At two points on a horizontal tube of varying cross-section the radii are 1 cm and $\mathbf{0 . 4 ~ c m}$, velocities of fluid are $\mathrm{V}_1, \mathrm{~V}_2$ and pressure difference ( $P_1-P_2$ ) between these points is 4.9 cm of water. The value of $\sqrt{V_2^2-V_1^2}$ is ( $\mathrm{g}=$ acceleration due to gravity $\mathrm{g}=980 \mathrm{~cm} / \mathrm{s}^2$ )

A simple harmonic progressive wave is given by equation $y=\operatorname{asin} 2 \pi\left(n t-\frac{x}{\lambda}\right)$. If the wave velocity is equal to $\frac{1}{4} \times$ (maximum particle velocity), then the wavelength ' $\lambda$ ' is (Given $\rightarrow \mathrm{a}=$ amplitude, $\mathrm{n}=$ frequency, $\mathrm{t}=$ time, $\mathrm{y}=$ displacement, $\mathrm{x}=$ distance )

$$ \text { In the following figure, the current } \mathrm{I} \text { is equal to } $$

The work function of a photosensitive metallic surface is $h v_0$. If photons of energy (2.5) $h v_0$ fall on this surface, the electrons come out with maximum velocity ' v '. When the photon energy is increased to $7 \mathrm{~h} v_0$, the maximum velocity of photoelectrons will be