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Units and Measurements

Section: Physics

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Question : 30 of 34

Marks: +1, -0

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

[JEE Adv 2011 P2]

0.9%
2.4%
3.1%
4.2%

Solution:

For the screw gauge, we have

Least count =

\frac{\text{Pitch}}{\text{Number of divisions on head scale}}

⇒ L.C. =

\frac{0.5}{50}

mm = 0.01 mm

Therefore, the diameter of the ball is

D = 2.5 + 0.01× 20 = 2.70 mm

The density of the ball is

ρ =

\frac{M}{V}

=

\frac{M}{\frac{4}{3}\pi r^3 \left(\frac{D}{2}\right)^3}

Therefore,

\frac{\Delta\rho}{\rho}

=

\frac{M}{M} + \frac{3\Delta D}{D}

Now,

\frac{\Delta\rho}{\rho}

= 0.02 and D = Least count. Therefore,

\frac{\Delta\rho}{\rho}

= 0.02 +

3\left(\frac{0.01}{2.7}\right)

= 0.031

or the percentage error in density is 3.1%.

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