Units and Measurement JEE Main Questions 2025: Important Practice Questions with PYQs

Units and Measurement JEE Main Questions 2025 - JEE Main Physics is considered the most important section of the exam. Units and Measurement is a fundamental JEE Main Physics exam. As per previous year's trends, Units and Measurement is seen to hold a weightage of  3 to 4%. A strong grasp of the topic is crucial for understanding and solving the problems of other topics of JEE Main Physics. Some of the key portions to focus on are SI units and their derived units, using dimensions to check the consistency of equations understanding different types of errors (systematic, random) and how to estimate them, and knowing how to express measurements with appropriate precision. 

In this article, we will discuss about the Important Practice Questions with PYQs related to Units and Measurement in JEE Main.

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Units and Measurement JEE Main Questions 2025

1. The diameter of a cylinder is measured using vernier callipers with no zero error. It is found that the zero of the vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The vernier scale has 50 divisions equivalent to 2.45 cm. The 24th division of the vernier scale exactly coincides with one of the main scale divisions. Then, the diameter of the cylinder is

a) 5.112 cm

b) 5. 124 cm

c) 5.136 cm

d) 5.148 cm

Answer: b) 5. 124 cm

Solution:

The least count of a vernier is given by

L.C. = 1 Main scale division/Number of divisions on the vernier scale

L.C. = 1 M.S.D./n

One main scale division = 0.05 cm

n = 50

L.C. = 0.05/50 = 0.001 cm

Diameter of the cylinder = Main scale reading + (Least count x Vernier scale reading)

= 5.10 + (24 x 0.001) = 5.124 cm

1. The density of a solid ball is to be determined in an experiment. The diameter of the ball is measured with a screw gauge, whose pitch is 0.5 mm, and there are 50 divisions on the circular scale. The reading on the main scale is 2.5 mm, and that on the circular scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is

a) 0.9%

b) 2.4 %

c) 3.1 %

d) 4.2 %

Answer: c) 3.1 %

Solution:

Given,

Pitch = 0.5 mm

Circular scale division = 50

Main scale division = 2.5 mm

Least count = Pitch/Circular scale division= 0.5/50 = 0.01 mm

Circular scale division reading = 20

Relative error = 2%

Screw gauge reading = Main scale reading + (Least count x Circular scale division reading)

= 2.5 + (0.01 x 20)

= 2.7 mm

Density, ρ = mass/volume

The relative percentage error in density is

1. The period of oscillation of a simple pendulum in the 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

Answer: b) 0.11 s

Solution:

Average value = (2.63 + 2.56 + 2.42 + 2.71 + 2.80)/5

= 2.62 sec

Now,

|△T1| = 2.63 – 2.62 = 0.01

|△T2| = 2.62- 2.56 = 0.06

|△T3| = 2.62- 2.42 = 0.20

|△T4| = 2.71- 2.62 = 0.09

|△T4| = 2.80- 2.62 = 0.18

Mean absolute error,

= (0.01 + 0.06 + 0.20 + 0.09 + 0.18)/5

= 0.54/5 = 0.108 = 0.11 sec

1.  A thin copper wire of length l metre increases in length by 2% when heated through 10°C. What is the percentage increase in the area when a square copper sheet of length l metre is heated through 10°C?

a) 4%

b) 8%

c) 16%

d) None of these

Answer: a) 4%

Solution:

△l = l αΔT

△l/l = 2/100 = α x 100

α = 2/1000

β = 2α = 4/1000

△A = A βΔT

△A/A = βΔT

= (4/1000) x 10

= 4/100

Percentage increase in area = (4/100) x 100

= 4%

1. The dimensional formula for the relative refractive index is

a) [M0L1T-1]

b) [M0L0T0]

c) [M0L1T1]

d) [MLT-1]

Answer: b) [M0L0T0]

Solution:

The relative refractive index is the ratio of the refractive index of the medium to the refractive index of the vacuum. Hence, it is a dimensionless quantity.

