11.5 \: g {cm}^{ - 3} \\ \\ \\ \pink{ \boxed{ \therefore \sf \: density \: of \: the \: object \: \: d = 11.5 \: g {cm}^{ - 3}}} \end{array}}}}}$