JEE Main 2020 · 4 September · evening · Single correct · +4 / −1

A cube of metal is subjected to a hydrostatic pressure of 4 GPa. The percentage change in the length of the side of the cube is close to :
(Given bulk modulus of metal, B = 8 × 10¹⁰ Pa)

  1. (a) 0.6
  2. (b) 20
  3. (c) 1.67
  4. (d) 5
Answer and solution

Answer (c) 1.67

Solution

Physics sir · 3 steps

Bulk modulus gives you the volume strain. The side changes by one third of that.

  1. Volume strain. Bulk modulus is pressure divided by volume strain, so (Δ V)/V = P/B. With P = 4 GPa = 4 × 10⁹ Pa, this gives (Δ V)/V = (4 × 10⁹)/(8 × 10¹⁰) = 0.05, which is a 5% decrease in volume.
  2. Link volume to side. For a cube V = L³. For small changes, take logs and differentiate. The fractional change in volume is three times the fractional change in side.

    (Δ V)/V = 3 (Δ L)/L ⇒ (Δ L)/L = 1/3 × 5% ≈ 1.67%

  3. So. The side of the cube shrinks by about 1.67%, which is option (c).

The idea to keep. Pressure over bulk modulus is the fractional change in volume. For a cube or sphere, divide by 3 to get the fractional change in length or radius.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 90 times in JEE Main since 2020. This question is from 2020.