How to handle negative zero error in a screw gauge reading?

True Reading = Observed Reading − Zero Error. For a negative zero error (e.g., −0.03 mm), subtracting a negative number means you ADD 0.03 mm. A negative zero error means the instrument under-reports, so you compensate by adding. Example: 3 + 0.35 − (−0.03) = 3.38 mm.

Which topics from Units and Measurements are most important for KVS NVS EMRS PGT Physics 2026?

Most important topics: (1) Dimensional Analysis with new fundamental units like F-v-T, P-A-T, E-v-F systems, (2) Percentage error using power rule for density of cube and cylinder, pendulum g-experiment, (3) Significant figures rules and arithmetic, (4) Screw gauge and Vernier caliper with zero error correction, (5) Matching dimensional formulas of viscosity, surface tension, impulse, permeability, magnetic field.

Why do moment of inertia and moment of force have different dimensions?

Moment of inertia = mr² has dimensions [ML²] — no time factor. Moment of force (Torque) = F×r has dimensions [ML²T⁻²] because force involves acceleration (T⁻²). They differ by T⁻², making this a classic PGT Physics MCQ trap. Note: Energy and Torque both have [ML²T⁻²] — they are dimensionally identical.

How to correctly calculate percentage error in g from a simple pendulum experiment?

From g = 4π²L/T²: % error in g = %ΔL + 2×(%ΔT). Critical rule: %ΔT = (Least Count of stopwatch / Total time measured for all oscillations) × 100. Do NOT divide least count by the time of one oscillation — always use total measured time. For 20 oscillations in 30s with 1s least count: %ΔT = (1/30)×100 = 3.33%.

Top 40 MCQs On Units & Measurements For KVS NVS EMRS PGT Physics Mains 2026 With Solutions

Q: What is a parsec and what does it measure?

A parsec is a unit of distance (not time) equal to approximately 3.086 × 10^16 metres or about 3.26 light-years. It stands for Parallax Second — the distance at which 1 AU subtends an angle of 1 arc-second. Despite the '-sec' suffix, it measures distance, not time.

Q: How do you calculate percentage error in a derived quantity like P = a³b²/cd?

Multiply each quantity's percentage error by its power in the formula, then add all terms. For P = a³b²/cd: % error = 3(%a) + 2(%b) + 1(%c) + 1(%d). Errors always add — never subtract. Powers in numerator and denominator both contribute positively.

Q: How many significant figures does 0.0003 have?

0.0003 has only 1 significant figure. Leading zeros are never counted as significant — they only indicate decimal position. Only the digit 3 is significant.

Q: What are the dimensions of Gravitational constant G?

From F = GM₁M₂/r², G = Fr²/M² = [M⁻¹L³T⁻²]. This is derived by substituting dimensions of Force (MLT⁻²), length squared (L²), and mass squared (M²).

Q: What does 1/√(μ₀ε₀) represent and what are its dimensions?

1/√(μ₀ε₀) equals the speed of light c in vacuum (≈ 3×10⁸ m/s). Its dimensions are [LT⁻¹]. This is a fundamental result from Maxwell's electromagnetic equations.

Q: What is the percentage error formula for density of a cylindrical wire?

For density ρ = M/(πr²L): % error in ρ = %ΔM + 2×(%Δr) + %ΔL. The radius error is always doubled because r is squared in the formula. Constants like π contribute zero error.

5/5 - (4 votes)

Welcome to your ultimate revision guide for the KVS, NVS, and EMRS PGT Physics Mains 2026. The “Units & Measurements” chapter is highly scoring if you know the right tricks. Below are the Top 40 MCQs extracted from premium study material, complete with unique, easy-to-understand solutions and special Cheat Points to save you crucial seconds in the exam hall.

📋 Table of Contents

Part 1 — Units and Basic Concepts (Q1–Q3)
Part 2 — Significant Figures & Error Analysis (Q4–Q19)
Part 3 — Dimensional Analysis Masterclass (Q20–Q39)
Part 4 — Experimental Skills & Instruments (Q40)
FAQs — Frequently Searched Questions

Units and Basic Concepts

parsec is the unit of

▼

(a) time(b) distance(c) frequency(d) angular acceleration

✓ Ans: (b)

Solution

A parsec is an astronomical unit used to measure massive distances outside our solar system.

💡 Cheat Point: Don’t let the suffix “-sec” fool you! Parsec stands for Parallax Second, which measures distance, not time.

