What are JEE Laboratory Experiments Class 11 questions?

JEE Laboratory Experiments Class 11 questions test conceptual understanding of standard Physics experiments, including measuring instruments, experimental procedures, graph interpretation, error analysis, and the physical principles behind experiments.

Are Laboratory Experiments important for JEE Main?

Yes. Laboratory Experiments form an important part of the experimental skills section and typically contribute 1–2 direct conceptual questions in JEE Main.

Which Laboratory Experiments are most important for JEE?

The simple pendulum, vernier caliper, screw gauge, resonance tube, and experiments related to elasticity and mechanics are among the most frequently tested practical topics.

Is the Laboratory Experiments section easy for JEE?

Yes. This section is generally considered easy to moderate because most questions focus on experimental procedures, instrument usage, and conceptual understanding rather than lengthy calculations.

How many Laboratory Experiments questions come in JEE Main and Advanced?

JEE Main usually includes 1–2 questions from experimental skills and laboratory-based concepts, while JEE Advanced may incorporate experimental reasoning into data interpretation or conceptual questions.

How can I prepare Laboratory Experiments for JEE?

To prepare effectively, study each experiment systematically, learn instrument least counts, understand experimental procedures, revise graphs and observations, and solve previous year questions.

What are the most common mistakes in JEE Laboratory Experiment questions?

Students often misread instrument scales, confuse systematic and random errors, forget least count calculations, and incorrectly interpret experimental graphs and observations.

Which concepts are important in Laboratory Experiments for JEE?

Key concepts include least count calculations, error analysis, graph interpretation, significant figures, experimental procedures, and the working principles of common laboratory instruments such as the vernier caliper, screw gauge, and resonance tube.

JEE Laboratory Experiments XI Questions & Solutions PDF

Question 1

Two thin wires, Wire-1 of diameter $$0.650\,\mathrm{mm}$$ and Wire-2 of unknown diameter $$d$$ are given. To obtain the value of $$d$$, the diameters of the two wires are measured with a screw gauge. The screw gauge has a pitch of $$0.5\,\mathrm{mm}$$ and there are $$100$$ divisions on the circular scale (CS). The smallest division on the linear scale (LS) is $$0.5\,\mathrm{mm}$$. The table shows the readings of LS and CS for the measurements. The value of $$d$$ (in $$\mu\mathrm{m}$$) is:

	Readings
	LS (mm)	CS
Wire-1	0.5	42
Wire-2	1.5	95

Question 2

In a screw gauge, the zero of the circular scale lies 3 divisions above the horizontal pitch line when their metallic studs are brought in contact. Using this instrument thickness of a sheet is measured. If pitch scale reading is 1 mm and the circular scale reading is 51 then the correct thickness of the sheet is_____ mm.
[Assume least count is 0.01 mm]

A

1.48

B

1.54

C

1.50

D

1.51

Question 3

In a vernier callipers, 50 vernier scale divisions are equal to 48 main scale divisions. If one main scale division= 0.05 mm, then the least count of the vernier callipers is__________ mm.

A

0.05

B

0.02

C

0.002

D

0.005

Question 4

Surface tension of two liquids (having same densities), $$T_{1}$$ and $$T_{2}$$, are measured using capillary rise method utilizing two tubes with inner radii of $$r_{1}$$ and $$r_{2}$$
where $$r_{1} > r_{2}$$. The measured liquid heights in these tubes are $$h_{1}$$ and $$h_{2}$$ respectively. [Ignore the weight of the liquid about the lowest point of miniscus]. The heights $$h_{1}$$ and $$h_{2}$$ and surfaces tensions $$T_{1}$$ and $$T_{2}$$ satisfy the relation :

A

$$h_{1} = h_{2}$$ and $$T_{1} = T_{2}$$

B

$$h_{1} > h_{2}$$ and $$T_{1} < T_{2}$$

C

$$h_{1} < h_{2}$$ and $$T_{1} = T_{2}$$

D

$$h_{1} > h_{2}$$ and $$T_{1} = T_{2}$$

Question 5

When both jaws of vernier callipers touch each other, zero mark of the vernier scale is right to zero mark of main scale $$4^{th}$$ mark on vernier scale coincides with certain mark on the main scale. while measuring the length of a cylinder, observer observes 15 divisions on main scale and $$5^{th}$$ division of vernier scale coincides with a main scale division. Measured length of cylinder is______mm.
(Least count of Vernier calliper= 0.1 mm)

A

15.5

B

15.1

C

15.9

D

15.4

Solution

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The reading shown by a vernier calliper has two parts:
• Main-scale reading (MSR) = value at the mark just left of the vernier zero.
• Vernier-scale reading (VSR) = (coinciding vernier division) $$\times$$ least count (LC).
Total reading $$=$$ MSR $$+$$ VSR.

Case 1: Determining the zero error (jaws closed)

When the jaws are in contact, the vernier zero is to the right of the main-scale zero; this is a positive zero error.
Here the $$4^{\text{th}}$$ vernier division coincides with a main-scale mark.

