Previous year questions: Units and Measurements

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar for this unit. Each heading states the paper and the marks. Attempt the question before reading the answer, and rewrite every equation on paper rather than reading it as prose. The answer keys used here are the official final keys for the papers.

Q1 (JEE Main 2026, 2 Apr Shift 1, 4 marks)

The dimensional formula of the energy density (1/2) epsilon0 E^2 of an electric field E is M^a L^b T^c. The value of 2a - b + c is

Answer: The quantity is energy per unit volume, so its dimensions are those of energy divided by volume: M L^2 T^-2 divided by L^3, which gives M L^-1 T^-2. Therefore a = 1, b = -1 and c = -2. Substitute: 2(1) - (-1) + (-2) = 2 + 1 - 2 = 1. Among the printed options 0, 1, -1 and 2, the correct choice is 1.

Hinglish: Energy density matlab energy per unit volume. Energy ki dimensions M L^2 T^-2 hain aur volume ki L^3, isliye divide karne par M L^-1 T^-2 milta hai. Iska matlab a = 1, b = -1, c = -2. Ab 2a - b + c = 2 + 1 - 2 = 1. Isliye answer 1 hai.

Q2 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

In a screw gauge, the circular scale is given five complete rotations and it moves linearly by 2.5 mm. If the circular scale has 100 divisions, the least count of the screw gauge is

Answer: Five complete rotations move the spindle by 2.5 mm, so one complete rotation moves it by the pitch, which is 2.5 / 5 = 0.5 mm. The least count is the pitch divided by the number of circular scale divisions: 0.5 / 100 = 0.005 mm, which is 5 x 10^-3 mm. Among the options 1 x 10^-2 mm, 1 x 10^-3 mm, 5 x 10^-2 mm and 5 x 10^-3 mm, the answer is 5 x 10^-3 mm.

Hinglish: Paanch complete rotations mein spindle 2.5 mm chalta hai, isliye ek rotation ka pitch 2.5 / 5 = 0.5 mm hai. Least count = pitch / divisions = 0.5 / 100 = 0.005 mm, matlab 5 x 10^-3 mm. Options mein 5 x 10^-3 mm hi sahi hai.

Q3 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

In a vernier caliper, when the jaws are closed the zero of the vernier scale is shifted to the right of the main scale zero, and the 7th vernier division coincides with a main scale reading. One main scale division equals 1 mm and there are 10 vernier scale divisions. The vernier caliper has

Answer: The least count is one main scale division minus one vernier scale division. With 10 vernier divisions spanning 9 main scale divisions, the least count is 0.1 mm, or 0.01 cm. The zero error is the coinciding division multiplied by the least count: 7 x 0.01 = 0.07 cm. Because the vernier zero sits to the right of the main scale zero when the jaws are closed, the instrument reads high, so the error is positive. The option 0.07 cm positive zero error is correct.

Hinglish: Least count = 1 MSD - 1 VSD = 0.1 mm = 0.01 cm. Zero error = 7 x 0.01 = 0.07 cm. Jaws band karne par vernier ka zero main scale ke zero ke right mein hai, matlab instrument zyada padhta hai, isliye error positive hai. Option 0.07 cm positive zero error sahi hai.

Q4 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

The density of a uniform cylinder is found from its mass m, length l and diameter d, measured as 97.42 plus or minus 0.02 g, 8.35 plus or minus 0.05 mm and 20.20 plus or minus 0.02 mm. The calculated percentage error in the density is

Answer: Write the density as rho = 4m / (pi d^2 l). The relative errors add, with the diameter error counted twice because of the square: (0.02/97.42) + 2(0.02/20.20) + (0.05/8.35). These are 0.000205, 0.001980 and 0.005988, whose sum is about 0.00817, that is 0.82 percent. The option 0.82 percent is correct. The length contributes the largest single share, which is worth noting when a student assumes the small-diameter reading is the weak point.

Hinglish: Density rho = 4m / (pi d^2 l). Diameter square mein hai, isliye us error ka double count hoga. Relative errors: 0.02/97.42 + 2(0.02/20.20) + 0.05/8.35 = 0.000205 + 0.001980 + 0.005988, total lagbhag 0.00817, matlab 0.82 percent. Option 0.82 percent sahi hai. Length ka error sabse bada hissa deta hai, yeh dhyan rakhne wali baat hai.

Q5 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

The percentage error in the calculated volume of a sphere is, if there is a 2 percent error in its diameter measurement,

Answer: The volume of a sphere is (pi/6)d^3, so it depends on the cube of the diameter. For a power, the relative error is the exponent times the relative error of the measured quantity: 3 x 2 percent = 6 percent. The option 6 is correct. Note that a percentage error quoted on the diameter is the same number as on the radius, so there is no extra factor of 2 to apply here.

Hinglish: Sphere ka volume (pi/6)d^3 hota hai, matlab diameter ke cube par depend karta hai. Power ke liye relative error ko power se multiply karte hain: 3 x 2 percent = 6 percent. Isliye 6 sahi hai. Diameter ka percentage error radius ke percentage error ke barabar hi hota hai, isliye 2 ka extra factor nahi lagana.

