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NEET Experimental Skills MCQs with answers

954 questions on Experimental Skills for NEET 2027. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. A student measures the depth of a flat-bottomed beaker using the depth probe of a vernier calliper. To avoid parallax and systematic errors, the probe should be placed:

  1. Perpendicular to the base, with the edge of the main scale beam resting squarely on the top rim
  2. Tilted diagonally across the bottom of the beaker
  3. At the center of the beaker without touching the bottom surface
  4. Inside the liquid contained in the beaker at an angle of 45°
Show answer and explanation

Answer: (A) Perpendicular to the base, with the edge of the main scale beam resting squarely on the top rim

To measure depth accurately, the flat shoulder or end of the main beam of the calliper must rest firmly and squarely on the upper rim of the vessel, while the sliding strip projects down vertically (perpendicular to the base) until it gently touches the bottom.

medium

2. In a laboratory experiment to verify Stokes' law, a steel ball is released from the top surface of a tall jar filled with glycerin. A student plots the viscous force F_v acting on the sphere as a function of depth y from the free surface. Which description best matches this curve?

  1. F_v increases linearly with depth y throughout the fall.
  2. F_v starts at zero, increases at a decreasing rate, and becomes constant.
  3. F_v is constant throughout the fall since gravity is constant.
  4. F_v starts at a maximum and decreases exponentially to zero.
Show answer and explanation

Answer: (B) F_v starts at zero, increases at a decreasing rate, and becomes constant.

At the free surface, the ball starts from rest, so initial velocity is zero and F_v = 0. As it falls, downward acceleration causes velocity to increase, which increases F_v. As F_v increases, the net force decreases, so the rate of increase of velocity diminishes until terminal velocity is attained, after which F_v remains constant.

hard

3. In an experiment to determine the Young's modulus of a wire of initial length 1.0 m and cross-sectional area 1.0 mm², a load of 5.0 kg is applied. The wire elongates by 0.25 mm. If the energy converted into heat during the stretching process is negligible, the work done by the gravitational field on the suspended mass and the elastic energy stored in the wire are, respectively: (take g = 10 m/s²)

  1. 12.5 mJ and 12.5 mJ
  2. 25.0 mJ and 12.5 mJ
  3. 6.25 mJ and 6.25 mJ
  4. 12.5 mJ and 6.25 mJ
Show answer and explanation

Answer: (D) 12.5 mJ and 6.25 mJ

The work done by gravity is W_grav = M g l = (5.0 kg)(10 m/s²)(0.25 × 10⁻³ m) = 1.25 × 10⁻² J = 12.5 mJ. The elastic potential energy stored in the wire is U = (1/2) M g l = (1/2)(12.5 mJ) = 6.25 mJ. The remaining half of the gravitational work is dissipated as heat or damped during oscillations.

medium

4. A student notes that when the jaws of a vernier calliper are in contact, the zero of the vernier scale lies to the left of the main scale zero, and the 7th division of the 10-division vernier scale coincides with a main scale division. If 1 MSD = 1 mm and 10 VSD = 9 MSD, what is the zero error of the instrument?

  1. -0.7 mm
  2. -0.3 mm
  3. +0.3 mm
  4. +0.7 mm
Show answer and explanation

Answer: (B) -0.3 mm

Least count = 1 mm / 10 = 0.1 mm. When vernier zero lies to the left of main scale zero, the error is negative. For a 10-division scale with the 7th mark coinciding, the negative zero error is -(10 - 7) × LC = -3 × 0.1 mm = -0.3 mm.

easy

5. Which of the following defines the relationship between the figure of merit (k) and the voltage sensitivity (Sᵥ) of a galvanometer having resistance G?

  1. Sᵥ = k × G
  2. Sᵥ = G / k
  3. Sᵥ = k / G
  4. Sᵥ = 1 / (k × G)
Show answer and explanation

Answer: (D) Sᵥ = 1 / (k × G)

Figure of merit is k = I / θ, so current sensitivity is Sᵢ = θ / I = 1 / k. Voltage sensitivity is given by Sᵥ = θ / V = θ / (I × G) = Sᵢ / G = 1 / (k × G).

hard

6. In an experiment using the principle of moments to determine an unknown mass, m = m_s × (d₂ / d₁), where m_s = 100 g (known with negligible error), d₁ = 25.0 cm, and d₂ = 50.0 cm. The positions on the metre scale are measured with an ordinary ruler of least count 0.1 cm. Since each arm length is obtained from the difference between two position readings, the absolute error in each arm length is Δd = 0.2 cm. What is the maximum percentage error in the calculated mass m?

  1. 0.8%
  2. 1.2%
  3. 1.6%
  4. 2.0%
Show answer and explanation

Answer: (B) 1.2%

The fractional error in m is given by Δm / m = Δd₁ / d₁ + Δd₂ / d₂. Given Δd₁ = Δd₂ = 0.2 cm, d₁ = 25.0 cm, and d₂ = 50.0 cm: Δm / m = 0.2 / 25.0 + 0.2 / 50.0 = 0.008 + 0.004 = 0.012. Therefore, the maximum percentage error is 0.012 × 100% = 1.2%.

easy

7. The height of capillary rise of a liquid in a tube is:

  1. Inversely proportional to the radius of the tube
  2. Directly proportional to the radius
  3. Independent of the radius
  4. Proportional to the square of the radius
Show answer and explanation

Answer: (A) Inversely proportional to the radius of the tube

h = 2T cos θ/(rρg), so a narrower tube gives a higher capillary rise.

