Question

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Answer 1

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \bf \to \: given \: : \\ \\ \sf \hookrightarrow \: mass \: of \: the \:o bject \: \: m = 90 \: g \\ \\ \sf \hookrightarrow \: initial \: volume \: of \: water \: v_ 1 = 40 \: ml \\ \\ \sf \hookrightarrow \: final \: volume \: of \: water \: \: v_2 = 65 \: ml \\ \\ \therefore \: \sf \: volume \: of \: the \: object \: \: v = v_2 - v_1 \\ \\ = 65 - 40 \\ \\ = 25 \: ml \\ \\ \sf \: \therefore \: v = 25 \: {cm}^{3} \: \: \: \: \: \: \{\pink{ \because \: \sf \: 1 \: ml = 1 {cm}^{3}} \} \\ \\ \\ \sf \hookrightarrow \: density \: of \: the \: object \: \: d = \: to \: fnid \\ \\ \\ \bf \: we \: know \: that : \\ \\ \blue{ \boxed{ \sf \: density(d) = \frac{mass(m)}{volume(v)}}} \\ \\ \sf \implies \:d = \frac{m}{v} \\ \\ \sf \implies \:d= \frac{90}{25} \\ \\ \sf \implies \:d = 3.6 \: g {cm}^{ - 3} \\ \\ \\ \pink{ \boxed{ \therefore \sf \: density \: of \: the \: object \: \: d = 3.6 \: g {cm}^{ - 3}}} \end{array}}}}}$