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9.1 , fluid mechanics — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics9.1 , fluid mechanics1 Mark
MCQ
Given below are two statements :

Statement $I$ : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation $\mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{g}\left(\mathrm{h}_2-\mathrm{h}_1\right)$

Statement $II$ : In ventury tube shown $2 \mathrm{gh}=v_1^2-v_2^2$

In the light of the above statements, choose the most appropriate answer from the options given below.

  • A
    Both Statement $I$ and Statement $II$ are correct.
  • B
    Statement $I$ is incorrect but Statement $II$ is correct.
  • C
    Both Statement $I$ and Statement $II$ are incorrect.
  • ✓
    Statement $I$ is correct but Statement $II$ is incorrect.
✓

Answer

Correct option: D.
Statement $I$ is correct but Statement $II$ is incorrect.
d
Applying Bernoulli's equation

$\mathrm{P}_1+\rho \mathrm{gh}_1+\frac{1}{2} \rho v_1^2=\mathrm{P}_2+\rho \mathrm{gh}_2+\frac{1}{2} \rho v_2^2$

$\left[h_1 \& h_2\right.$ are height of point from any reference level]

$\text { Given } \mathrm{V}_1=\mathrm{V}_2=0 \text { (for statement-1) }$

$\therefore \mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{g}\left(\mathrm{h}_2-\mathrm{h}_2\right)$

For statement-2

$\mathrm{P}_1+\frac{1}{2} \rho v_1^2=\mathrm{P}_2+\frac{1}{2} \rho v_2^2$

$\mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{gh}$

$\mathrm{P}_1-\mathrm{P}_2=\frac{1}{2} \rho v_2^2-\frac{1}{2} \rho v_1^2$

$\rho \mathrm{gh}=\frac{1}{2} \rho v_2^2-\frac{1}{2} \rho v_1^2$

$2 \mathrm{gh}=\mathrm{v}_2^2-\mathrm{v}_1^2$

Hence answer $(4)$

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