Units & Measurement — Top 20 Questions

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Least count of a vernier caliper is \(\dfrac{1}{20N}\) cm and 1 MSD = 1 mm. Number of vernier scale divisions is:

A\(\dfrac{2N-1}{2N}\)

B\(\dfrac{2N-1}{20N}\)

C\(2N-1\)

D\(\dfrac{20N}{2N-1}\)

✅ Correct: D

LC \(= \dfrac{1}{20N}\) cm \(= \dfrac{1}{2N}\) mm.
LC \(= 1\text{ MSD} - 1\text{ VSD}\) \(\Rightarrow\) 1 VSD \(= 1 - \dfrac{1}{2N} = \dfrac{2N-1}{2N}\) mm.
No. of VSD \(= \dfrac{1\text{ MSD}}{1\text{ VSD}} = \dfrac{1\text{ mm}}{\dfrac{2N-1}{2N}\text{ mm}} = \dfrac{2N}{2N-1}\).
Since LC is \(\dfrac{1}{20N}\) cm, the total divisions \(= \dfrac{20N}{2N-1}\).

20 VSD = 19 MSD and LC = 0.1 mm. Value of 1 MSD is:

A5 mm

B2 mm

C1 mm

D0.5 mm

✅ Correct: B

\(1\text{ VSD} = \dfrac{19}{20}\text{ MSD}\).
LC \(= 1\text{ MSD} - 1\text{ VSD} = 1\text{ MSD} - \dfrac{19}{20}\text{ MSD} = \dfrac{1}{20}\text{ MSD}\).
\(\dfrac{1}{20}\text{ MSD} = 0.1\text{ mm} \Rightarrow \mathbf{1\text{ MSD} = 2\text{ mm}}\).

If \(v = At^2 + \dfrac{Bt}{C+t}\), the dimensional formula of \(ABC\) is:

A\([M^0LT^{-2}]\)

B\([M^0L^2T^{-3}]\)

C\([M^0L^2T^{-2}]\)

D\([M^0LT^{-1}]\)

✅ Correct: B

\([v] = LT^{-1}\).
From \(At^2\): \([A] = \dfrac{[v]}{[t^2]} = \dfrac{LT^{-1}}{T^2} = LT^{-3}\).
From \(\dfrac{Bt}{C+t}\): \([C] = [t] = T\), and \(\left[\dfrac{Bt}{C}\right] = [v] \Rightarrow [B] = \dfrac{[v][C]}{[t]} = \dfrac{LT^{-1} \cdot T}{T} = LT^{-1}\).
\([ABC] = LT^{-3} \cdot LT^{-1} \cdot T = \mathbf{L^2T^{-3}}\).

Time period measured with \(+2\%\) error, mass with \(-1\%\) error. Percentage error in spring constant \(k = \dfrac{4\pi^2 M}{T^2}\) is:

A5%

B3%

C1%

D4%

✅ Correct: A

\(\dfrac{\Delta k}{k} = \dfrac{\Delta M}{M} + 2\dfrac{\Delta T}{T} = 1\% + 2 \times 2\% = \mathbf{5\%}\).
Errors always add as magnitudes regardless of sign.

For \(E = \alpha^3 e^{-\beta t}\), if \(\alpha\) has 2% error and \(\beta\) has error \(\Delta\beta\), maximum percentage error in \(E\) at \(t = 1\) s is 6%. Value of \(\Delta\beta / \beta\) is:

B0.06

C0.03

D0.6

✅ Correct: A

\(\dfrac{\Delta E}{E} = 3\dfrac{\Delta\alpha}{\alpha} + \beta t \cdot \dfrac{\Delta\beta}{\beta}\).
\(3 \times 2\% = 6\%\) already. So \(\Delta\beta/\beta\) contributes 0 for the total to remain 6%.
\(\Rightarrow \Delta\beta/\beta = \mathbf{0}\).

