Explain why

**(a)** To keep a piece of paper horizontal, you should blow over, not under, it

**(b)** When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers

**(c)** The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection

**(d)** A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel

**(e)** A spinning cricket ball in air does not follow a parabolic trajectory

Asked by Pragya Singh | 1 year ago | 73

**(a)** If we blow over a piece of paper, velocity of air above the paper becomes more than that below it. As K.E. of air above the paper increases, so in accordance with Bernoulli’s theorem its pressure energy and hence its pressure decreases.

Due to greater value of pressure below the piece of paper = atmospheric pressure, it remains horizontal and does not fall.

**(b)** As per the equation of continuity area × velocity = constant. When we try to close a water tap with our fingers, the area of cross-section of the outlet of water jet is reduced considerably as the openings between our fingers provide constriction (regions of smaller area)

Thus, velocity of water increases greatly and fast jets of water come through the openings between our fingers.

**(c)** The size of the needle controls the velocity of flow and the thumb pressure controls pressure.

According to the Bernoulli’s theorem \( P + \dfrac{ 1 }{ 2 } pv^{ 2 }\) = Constant

In this equation, the pressure P occurs with a single power whereas the velocity occurs with a square power. Therefore, the velocity has more effect compared to the pressure. It is for this reason that needle of the syringe controls flow rate better than the thumb pressure exerted by the doctor.

**(d)** This is because of principle of conservation of momentum. While the flowing fluid carries forward momentum, the vessel gets a backward momentum.

**(e)** A spinning cricket ball would have followed a parabolic trajectory has there been no air. But because of air the Magnus effect takes place. Due to the Magnus effect the spinning cricket ball deviates from its parabolic trajectory.

**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 = \( 9.8 m/s^2\)