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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 ?

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 ?

Grade:11

2 Answers

Arun
25763 Points
one year ago
Given
 
Screw gauge readings
 
Pitch =0.5mm
 
Circular scale division =50
 
Main scale reading =2.5mm
 
Least count =500.5=0.01mm
 
Circular scale division reading =20divisions
 
Relative error =2%
 
So,
Then dimensions reading from screw gauge= Diameter of a solid ball 
 
S.G. Reading = main scale reading +(circular scale divisionpitch)× circular scale division reading
 
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
one year ago
Screw gauge readings
Pitch =0.5mm
Circular scale division=50
Main scale reading =2.5 mm
Circular scale division reading =20 divisions
Relative error =2%
So
Then dimensions reading from screw gauge= Dia of ball
S.G. Reading = main scale reading +[(pitch/circular scale division) x circular scale division reading]
=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%

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