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

Units and Measurements PYQ — JEE Advanced 2011 · Shift 2 · Q48 with Solution

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Correct answer: C

Given:

So, least count:

\text{L.C.} = \frac{0.5}{50} = 0.01\text{ mm}

Main scale reading $= 2.5\text{ mm}$

Circular scale reading $= 20$ divisions

Circular scale contribution:

20 \times 0.01 = 0.20\text{ mm}

Hence diameter:

d = 2.5 + 0.20 = 2.70\text{ mm}

Take the absolute error in diameter as the least count:

\Delta d = 0.01\text{ mm}

Relative error in diameter:

\frac{\Delta d}{d} = \frac{0.01}{2.70}

Percentage error in diameter:

\frac{0.01}{2.70} \times 100 \approx 0.37\%

Density of a sphere:

\rho = \frac{m}{V}

For a sphere,

V = \frac{\pi d^3}{6}

So,

\rho \propto \frac{m}{d^3}

Therefore, relative error in density is:

\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + 3\frac{\Delta d}{d}

Given relative percentage error in mass $= 2\%$

Thus,

\text{Percentage error in density} = 2 + 3(0.37)

= 2 + 1.11 = 3.11\%

So the relative percentage error in density is approximately:

3.1\%

Hence the correct option is C.