Given that $y = A \sin \left[ \frac{2\pi}{\lambda} (ct - x) \right]$ , where $y$ and $x$ are measured in metres. Which of the following statements is correct?

▼

(a) the unit of $(ct - x)$

is same as that of $\frac{2\pi}{\lambda}$

(b) the unit of $c$

is same as that of $\frac{2\pi}{\lambda}$

is same as that of $x$

but not of $A$

(d) the unit of $\lambda$

is same as that of $x$

and $A$

✓ Ans: (d)

Solution

The physical quantities $y$

, $x$

, and Amplitude $A$

are all lengths, so they are in meters. The argument inside the sine function must be dimensionless, so $\lambda$

must match the length unit of $x$

💡 Cheat Point: In wave equations, variables representing physical spread (like $y, x, \lambda, A$

) always share the exact same spatial unit!

Match List-I with List-II.

▼

List-I	List-II
A. Surface Tension	I. $kg\ m^{-1} s^{-1}$
B. Pressure	II. $kg\ m\ s^{-1}$
C. Viscosity	III. $kg\ s^{-2}$
D. Impulse	IV. $kg\ m^{-1} s^{-2}$

(a) A-III, B-IV, C-I, D-II(b) A-II, B-I, C-III, D-IV
(c) A-IV, B-III, C-I, D-II(d) A-IV, B-III, C-II, D-I

✓ Ans: (a)

Solution

Surface tension $= F/L = MT^{-2}\ (kg\ s^{-2})$

. Pressure $= F/A = ML^{-1}T^{-2}\ (kg\ m^{-1}s^{-2})$

. Viscosity $= ML^{-1}T^{-1}\ (kg\ m^{-1}s^{-1})$

. Impulse $= MLT^{-1}\ (kg\ ms^{-1})$

💡 Cheat Point: Just match Impulse = Force $\times$

Time ( $kg\ m s^{-1}$

). D matches II. Only option (a) and (c) have D-II. Check Surface Tension next to crack it instantly!

Significant Figures & Error Analysis

The respective number of significant figures for the numbers 23.023, 0.0003 and $2.1 \times 10^{-3}$ are

▼

(a) 4, 4, 2(b) 5, 1, 2(c) 5, 1, 5(d) 5, 5, 2

✓ Ans: (b)

Solution

All non-zero digits and trapped zeros count (23.023 $\rightarrow$

5). Leading zeros never count (0.0003 $\rightarrow$

1). Powers of 10 are ignored ( $2.1 \times 10^{-3} \rightarrow$

2).

💡 Cheat Point: Ignore $10^x$

completely. Only look at the base number!

In an experiment four quantities $a, b, c$ and $d$ are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity P is calculated as $P = \frac{a^3 b^2}{cd}$ . % error in P is

▼

(a) 10%(b) 7%(c) 4%(d) 14%

✓ Ans: (d)

Solution

Error $= 3(\%a) + 2(\%b) + 1(\%c) + 1(\%d) = 3(1) + 2(2) + 3 + 4 = 14\%$

💡 Cheat Point: Powers jump to the front as multipliers. Divisions turn into additions. Errors always add up; they never subtract!

A physical quantity $y = \frac{a^4 b^2}{(cd^4)^{1/3}}$ has four observables $a, b, c$ and $d$ . The percentage errors are 2%, 3%, 4% and 5%. The error in $y$ will be

▼

(a) 6%(b) 11%(c) 12%(d) 22%

✓ Ans: (d)

Solution

Error $= 4(2\%) + 2(3\%) + \frac{1}{3}(4\%) + \frac{4}{3}(5\%) = 8 + 6 + 1.33 + 6.67 = 22\%$

The density of a cube is measured by measuring its mass and length. If the maximum errors in mass and lengths are 3% and 2% respectively, the maximum error in density would be

▼

(a) 12%(b) 14%(c) 7%(d) 9%

✓ Ans: (d)

Solution

Density $\rho = \frac{M}{L^3}$

. Error $= \%M + 3(\%L) = 3\% + 3(2\%) = 9\%$

The density of a material (cube) is determined. Relative error in measuring mass and length are 1.5% and 1%. Maximum error in density is

▼

(a) 2.5%(b) 3.5%(c) 4.5%(d) 6%

✓ Ans: (c)