MSR = $$0 \text{ mm}$$ because the vernier zero has not crossed the first main-scale division.
VSR $$= 4 \times LC = 4 \times 0.1 \text{ mm} = 0.4 \text{ mm}$$.

Observed zero reading $$= 0 + 0.4 = 0.4 \text{ mm}$$, so
zero error $$= +0.4 \text{ mm}$$ and the zero correction (to be applied later) is $$-0.4 \text{ mm}$$.

Case 2: Measuring the cylinder

The observer notes:
• MSR = $$15 \text{ mm}$$ (15 main-scale divisions lie to the left of the vernier zero).
• The $$5^{\text{th}}$$ vernier division coincides with a main-scale mark.

VSR $$= 5 \times 0.1 \text{ mm} = 0.5 \text{ mm}$$.

Raw reading (uncorrected) $$= 15 + 0.5 = 15.5 \text{ mm}$$.

Applying the zero correction
True length $$= 15.5 \text{ mm} - 0.4 \text{ mm} = 15.1 \text{ mm}$$.

Therefore, the measured length of the cylinder is $$\mathbf{15.1 \text{ mm}}$$ (Option B).

Question 1

Given below are two statements : Statement I : In a vernier callipers, one vernier scale division is always smaller than one main scale division. Statement II : The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions. In the light of the above statements, choose the correct answer from the options given below.

A

Statement I is true but Statement II is false

B

Statement I is false but Statement II is true

C

Both Statement I and Statement II are false

D

Both Statement I and Statement II are true

Solution

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We need to evaluate two statements about vernier callipers.

Statement I: "In a vernier callipers, one vernier scale division is always smaller than one main scale division."

In a standard vernier callipers, n vernier scale divisions (VSD) coincide with (n-1) main scale divisions (MSD). So:

$$n \times VSD = (n-1) \times MSD$$

$$VSD = \frac{(n-1)}{n} \times MSD$$

This means one VSD is smaller than one MSD. However, the statement says "always." In some vernier callipers, n VSD = (n+1) MSD, which would make VSD larger than MSD. So the word "always" makes Statement I false, as there exist vernier callipers where VSD > MSD.

Statement II: "The vernier constant is given by one main scale division multiplied by the number of vernier scale divisions."

The vernier constant (least count) is given by:

$$VC = 1 \text{ MSD} - 1 \text{ VSD} = MSD - \frac{(n-1)}{n} MSD = \frac{MSD}{n}$$

This equals one MSD divided by (not multiplied by) the number of vernier scale divisions. So Statement II is also false.

Both statements are false.

The correct answer is Option 3: Both Statement I and Statement II are false.

Question 2

A tiny metallic rectangular sheet has length and breadth of 5 mm and 2.5 mm , respectively. Using a specially designed screw gauge which has pitch of 0.75 mm and 15 divisions in the circular scale, you are asked to find the area of the sheet. In this measurement, the maximum fractional error will be $$\frac{x}{100}$$ where x is __________.

Question 3

The least count of a screw gauge is $$0.01\,mm.$$ If the pitch is increased by $$75\%$$ and number of divisions on the circular scale is reduced by $$50\%,$$ the new least count will be $$\underline{\hspace{2cm}}\times10^{-3}\,mm.$$

Question 1

Identify compound (Z) in the following reaction sequence.

A

B

C

D

Question 2

Identify the product (P) in the following reaction:

A

B

C

D

Question 3

Match List - I with List - II.

Choose the correct answer from the options given below :

A

(A)-(IV), (B)-(III), (C)-(I), (D)-(II)

B

(A)-(II), (B)-(III), (C)-(IV), (D)-(I)

C

(A)-(IV), (B)-(I), (C)-(II), (D)-(III)

D

(A)-(III), (B)-(I), (C)-(IV), (D)-(II)

Question 4

Which among the following aldehydes is most reactive towards nucleophilic addition reactions?

A

B

C

D

Question 5

Identify major product "X" formed in the following reaction :

A

B

C

D

Question 6

Two moles of benzaldehyde and one mole of acetone under alkaline conditions using aqueous NaOH after heating gives $$x$$ as the major product. The number of $$\pi$$ bonds in the product $$x$$ is _____

Question 1

A spectrometer gives the following reading when used to measure the angle of a prism. Main scale reading: $$58.5$$ degree; Vernier scale reading: $$09$$ divisions. Given that $$1$$ division on main scale corresponds to $$0.5$$ degree. Total divisions on the vernier scale are $$30$$ and match with $$29$$ divisions of the main scale. The angle of the prism from the above data

A

$$58.59^\circ$$

B

$$58.77^\circ$$

C

$$58.65^\circ$$

D

$$59^\circ$$

Question 2

$$N$$ divisions on the main scale of a vernier calliper coincide with $$(N+1)$$ divisions of the vernier scale. If each division of main scale is $$a$$ units, then the least count of the instrument is

