# The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2 %, the relative percentage error in the density is ?

Arun
25757 Points
3 years ago
Given

Pitch =0.5mm

Circular scale division =50

Least count =500.5=0.01mm

Relative error =2%

So,
Then dimensions reading from screw gauge= Diameter of a solid ball

Diameter of solid ball =2.5+500.5×20=2.7mm

As density ρ=Volumemass=34π(2D)3M

The relative percentage error in the density is
ρΔρ×100=(MΔM+D3ΔD)×100

% error =(MΔM×100+D3ΔD)×100

=(2+3×2.70.01)×100

=3.1 %
Vikas TU
14149 Points
3 years ago
Pitch =0.5mm
Circular scale division=50
Circular scale division reading =20 divisions
Relative error =2%
So
Then dimensions reading from screw gauge= Dia of ball
=2.5+[(0.5/50)x20)]
=2.5=0.2
=2.7mm
Density = mass/ volume
Volume =4/3 pi r³
=10.3065mm³

Density (§ using this symbol for rho)
Mass=m
Volume =v
§= m/v

% error in density = dm/m x100 + (dD/D) 3 x 100
=3.1%