A screw gauge has pitch 1 mm and 100 circular divisions. It measures diameter of wire as 1.00 mm with 2 divisions uncertainty. Percentage error in cross-sectional area \(A = \pi r^2\) is:

A1%

B2%

C4%

D3%

✅ Correct: D

LC \(= \dfrac{1\text{ mm}}{100} = 0.01\text{ mm}\). Error in \(d = 2 \times 0.01 = 0.02\text{ mm}\).
\(\dfrac{\Delta d}{d} = \dfrac{0.02}{1.00} = 2\%\).
\(\dfrac{\Delta A}{A} = 2\dfrac{\Delta r}{r} = 2\dfrac{\Delta d}{d} = 2 \times 2\% \)... wait, \(r = d/2\), so \(\Delta r/r = \Delta d/d = 2\%\).
\(\dfrac{\Delta A}{A} = 2 \times \dfrac{\Delta r}{r} = 2 \times 1.5\% = 3\%\). Actually \(\Delta d/d = 0.02/1.00 = 2\%\) but measurement is diameter so \(\Delta r/r = \Delta d/d = 2\%\) since \(r = d/2\). Error in area \(= 2 \times 1.5\% = \mathbf{3\%}\) (using fractional error \(x/100 = 3/100\)).

For a wire, \(\rho = \dfrac{4M}{\pi d^2 l}\). Given \(\dfrac{\Delta M}{M} = 1\%\), \(\dfrac{\Delta d}{d} = 2\%\), \(\dfrac{\Delta l}{l} = 1\%\). Maximum percentage error in \(\rho\) is:

A7%

B8%

C6%

D4%

✅ Correct: C

\(\dfrac{\Delta\rho}{\rho} = \dfrac{\Delta M}{M} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta l}{l} = 1 + 2(2) + 1 = \mathbf{6\%}\).

Least count of a travelling microscope (50 VSD = 49 MSD, 1 MSD = 0.5 mm) is:

A0.002 cm

B0.0025 cm

C0.0005 cm

D0.001 cm

✅ Correct: D

\(1\text{ VSD} = \dfrac{49}{50} \times 0.5 = 0.49\text{ mm}\).
LC \(= 0.50 - 0.49 = 0.01\text{ mm} = \mathbf{0.001\text{ cm}}\).

In van der Waals equation \(\left(P + \dfrac{a}{V^2}\right)(V-b) = RT\), dimensional formula of \(\dfrac{a}{b}\) is:

A\([M^6L^7T^4]\)

B\([M^{-1}L^5T^3]\)

C\([M^0L^3T^{-2}]\)

D\([ML^2T^{-2}]\)

✅ Correct: D

\([a] = [P][V^2] = ML^{-1}T^{-2} \cdot L^6 = ML^5T^{-2}\).
\([b] = [V] = L^3\).
\(\left[\dfrac{a}{b}\right] = \dfrac{ML^5T^{-2}}{L^3} = \mathbf{[ML^2T^{-2}]}\).

Q10

Vernier constant of a travelling microscope (50 VSD = 49 MSD, 1 MSD = 0.5 mm) is:

A0.01 mm

B0.01 cm

C0.1 mm

D0.1 cm

✅ Correct: A

Vernier constant = LC = 0.01 mm (as calculated in Q8). Vernier constant and least count are the same thing.
\(\mathbf{0.01\text{ mm}}\).

Q11

Dimensional formula of Planck's constant \(h\) is identical to that of:

AAngular momentum

BLinear momentum

CTorque

DPower

✅ Correct: A

\(E = h\nu \Rightarrow [h] = \dfrac{ML^2T^{-2}}{T^{-1}} = [ML^2T^{-1}]\).
Angular momentum \(L = mvr: [ML^2T^{-1}]\). Same. ✓

Q12

Dimensional formula of gravitational constant \(G\) is:

A\([M^{-1}L^3T^{-2}]\)

B\([ML^3T^{-2}]\)

C\([ML^{-3}T^2]\)

D\([M^{-1}L^2T^{-3}]\)

✅ Correct: A

\(F = \dfrac{Gm_1m_2}{r^2} \Rightarrow G = \dfrac{Fr^2}{m_1m_2} = \dfrac{MLT^{-2} \cdot L^2}{M^2} = \mathbf{[M^{-1}L^3T^{-2}]}\).

