JEE Physics Units and Measurements MCQs Set 13

Physics Objective Questions and Answers: Units and Measurements

Review structured MCQ sets for JEE Physics Units and Measurements. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Download Units and Measurements MCQs with Answers

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Multiple Choice Questions

Question 1: If the time period (T) of vibration of a liquid drop depends on surface tension (S), radius (r) of the drop and density (ρ) of liquid, then the expression of T is
(a) \( T=k\sqrt{\rho r^3/S} \)
(b) \( T=k\sqrt{\rho^{1/2}r^3/S} \)
(c) \( T=k\sqrt{\rho r^3/S^{1/2}} \)
(d) none of these
Show Answer & Explanation

Answer: (a) \( T=k\sqrt{\rho r^3/S} \)

Explanation:
1. Surface tension has dimensions \( [MT^{-2}] \).
2. Try \( \rho r^3/S \): \( [ML^{-3}][L^3]/[MT^{-2}] = [T^2] \).
3. So \( T = k\sqrt{\rho r^3/S} \).

Teacher's Note:
1. A product of powers must match dimensions of T exactly.
2. Check by substituting dimensions of each quantity.

Question 2: The electric field is given by \( \vec E = \dfrac{A}{x^3}\hat i + By\hat j + Cz^2\hat k \). The SI units of A, B and C respectively are
(a) N-m³/C, V/m², N/m²-C
(b) V-m², V/m, N/m²-C
(c) V/m², V/m, N-C/m²
(d) V/m, N-m³/C, N-C/m
Show Answer & Explanation

Answer: (a) N-m³/C, V/m², N/m²-C

Explanation:
1. Unit of E is V/m = N/C.
2. \( A = E x^3 \) has unit N-m³/C.
3. \( B = E/y \) has unit (N/C)/m = V/m².
4. \( C = E/z^2 \) has unit N/(C-m²).

Teacher's Note:
1. Each term in E must have the units of E.
2. Divide E by the coordinate factor to get each constant.

Question 3: What are the dimensions of A/B in the relation \( F = A\sqrt{x} + Bx^2 \), where F is the force and x is the distance?
(a) \( L^{1/2} \)
(b) \( L^{3/2} \)
(c) \( L^{-3/2} \)
(d) \( L^{-1/2} \)
Show Answer & Explanation

Answer: (b) \( L^{3/2} \)

Explanation:
1. Both terms have the dimensions of force.
2. \( [A] = [F]L^{-1/2} \) and \( [B] = [F]L^{-2} \).
3. \( [A/B] = L^{-1/2+2} = L^{3/2} \).

Teacher's Note:
1. Terms added in a formula share the same dimensions.
2. Divide the two dimensional formulas carefully.

Question 4: The potential energy of a particle is given by \( U(x) = -\alpha x + \beta\sin\dfrac{x}{\gamma} \). A dimensionless combination of the constants \( \alpha, \beta, \gamma \) is
(a) \( \alpha/(\beta\gamma) \)
(b) \( \alpha^2/(\beta\alpha) \)
(c) \( \gamma/(\alpha\beta) \)
(d) \( \alpha\gamma/\beta \)
Show Answer & Explanation

Answer: (d) \( \alpha\gamma/\beta \)

Explanation:
1. The argument x/γ is dimensionless, so γ has the dimensions of length.
2. \( \alpha x \) is an energy, so \( [\alpha] = [U]/L \).
3. \( \beta \) has the dimensions of energy.
4. \( \alpha\gamma/\beta = ([U]/L)(L)/[U] = 1 \).

Teacher's Note:
1. The argument of sin must be dimensionless.
2. Use that to find the dimensions of γ first.

Question 5: The dimensions of \( \dfrac{1}{\sqrt{\epsilon_0\mu_0}} \) is that of
(a) velocity
(b) time
(c) capacitance
(d) distance
Show Answer & Explanation

Answer: (a) velocity

Explanation:
1. The speed of light in vacuum is \( c = 1/\sqrt{\mu_0\epsilon_0} \).
2. So the quantity has the dimensions of velocity.

Teacher's Note:
1. This is the famous relation linking electricity, magnetism and light.
2. Remember it as c.