Solution

Using $\rho = \frac{M}{L^3}$

, error $= 1.5\% + 3(1\%) = 4.5\%$

A wire has a mass (0.3 ± 0.003) g, radius (0.5 ± 0.005) mm and length (6 ± 0.06) cm. The maximum percentage error in density is

▼

(a) 1(b) 2(c) 3(d) 4

✓ Ans: (d)

Solution

Convert values to percentages: Mass error $= (\frac{0.003}{0.3}) \times 100 = 1\%$

. Radius error $= 1\%$

. Length error $= 1\%$

. Density of a cylinder $\rho = \frac{M}{\pi r^2 L}$

. Total error $= \%M + 2(\%r) + \%L = 1\% + 2(1\%) + 1\% = 4\%$

💡 Cheat Point: For wires, formula is always cylinder ( $\pi r^2 L$

). The radius error is always doubled!

A quantity $X = \frac{A^2 B^{1/2}}{C^{1/3} D^3}$ . Percentage errors in A, B, C, D are 1%, 2%, 3%, 4%. Max error in X is

▼

(a) (3/13)%(b) 16%(c) -10%(d) 10%

✓ Ans: (b)

Solution

Error $= 2(1\%) + \frac{1}{2}(2\%) + \frac{1}{3}(3\%) + 3(4\%) = 2 + 1 + 1 + 12 = 16\%$

In the density measurement of a cube, mass and edge length are (10.00 ± 0.10) kg and (0.10 ± 0.01) m. Error in density is:

▼

(a) $0.10\ kg/m^3$

(b) $0.31\ kg/m^3$

(d) $0.10\ kg/m^3$

*(Note: Source prints options identically; actual calculated absolute error differs in magnitude scale, but base value aligns with 0.31)*

✓ Ans: (b)

Solution

Relative error $\frac{\Delta \rho}{\rho} = \frac{0.10}{10} + 3 \times \frac{0.01}{0.10} = 0.01 + 0.3 = 0.31$

Simple pendulum g-determination: 20 oscillations take 30s (measured with 1s least count watch). Length is 55.0 cm (1 mm least count scale). Percentage error in g is close to:

▼

(a) 0.7%(b) 0.2%(c) 3.5%(d) 6.8%

✓ Ans: (d)

Solution

$g = \frac{4\pi^2 L}{T^2}$

. % error in $g = \%L + 2(\%T)$

. Error in $L = (\frac{0.1}{55.0}) \times 100 \approx 0.18\%$

. Error in $T$

for total time $= (\frac{1}{30}) \times 100 \approx 3.33\%$

. Total $= 0.18 + 2(3.33) \approx 6.8\%$

💡 Cheat Point: Total time error is $\frac{\text{Least Count}}{\text{Total Time}}$

. Don’t divide by the number of oscillations for the percentage!

$Z = \frac{a^2 b^{2/3}}{\sqrt{c} d^3}$ . Percentage errors in a, b, c, d are 2%, 1.5%, 4% and 2.5%. Percentage error in Z is

▼

(a) 12.25%(b) 16.5%(c) 13.5%(d) 14.5%

✓ Ans: (d)

Solution

Error $= 2(2) + \frac{2}{3}(1.5) + \frac{1}{2}(4) + 3(2.5) = 4 + 1 + 2 + 7.5 = 14.5\%$

Taking into account significant figures, what is the value of 9.99 m – 0.0099 m

▼

(a) 9.98 m(b) 9.980 m(c) 9.9 m(d) 9.9801 m

✓ Ans: (a)

Solution

$9.99 - 0.0099 = 9.9801$

m. In addition/subtraction, the result retains the least number of decimal places of the given terms (two decimal places from 9.99). So, round to 9.98 m.

Physical quantity $y = m^2 r^{-4} g^x l^{-3/2}$ . Percentage errors in y, m, r, l and g are 18, 1, 0.5, 4 and p respectively. Find x and p.

▼

(a) 5 and ± 2(b) 4 and ± 3
(c) 8 and ± 2(d) 16/3 and ± 3/2

✓ Ans: (d)

Solution

$18 = 2(1) + 4(0.5) + x(p) + 1.5(4) \Rightarrow 18 = 2 + 2 + xp + 6 \Rightarrow xp = 8$

. Checking options, (16/3) × (3/2) = 8. So (d) is correct.

💡 Cheat Point: Work backwards from the options! Find which product $xp$

equals 8.

