The density of a solid metal sphere is determ | vedclass

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Question

$\rho=\frac{M}{V}=\frac{M}{\frac{4}{3} \pi\left(\frac{D}{2}\right)^{3}}$

$\rho=\frac{6}{\pi} M D ^{-3}$

taking log

$\ell n \rho=\ell n \left(

Vedclass · Accepted Answer

$1050$

Vedclass · Answer

$ho=\frac{M}{V}=\frac{M}{\frac{4}{3} \pi\left(\frac{D}{2}ight)^{3}}$

$ho=\frac{6}{\pi} M D ^{-3}$

taking log

$\ell n ho=\ell n \left(\frac{6}{\pi}ight)+\ell n M -3 \ell m D$

Differentiates

$\frac{ d p}{ho}=0+\frac{ d M }{ M }-3 \frac{ d ( D )}{ D }$

for maximum error

$100 	imes \frac{ d ho}{ho}=\frac{ dM }{ M } 	imes 100+\frac{3 d D }{ D } 	imes 100$

$=6+3 	imes 1.5$

$=10.5 \%$

$=\frac{1050}{100} \%$

$x=1050.00$

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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