Properties of Bulk Matter

20 NEET questions on Properties of Bulk Matter. A sample is shown below with full explanations; the rest are available free after signing up.

Properties of Bulk Matter2026medium

Q1.The temperature of a metallic sphere of radius R is increased by a small amount ΔT. If the linear coefficient of thermal expansion of the metal is α, the approximate increase in the volume of the sphere is:

  • A2πR³αΔT
  • B3πR³αΔT
  • C4πR³αΔTAnswer
  • D6πR³αΔT

Explanation

For small changes the fractional volume change is three times the fractional linear change: ΔV/V = 3ΔR/R = 3αΔT. With V = (4/3)πR³ this gives ΔV = 3αΔT × (4/3)πR³ = 4πR³αΔT.

Source: NEET 2026 re-exam (21 June), Code 50, Q24

Properties of Bulk Matter2024medium

Q2.A thin flat circular disc of radius 4.5 cm is placed gently over the surface of water. If the surface tension of water is 0.07 N m⁻¹, then the excess force required to take it away from the surface is

  • A198 N
  • B1.98 mN
  • C99 N
  • D19.8 mNAnswer

Explanation

The surface tension acts along the circumference of the disc, so the extra force needed is F = T × 2πr = 0.07 × 2 × 3.14 × 0.045 ≈ 0.0198 N, that is 19.8 mN. Note the circumference and not the area is what matters, and the answer is in millinewtons — the two large distractors come from using the wrong length or dropping the metre-to-centimetre conversion.

Source: NEET 2024 (5 May), Code R3, Q20

Properties of Bulk Matter2024medium

Q3.The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young’s modulus are 8 × 10⁸ N m⁻² and 2 × 10¹¹ N m⁻² respectively is

  • A0.4 mm
  • B40 mm
  • C8 mm
  • D4 mmAnswer

Explanation

Young’s modulus Y = stress/strain, so the maximum strain is the elastic limit divided by Y: 8 × 10⁸ / 2 × 10¹¹ = 4 × 10⁻³. The elongation is that strain times the original length, 4 × 10⁻³ × 1 m = 4 × 10⁻³ m, which is 4 mm. Beyond this the wire would no longer return to its original length.

Source: NEET 2024 (5 May), Code R3, Q21

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