Also Read: JEE Main Previous Year Question Paper

Unsolved Units and Measurement JEE Main Questions

1. Two full turns of the circular scale of a screw gauge cover a distance of 1 mm on its main scale. The total number of divisions on the circular scale is 50. Further, it is found that the screw gauge has a zero error of –0.03 mm. While measuring the diameter of a thin wire, a student notes the main scale reading of 3 mm and the number of circular scale divisions in line with the main scale as 35. The diameter of the wire is

a) 3.73 mm

b) 3.32 mm

c) 3.38 mm

d) 3.67 mm

1. The current-voltage relation of the diode is given by I = (e1000V/T – 1) mA, where the applied V is in volts, and the temperature T is in degrees, Kelvin. If a student makes an error measuring ±0.01 V while measuring the current of 5 mA at 300K, what will be the error in the value of current in mA?

a) 0.05 mA

b) 0.02 mA

c) 0.5 mA

d) 0.2 mA

1. The energy (E), angular momentum (L) and universal gravitational constant (G) are chosen as fundamental quantities. The dimensions of the universal gravitational constant in the dimensional formula of Planck’s constant (h) is

a) zero

b) 1

c) 5/3

d) -1 

1. To find the distance d over which a signal can be seen clearly in foggy conditions, a Railway Engineer uses dimensional analysis and assumes that the distance depends on the mass density ρ of the fog, intensity (power/area) S of the light from the signal and its frequency f. The Engineer finds that d is proportional to S1/n. The value of n is

a) 3

b) 2

c) 1

d) 4

1. A student experiments to determine Young’s modulus of a wire, exactly 2 m long, by Searle’s method. In a particular reading, the student measures the extension in the length of the wire to be 0.8 mm with an uncertainty of ± 0.05 mm at a load of exactly 1.0 kg. The student also measures the diameter of the wire to be 0.4 mm with an uncertainty of ± 0.01 mm. Take g = 9.8 ms-2 (exact). The Young’s modulus obtained from the reading is

a) (2 ± 0.2) x 1011 N/m2

b) (2 ± 0.3) x 1011 N/m2

c) (2 ± 0.05) x 1011 N/m2

d)(2 ± 0.1) x 1011 N/m2

1. The density of a substance in the shape of a cube is found by measuring the cube's three sides and mass. Calculate the maximum error in predicting density if the relative errors in measuring mass and length are 1.5% and 1%, respectively. 
2. A copper wire is stretched to lengthen it by 0.5 percent. If its volume remains constant, the percentage change in its electrical resistance is ?

Previous Years Units and Measurement JEE Main Questions

1. If E, L, M and G denote the quantities as energy, angular momentum, mass and constant of gravitation respectively, then the dimensions of P in the formula - 

{JEE Main 2018}

1. [M⁰L¹Tº]
2. [M⁻¹L⁻¹T²]
3. [M¹L¹T⁻²]
4. [M⁰L⁰T⁰]

Given the formula, P=EL²M⁻⁵G⁻²

In order to find the dimension of P, we have to use the dimensional formula of quantities E, L, M and G which are energy, angular momentum, mass and constant of gravitation.

Dimensional Formula of energy, E=[M¹L²T⁻²] 

Dimensional Formula of angular momentum, L=[M¹L²T⁻¹]

Dimensional Formula of mass,M=[M¹] 

Dimensional Formula for the constant of gravitation, G=[M⁻¹L³T²] 

After putting the dimensional formula of all the quantities in the given formula we can obtain the dimension of P as,

P=[M¹L²T⁻²][M²L⁴T⁻²][M⁻⁵][M²L⁻⁶T⁴]

After adding and subtracting the powers of M,L and T, we get;

P=[M⁰LºTº]

Answer - [M⁰L⁰T⁰]

Trick: Recalling the dimensional formula for different standard quantities like force, momentum, energy, moment of inertia etc. is necessary to crack such a problem.

2. Resistance of a given wire is obtained by measuring the current flowing in it and the voltage difference applied across it. If the percentage errors in the measurement of the current and the voltage difference are 3% each, then the error in the value of resistance of the wire is (JEE Main 2017)

1. 3 %
2. 6 %
3. Zero
4. 1 %

Percentage error in measurement of current, ΔII×100=3%

Percentage error in measurement of potential difference, ΔVV×100=3%

We know that the formula between potential difference and current is given as,

V=IR

In the form of percentage error, we can write the above relation as,

ΔVV×100=ΔII×100+ΔRR×100

From the above relation, the error in the measurement of resistance is given as,

ΔRR×100=ΔVV×100+ΔII×100 

After putting the values of known quantities we get;

ΔRR×100=3%+3%=6%

Hence error in the measurement of resistance is 6%

∴ Option C is correct.

Trick: To crack questions like this, we must keep knowledge about the percentage error formula and combinational error.

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