- A$\frac{\lambda }{{2\pi }}\,\phi $
- B$\frac{\lambda }{{2\pi }}\,\left( {\phi - \frac{\pi }{2}} \right)$
- ✓$\frac{\lambda }{{2\pi }}\,\left( {\phi + \frac{\pi }{2}} \right)$
- D$\frac{{2\pi }}{\lambda }\,\phi $
$\mathrm{y}_{2}=\mathrm{a}_{2} \sin \left(\omega \mathrm{t}-\frac{2 \pi \mathrm{x}}{\lambda}+\frac{\pi}{2}+\phi\right)$
$\Delta \phi=\frac{\pi}{2}+\phi$
So, $\frac{\Delta \phi}{2 \pi}=\frac{\Delta x}{\lambda} \Rightarrow \Delta x=\frac{\lambda}{2 \pi}\left(\frac{\pi}{2}+\phi\right)$
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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.
| List - I | List - II |
|---|---|
| $(A)$ Distance between earth and stars | $(1)$ Microns |
| $(B)$ Inter-atomic distance in a solid | $(2)$ Angstroms |
| $(C)$ Size of the nucleus | $(3)$ Light years |
| $(D)$ Wavelength of infrared laser | $(4)$ Fermi |
| $(5)$ Kilometres |