Area of rectangular field of length 55.3 m and breadth 25 m after rounding off for correct significant digits is

▼

(a) 1382(b) 1382.5
(c) $14 \times 10^2$

(d) $138 \times 10^1$

✓ Ans: (c)

Solution

Area $= 55.3 \times 25 = 1382.5\ m^2$

. Since ’25’ has only 2 significant figures, the answer must be rounded to 2 significant figures, which gives $1400$

or $14 \times 10^2\ m^2$

If $Z = \frac{A^2 B^3}{C^4}$ , then relative error in Z will be

▼

(a) $\frac{\Delta A}{A} + \frac{\Delta B}{B} + \frac{\Delta C}{C}$

(b) $2\frac{\Delta A}{A} + 3\frac{\Delta B}{B} + 4\frac{\Delta C}{C}$

(d) $\frac{\Delta A}{A} + \frac{\Delta B}{B} - \frac{\Delta C}{C}$

✓ Ans: (b)

Solution

Powers become coefficients and negative powers become positive because maximum relative errors are always added.

A cylindrical wire of mass (0.4 ± 0.01) g has length (8 ± 0.04) cm and radius (6 ± 0.03) mm. Maximum error in density is

▼

(a) 3.5%(b) 1%(c) 4%(d) 5%

✓ Ans: (c)

Solution

Error $= (\frac{0.01}{0.4} \times 100) + 2(\frac{0.03}{6} \times 100) + (\frac{0.04}{8} \times 100) = 2.5\% + 1\% + 0.5\% = 4\%$

A metal wire has mass (0.4 ± 0.002) g, radius (0.3 ± 0.001) mm and length (5 ± 0.02) cm. Maximum percentage error in density is

▼

(a) 1.2%(b) 1.3%(c) 1.6%(d) 1.4%

✓ Ans: (c)

Solution

Error $= (\frac{0.002}{0.4} \times 100) + 2(\frac{0.001}{0.3} \times 100) + (\frac{0.02}{5} \times 100) = 0.5\% + 0.66\% + 0.4\% = 1.6\%$

Dimensional Analysis Masterclass

If force (F), velocity (v) and time (T) are taken as fundamental units, dimensions of mass are

▼

(a) $[F v^{-1} T^{-1}]$

(b) $[F v^{-1} T]$

(d) $[F v T^{-2}]$

✓ Ans: (b)

Solution

Force $= \text{Mass} \times \text{Acceleration} = \frac{M \times v}{T}$

. Rearranging, $M = \frac{F \times T}{v} = [F v^{-1} T]$

💡 Cheat Point: Don’t write full MLT equations for these. Use basic physics formulas (like $F = ma = m(v/t)$

) to isolate the unknown!

If momentum (P), area (A) and time (T) are fundamental, dimensional formula of energy is

▼

(a) $[P A^{-1} T]$

(b) $[P^2 A T]$

(d) $[P A^{1/2} T^{-1}]$

✓ Ans: (d)

Solution

$P = MLT^{-1}$

, $A = L^2 \Rightarrow L = A^{1/2}$

. Energy $E = ML^2T^{-2}$

. Multiply and divide P by L/T logic: $E = P \times \frac{L}{T} = P \times A^{1/2} \times T^{-1}$

If energy (E), velocity (v) and force (F) are fundamental, dimensions of mass will be

▼

(a) $[F v^{-2}]$

(b) $[F v^{-1}]$

(d) $[E v^2]$

✓ Ans: (c)

Solution

Kinetic Energy $E = \frac{1}{2} M v^2$

. Constants vanish in dimensions, so $M = \frac{E}{v^2} = [E v^{-2}]$

💡 Cheat Point: $K.E. = \frac{1}{2}mv^2$

finishes this in 3 seconds!

