The density of a solid metal sphere is determ | vedclass

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(D) The density $\rho$ of a sphere is given by $\rho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi (D/2)^3} = \frac{6M}{\pi D^3}$.Taking the natural logari

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$1050$

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(D) The density $ho$ of a sphere is given by $ho = \frac{M}{V} = \frac{M}{\frac{4}{3} \pi (D/2)^3} = \frac{6M}{\pi D^3}$.Taking the natural logarithm on both sides: $\ln ho = \ln(6/\pi) + \ln M - 3 \ln D$.Differentiating both sides to find the relative error: $\frac{dho}{ho} = \frac{dM}{M} - 3 \frac{dD}{D}$.For the maximum percentage error,we add the absolute values of the relative errors: $\left( \frac{dho}{ho} 	imes 100 ight)_{	ext{max}} = \left( \frac{dM}{M} 	imes 100 ight) + 3 \left( \frac{dD}{D} 	imes 100 ight)$.Given $\frac{dM}{M} 	imes 100 = 6.0 \%$ and $\frac{dD}{D} 	imes 100 = 1.5 \%$.Substituting these values: $	ext{Max error} = 6.0 + 3(1.5) = 6.0 + 4.5 = 10.5 \%$.We are given the error as $\left(\frac{x}{100}ight) \% = 10.5 \%$.Therefore,$\frac{x}{100} = 10.5 \implies x = 1050$.

The density of a solid metal sphere is determined by measuring its mass and its diameter. The maximum error in the density of the sphere is $\left(\frac{x}{100}\right) \%$. If the relative errors in measuring the mass and the diameter are $6.0 \%$ and $1.5 \%$ respectively,the value of $x$ is

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