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Properties of Matter question

2024 · 8 Apr · Shift 1 · Q77
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Properties of Matter question

2024 · 8 Apr · Shift 1 · Q77

JEE MainPhysicsProperties of MatterMCQ+4 / −1
Correct Bernoulli's equation is (symbols have their usual meaning) :
  1. A
    P+12ρgh+12ρv2=P+\frac{1}{2} \rho g h+\frac{1}{2} \rho v^2=P+21​ρgh+21​ρv2= constant
  2. B
    P+mgh+12mv2=P+m g h+\frac{1}{2} m v^2=P+mgh+21​mv2= constant
  3. C
    P+ρgh+ρv2=P+\rho g h+\rho v^2=P+ρgh+ρv2= constant
  4. D
    P+ρgh+12ρv2=P+\rho g h+\frac{1}{2} \rho v^2=P+ρgh+21​ρv2= constant
View written solutionFree

Correct answer: D

  1. Bernoulli’s equation for an ideal fluid flowing steadily is

P+ρgh+12ρv2=constantP + \rho g h + \frac{1}{2}\rho v^2 = \text{constant}P+ρgh+21​ρv2=constant

where:

  • PPP = pressure energy per unit volume
  • ρgh\rho g hρgh = potential energy per unit volume
  • 12ρv2\frac{1}{2}\rho v^221​ρv2 = kinetic energy per unit volume
  1. Check each option:
  • Option A: P+12ρgh+12ρv2=constantP+\frac{1}{2}\rho g h+\frac{1}{2}\rho v^2=\text{constant}P+21​ρgh+21​ρv2=constant This is incorrect because the gravitational term should be ρgh\rho g hρgh, not 12ρgh\frac{1}{2}\rho g h21​ρgh.

  • Option B: P+mgh+12mv2=constantP+mgh+\frac{1}{2}mv^2=\text{constant}P+mgh+21​mv2=constant This is incorrect because Bernoulli’s equation is written per unit volume, so density ρ\rhoρ appears, not mass mmm.

  • Option C: P+ρgh+ρv2=constantP+\rho g h+\rho v^2=\text{constant}P+ρgh+ρv2=constant This is incorrect because the kinetic term should be 12ρv2\frac{1}{2}\rho v^221​ρv2, not ρv2\rho v^2ρv2.

  • Option D: P+ρgh+12ρv2=constantP+\rho g h+\frac{1}{2}\rho v^2=\text{constant}P+ρgh+21​ρv2=constant This matches the correct Bernoulli’s equation.

  1. Final answer:

Therefore, the correct option is D.

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