Given:

Number of holes, n = 40

Cross-sectional area of the spray pump, A_{1 }= 8 cm ^{-2}

= 8 × 10^{-4} m^{-2}

Radius of each hole, r = 0.5 × 10^{-3} m

Cross-sectional area of each hole, a = πr^{2}

= π (0.5 × 10^{-3} )^{2} m^{2}

Total area of 40 holes, A_{2}= n × a

= \(40 \times \pi \left ( 0.5 \times 10^{ -3 } \right )^{ 2 }\)

= 3.14 x 10^{-5} m^{2}

Speed of flow of water inside the tube, V_{1} = 1.5 m/min = 0.025 m/s

Let, the water ejected through the holes at a speed = V_{2}

Using the law of continuity:

A_{1}V_{1} = A_{2 }V_{2}

V_{2} = \( \dfrac{ A_{ 1 } V_{ 1 } }{ A_{ 2 } } = \dfrac{ 8 \times 10^{ -4 }\times 0.025 }{ 3.14 \times 10^{ -5 } }\)

Therefore, V_{2 }= 0.636 m/s

**Answer the following questions:-**

**(a)** According to the law of atmospheres density of air decreases with increase in height y as \( ρ=\rho _{0}e^{\dfrac{-y}{y_{0}}}\). Where ρ_{0} =1.25 kg m^{-3} is the density of air at sea level and y_{0 }is a constant. Derive this equation/law considering that the atmosphere and acceleration due to gravity remain constant.

**(b)** A zeppelin of volume 1500 m^{3} is filled with helium and it is lifting a mass of 400 kg. Assuming that the radius of the zeppelin remains constant as it ascends. How high will it rise?[ y_{0} = 8000 m and ρH_{e} =0.18 kg m^{-3}].

Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10^{–2} N m^{–1}. Take the angle of contact to be zero and density of water to be 1.0 × 10^{3} kg m^{–3} (g = 9.8 m s^{–2}).

Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside ? Surface tension of mercury at the temperature of the experiment is 0.465 N m^{–1}. Density of mercury = 13.6 × 10^{3} kg m^{–3}

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius \( 2.0 × 10^{–5} m\) and density \( 1.2 × 10^3 kg m^{–3}\). Take the viscosity of air at the temperature of the experiment to be \( 1.8 × 10^{–5} Pa \). How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

A plane is in level flight at constant speed and each of its wings has an area of \( 25 m^2\). If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be \( 1 kg/m^3\)), g = \( 9.8 m/s^2\)