NCERT Class 11

medium

8. In a metre-bridge experiment, a 4 Ω resistor is in the left gap and the null point is 55 cm from the left end. The resistance in the right gap is:

  1. 2.2 Ω
  2. 3.27 Ω
  3. 4.89 Ω
  4. 1.8 Ω
Show answer and explanation

Answer: (B) 3.27 Ω

At balance R/S = l/(100 − l), so S = R(100 − l)/l = 4 × 45/55 = 3.27 Ω.

medium

9. A screw gauge has a pitch of 0.5 mm and 50 circular-scale divisions. The main-scale reading is 4.5 mm and the 40th circular division coincides with the reference line (no zero error). The diameter of the wire is:

  1. 4.9 mm
  2. 5.3 mm
  3. 8.5 mm
  4. 6.5 mm
Show answer and explanation

Answer: (A) 4.9 mm

Least count = pitch/divisions = 0.5/50 = 0.01 mm. Diameter = 4.5 + 40 × 0.01 = 4.9 mm.

medium

10. A screw gauge has a pitch of 1 mm and 100 circular-scale divisions. The main-scale reading is 9 mm and the 49th circular division coincides with the reference line (no zero error). The diameter of the wire is:

  1. 0.49 mm
  2. 9.49 mm
  3. 18.98 mm
  4. 13.9 mm
Show answer and explanation

Answer: (B) 9.49 mm

Least count = pitch/divisions = 1/100 = 0.01 mm. Diameter = 9 + 49 × 0.01 = 9.49 mm.

hard

11. In a modified resonance tube experiment, the tube is filled with a gas of molar mass M at absolute temperature T. If the fundamental resonant length is l₁ and the second resonant length is l₂, which expression correctly gives the ratio of molar heat capacities γ = C_p/C_v of the gas? (R is the universal gas constant, f is the frequency of the fork)

  1. f² M (l₂ - l₁)² / (4 R T)
  2. 16 f² M (l₂ - l₁)² / (R T)
  3. 4 f² M (l₂ - l₁)² / (R T)
  4. 2 f² M (l₂ - l₁)² / (R T)
Show answer and explanation

Answer: (C) 4 f² M (l₂ - l₁)² / (R T)

The speed of sound obtained from the resonance lengths is v = 2f(l₂ - l₁). For an ideal gas, the Laplace formula states that v = √(γRT/M). Squaring both sides gives v² = γRT/M, so γ = v² M / (RT). Substituting v = 2f(l₂ - l₁) gives γ = [2f(l₂ - l₁)]² M / (RT) = 4 f² M (l₂ - l₁)² / (RT).

hard

12. The external diameter of a hollow metal cylinder is measured as (4.24 ± 0.01) cm, and its internal diameter is measured as (3.82 ± 0.01) cm using the same vernier calliper. What is the thickness of the cylinder wall with the appropriate absolute uncertainty?

  1. (0.42 ± 0.01) cm
  2. (0.42 ± 0.02) cm
  3. (0.21 ± 0.02) cm
  4. (0.21 ± 0.01) cm
Show answer and explanation

Answer: (D) (0.21 ± 0.01) cm

The wall thickness is t = (D_ext - D_int) / 2 = (4.24 - 3.82) / 2 = 0.42 / 2 = 0.21 cm. The absolute error in the difference (D_ext - D_int) is Δ(D_ext) + Δ(D_int) = 0.01 + 0.01 = 0.02 cm. Therefore, the absolute error in wall thickness t is Δt = 0.02 / 2 = 0.01 cm. Hence, thickness = (0.21 ± 0.01) cm.

hard

13. A non-uniform metre scale of length 100 cm rests horizontally on a knife-edge at the 40 cm mark when a mass of 20 g is suspended from the 10 cm mark. When the suspended mass is changed to 60 g at the 10 cm mark, the knife-edge must be moved to the 30 cm mark for horizontal balance. What is the mass of the non-uniform metre scale?

  1. 80 g
  2. 40 g
  3. 60 g
  4. 50 g
Show answer and explanation

Answer: (C) 60 g

Let the mass of the scale be M and its centre of gravity be at x cm. Case 1: 20 × (40 - 10) = M × (x - 40) → 600 = M(x - 40) = Mx - 40M. Case 2: 60 × (30 - 10) = M × (x - 30) → 1200 = M(x - 30) = Mx - 30M. Subtracting Case 1 from Case 2: 600 = 10M, hence M = 60 g.

easy

14. A student records the following three readings on the vertical scale of a travelling microscope: Reading 1 (mark on paper): 1.150 cm Reading 2 (mark viewed through slab): 2.450 cm Reading 3 (chalk dust on top of slab): 5.050 cm What is the refractive index of the slab?

  1. 1.35
  2. 1.42
  3. 1.50
  4. 1.60
Show answer and explanation

Answer: (C) 1.50

Real thickness = Reading 3 - Reading 1 = 5.050 - 1.150 = 3.900 cm. Apparent thickness = Reading 3 - Reading 2 = 5.050 - 2.450 = 2.600 cm. Refractive index μ = 3.900 / 2.600 = 1.50.

easy

15. While identifying terminals of an electrolytic capacitor before using it in a rectification filter circuit, a student should note that:

  1. the metallic body is universally connected to the positive terminal
  2. the longer lead always indicates the negative terminal
  3. both leads are identical and the component can be connected in either direction
  4. the terminal near the broad stripe containing '-' signs is the cathode (negative terminal)
Show answer and explanation

Answer: (D) the terminal near the broad stripe containing '-' signs is the cathode (negative terminal)

Electrolytic capacitors are polarized. The negative terminal is typically identified by a prominent vertical stripe on the casing marked with minus signs ('-') and possesses the shorter lead.

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