A syringe is used to exert 1.5 atmospheric pressure to release water horizontally. The speed of water immediately after ejection is ............ m s$^{-1}$. (Take 1 atmospheric pressure = $10^5$ Pa, density of water = $10^3$ kg m$^{-3}$.) (Specify your answer in m s$^{-1}$ as an integer.)
Step 1: Use Bernoulli's equation for fluid exit velocity.
The exit speed from a pressurized container is given by
$v = \sqrt{\frac{2\Delta P}{\rho}}$.
Step 2: Compute pressure difference.
$\Delta P = 1.5 \times 10^5$ Pa.
Step 3: Substitute values.
$v = \sqrt{\frac{2(1.5\times10^5)}{10^3}} = \sqrt{300}$.
Step 4: Final value.
$\sqrt{300} \approx 17.3$, so the answer is $17$ m/s (integer).
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)

