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Metre scale: mass by principle of moments: NEET MCQs with answers

47 questions on this topic of Experimental Skills (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. In an experiment to determine the mass of an unknown object using a metre scale, the system is balanced horizontally. What is the normal reaction force exerted by the knife-edge support on the scale, if the mass of the scale is M and two masses m₁ and m₂ are suspended from it? (Take acceleration due to gravity as g)

  1. (M + m₁ + m₂)g
  2. (m₁ + m₂)g
  3. (M + m₁ - m₂)g
  4. (m₁ - m₂)g
Show answer and explanation

Answer: (A) (M + m₁ + m₂)g

For vertical translational equilibrium, the net vertical force on the scale must be zero. The downward forces are the weight of the metre scale (Mg) and the weights of the two attached masses (m₁g and m₂g). Hence, the upward normal reaction force exerted by the knife-edge is N = (M + m₁ + m₂)g.

medium

2. Consider the following statements regarding the determination of mass using a metre scale: Statement I: The principle of moments is strictly independent of the local value of acceleration due to gravity, provided 'g' is uniform across the scale. Statement II: If the knife-edge is blunt instead of sharp, the effective point of support is indeterminate, leading to an error in distance measurements. Choose the correct option:

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (A) Both Statement I and Statement II are correct

Statement I is true because g cancels out from both sides of the moment equation (m₁·g·d₁ = m₂·g·d₂) when g is uniform. Statement II is also true because a blunt edge provides a contact area rather than a line pivot, introducing uncertainty in the exact position of the fulcrum.

hard

3. In a metre scale experiment, an unknown mass m is balanced against a standard mass m_s = 50 g. The position of the knife-edge is at x_f = 50.0 cm. The standard mass is at x₁ = 20.0 cm and m is at x₂ = 75.0 cm. If the position of each load has an absolute reading uncertainty of ±0.1 cm, what is the maximum percentage error in the determination of m?

  1. 0.73%
  2. 0.43%
  3. 1.25%
  4. 2.00%
Show answer and explanation

Answer: (A) 0.73%

The arm lengths are d₁ = 50.0 - 20.0 = 30.0 cm and d₂ = 75.0 - 50.0 = 25.0 cm. From m = m_s(d₁ / d₂), the fractional error is Δm/m = Δd₁/d₁ + Δd₂/d₂. Since the knife-edge is fixed, the uncertainty in each arm length is Δd = 0.1 cm. Thus Δm/m = (0.1/30.0) + (0.1/25.0) = 0.00333 + 0.00400 = 0.00733, which corresponds to approximately 0.73%.

medium

4. A uniform metre scale of mass 100 g is balanced horizontally on a knife-edge placed at the 65 cm mark. If a single unknown mass m is suspended from the 90 cm mark, what is the value of m?

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

Answer: (D) 60 g

The scale is uniform, so its weight acts at its centre of gravity at the 50 cm mark. The distance of the centre of gravity from the knife-edge (at 65 cm) is 65 - 50 = 15 cm (anticlockwise moment). The mass m is suspended at 90 cm, at a distance of 90 - 65 = 25 cm from the knife-edge (clockwise moment). By the principle of moments: 100 × 15 = m × 25, giving m = 1500 / 25 = 60 g.

easy

5. A uniform metre scale is balanced horizontally on a knife-edge placed at the 50 cm mark. When a mass of 45 g is suspended at the 10 cm mark, at what mark on the scale must a 75 g mass be suspended to maintain horizontal equilibrium?

  1. 66 cm mark
  2. 74 cm mark
  3. 80 cm mark
  4. 84 cm mark
Show answer and explanation

Answer: (B) 74 cm mark

Taking moments about the knife-edge at the 50 cm mark: Clockwise moment = Anticlockwise moment. The 45 g mass is at a distance of (50 - 10) = 40 cm to the left. Let the 75 g mass be at a distance x to the right. Then 45 × 40 = 75 × x, which gives x = 1800 / 75 = 24 cm. Therefore, the mark on the scale is 50 + 24 = 74 cm.

hard

6. In a metre scale experiment, an unknown mass m suspended at 15 cm balances a known mass of 60 g at 75 cm when the fulcrum is at the 45 cm mark. If both the scale and the masses are completely immersed in a large container of water, assuming all components have the same uniform density greater than water, the scale will:

  1. Tilt towards the 75 cm mark
  2. Tilt towards the 15 cm mark
  3. Remain in horizontal balance
  4. Oscillate continuously without balancing
Show answer and explanation

Answer: (C) Remain in horizontal balance

When fully immersed, the effective gravitational weight of each element is scaled by (1 - ρ_water / ρ_body). If the scale and suspended masses share the same uniform density, the buoyant reduction factor is identical for every term in the moment equation, keeping the balance condition unchanged.

medium

7. A uniform metre rule of mass 150 g is balanced on a wedge at the 35 cm mark. At which position should an external mass of 75 g be suspended to maintain horizontal equilibrium?

