If 1 % and 2 % are the errors in the measurem | vedclass

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(A) Given,error in measurement of mass,$\frac{\Delta m}{m} = 1 \%$. Error in measurement of density,$\frac{\Delta d}{d} = 2 \%$. We know that density

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(A) Given,error in measurement of mass,$\frac{\Delta m}{m} = 1 \%$. Error in measurement of density,$\frac{\Delta d}{d} = 2 \%$. We know that density $d = \frac{m}{V}$,where $V$ is the volume of the cube. Since $V = l^3$,where $l$ is the length of the side of the cube,we have $d = \frac{m}{l^3}$. Rearranging for $l$,we get $l = (m \cdot d^{-1})^{1/3}$. The relative error in length is given by $\frac{\Delta l}{l} = \frac{1}{3} \left( \frac{\Delta m}{m} + \frac{\Delta d}{d} \right)$. Substituting the given values: $\frac{\Delta l}{l} = \frac{1}{3} (1 \% + 2 \%) = \frac{1}{3} (3 \%) = 1 \%$. Thus,the error in the measurement of length is $1 \%$.

If $1 \%$ and $2 \%$ are the errors in the measurement of mass and density of a cube respectively,then the error in the measurement of length is (in $\%$)

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