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Units and Measurement Questions and Answers

31.

Of the given numbers,
(1) 25300
(2) 5.600
(3) 0.07200
the number of significant figures for

Exam Year(s)

KEAM - 2026,

Answer: E

The correct statement is (2) and (3) are four.

In 5.600 all digits count: the 5, the 6 and the two trailing zeros after the decimal point, which are significant. In 0.07200 the leading zeros only fix the decimal point and do not count, leaving 7, 2, 0, 0: four significant figures.

For 25300 the trailing zeros of a whole number without a decimal point are not taken as significant, so it has three, not five. (2) is not two and (3) is not three, and (1) does not have four.

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Answer: B

The mismatch is Torque : \(\mathrm{ML^{2}T^{-1}}\).

Torque is force times distance, so its dimensions are \(\mathrm{MLT^{-2}} \times \mathrm{L} = \mathrm{ML^{2}T^{-2}}\), the same as energy. \(\mathrm{ML^{2}T^{-1}}\) is the dimension of angular momentum.

The others are correct: force \(\mathrm{MLT^{-2}}\); pressure, force per area, \(\mathrm{ML^{-1}T^{-2}}\); heat energy \(\mathrm{ML^{2}T^{-2}}\); thermal conductivity, from \(\frac{Q}{t} = kA\frac{\Delta T}{x}\), \(\mathrm{MLT^{-3}K^{-1}}\).

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33.

The unit of measurement of EMF (Electromotive Force) is:

Exam Year(s)

Northern Coalfields Limited - 2020, NCL Technician - 2020,

Answer: B

EMF is measured in volts.

Electromotive force is the energy a source such as a cell or generator gives each coulomb of charge that passes through it. Energy per unit charge, joules per coulomb, is the volt, the same unit as potential difference.

The ohm is the unit of resistance, the ampere of current and the watt of power, so none of them can measure an electromotive force.

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34.

In Millikan's oil drop experiment the electronic charge e could be written as \(K\eta^{1.5}\), where K was a function of other experimental parameters with negligible error, while the viscosity of air \(\eta\) assumed was 0.4% too low. What was the error in e ?

Exam Year(s)

MH SET - 2020, SET Exam - 2020,

Answer: C

The error in e was 0.6%.

For a power law \(e = K\eta^{n}\) with K exact, relative errors scale by the exponent:

\[ \begin{aligned} \frac{\delta e}{e} &= n\frac{\delta\eta}{\eta} \\ &= 1.5 \times 0.4\% = 0.6\% \end{aligned} \]

Since \(\eta\) was too low, e came out about 0.6% too low. 0.4% ignores the exponent, 1.5% confuses the exponent with the error, and 0.7% has no basis.

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35.

The percentage error in the measurement of kinetic energy of a moving body is calculated to be 7%. If the percentage error in the measurement of mass of the body is 3%, then the percentage error in its velocity is

Exam Year(s)

Tripura Public Service Commission - 2026,

Answer: B

The percentage error in velocity is 2%.

For a product of powers, the relative errors add, each multiplied by its power. With K = \(\frac{1}{2}mv^{2}\):

\[ \begin{aligned} \frac{\Delta K}{K} &= \frac{\Delta m}{m} + 2\frac{\Delta v}{v} \\ 7 &= 3 + 2x \\ x &= 2\% \end{aligned} \]

4% is the result of forgetting to divide by the power 2, and 1% or 3% do not satisfy the equation.

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Tags: Units and Measurement Questions and Answers