If E = energy, G = gravitational constant, I = impulse and M = mass, dimensions of $\frac{GIM^2}{E^2}$ are same as

▼

(a) mass(b) length(c) time(d) force

✓ Ans: (c)

Solution

$\frac{[M^{-1}L^3T^{-2}] [MLT^{-1}] [M^2]}{[ML^2T^{-2}]^2} = \frac{M^2L^4T^{-3}}{M^2L^4T^{-4}} = [T]$

In a system where force (F), acceleration (A) and time (T) are fundamental, formula of energy is

▼

(a) $[F A T^2]$

(b) $[F A^2 T]$

(d) $[F A T]$

✓ Ans: (a)

Solution

Energy = Work = Force $\times$

Distance. Distance from kinematics $d = \frac{1}{2}AT^2$

. So, Energy $= F \times AT^2 = [FAT^2]$

Dimensional formula of Gravitational constant is

▼

(a) $[M^{-1} L^2 T^{-2}]$

(b) $[M^{-1} L^{-1} T^{-2}]$

(d) $[M L^3 T^{-2}]$

✓ Ans: (c)

Solution

$F = \frac{GM_1M_2}{r^2} \Rightarrow G = \frac{Fr^2}{M^2} = \frac{MLT^{-2} \times L^2}{M^2} = [M^{-1}L^3T^{-2}]$

Match Column-I with Column-II:

▼

Column-I	Column-II
(A) Torque	(p) $[M^1 L^1 T^{-2} A^{-2}]$
(B) Magnetic Field	(q) $[L^2 A^1]$
(C) Magnetic moment	(r) $[M^1 T^{-2} A^{-1}]$
(D) Permeability	(s) $[M^1 L^2 T^{-2}]$

✓ Ans: (a) A→s ; B→r ; C→q ; D→p

Solution

Torque = Force $\times$

distance $= [ML^2T^{-2}]$

. Matches (s). Magnetic moment = Current $\times$

Area $= [AL^2]$

. Matches (q).

Dimensions of $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$ are

▼

(a) $[L^2 T^{-2}]$

(b) $[L^{-2} T^2]$

(d) $[L^{-1} T]$

✓ Ans: (c)

Solution

The expression $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$

is exactly the formula for the speed of light $c$

in a vacuum. Speed $= [LT^{-1}]$

💡 Cheat Point: Memorize EM constants! $c = 1/\sqrt{\mu_0\epsilon_0}$

. E/B = c. Both are velocity.

E, m, l and G denote energy, mass, angular momentum and gravitational constant. $\frac{El^2}{m^5G^2}$ has dimensions of:

▼

(a) angle(b) length(c) mass(d) time

✓ Ans: (a)

Solution

Substitute: $E = ML^2T^{-2}$

, $l = ML^2T^{-1}$

, $m = M$

, $G = M^{-1}L^3T^{-2}$

. Solving $\frac{(ML^2T^{-2})(M^2L^4T^{-2})}{M^5(M^{-2}L^6T^{-4})}$

yields $[M^0L^0T^0]$

. It is dimensionless, matching an angle.

Speed of light (c), G, and Planck’s constant (h) are fundamental. Dimensions of time should be

▼

(a) $G^{1/2} h^{1/2} c^{-5/2}$

(b) $G^{1/2} h^{1/2} c^{1/2}$

(d) $G^{1/2} h^{1/2} c^{1/2}$

✓ Ans: (a)

Solution

Assume $Time = c^x G^y h^z$

. Solving powers of M, L, T yields $x = -5/2, y = 1/2, z = 1/2$

In which pair do the quantities have different dimensions?

▼

(a) Planck’s constant and angular momentum
(b) Impulse and linear momentum
(c) Moment of inertia and moment of force
(d) Energy and torque

✓ Ans: (c)

Solution

Moment of inertia $= MR^2 = [ML^2]$

. Moment of force (Torque) $= F \times r = [ML^2T^{-2}]$

. They do not match.

Equation of state is $(P + \frac{a}{V^3})(V - b^2) = cT$ . Dimensions of a and b are:

▼

(a) $ML^8T^{-2}$

and $L^{3/2}$

(b) $ML^5T^{-2}$

and $L^3$

and $L^6$

(d) $ML^6T^{-2}$

and $L^{3/2}$

✓ Ans: (a)

Solution

By principle of homogeneity, $b^2$

must have dimensions of volume $V = L^3 \Rightarrow b = L^{3/2}$

. $\frac{a}{V^3}$

must match pressure $P = ML^{-1}T^{-2}$

. Therefore $a = ML^{-1}T^{-2} \times (L^3)^3 = ML^8T^{-2}$

h, c, and G are fundamental. Which combination has dimension of length?

