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Measurement MCQs for NEET — Physics Questions with Answers

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In a vernier calliper N divisions of vernier scale coincides with N-1 divisions of main scale (in which length of one division is 1 mm). The least count of the instrument should be

$N mm$

$N - 1 mm$

$\frac{1}{10 N} cm$

$\frac{1}{N - 1} mm$

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Explanation

Least Count = 1MSD - 1VSD

$N VSD = (N-1) MSD 1 V S D = \frac{N - 1}{N} M S D L C = 1 M S D - \frac{N - 1}{N} M S D = \frac{1}{N} M S D = \frac{1}{10 N} c m$

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In certain vernier callipers 25 divisions on vernier scale have same length as 24 divisions on main scale. One division on main scale is 1 mm long. The least count of the instrument is

0.04 mm

0.01 mm

0.02 mm

0.08 mm

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Explanation

L.C. = 1MSD -1VSD

25 VSD = 24 MSD

$1 V S D = \frac{24}{25} M S D L . C . = 1 M S D - \frac{24}{25} M S D = \frac{1}{25} M S D = \frac{1 m m}{25} = 0.04 m m$

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One centimeter on the main scale of Vernier calliper is divided into ten equal parts. If 10 divisions of Vernier scale coincide with 8 small divisions of the main scale, the least count of the callipers is:

0.01 cm

0.02 cm

0.05 cm

0.07 cm

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Explanation

L.C. = 1MSD - 1VSD

10 VSD = 8 MSD

1 VSD = 0.8 MSD

L.C. = 1 MSD - 0.8 MSD

= 0.2 MSD

= 0.2 mm or 0.02 cm

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If a screw gauge has a pitch of 1.5 mm and 300 divisions on circular scale, which of the following reading can be made from this screw gauge?

1.05 mm

3.01 cm

0.030000 m

2.1185 dm

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Explanation

Let us first find the least count.

L.C. = $\frac{P i t c h}{N o . o f C i r c u l a r s c a l e d i v i s i o n s}$ = $\frac{1.5 m m}{300} = 0.005 m m$

In (C) option, 0.030000m = 30.000 mm

The instrument from which reading has been taken has a least count of 0.001 mm or more.

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A Screw Guage gives the following readings when used to measure the diameter of a wire.
Main scale reading = 0mm
Circular scale reading = 52 divisions
Given that: 1 mm on main scale corresponds to 100 divisions of the circular scale.
The diameter of wire from the above data is:

0.026 cm

0.005 cm

0.52 cm

0.052 cm

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Explanation

$L e a s t C o u n t = \frac{P i t c h}{N o . o f c i r c u l a r s c a l e d i v i s i o n s} = \frac{1 m m}{100} = 0.01 m m$

Reading = Main scale reading (MSR) + Circular scale reading (CSR)

= 0 + 0.01 mm X 52

= 0.52 mm or 0.052 cm

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Two full turns of the circular scale of gauge cover a distance of 1 mm on scale. The total number of divisions on circular scale is 50. Further, it is found that screw gauge has a zero error of -0.03 mm. While measuring the diameter of a thin wire a student notes the main scale reading of 3 mm and the number of circular scale division in line, with the main scale as 35. The diameter of the wire is

3.32 mm

3.73 mm

3.67 mm

3.38 mm

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Explanation

Pitch = 0.5 mm

$L e a s t c o u n t = \frac{P i t c h}{N o . o f C S D}$

$= \frac{0.5 m m}{50} = 0.01 m m$

Observed Reading = MSR + CSR

= 3 mm + 35 X 0.01mm

= 3.35 mm

Correct Reading = Observed reading- Error

= 3.35 mm - (-0.03 mm)

= 3.38 mm

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The pitch of a screw gauge is 1mm and there are 100 divisions on the circular scale. While measuring the diameter of a wire, the linear scale reads 1 mm and 47^th division on the circular scale coincides with the reference line. The length of the wire is 5.6 cm. Find the curved surface area (in cm²) of the wire in appropriate number of significant figures.

2.4 cm²

2.56 cm²

2.6 cm²

2.8 cm²

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Explanation

$L . C . = \frac{1 m m}{100} = 0.01 m m$

Diameter (d) = 1mm + 0.01 mm X 47

= 1.47 mm

Length (l) = 5.6 cm

Curved surface area A = 2 $π r l$ = 2.586 $c m^{2}$

The final answer should have the same no. of significant figures (S.F.) as quantity having least no. of significant figures.

d = 1.47 mm ( 3 S.F.)

l = 5.6 cm (2 S.F.)

Ans : 2.6 $c m^{2}$ (rounded to 2 decimal places)

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One cm on the main scale of vernier callipers is divided into ten equal parts. If 20 divisions of vernier scale coincide with 8 small divisions of the main scale. What will be the least count of callipers?

0.06 cm

0.6 cm

0.5 cm

0.7 cm

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Explanation

(A)

20 VSD = 8 MSD

1 VSD = 0.4 MSD

L.C. = 1 MSD - 0.4 MSD

= 0.6 MSD = 0.6 X 1 mm

= 0.6 mm or 0.06 cm

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In Vander Waal's equation $[P + \frac{a}{V^{2}}] (V - b) = RT,$ the dimensions of a are

$M^{1} L^{5} T^{- 2}$

$M^{0} L^{2} T^{- 3}$

$M^{1} L^{3} T^{- 2}$

$M^{1} L^{1} T^{- 2}$

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Explanation

1. According to the principle of homogeneity dimensional of $P$ = Dimension of $\frac{a}{V^{2}}$

$\Rightarrow \frac{a}{{[L^{3}]}^{2}} = [{ML}^{- 1} T^{- 1}] \Rightarrow a = [{ML}^{5} T^{- 2}]$

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The dimension of the magnetic field intensity B is:

${MLT}^{- 2} A^{- 1}$

${MT}^{- 2} A^{- 1}$

${ML}^{2} {TA}^{- 2}$

$M^{2} {LT}^{- 2} A^{- 1}$

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Explanation

(b) F = Bil $\Rightarrow [B] = \frac{[F]}{[i] [L]} = \frac{{MLT}^{- 2}}{AL} = {MT}^{- 2} A^{- 1}$

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