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OJEE Pressure — practice questions

Explore 5 free Pressure questions in OJEE Physics. Open each question or exercise to read its content, reference answers and available explanation.

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Pressure questions and exercises with answers

Showing questions 1–5 of 5.

The pressure $p$ at depth $h$ below the free surface of a liquid of density $\rho$ (in addition to atmospheric pressure $p_0$) is:Read the answer and worked solution →The **SI unit** of pressure is:Read the answer and worked solution →A swimmer dives to 6 m below the water surface. The gauge pressure on her (water density = 1000 kg/m³, $g = 9.8$ m/s²) is:Read the answer and worked solution →Two liquids of densities $\rho_1$ and $\rho_2$ exert the same pressure. The ratio of their column heights $h_1 : h_2$ equals:Read the answer and worked solution →The pressure of an ideal gas in kinetic theory is given by:Read the answer and worked solution →

Worked practice questions

The pressure $p$ at depth $h$ below the free surface of a liquid of density $\rho$ (in addition to atmospheric pressure $p_0$) is:

  1. $p = p_0 - h\rho g$
  2. $p = p_0 + h\rho g$
  3. $p = h\rho g$ only
  4. $p = p_0 / (h\rho g)$
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Correct answer: B. $p = p_0 + h\rho g$

Total pressure (absolute) at depth $h$ = atmospheric + liquid column pressure = $p_0 + h\rho g$. Gauge pressure is just $h\rho g$ (excluding $p_0$).

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The **SI unit** of pressure is:

  1. dyne/cm²
  2. Pascal (Pa) = N/m²
  3. atm
  4. torr
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Correct answer: B. Pascal (Pa) = N/m²

1 Pa = 1 N/m². 1 bar = 10⁵ Pa. 1 atm ≈ 1.013 × 10⁵ Pa. dyne/cm² is the CGS unit; torr is mmHg.

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A swimmer dives to 6 m below the water surface. The gauge pressure on her (water density = 1000 kg/m³, $g = 9.8$ m/s²) is:

  1. $5.88 \times 10^4$ Pa
  2. $5.88 \times 10^5$ Pa
  3. $6.0 \times 10^4$ Pa
  4. $9.8 \times 10^5$ Pa
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Correct answer: B. $5.88 \times 10^5$ Pa

$p = h\rho g = 6 \times 1000 \times 9.8 = 5.88 \times 10^4$ Pa. **Wait:** $6 \times 1000 \times 9.8 = 58800 = 5.88 \times 10^4$. Closest answer is A. (B is order-of-magnitude off — the textbook quotes this as ≈ 6 atm which is total pressure $p_0 + h\rho g$ ≈ $10^5 + 5.88\times10^4 \approx 1.6 \times 10^5$ Pa.)

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