▼

(a) $\sqrt{\frac{Gc}{h^{3/2}}}$

(b) $\frac{\sqrt{hG}}{c^{3/2}}$

(d) $\sqrt{\frac{hc}{G}}$

✓ Ans: (b)

Solution

Length $\propto h^a c^b G^c$

. Solving exponents gives $L = K \frac{\sqrt{hG}}{c^{3/2}}$

A length dimension can be formed out of c, G and $\frac{e^2}{4\pi\epsilon_0}$ as:

▼

(a) $\frac{1}{c^2}[G(\frac{e^2}{4\pi\epsilon_0})]^{1/2}$

(b) $c^2[G(\frac{e^2}{4\pi\epsilon_0})]^{1/2}$

(d) $\frac{1}{c}G(\frac{e^2}{4\pi\epsilon_0})$

✓ Ans: (a)

Solution

Note that $\frac{e^2}{4\pi\epsilon_0} = F \times r^2 = [ML^3T^{-2}]$

. Equating $L \propto c^x G^y (\frac{e^2}{4\pi\epsilon_0})^z$

leads to $x=-2, y=1/2, z=1/2$

$F = \alpha \beta \exp \left( -\frac{x^2}{\alpha k T} \right)$ . Dimension of $\beta$ is :

▼

(a) $M^2 L^2 T^{-2}$

(b) $M^2 L T^{-4}$

(d) $M L T^{-2}$

✓ Ans: (b)

Solution

The power in exponent $\frac{x^2}{\alpha k T}$

is dimensionless. $k T$

has dimensions of Energy ( $ML^2T^{-2}$

). So $\alpha = \frac{x^2}{kT} = \frac{L^2}{ML^2T^{-2}} = M^{-1}T^2$

. Now, $F = \alpha \beta \Rightarrow \beta = \frac{F}{\alpha} = \frac{MLT^{-2}}{M^{-1}T^2} = M^2LT^{-4}$

In $X = 5YZ^2$ , X is capacitance and Z is magnetic field. Dimensions of Y are:

▼

(a) $[M^{-2} L^{-2} T^6 A^3]$

(b) $[M^{-1} L^{-2} T^4 A^2]$

(d) $[M^{-2} L^0 T^{-1} A^{-2}]$

✓ Ans: (c)

Solution

$Y = \frac{X}{Z^2}$

. $X = [M^{-1}L^{-2}T^4A^2]$

, $Z = [MT^{-2}A^{-1}]$

. Then $Z^2 = [M^2T^{-4}A^{-2}]$

. Dividing them yields $Y = [M^{-3}L^{-2}T^8A^4]$

If speed (V), acceleration (A) and force (F) are fundamental, dimension of Young’s modulus is:

▼

(a) $V^{-2} A^2 T^{-2}$

(b) $V^{-4} A^2 F$

(d) $V^{-2} A^2 F^{-2}$

✓ Ans: (b)

Solution

Young’s Modulus $Y = \frac{F}{\text{Area}}$

. Area $= L^2$

. From $V = LT^{-1}$

and $A = LT^{-2}$

, we get $L = \frac{V^2}{A}$

. Thus, Area $= \frac{V^4}{A^2}$

. Substituting this back, $Y = F \times A^2 V^{-4}$

$F = A \cos Bx + C \sin Dt$ . Dimensional formula of $\frac{AD}{B}$ is

▼

(a) $M^2 L^2 T^{-3}$

(b) $M L T^{-2}$

(d) $M L^2 T^{-3}$

✓ Ans: (d)

Solution

Arguments of trig functions are dimensionless: $Bx = 1 \Rightarrow B = L^{-1}$

. $Dt = 1 \Rightarrow D = T^{-1}$

. F and A must match: $A = MLT^{-2}$

. Thus $\frac{AD}{B} = \frac{(MLT^{-2})(T^{-1})}{L^{-1}} = ML^2T^{-3}$

💡 Cheat Point: Cosines/Sines eat dimensions! Anything multiplying the variable inside ( $x$

or $t$

) is just the inverse of that variable’s dimension.

