If (1 \%) and (2 \%) are the errors in the measurement of mass and...

Given, error in measurement of mass, \$(\frac{\Delta m}{m}=$1 \%) and error in measurement of density, \$(\frac{\Delta d}{d}=$2 \%) Now, density, \$(d=\frac{\text { mass }}{\text { volume }} \Rightarrowd=\frac{m}{V}$) (\therefore) Volume of cube, $V=l^3$ Then, \$(d=\frac{m}{l^3} \Rightarrowl=\left(\frac{m}{d}\right)^{\frac{1}{3}}$) Then the error in measurement of length, ($$\begin{aligned} \frac{\Delta l}{l} \times100 & =\frac{1}{3}\left(\frac{\Delta m}{m} \times₁₀₀⁺\frac{\Delta d}{d} \times100\right) \\ \frac{\Delta l}{l} \times100 & =\frac{1}{3}(1+2) \times 100 \\ \frac{\Delta l}{l}& =1 \% \end{aligned}$$)