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JEE Main 2020 (4 Sep, Shift 2) · Bulk Modulus & Compressibility

Bulk Modulus & Compressibility question from JEE Main 2020 (4 Sep, Shift 2)

Q17JEE Main 2020 · 4 SEP · SHIFT 2Medium

Bulk Modulus & Compressibility

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 ×\times 1010 Pa)

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Answer: (c)

  1. ΔVV=PB=4×1098×1010=0.05=5%\frac{\Delta V}{V} = \frac{P}{B} = \frac{4\times10^{9}}{8\times10^{10}} = 0.05 = 5\%
  2. For a cube V=L3⇒ΔLL=13ΔVVV = L^3 \Rightarrow \frac{\Delta L}{L} = \frac{1}{3}\frac{\Delta V}{V}
  3. ΔLL=53%≈1.67%\frac{\Delta L}{L} = \frac{5}{3}\% \approx 1.67\%
  4. So the answer is (c).
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