JEE Advanced (Subjective Type Questions): Units & Measurements

Q.1. Give the MKS units for each of the following  quantities.      (1980) 
(i) Young's modulus
(ii) Magnetic Induction
(iii) Power of a lens

Ans. (i) N/m²;  
(ii) Tesla;  
(iii) Dioptre;

Solution.  Young's modulus is defined as stress/strain. Stress has units of force per unit area, i.e. N/m², while strain is dimensionless. Hence Young's modulus has the unit N/m².
The unit of magnetic induction is the Tesla. One tesla may be written in base MKS units as kg s^-2 A^-1, but it is customary to use the name Tesla for this unit.
The power (optical power) of a lens is defined as the reciprocal of its focal length (in metres). Its SI unit is metre^-1, commonly expressed as the dioptre.

Q.2. A gas bubble, from an explosion under water, oscillates with a period T proportional to p^ad^bE^c. Where 'P' is the static pressure, 'd' is the density of water and 'E' is the total energy of the explosion. Find the values of a, b and c.

Ans.  a = - 5/6, b = 1/2, c = 1/3

Solution. Given T ∝ P^ad^bE^c.
Write dimensions of each quantity (using M, L, T):
[T] = [M⁰L⁰T¹]
[P] = [M L^-1 T^-2], [d] = [M L^-3], [E] = [M L² T^-2].
Therefore,
[T] = [P]^a[d]^b[E]^c = [M^a+b+c L^{-a -3b +2c} T^{-2a -2c}].
Equate exponents for M, L and T:
a + b + c = 0
-a - 3b + 2c = 0
-2a - 2c = 1
From the third equation: a + c = -1/2 ⇒ a = -1/2 - c.
Substitute in the first: (-1/2 - c) + b + c = 0 ⇒ b = 1/2.
Substitute b = 1/2 into the second:
-a - 3(1/2) + 2c = 0 ⇒ -a - 3/2 + 2c = 0.
Replace a = -1/2 - c: -(-1/2 - c) - 3/2 + 2c = 0 ⇒ 1/2 + c - 3/2 + 2c = 0 ⇒ 3c - 1 = 0 ⇒ c = 1/3.
Hence a = -1/2 - 1/3 = -5/6, b = 1/2, c = 1/3, as required.

Q.3. Write the dimensions of the following in terms of mass, time, length and charge      (1982 - 2 Marks) 
(i) magnetic flux 
(ii) rigidity modulus

Ans. (i) [M¹L²T^-1Q^-1]  (ii) [ML^-1T^-2]

Solution. Magnetic flux Φ = magnetic induction B × area.
From Q.1, B (Tesla) has dimensions [M T^-1 Q^-1] when expressed using charge Q (since current I = Q/T). Area has [L²].
Therefore Φ: [M L² T^-1 Q^-1].
Modulus of rigidity (shear modulus) has the same dimensions as pressure (force per unit area): N/m² ⇒ [M L^-1 T^-2].

Q.4. Match the ph ysical quan tities given in column I with dimensions expressed in terms of mass (M), length (L), time (T), and charge (Q) given in column II and write the correct answer against the matched quantity in a tabular form in your answer book.

Solution. Match each quantity with its correct dimensional form and give a brief reason:
Angular momentum: [M L² T^-1]. (Angular momentum = moment of linear momentum ⇒ M L T^-1 × L.)
Latent heat: [L² T^-2]. (Latent heat per unit mass has dimensions of energy per mass ⇒ (M L² T^-2)/M.)
Torque: [M L² T^-2]. (Torque = force × distance ⇒ N·m ⇒ M L² T^-2.)
Capacitance: [M^-1 L^-2 T² Q²]. (Capacitance = Q/V and V has dimensions M L² T^-2 Q^-1, so C = Q/(M L² T^-2 Q^-1) = M^-1 L^-2 T² Q².)
Inductance: [M L² Q^-2]. (Inductance L = magnetic flux / current; using Φ dimensions from Q.3 and I = Q/T gives this form.)
Resistivity: [M L³ T^-1 Q^-2]. (Resistivity = ohm·m and ohm = V/A → substitute V and A in base units.)

Q.5. Column -I gives three physical quant ities. Select the appropriate units for the choices given in Column-II. Some of the physical quantities may have more than one choice correct :

Column I                                     Column II 

Capacitance                              (i) ohm-second 
In ductan ce                             (ii) coulomb²-joule^-1 
Magnetic Induction              (iii) coulomb (volt)^-1 
                                                    (iv) newton (amp-metre)^-1 
                                                    (v) volt-second (ampere)^-1

Ans. Capacitance                   coulomb-volt^-1, coulomb²-joule^-1
In ductance ohm-sec, volt-second (ampere)^-1
Magnetic Induction newton (ampere-metre)^-1

Solution. Brief justification for each choice:
Capacitance: 1 C = charge, V = energy/charge = J/C, so capacitance C = Q/V has units C·V^-1. Using V = J/C, C = Q/(J/C) = Q² J^-1; hence (ii) and (iii) are correct for capacitance.
Inductance: 1 H = Wb/A and Wb = V·s. Since ohm = V/A, ohm·s = (V/A)·s = V·s / A = H, so (i) and (v) represent inductance units.
Magnetic induction B: dimensionally B = force/(current × length) so N (A·m)^-1 is correct; hence (iv) is the correct choice for magnetic induction.

