We apply the principle of conservation of energy and the equation for fluid flow to determine the flow rate from a hole in the mesh.
The Bernoulli equation for the flow through the pipe gives the speed of the fluid at the hole as:
\[ v = \sqrt{v_0^2 + 2gh} \]
Here:
The flow rate \( Q \) for an individual hole is given by the area \( A_h \) multiplied by the velocity \( v \):
\[ Q = A_h v = A_h \sqrt{v_0^2 + 2gh} \]
The total flow rate through all \( n \) holes is:
\[ Q_{\text{total}} = n \times A_h \times \sqrt{v_0^2 + 2gh} \]
Since the total cross-sectional area of the pipe is \( A_0 \), we use \( A_0 \) in place of \( A_h \) for the total flow rate per hole. The final expression for the volume flow rate from an individual hole is:
\[ \frac{A_0}{n} \]
Thus, the correct answer is option (A).
The velocity of fluid flow is given by:
\[ v = \sqrt{v_0^2 + 2gh} \]
In order to achieve the static equilibrium of the see-saw about the fulcrum \( P \), shown in the figure, the weight of Box B should be _________ kg, if the weight of Box A is 50 kg.
A particle of mass 1kg, initially at rest, starts sliding down from the top of a frictionless inclined plane of angle \(\frac{π}{6}\)\(\frac{\pi}{6}\) (as schematically shown in the figure). The magnitude of the torque on the particle about the point O after a time 2seconds is ______N-m. (Rounded off to nearest integer)
(Take g = 10m/s2)
At a particular temperature T, Planck's energy density of black body radiation in terms of frequency is \(\rho_T(\nu) = 8 \times 10^{-18} \text{ J/m}^3 \text{ Hz}^{-1}\) at \(\nu = 3 \times 10^{14}\) Hz. Then Planck's energy density \(\rho_T(\lambda)\) at the corresponding wavelength (\(\lambda\)) has the value \rule{1cm}{0.15mm} \(\times 10^2 \text{ J/m}^4\). (in integer)
[Speed of light \(c = 3 \times 10^8\) m/s]
(Note: The unit for \(\rho_T(\nu)\) in the original problem was given as J/mΒ³, which is dimensionally incorrect for a spectral density. The correct unit J/(mΒ³Β·Hz) or JΒ·s/mΒ³ is used here for the solution.)