Question 6: The dimensional formulae for Planck's constant h and the gravitational constant G respectively are
(a) \( [ML^3T^{-2}], [M^{-1}L^2T^{-3}] \)
(b) \( [ML^2T^{-1}], [M^{-1}L^3T^{-2}] \)
(c) \( [ML^3T^{-2}], [M^{-1}L^2T^{2}] \)
(d) \( [MLT^{-3}], [M^{-1}L^3T^{-3}] \)
Show Answer & Explanation

Answer: (b) \( [ML^2T^{-1}], [M^{-1}L^3T^{-2}] \)

Explanation:
1. From \( E = h\nu \), \( [h] = [ML^2T^{-2}]/[T^{-1}] = [ML^2T^{-1}] \).
2. From \( F = Gm_1m_2/r^2 \), \( [G] = [MLT^{-2}][L^2]/[M^2] = [M^{-1}L^3T^{-2}] \).

Teacher's Note:
1. Derive each constant from a defining formula.
2. Do not rely on memory for exponents.

Question 7: If energy (E), velocity (V) and time (T) are chosen as the fundamental quantities, the dimensional formula of surface tension will be
(a) \( [EV^{-1}T^{-2}] \)
(b) \( [EV^{-2}T^{-2}] \)
(c) \( [E^{-2}V^{-1}T^{-3}] \)
(d) \( [EV^{-2}T^{-1}] \)
Show Answer & Explanation

Answer: (b) \( [EV^{-2}T^{-2}] \)

Explanation:
1. Surface tension \( = [MT^{-2}] \).
2. \( E/V^2 = [ML^2T^{-2}]/[L^2T^{-2}] = [M] \).
3. So surface tension \( = [EV^{-2}T^{-2}] \).

Teacher's Note:
1. Express M in terms of the new base quantities first.
2. Then substitute in the old formula.

Question 8: If the dimension of critical velocity \( v_c \) of a liquid flowing through a tube is expressed as \( \eta^x\rho^y r^z \), where \( \eta, \rho \) and r are the coefficient of viscosity of the liquid, density of the liquid and radius of the tube, then the values of x, y and z are given by
(a) 1, 1, 1
(b) 1, −1, −1
(c) −1, −1, 1
(d) −1, −1, −1
Show Answer & Explanation

Answer: (b) 1, −1, −1

Explanation:
1. \( [\eta] = ML^{-1}T^{-1} \), \( [\rho] = ML^{-3} \), \( [r] = L \).
2. Time: \( -x = -1 \), so x = 1.
3. Mass: \( x + y = 0 \), so y = −1.
4. Length: \( -x - 3y + z = 1 \) gives z = −1.

Teacher's Note:
1. Equate powers of M, L and T separately.
2. Start with the quantity that has only one base unit, which is time here.

Question 9: A physical quantity of the dimensions of length that can be formed out of c, G and \( e^2/(4\pi\epsilon_0) \) is [c is velocity of light, G is the universal constant of gravitation and e is charge]
(a) \( c^2\left[G\dfrac{e^2}{4\pi\epsilon_0}\right]^{1/2} \)
(b) \( \dfrac{1}{c^2}\left[\dfrac{e^2}{G\,4\pi\epsilon_0}\right]^{1/2} \)
(c) \( \dfrac{1}{c}\left[G\dfrac{e^2}{4\pi\epsilon_0}\right]^{1/2} \)
(d) \( \dfrac{1}{c^2}\left[G\dfrac{e^2}{4\pi\epsilon_0}\right]^{1/2} \)
Show Answer & Explanation

Answer: (d) \( \dfrac{1}{c^2}\left[G\dfrac{e^2}{4\pi\epsilon_0}\right]^{1/2} \)

Explanation:
1. \( e^2/(4\pi\epsilon_0) \) has dimensions of force × length² = \( ML^3T^{-2} \).
2. \( G\,e^2/(4\pi\epsilon_0) = L^6T^{-4} \), whose square root is \( L^3T^{-2} \).
3. Dividing by \( c^2 = L^2T^{-2} \) gives L.

Teacher's Note:
1. Find the dimensions of the charge term from Coulomb's law.
2. Combine the three quantities to cancel M and T.