(b) Refer to solution of Q. 3, type D

Q.6. If nth division of main scale coincides with (n+1)^th divisions of vernier scale. Given one main scale division is equal to 'a' units. Find the least count of the vernier.       (2003 - 2 Marks)

Ans. a/(n+1) units

Solution. (n + 1) divisions of vernier scale = n divisions of main scale.
So one vernier division = n a /(n + 1).
Least count = value of one main scale division - value of one vernier division = a - n a /(n + 1) = a/(n + 1).
Therefore the least count of the vernier is a/(n+1) units.

Q.7. A screw gauge having 100 equal divisions and a pitch of length 1 mm is used to measure the diameter of a wire of length 5.6 cm. The main scale reading is 1 mm and 47^th circular division coincides with the main scale. Find the curved surface area of wire in cm² to appropriate significant figure. 

(use π = 22/7)

Ans. 2.6 cm²

Solution.  Least count of the screw gauge = pitch / number of divisions = 1 mm / 100 = 0.01 mm.
Diameter D = MSR + (circular division) × (least count) = 1 mm + 47 × 0.01 mm = 1.47 mm.
Convert diameter to cm: D = 1.47 mm = 0.147 cm.
Length of wire l = 5.6 cm.
Curved surface area = π D l = (22/7) × 0.147 × 5.6 = 2.587... cm².
Rounding to appropriate significant figures (measurements limited by least count 0.01 mm and given data), the area ≈ 2.6 cm².

Q.8. In Searle's exper iment, which is used to find Young's Modulus of elasticity, the diameter of experimental wire is D = 0.05 cm (measured by a scale of least count 0.001 cm) and length is L = 110 cm (measured by a scale of least count 0.1 cm). A weight of 50 N causes an extension of  X = 0.125 cm (measured by a micrometer of least count 0.001cm). Find maximum possible error in the values of Young's modulus.

Screw gauge and meter scale are free from error.        (2004 - 2 Marks)

Ans. 1.09 × 10¹⁰ Nm ^-2

Solution. Expression for Young's modulus for a wire: Y = (4 W L) / (π D² X).
Take relative (maximum possible) error by adding magnitudes of fractional errors according to the powers of each measured quantity:
ΔY / Y = ΔW / W + ΔL / L + 2 ΔD / D + ΔX / X.
Given values (convert to SI): W = 50 N (assumed exact here), D = 0.05 cm = 0.05 × 10^-2 m = 5.0 × 10^-4 m with least count 0.001 cm = 1.0 × 10^-5 m ⇒ ΔD = 1.0 × 10^-5 m.
X = 0.125 cm = 0.125 × 10^-2 m = 1.25 × 10^-3 m with micrometer least count 0.001 cm = 1.0 × 10^-5 m ⇒ ΔX = 1.0 × 10^-5 m.
L = 110 cm = 1.10 m with least count 0.1 cm = 1.0 × 10^-3 m ⇒ ΔL = 1.0 × 10^-3 m.
Compute fractional errors:
ΔW/W ≈ 0 (weight taken as exact), ΔL/L = 1.0×10^-3 / 1.10 ≈ 9.09×10^-4.
ΔD/D = 1.0×10^-5 / 5.0×10^-4 = 0.02 ⇒ 2 ΔD/D = 0.04.
ΔX/X = 1.0×10^-5 / 1.25×10^-3 = 8.0×10^-3.
Total relative error ΔY/Y ≈ 0 + 9.09×10^-4 + 0.04 + 8.0×10^-3 ≈ 0.0489.
Now calculate Y itself: Y = (4 W L) / (π D² X) = (4 × 50 × 1.10) / [π × (5.0×10^-4)² × 1.25×10^-3] ≈ 2.24 × 10¹¹ N/m².
Maximum possible error ΔY = (ΔY/Y) × Y ≈ 0.0489 × 2.24 × 10¹¹ ≈ 1.09 × 10¹⁰ N/m².

Q.9. The side of a cube is measured by vernier callipers (10 divisions of a vernier scale coincide with 9 divisions of main scale, where 1 division of main scale is 1 mm). The main scale reads 10 mm and first division of vernier scale coincides with the main scale. Mass of the cube is 2.736 g. Find the density of the cube in appropriate significant figures.

Ans. 2.66 gm/cm³

Solution.Given: 10 VSD = 9 MSD, with 1 MSD = 1 mm.
So one VSD = 9/10 mm = 0.9 mm. Least count = MSD - VSD = 1.0 - 0.9 = 0.1 mm.
Reading: side s = MSR + (vernier divisions coinciding) × (least count) = 10 mm + 1 × 0.1 mm = 10.1 mm = 1.01 cm.
Volume = s³ = (1.01 cm)³ = 1.030301 cm³.
Mass = 2.736 g.
Density ρ = mass / volume = 2.736 / 1.030301 ≈ 2.655... g/cm³.
Rounding to appropriate significant figures (measurements given to three significant figures), ρ ≈ 2.66 g/cm³.