If momentum (P), area (A) and time (T) are fundamental, dimensional formula for coefficient of viscosity is

▼

(a) $[P A^{-1} T^0]$

(b) $[P A T^{-1}]$

(d) $[P A^{-1} T^{-1}]$

✓ Ans: (a)

Solution

Viscosity $\eta = [ML^{-1}T^{-1}]$

. $M = \frac{P}{V} = PL^{-1}T$

. $L = A^{1/2}$

. $\eta = (PA^{-1/2}T)(A^{-1/2})(T^{-1}) = PA^{-1}$

Speed of wave $v = \lambda^a g^b \rho^c$ . Values of a, b, c respectively are

▼

(a) $1/2, 0, 1/2$

(b) $1, -1, 0$

(d) $1/2, 1/2, 0$

✓ Ans: (d)

Solution

$[LT^{-1}] = [L]^a [LT^{-2}]^b [ML^{-3}]^c = M^c L^{a+b-3c} T^{-2b}$

. Comparing powers: $c=0$

, $-2b=-1 \Rightarrow b=1/2$

, $a+b=1 \Rightarrow a=1/2$

Experimental Skills & Instruments

Screw gauge: Two full turns cover 1 mm. Circular scale has 50 divisions. Zero error is -0.03 mm. Main scale reads 3 mm and circular scale reads 35. Diameter of the wire is

▼

(a) 3.38 mm(b) 3.32 mm
(c) 3.73 mm(d) 3.67 mm

✓ Ans: (a)

Solution

Pitch $= \frac{1\ mm}{2} = 0.5\ mm$

. Least count $= \frac{0.5}{50} = 0.01\ mm$

. Reading $= \text{Main Scale} + (\text{Circular Scale} \times LC) - \text{Zero Error}$

. Reading $= 3 + (35 \times 0.01) - (-0.03) = 3 + 0.35 + 0.03 = 3.38\ mm$

💡 Cheat Point: Negative zero error always ADDS to your final reading! Think of it like a clock running slightly slow—you must add time to get the true reading.

Frequently Asked Questions

Most searched questions on Units & Measurements for KVS NVS EMRS PGT Physics 2026 — answered in detail.

What is a parsec and what does it measure? +

A parsec is a unit of distance (not time) equal to approximately 3.086 × 10¹⁶ metres or about 3.26 light-years. It is derived from the parallax method: one parsec is the distance at which one astronomical unit (AU) subtends an angle of one arc-second. Despite containing “-sec,” parsec stands for Parallax Second — it uses arc-seconds as an angular measure, not as time. The suffix simply refers to the angle (arc-second) used in the parallax calculation. It is the standard unit for measuring distances to stars and galaxies outside our solar system.

How do you calculate percentage error in a derived quantity? What is the rule? +

The rule is: multiply each quantity’s percentage error by its power in the formula, then add all terms. For $P = \frac{a^3 b^2}{cd}$

: percentage error $= 3(\%a) + 2(\%b) + 1(\%c) + 1(\%d)$

. Key points — (1) Powers in numerator and denominator both contribute positively; errors always add, never subtract. (2) Fractional powers like ½ or ⅓ also become multipliers. (3) Constants like π or numerical factors contribute zero error. This rule is derived from taking the logarithm of the formula and differentiating — the absolute values ensure errors always add regardless of the sign of the power.

How many significant figures does 0.0003 have? What are the complete rules? +

0.0003 has only 1 significant figure. The complete rules: (1) All non-zero digits are always significant. (2) Leading zeros (before the first non-zero digit) are never significant — they only show decimal position. (3) Trapped zeros (between non-zero digits) are always significant — e.g., 23.023 → 5 sig figs. (4) Trailing zeros after a decimal point are significant — e.g., 9.980 → 4 sig figs. (5) In scientific notation $2.1 \times 10^{-3}$

, only the coefficient (2.1) counts — the power of 10 is ignored → 2 sig figs. For arithmetic: in addition/subtraction keep the least decimal places; in multiplication/division keep the least significant figures.

What are the dimensions of Gravitational constant G and how to derive them? +

From Newton’s law: $F = \frac{GM_1M_2}{r^2}$

, rearrange to $G = \frac{Fr^2}{M^2}$

. Substituting dimensions: $G = \frac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$

. This is one of the most tested dimensional formulas in PGT Physics. The negative power of M (i.e., M⁻¹) is the key identifier. SI value of G is $6.674 \times 10^{-11}\ Nm^2kg^{-2}$

, consistent with this dimension.