Question 10: Which one of the following does not have the same dimensions?
(a) work and energy
(b) angle and strain
(c) relative density and refractive index
(d) Planck constant and energy
Show Answer & Explanation

Answer: (d) Planck constant and energy

Explanation:
1. Work and energy are both \( ML^2T^{-2} \).
2. Angle, strain, relative density and refractive index are all dimensionless.
3. Planck's constant is \( ML^2T^{-1} \), not the dimensions of energy.

Teacher's Note:
1. Ratios of like quantities are dimensionless.
2. Energy and Planck's constant differ by a factor of frequency.

Question 11: If the unit of length and force are increased four times, then the unit of energy
(a) is increased by 4 times
(b) is increased by 16 times
(c) is increased by 8 times
(d) remains unchanged
Show Answer & Explanation

Answer: (b) is increased by 16 times

Explanation:
1. Energy = force × length.
2. New unit = 4 × 4 = 16 times the old unit.

Teacher's Note:
1. Energy has dimensions of force times length.
2. Scale each factor separately and multiply.

Question 12: The density of a material in CGS system is 8 g/cm³. In a system of units in which unit of length is 5 cm and unit of mass is 20 g, the density of material is
(a) 8
(b) 20
(c) 50
(d) 80
Show Answer & Explanation

Answer: (c) 50

Explanation:
1. \( 8\ \text{g/cm}^3 = \dfrac{8/20\ \text{(new mass units)}}{(1/5)^3\ \text{(new length units)}^3} \).
2. \( = 0.4 \times 125 = 50 \).

Teacher's Note:
1. A larger unit gives a smaller number.
2. Mass units divide, length units multiply (cubed).

Question 13: In a new system the unit of mass is α kg, unit of length is β m and unit of time is γ s. The value of 1 J in this new system is
(a) \( \gamma^2/(\alpha\beta^2) \)
(b) \( \gamma\alpha/\beta^2 \)
(c) \( \alpha\beta\gamma \)
(d) \( \alpha\gamma^2/\beta^2 \)
Show Answer & Explanation

Answer: (a) \( \gamma^2/(\alpha\beta^2) \)

Explanation:
1. \( 1\,\text{J} = 1\,\text{kg m}^2\text{s}^{-2} \).
2. 1 kg = 1/α new mass units, 1 m = 1/β, 1 s = 1/γ.
3. \( n = \alpha^{-1}\beta^{-2}\gamma^{2} = \gamma^2/(\alpha\beta^2) \).

Teacher's Note:
1. Replace each SI unit by its equivalent in the new unit.
2. Mass, length and time exponents are 1, 2 and −2.

Question 14: Which of the following has the highest number of significant figures?
(a) 0.007 m²
(b) 2.64 × 10²⁴ kg
(c) 0.0006032 m²
(d) 6.3200 J
Show Answer & Explanation

Answer: (d) 6.3200 J

Explanation:
1. 0.007 has 1 significant figure.
2. 2.64 × 10²⁴ has 3.
3. 0.0006032 has 4.
4. 6.3200 has 5, since trailing zeros after the decimal point count.

Teacher's Note:
1. Leading zeros never count.
2. Trailing zeros after a decimal point do count.

Question 15: The density of a material in CGS system of units is 4 g/cm³. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be
(a) 0.04
(b) 0.4
(c) 40
(d) 400
Show Answer & Explanation

Answer: (c) 40

Explanation:
1. \( 4\ \text{g/cm}^3 = \dfrac{4/100}{(1/10)^3} = 0.04 \times 1000 = 40 \).

Teacher's Note:
1. The new unit of density is much larger in length but not in mass.
2. Convert mass and length separately.

Question 16: The least count of a stop watch is \( \tfrac15 \) s. The time of 20 oscillations of a pendulum is measured to be 25 s. What is the maximum percentage error in this measurement?
(a) 8%
(b) 1%
(c) 0.8%
(d) 16%
Show Answer & Explanation

Answer: (c) 0.8%

Explanation:
1. Maximum error equals least count = 0.2 s.
2. Percentage error \( = (0.2/25)\times100 = 0.8\% \).

Teacher's Note:
1. Least count is the maximum possible error of one reading.
2. Divide by the measured time.