  1. 5 cm mark
  2. 10 cm mark
  3. 15 cm mark
  4. 20 cm mark
Show answer and explanation

Answer: (A) 5 cm mark

The weight of the scale acts at its centre of mass (50 cm). Clockwise torque about 35 cm = 150 × (50 - 35) = 150 × 15 = 2250 g·cm. To balance this, the 75 g mass must provide an anticlockwise torque: 75 × (35 - x) = 2250, so 35 - x = 30, giving x = 5 cm.

easy

8. A uniform metre scale is balanced on a knife-edge placed at the 55 cm mark when a mass of 40 g is suspended from the 75 cm mark. What is the mass of the metre scale?

  1. 160 g
  2. 80 g
  3. 120 g
  4. 40 g
Show answer and explanation

Answer: (A) 160 g

Since the scale is uniform, its centre of gravity is at 50 cm. Taking moments about the knife-edge at 55 cm: M × (55 - 50) = 40 × (75 - 55). Thus, M × 5 = 40 × 20 = 800, which gives M = 160 g.

medium

9. A uniform metre scale of mass 100 g is balanced horizontally on a knife-edge at the 40 cm mark. An unknown mass m is hung at the 10 cm mark, and a known mass of 25 g is hung at the 90 cm mark. What is the value of m?

  1. 60 g
  2. 85 g
  3. 75 g
  4. 95 g
Show answer and explanation

Answer: (C) 75 g

Taking torques about the fulcrum at 40 cm: Anticlockwise torque = m × (40 - 10) = 30m. Clockwise torque is due to the scale's weight (acting at 50 cm) and the 25 g mass (at 90 cm): Clockwise torque = 100 × (50 - 40) + 25 × (90 - 40) = 100 × 10 + 25 × 50 = 1000 + 1250 = 2250 g·cm. Equating both torques: 30m = 2250, which yields m = 75 g.

easy

10. A uniform metre scale is balanced on a knife-edge placed at the 50 cm mark. When a small wooden block of mass m is suspended at the 10 cm mark, the scale can be rebalanced horizontally by placing a known mass of 120 g at the 80 cm mark. What is the mass m of the wooden block?

  1. 120 g
  2. 80 g
  3. 60 g
  4. 90 g
Show answer and explanation

Answer: (D) 90 g

Taking torque about the knife-edge at the 50 cm mark: Clockwise torque = Counter-clockwise torque. The distance of mass m from the pivot is (50 - 10) = 40 cm. The distance of the 120 g mass from the pivot is (80 - 50) = 30 cm. Thus, m × 40 = 120 × 30, which gives m = 3600 / 40 = 90 g.

medium

11. A metre scale is balanced at its centre. A 50 g mass at 20 cm from the fulcrum balances an unknown mass at 25 cm on the other side. The unknown mass is:

  1. 50 g
  2. 30 g
  3. 40 g
  4. 62.5 g
Show answer and explanation

Answer: (C) 40 g

By the principle of moments, m × 25 = 50 × 20, so m = 1000/25 = 40 g.

NCERT Class 11

hard

12. In a metre scale experiment, an unknown mass m is balanced against a standard mass m₀ = 40.0 g. The knife-edge is at the 50.0 cm mark. The mass m is suspended at (10.0 ± 0.1) cm and the mass m₀ is suspended at (80.0 ± 0.1) cm. The percentage error in the calculated value of m is:

  1. 0.25%
  2. 0.33%
  3. 0.58%
  4. 0.75%
Show answer and explanation

Answer: (C) 0.58%

Distance of unknown mass from fulcrum is d₁ = 50.0 - 10.0 = 40.0 cm with error Δd₁ = 0.1 cm. Distance of standard mass is d₂ = 80.0 - 50.0 = 30.0 cm with error Δd₂ = 0.1 cm. From m = m₀(d₂/d₁), assuming negligible error in m₀, the fractional error is Δm/m = Δd₁/d₁ + Δd₂/d₂ = 0.1/40.0 + 0.1/30.0 = 1/400 + 1/300 = 7/1200 ≈ 0.00583. Therefore, percentage error is 0.58%.

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