The density of a solid ball is to be determin | vedclass

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(C) Given:Pitch $= 0.5 \ mm$Circular scale divisions $= 50$Main scale reading $= 2.5 \ mm$Circular scale reading $= 20$Relative error in mass $(\Delta

Vedclass · Accepted Answer

$3.1$

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(C) Given:Pitch $= 0.5 \ mm$Circular scale divisions $= 50$Main scale reading $= 2.5 \ mm$Circular scale reading $= 20$Relative error in mass $(\Delta M/M) 	imes 100 = 2 \%$Least count $(LC) = \frac{	ext{Pitch}}{	ext{Circular scale divisions}} = \frac{0.5 \ mm}{50} = 0.01 \ mm$Diameter of the ball $(D) = 	ext{Main scale reading} + (LC 	imes 	ext{Circular scale reading})$$D = 2.5 \ mm + (0.01 \ mm 	imes 20) = 2.5 \ mm + 0.2 \ mm = 2.7 \ mm$Density $ho = \frac{M}{V} = \frac{M}{\frac{4}{3}\pi (D/2)^3} = \frac{6M}{\pi D^3}$The relative percentage error in density is given by:$\frac{\Delta ho}{ho} 	imes 100 = \left( \frac{\Delta M}{M} + 3 \frac{\Delta D}{D} ight) 	imes 100$Here,$\Delta D = LC = 0.01 \ mm$ and $D = 2.7 \ mm$.$\frac{\Delta ho}{ho} 	imes 100 = 2 \% + 3 	imes \left( \frac{0.01}{2.7} ight) 	imes 100$$\frac{\Delta ho}{ho} 	imes 100 = 2 \% + 3 	imes 0.37 \% = 2 \% + 1.11 \% = 3.11 \% \approx 3.1 \%$

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