Question 17: The refractive index of water measured by the relation \( \mu = \dfrac{\text{real depth}}{\text{apparent depth}} \) is found to have values of 1.34, 1.38, 1.32 and 1.36. The mean value of refractive index with percentage error is
(a) 1.35 ± 1.48%
(b) 1.35 ± 0%
(c) 1.36 ± 6%
(d) 1.36 ± 0%
Show Answer & Explanation

Answer: (a) 1.35 ± 1.48%

Explanation:
1. Mean = (1.34 + 1.38 + 1.32 + 1.36)/4 = 1.35.
2. Absolute deviations: 0.01, 0.03, 0.03, 0.01; mean = 0.02.
3. Percentage error \( = 0.02/1.35 \times 100 = 1.48\% \).

Teacher's Note:
1. Compute the mean first, then deviations from it.
2. Percentage error uses the mean absolute error.

Question 18: Choose the incorrect statement out of the following.
(a) Every measurement made by any measuring instrument has some error.
(b) Every calculated physical quantity that is based on measured values has some error.
(c) A measurement can have more accuracy but less precision and vice versa.
(d) The percentage error is different from relative error.
Show Answer & Explanation

Answer: (d) The percentage error is different from relative error.

Explanation:
1. Statements 1 to 3 are true statements about measurement.
2. Percentage error is simply relative error × 100, so the statement that they are unrelated to each other is not correct.

Teacher's Note:
1. Percentage error and relative error carry the same information.
2. One is the other multiplied by 100.

Question 19: A student measures the distance traversed in free fall of a body, initially at rest, in a given time. He uses this to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e₁ and e₂ respectively, the percentage error in the estimation of g is
(a) e₁ + 2e₂
(b) e₁ + e₂
(c) e₁ − 2e₂
(d) e₂ − e₁
Show Answer & Explanation

Answer: (a) e₁ + 2e₂

Explanation:
1. \( h = \tfrac12 g t^2 \), so \( g = 2h/t^2 \).
2. \( \Delta g/g = \Delta h/h + 2\Delta t/t \), i.e. \( e_1 + 2e_2 \).

Teacher's Note:
1. Errors in a quotient add.
2. The power of t doubles its contribution.

Question 20: If a set of defective weights is used by a student to find the mass of an object using a physical balance, a large number of readings will reduce
(a) random error
(b) random as well as systematic error
(c) systematic error
(d) neither random nor systematic error
Show Answer & Explanation

Answer: (a) random error

Explanation:
1. Averaging many readings reduces random error.
2. Defective weights give the same bias every time, a systematic error, which averaging cannot remove.

Teacher's Note:
1. Systematic errors have a fixed direction.
2. Only random errors average out.

Question 21: A quantity is represented by \( X = M^a L^b T^c \). The percentage error in measurement of M, L and T are α%, β% and γ% respectively. The percentage error in X would be
(a) (αa + βb + γc)%
(b) (αa − βb + γc)%
(c) (αa − βb − γc)%
(d) None of these
Show Answer & Explanation

Answer: (a) (αa + βb + γc)%

Explanation:
1. For a product of powers, relative errors add, each multiplied by its power.
2. So the percentage error is (aα + bβ + cγ)%.

Teacher's Note:
1. Maximum error is always the sum of magnitudes.
2. Do not subtract for negative powers.

Question 22: The period of oscillation of a simple pendulum in an experiment is recorded as 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s respectively. The average absolute error is
(a) 0.1 s
(b) 0.11 s
(c) 0.01 s
(d) 1.0 s
Show Answer & Explanation

Answer: (b) 0.11 s

Explanation:
1. Mean = 13.12/5 = 2.624 s.
2. Absolute deviations: 0.006, 0.064, 0.204, 0.086, 0.176.
3. Sum = 0.536, so the mean = 0.107 ≈ 0.11 s.

Teacher's Note:
1. Average absolute error is the mean of |reading − mean|.
2. Round sensibly to the precision of the data.

Question 23: The length of a cylinder is measured with a metre rod having least count 0.1 cm. Its diameter is measured with vernier callipers having least count 0.01 cm. Given that length is 5.0 cm and radius is 2.00 cm, the percentage error in the calculated value of the volume will be
(a) 2%
(b) 1%
(c) 2.5%
(d) 4%
Show Answer & Explanation

Answer: (c) 2.5%

Explanation:
1. \( V = \pi r^2 l \), so \( \Delta V/V = 2\Delta r/r + \Delta l/l \).
2. \( \Delta r = 0.01/2 = 0.005 \) cm, so \( 2\Delta r/r = 0.5\% \).
3. \( \Delta l/l = 0.1/5 = 2\% \). Total = 2.5%.