A

$$a$$

B

$$\frac{a}{N}$$

C

$$\frac{N}{N+1}\times a$$

D

$$\frac{a}{N+1}$$

Question 1

A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading: $$0 \, \text{mm}$$. Circular scale reading: 52 divisions. Given that $$1 \, \text{mm}$$ on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is:

A

$$0.052 \, \text{cm}$$

B

$$0.026 \, \text{cm}$$

C

$$0.005 \, \text{cm}$$

D

$$0.52 \, \text{cm}$$

Question 1

In an experiment the angles are required to be measured using an instrument. $$29$$ divisions of the main scale exactly coincide with the $$30$$ divisions of the vernier scale. If the smallest division of the main scale is half-a-degree $$(= 0.5°)$$, then the least count of the instrument is

A

one minute

B

half minute

C

one degree

D

half degree

Question 1

An experiment is performed to find the refractive index of glass using a travelling microscope. In this experiment distance are measured by

A

a vernier scale provided on the microscope

B

a standard laboratory scale

C

a meter scale provided on the microscope

D

a screw gauage provided on the microscope

Question 2

Shown in the figure below is a meter-bridge set up with null deflection in the galvanometer.
The value of the unknown resistor $$R$$ is

A

$$13.75\ \Omega$$

B

$$220\ \Omega$$

C

$$110\ \Omega$$

D

$$55\ \Omega$$

Question 3

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

3.32 mm

B

3.73 mm

C

3.67 mm

D

3.38 mm

Solution

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Solution & Explanation

1. Calculate the Pitch and Least Count (LC) of the Screw Gauge

The pitch of a screw gauge is defined as the distance advanced by the spindle per complete rotation of the circular scale. We are given that 2 full turns cover a distance of $$1 \,\, \text{mm}$$:

$$\text{Pitch} = \frac{\text{Distance covered on main scale}}{\text{Number of full rotations}} = \frac{1 \,\, \text{mm}}{2} = 0.5 \,\, \text{mm}$$

The Least Count (LC) is the smallest measurement that can be accurately taken using the instrument. It is calculated by dividing the pitch by the total number of divisions on the circular scale ($$50$$):

$$\text{Least Count (LC)} = \frac{\text{Pitch}}{\text{Total number of circular divisions}} = \frac{0.5 \,\, \text{mm}}{50} = 0.01 \,\, \text{mm}$$

2. Determine the Observed Reading

The total observed reading before adjusting for any systematic tool misalignment is calculated using the main scale reading (MSR) and the circular scale reading (CSR) matching line:

Main Scale Reading ($$\text{MSR}$$): $$3 \,\, \text{mm}$$
Circular Scale Reading ($$\text{CSR}$$): $$35 \,\, \text{divisions}$$

$$\text{Observed Reading} = \text{MSR} + (\text{CSR} \times \text{LC})$$

$$\text{Observed Reading} = 3 \,\, \text{mm} + (35 \times 0.01 \,\, \text{mm}) = 3 + 0.35 = 3.35 \,\, \text{mm}$$

3. Correct for Zero Error to Find True Diameter

The instrument is given to have a built-in zero error of $$-0.03 \,\, \text{mm}$$. To find the true, accurate thickness parameter, we must subtract this zero error from our raw observed calculation:

$$\text{True Diameter} = \text{Observed Reading} - (\text{Zero Error})$$

$$\text{True Diameter} = 3.35 \,\, \text{mm} - (-0.03 \,\, \text{mm})$$

$$\text{True Diameter} = 3.35 \,\, \text{mm} + 0.03 \,\, \text{mm} = 3.38 \,\, \text{mm}$$

Concept Check: Because the tool has a negative zero error, it sits slightly behind the true zero index mark when completely closed. As a result, it consistently under-reports measurements by $$0.03 \,\, \text{mm}$$. We mathematically add this value back to our observed measurement to calculate the correct structural thickness.

Correct Option Key: Option D ($$3.38 \,\, \text{mm}$$)

Question 1

In a metre bridge experiment null point is obtained at $$20$$ cm from one end of the wire when resistance $$X$$ is balanced against another resistance $$Y$$. If $$X < Y$$, then where will be the new position of the null point from the same end, if one decides to balance a resistance of $$4X$$ against $$Y$$?

A

$$50$$ cm

B

$$80$$ cm

C

$$40$$ cm

D

$$70$$ cm

JEE Laboratory Experiments - XI Questions

Solution & Explanation

1. Calculate the Pitch and Least Count (LC) of the Screw Gauge

2. Determine the Observed Reading

3. Correct for Zero Error to Find True Diameter

Frequently Asked Questions

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JEE Laboratory Experiments - XI Questions

Solution & Explanation

1. Calculate the Pitch and Least Count (LC) of the Screw Gauge

2. Determine the Observed Reading

3. Correct for Zero Error to Find True Diameter

Frequently Asked Questions

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