Q6 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

Match List I with List II. List I: A. Boltzmann constant, B. Stefan constant, C. Planck constant, D. Gravitational constant. List II: I. [M^-1 L^3 T^-2], II. [M L^2 T^-1], III. [M L^2 T^-2 K^-1], IV. [M L^0 T^-3 K^-4].

Answer: The Boltzmann constant is energy per kelvin, so [M L^2 T^-2 K^-1], which is III. The Stefan constant appears in energy per unit area per unit time per kelvin to the fourth, so [M L^0 T^-3 K^-4], which is IV. The Planck constant is energy multiplied by time, [M L^2 T^-1], which is II. The gravitational constant follows from F = G m1 m2 / r^2, giving [M^-1 L^3 T^-2], which is I. The correct match is A-III, B-IV, C-II, D-I.

Hinglish: Boltzmann constant energy per kelvin hai, [M L^2 T^-2 K^-1], matlab III. Stefan constant energy per area per time per K^4 hai, [M L^0 T^-3 K^-4], matlab IV. Planck constant energy guna time, [M L^2 T^-1], matlab II. G, F = G m1 m2 / r^2 se [M^-1 L^3 T^-2] aata hai, matlab I. Match A-III, B-IV, C-II, D-I.

Q7 (NCERT Exemplar, 1 mark)

The number of significant figures in 0.06900 is

Answer: The zeros between the decimal point and the 6 are place holders and are not significant. The two zeros after 69 are trailing zeros after a decimal point and are significant, as are the 6 and the 9. The count is four, so the answer is option (b). A good check is to write the value in scientific notation, 6.900 x 10^-2, where all four digits of the leading number count.

Hinglish: Decimal point aur 6 ke beech ke zeros place holder hain, woh significant nahi. 69 ke baad wale do zeros trailing zeros hain decimal ke baad, isliye significant hain. Total chaar significant figures, option (b). Scientific notation 6.900 x 10^-2 se check bhi kar sakte ho.

Q8 (NCERT Exemplar, 1 mark)

The sum of the numbers 436.32, 227.2 and 0.301 in appropriate significant figures is

Answer: For addition the rule is decimal places, not significant figures. The raw sum is 663.821. The least precise input in decimal places is 227.2, which has one decimal place, so the answer must be rounded to one decimal place: 663.8. That is option (c). Options 664, 663.82 and 663.821 all claim more decimal precision than the 227.2 measurement supports.

Hinglish: Addition mein rule decimal places ka hai, significant figures ka nahi. Raw sum 663.821 aata hai. Sabse kam decimal places wala input 227.2 hai, jisme ek decimal place hai, isliye answer 663.8 hoga, option (c). 664, 663.82 aur 663.821 sab zyada precision claim karte hain.

Q9 (NCERT Exemplar, 1 mark)

Which of the following pairs of physical quantities does not have the same dimensional formula? (a) Work and torque, (b) Angular momentum and Planck constant, (c) Tension and surface tension, (d) Impulse and linear momentum.

Answer: Work is force times distance, [M L^2 T^-2], and torque is also force times perpendicular distance, so the pair in (a) matches. Angular momentum is [M L^2 T^-1] and the Planck constant is energy times time with the same dimensions, so (b) matches. Tension is a force, [M L T^-2], while surface tension is force per unit length, [M T^-2], so (c) does not match. Impulse and change in momentum are the same quantity dimensionally, [M L T^-1], so (d) matches. The answer is (c).

Hinglish: Work ML^2 T^-2 hai aur torque bhi force guna distance, isliye (a) match. Angular momentum ML^2 T^-1 aur Planck constant bhi energy guna time, isliye (b) match. Tension force hai MLT^-2, jabki surface tension force per length MT^-2, isliye (c) match nahi. Impulse aur momentum dono MLT^-1, isliye (d) match. Answer (c).

Q10 (NCERT Exemplar, 3 marks)

The vernier scale of a travelling microscope has 50 divisions which coincide with 49 main scale divisions. If each main scale division is 0.5 mm, calculate the minimum inaccuracy in the measurement of distance.

Answer: The least count of a vernier instrument is one main scale division minus one vernier scale division. Here 50 vernier divisions span 49 main scale divisions, so one vernier scale division is (49/50) x 0.5 mm = 0.49 mm. The least count is therefore 0.5 - 0.49 = 0.01 mm. That is the minimum inaccuracy of a single measurement, so the microscope can be read to 0.01 mm. In centimetres it is 0.001 cm.

Hinglish: Least count = 1 MSD - 1 VSD. 50 vernier divisions 49 main scale divisions ke barabar hain, isliye 1 VSD = (49/50) x 0.5 = 0.49 mm. Ab least count = 0.5 - 0.49 = 0.01 mm. Matlab microscope ka minimum inaccuracy 0.01 mm hai, ya 0.001 cm.