Units and Measurements JEE Main PYQs (2020–2026)

Solved Units and Measurements PYQs

3 recent questions with answers. The full set of 180 is free to practise.

1. JEE Main 2026EasyUnits of Measurement and Dimensions of Physical Quantities
   Q1

   A new unit ( $\alpha$ ) of length is chosen such that it is equal to the speed of light in vacuum. What is the distance between Venus and Earth in terms of $\alpha$ units if light takes 6 min. 40 s to cover this distance?

   • A$200\alpha$
   • B$400\alpha$
   • C$300\alpha$
   • D$500\alpha$

   Show answer & solution

   Correct answer: (B)

   To find the distance between two points when an object moves at a constant speed, we use formula,
   
   Distance $=$ Speed × Time
   
   $\alpha$ is a new unit of length such that 1 unit $(\alpha )=$ speed of light in vacuum (c).
   
   The speed of light c is the distance light travels per unit of time (usually meters per second).
   
   $$1\alpha =c\times (1\text{ second })$$
   
   The time taken for light to travel from Venus to Earth is 6 minutes and 40 seconds.
   
   1 minute $=60$ seconds
   
   6 minutes $=6\times 60=360$ seconds
   
   Total time $t=360s+40s=400$ seconds
   
   Using distance formula:
   
   Distance $=$ Speed of light × Time
   
   Distance $=c\times 400$ seconds
   
   Since $1\alpha$ is the distance light travels in 1 second ( $c\times 1s$ ), we can substitute c with our new unit,
   
   Distance $=400\times (c\times 1s)$
   
   Distance $=400\alpha$
   
   Therefore, the distance between Venus and Earth is $400\alpha$ . Hence, the correct option is (B).
2. JEE Main 2026MediumErrors in Measurement and Error Propagation
   Q2

   The percentage error in the calculated volume of a sphere, if there is $2\%$ error in its diameter measurement, is $\_\_\_\_$ .

   • A1
   • B2
   • C6
   • D8

   Show answer & solution

   Correct answer: (C)

   The volume $V$ of a sphere is given by the formula $=\frac{4}{3}\pi r^3$ , where $r$ is the radius of the sphere.
   
   Since the error is given in terms of diameter (D), we express the radius as $=\frac{D}{2}$ .
   
   Substituting this into the volume formula:
   
   $$V=\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3$$
   
   $$\Rightarrow V=\frac{4}{3}\pi \left(\frac{D^3}{8}\right)$$
   
   $$\Rightarrow V=\frac{1}{6}\pi D^3$$
   
   To find the relative error, we take the natural logarithm $(ln)$ on both sides of the equation :
   
   $$ln(V)=ln\left(\frac{\pi }{6}\right)+ln\left(D^3\right)$$
   
   $$\Rightarrow ln(V)=ln\left(\frac{\pi }{6}\right)+3ln(D)$$
   
   Now, on differentiating both sides.
   
   $$\frac{dV}{V}=0+3\left(\frac{dD}{D}\right)$$
   
   $$\Rightarrow \frac{dV}{V}=3\frac{dD}{D}$$
   
   To express this as a percentage error, we multiply both sides by 100 :
   
   $$\left(\frac{dV}{V}\times 100\right)=3\times \left(\frac{dD}{D}\times 100\right)$$
   
   It is given that the percentage error in the diameter is $2\%$ .
   
   $$\frac{dD}{D}\times 100=2\%$$
   
   So, percentage error in Volume $=3\times 2\%$
   
   Percentage error in Volume $=6\%$
   
   Therefore, the correct option is (C).
3. JEE Main 2026EasyMeasuring Instruments (Vernier Calipers and Screw Gauge)
   Q3

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

   • A$1\times 10^{-2}$
   • B$1\times 10^{-3}$
   • C$5\times 10^{-2}$
   • D$5\times 10^{-3}$

   Show answer & solution

   Correct answer: (D)

   The pitch of a screw gauge is defined as the linear distance moved by the screw on the main scale when the circular head is given one complete rotation.
   
   The screw moves linearly by 2.5 mm when given five complete rotations. Therefore, the distance moved in a single rotation is :
   
   $$\text{ Pitch }=\frac{\text{ Total linear distance moved }}{\text{ Number of full rotations }}$$
   
   $$\text{ Pitch }=\frac{2.5mm}{5}=0.5mm$$
   
   The least count is the smallest distance that can be measured accurately using the instrument. It is the distance moved by the screw when the circular scale is rotated through only one of its divisions.
   
   So, the least count of screw gauge is the ratio of the pitch to the total number of divisions on the circular scale:
   
   It is given that the circular scale has 100 divisions,
   
   L. $C.=\frac{0.5mm}{100}=0.005mm=5\times 10^{-3}mm$
   
   Therefore, the least count of given screw gauge is $5\times 10^{-3}mm$ .
   
   Thus, the correct option is (D).

Topic-wise question count

Where JEE Main has asked from Units and Measurements, most frequent first.

• Units of Measurement and Dimensions of Physical Quantities55
• Measuring Instruments (Vernier Calipers and Screw Gauge)42
• Errors in Measurement and Error Propagation41
• Principle of Homogeneity and Applications of Dimensional Analysis36
• Significant Figures, Rounding Off and Order of Magnitude6

Year-wise Units and Measurements questions in JEE Main

Number of questions from this chapter in each year's paper(s).