Question

$\sf {\bold {\underline{find \: density \: of \: cube}}} \: \\ \\ \sf \: where \: mass \: = \small6 \frac{+}{-} 0.3kg \\ \\ \sf \: and \: volume \: = \small{ \sf{ \rm{ \pink{2 \frac{ + }{ - } 0.2kgl ^{ - 3} }}}}$
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Answer 1

Explanation:

$\sf\large\underline\purple{Given:-}$

${ \rm{ \implies{mass \: of \: cube \: = 6 \frac{ + }{ - } 0.3kg}}} \: \: \: \: \: \: \: \: \: \: \: \\ \\ { \rm{ \implies{volume \: of \: cube \: = 2 \frac{ + }{ - } 0.2kg {l}^{ - 3} }}}$

$\sf\large\underline\purple{Solution:-}$

$\displaystyle \sf \longrightarrow \: \small{\sf{ \green{ \boxed{density = \frac{mass}{volume} }}}} \: \\ \\ \displaystyle \longrightarrow \small{ \rm{ \pink{density = \frac{6}{2} = 3kg {l}^{ - 3} }}}$

$\sf\large\underline\purple{Find error:-}$

$\displaystyle \longrightarrow \sf{ \bold{ \boxed{ \frac{ \delta \: d}{d} = \frac{ \delta \: m}{m} + \frac{ \delta \: v}{v}}}} \: \: \: \\ \\ \displaystyle \sf \longrightarrow \: \delta \: d \: = (\frac{ \delta \: m}{m} + \frac{ \delta \: v}{v})d\\ \\ \displaystyle \sf \longrightarrow \: \delta \: d \: = 0.45 \\ \\ \\ \sf \:\bigstar\: density \: = 3 \frac{ + }{ - } 0.45$