What does 1/√(μ₀ε₀) represent and what are its dimensions? +

$\frac{1}{\sqrt{\mu_0 \epsilon_0}}$

equals the speed of light c in vacuum (≈ 3 × 10⁸ m/s). This is a fundamental result from Maxwell’s electromagnetic equations. Its dimensions are $[LT^{-1}]$

— the dimension of velocity. Related facts: $\frac{E}{B} = c$

(ratio of electric to magnetic field in an EM wave) also has the same $[LT^{-1}]$

dimension. For MCQs: $[\mu_0\epsilon_0] = [L^{-2}T^2]$

, so $[1/\sqrt{\mu_0\epsilon_0}] = [LT^{-1}]$

How to handle negative zero error in a screw gauge? +

The formula is: True Reading = Observed Reading − Zero Error. For a negative zero error (e.g., −0.03 mm) — the gauge reads below zero when fully closed — subtracting a negative value means adding: True = Observed − (−0.03) = Observed + 0.03. For a positive zero error (+0.03 mm), you subtract: True = Observed − 0.03. Memory trick: a negative zero error means the instrument under-reports the size, so you must compensate by adding. Always note the sign carefully — it completely determines whether you add or subtract from the reading.

What is the percentage error formula for density of a cylinder (wire) vs a cube? +

For a cylinder/wire, density $\rho = \frac{M}{\pi r^2 L}$

: % error $= \%\Delta M + 2 \times \%\Delta r + \%\Delta L$

. The radius error is doubled because r is squared. For a cube, density $\rho = M/L^3$

: % error $= \%\Delta M + 3 \times \%\Delta L$

. The length error is tripled because L is cubed. Constants like π contribute zero error in both cases. This pair is among the most repeated question types in KVS, NVS and EMRS PGT Physics papers.

Which topics from Units & Measurements have highest weightage in KVS NVS EMRS PGT Physics 2026? +

Based on previous years’ KVS, NVS, and EMRS PGT Physics paper analysis, the highest-weightage subtopics are: (1) Dimensional Analysis with new fundamental unit systems — expressing quantities when F-v-T, P-A-T, E-v-F or h-c-G are taken as fundamental units. (2) Percentage error using power rule — especially density of cube/cylinder, pendulum g, and multi-variable formulas with fractional powers. (3) Significant figures — counting rules and arithmetic (subtraction and multiplication). (4) Screw gauge and Vernier caliper — pitch, least count, zero error correction. (5) Matching dimensional formulas — viscosity, surface tension, impulse, permeability, magnetic field. Astronomical units (parsec vs light-year vs AU) and wave equation dimensional consistency also appear regularly.

Why do moment of inertia and moment of force (torque) have different dimensions? +

Moment of inertia $I = mr^2$

has dimensions $[ML^2]$

— purely a geometric-mass property with no time dependence. Moment of force (Torque) $\tau = F \times r = ma \times r$

has dimensions $[ML^2T^{-2}]$

— it includes time because force involves acceleration ( $T^{-2}$

). They differ by $[T^{-2}]$

. This is a classic MCQ trap — both involve mass × length² but they are dimensionally different because one involves force and the other does not. For contrast: Energy and Torque both have dimensions $[ML^2T^{-2}]$

— they are dimensionally identical despite being physically different.

How to correctly find percentage error in g from a simple pendulum experiment? +

From $g = \frac{4\pi^2 L}{T^2}$

: % error in $g = \%\Delta L + 2 \times \%\Delta T$

. The most common mistake: using $\%\Delta T = \frac{\text{Least Count}}{\text{time for one oscillation}} \times 100$

. This is wrong. Correct method: $\%\Delta T = \frac{\text{Least Count of stopwatch}}{\text{Total time measured}} \times 100$

. The total time is what was directly measured (e.g., 30 s for 20 oscillations). Absolute error (1 s) applies to 30 s total → relative error = 1/30 = 3.33%, not 1/1.5. The factor of 2 multiplying $\%\Delta T$

comes from T being squared in the g formula.

Best of luck with your KVS, NVS, and EMRS PGT Physics preparation! Keep revising these cheat points daily.

Top 40 MCQs on Units & Measurements for KVS NVS EMRS PGT Physics Mains 2026 with Solutions

Units and Basic Concepts

Significant Figures & Error Analysis

Dimensional Analysis Masterclass

Experimental Skills & Instruments

Amit Patel

Leave a ReplyCancel Reply

About : TotheScience

Units and Basic Concepts

Significant Figures & Error Analysis

Dimensional Analysis Masterclass

Experimental Skills & Instruments

Amit Patel

Leave a ReplyCancel Reply

Related Posts

UP TGT Science Hand written Notes [Free]| Unit (i): Units, Measurements and Vectors Complete Notes 100% Free For UP TGT & PGT

Trending now

About : TotheScience