Teacher's Note:
1. Radius error is half of the diameter error.
2. The square on r doubles its fractional error.

Question 24: A physical quantity ε is calculated by using the formula \( \varepsilon = \dfrac{xy^2}{10z^{1/3}} \), where x, y and z are experimentally measured quantities. If the fractional percentage error in the measurements of x, y and z are 2%, 1% and 3% respectively, then the fractional percentage error in ε will be
(a) 0.5%
(b) 5%
(c) 6%
(d) 7%
Show Answer & Explanation

Answer: (b) 5%

Explanation:
1. \( \Delta\varepsilon/\varepsilon = \Delta x/x + 2\Delta y/y + \tfrac13\Delta z/z \).
2. \( = 2 + 2(1) + \tfrac13(3) = 5\% \).

Teacher's Note:
1. The constant 10 has no error.
2. Fractional powers also scale the errors.

Question 25: 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 the measurement of its density is
(a) 1
(b) 2
(c) 3
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. \( \rho = m/(\pi r^2 l) \).
2. Errors: m 1%, r 1% (counted twice = 2%), l 1%.
3. Total = 1 + 2 + 1 = 4%.

Teacher's Note:
1. Density is mass over volume, and volume contains r².
2. Add all fractional errors.

Question 26: The resistance of a metal is given by R = V/I, where V is the potential difference and I is the current. In a circuit the potential difference across a resistance is V = (10 ± 0.5) V and current in resistance is I = (2 ± 0.2) A. The value of resistance in Ω with percentage error is
(a) 5 ± 10%
(b) 5 ± 15%
(c) 5 ± 20%
(d) 5 ± 25%
Show Answer & Explanation

Answer: (b) 5 ± 15%

Explanation:
1. R = 10/2 = 5 Ω.
2. Percentage error = 0.5/10 × 100 + 0.2/2 × 100 = 5% + 10% = 15%.

Teacher's Note:
1. Errors in a quotient add.
2. Compute each percentage separately.

Question 27: The period of oscillation of a simple pendulum is given by \( T = 2\pi\sqrt{\ell/g} \), where \( \ell \) is about 100 cm and is known to have 1 mm accuracy. The period is about 2 s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is
(a) 0.1%
(b) 1%
(c) 0.2%
(d) 0.8%
Show Answer & Explanation

Answer: (c) 0.2%

Explanation:
1. \( g = 4\pi^2\ell/T^2 \), so \( \Delta g/g = \Delta\ell/\ell + 2\Delta T/T \).
2. \( \Delta\ell/\ell = 0.1/100 = 0.1\% \).
3. 100 oscillations take 200 s, so \( \Delta T/T = 0.1/200 = 0.05\% \).
4. Total = 0.1 + 2(0.05) = 0.2%.

Teacher's Note:
1. Measuring many oscillations reduces the timing error.
2. The time error is on 200 s, not on one period.

Question 28: In a vernier callipers, one main scale division is x cm and n divisions of the vernier scale coincide with (n − 1) divisions of the main scale. The least count (in cm) of the callipers is
(a) \( \left(\dfrac{n-1}{n}\right)x \)
(b) \( \dfrac{nx}{n-1} \)
(c) \( \dfrac{x}{n} \)
(d) \( \dfrac{x}{n-1} \)
Show Answer & Explanation

Answer: (c) \( \dfrac{x}{n} \)

Explanation:
1. 1 vernier division = (n − 1)x/n.
2. Least count = x − (n − 1)x/n = x/n.

Teacher's Note:
1. Least count is one main division minus one vernier division.

Question 29: A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of −0.004 cm, the correct diameter of the ball is
(a) 0.521 cm
(b) 0.525 cm
(c) 0.053 cm
(d) 0.529 cm
Show Answer & Explanation

Answer: (d) 0.529 cm

Explanation:
1. Observed reading = 0.5 cm + 25 × 0.001 cm = 0.525 cm.
2. Zero error is −0.004 cm, so the correction is +0.004 cm.
3. Correct diameter = 0.525 + 0.004 = 0.529 cm.

Teacher's Note:
1. Correction = minus the zero error.
2. A negative zero error increases the true reading.

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