Question

calculate the volume of wood of mass 6000 kg,when density of wood iron 0.8 gcm-3​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Volume\:of\:wood=75\times 10^{5}\:cm^{3}\:or\:7500\:L}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given : }} \\ \tt: \implies Mass \: of \: wood = 6000 \: kg \\ \\ \tt: \implies Density\: of \: wood = 0.8 \:g/ {cm}^{3} \\ \\ \red{\underline \bold{To \: Find : }} \\ \tt: \implies Volume \: of \:wood = ?$

• According to given question :

$\tt \circ \: Mass = 6000 \times 1000 = 6 \times {10}^{6} \: g \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies Density = \frac{Mass}{Volume} \\ \\ \tt: \implies 0.8 \times \frac{g}{cm ^{3} } = \frac{6 \times {10}^{6} }{Volume} \times g \\ \\ \tt: \implies Volume = \frac{6 \times {10}^{6} }{0.8} \times \frac{g \times {cm}^{3} }{g} \\ \\ \tt: \implies Volume = \frac{6 \times 10^{7} }{8} \times {cm}^{3} \\ \\ \tt: \implies Volume = 0.75 \times {10}^{7} \: cm^{3} \\ \\ \green{\tt: \implies Volume = 75 \times {10}^{5} \: {cm}^{3} } \\ \\ \green{\tt\therefore Volume\:of\:wood\:is\:75\times 10^{5}\:cm^{3}\:or\:7500\:litre}\\\\ \blue {\boxed{ \bold{Some \: related \: formula}}} \\ \orange{ \tt \circ \: Kinetic \: energy = \frac{1}{2} {mv}^{2} } \\ \\ \orange{ \tt \circ \:Potential \: energy = mgh } \\ \\ \orange{ \tt \circ \: Linear \: momentum=mv} \\ \\ \orange{\tt\circ\: Pressure=\frac{Force}{Area}}$