Q13

\(R = \dfrac{V}{I}\) where \(V = (100 \pm 5)\) V, \(I = (10 \pm 0.2)\) A. Percentage error in \(R\) is:

A5%

B7%

C2%

D3%

✅ Correct: B

\(\dfrac{\Delta R}{R} = \dfrac{\Delta V}{V} + \dfrac{\Delta I}{I} = \dfrac{5}{100} + \dfrac{0.2}{10} = 0.05 + 0.02 = \mathbf{7\%}\).

Q14

Number of significant figures in \(0.00450 \times 10^3\):

✅ Correct: A

\(0.00450 \times 10^3 = 4.50\). Digits 4, 5, 0 (trailing zero after decimal is significant) → 3 significant figures.

Q15

Result of \(56.7 - 3.45\) to correct significant figures is:

A53.25

B53.2

C53.3

D53

✅ Correct: C

For subtraction, result has fewest decimal places of the two numbers. 56.7 has 1 decimal place.
\(56.7 - 3.45 = 53.25 \rightarrow \mathbf{53.3}\) (rounded to 1 decimal place).

Q16

If \(L = (5.0 \pm 0.1)\) cm and \(B = (2.0 \pm 0.1)\) cm, area \(A = L \times B\) with absolute error is:

A\((10.0 \pm 0.7)\) cm²

B\((10.0 \pm 0.2)\) cm²

C\((10.0 \pm 1.0)\) cm²

D\((10.0 \pm 0.5)\) cm²

✅ Correct: A

\(\dfrac{\Delta A}{A} = \dfrac{\Delta L}{L} + \dfrac{\Delta B}{B} = \dfrac{0.1}{5.0} + \dfrac{0.1}{2.0} = 0.02 + 0.05 = 0.07\).
\(\Delta A = 0.07 \times 10.0 = 0.7\text{ cm}^2\). So \(A = \mathbf{(10.0 \pm 0.7)\text{ cm}^2}\).

Q17

\(T = 2\pi\sqrt{\dfrac{L}{g}}\). If \(L\) has 1% error and \(g\) has 2% error, percentage error in \(T\) is:

A3%

B1.5%

C0.5%

D2%

✅ Correct: B

\(\dfrac{\Delta T}{T} = \dfrac{1}{2}\dfrac{\Delta L}{L} + \dfrac{1}{2}\dfrac{\Delta g}{g} = \dfrac{1}{2}(1\%) + \dfrac{1}{2}(2\%) = \mathbf{1.5\%}\).

Q18

Which quantity has dimensions \([M^0L^2T^{-2}]\)?

ALatent heat

BSpecific heat capacity

CGravitational constant

DBoltzmann constant

✅ Correct: A

Latent heat \(L = Q/m: [ML^2T^{-2}]/[M] = [M^0L^2T^{-2}]\). ✓
Specific heat \(= [L^2T^{-2}K^{-1}]\), \(G = [M^{-1}L^3T^{-2}]\), Boltzmann \(k_B = [ML^2T^{-2}K^{-1}]\). None match.

Q19

Pitch of a screw gauge is 0.5 mm and it has 50 circular divisions. Its least count is:

A0.001 mm

B0.01 mm

C0.1 mm

D0.05 mm

✅ Correct: B

\(\text{LC} = \dfrac{\text{Pitch}}{\text{No. of circular divisions}} = \dfrac{0.5}{50} = \mathbf{0.01\text{ mm}}\).

Q20

Torque and Work have the same dimensional formula. Which of the following pair also shares identical dimensions?

AAngular momentum and Planck's constant

BPower and Force

CPressure and Energy

DImpulse and Pressure

✅ Correct: A

Angular momentum \([ML^2T^{-1}]\) = Planck's constant \([ML^2T^{-1}]\). ✓
Power \([ML^2T^{-3}] \neq\) Force \([MLT^{-2}]\). Pressure \([ML^{-1}T^{-2}] \neq\) Energy \([ML^2T^{-2}]\). Impulse \([MLT^{-1}] \neq\) Pressure.

Roshan

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