Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density `984 kg m^-3`. Determine the height of the wine column for normal atmospheric pressure.

What is the absolute and gauge pressure of the gas above the liquid surface in the tank shown in the following figure? Density of oil = 820 kg m^-3 , density of mercury = 13.6 xx 10^3 kg m^-3 , 1 atmosphere pressure = 1.01 xx 10^5 pa .

A monometer contains a liquid of density 5.44g//cm^(3) is attached to a flask containing gas A' as follows If the same liquid is used in barometer to measure the atmospheric pressure, then what will be the length of the liquid column, which exerts pressure equal to 1 atm ? (density of Hg = 13.6 g//cm^(3) )

A monometer contains a liquid of density 5.44g//cm^(3) is attached to a flask containing gas 'A' as follows If the same liquid is used in barometer to measure the atmospheric pressure, then what will be the lengh of the liquid columnm which exerts pressure equal to 1 atm ? (density of Hg=13.6g//cm^(3) )

The density of air in atmoshpere decreases with height and can be expressed by the relaton rho = rho_0e^(-alphah where rho_0 is the density at sea level , alpha is a constant and h is the height, Calculate the atmospheric pressure at sea level. Assume g to be constatn. The numerical values of constants are : g = 9.8 ms^-2 , rho_0 = 1.3 kg m^-3 , alpha = 1.2 xx 10^-4 m^-1

Estimate the average mass density of a sodium atom assuming its size to be about 2.5 overset@A . (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase : 970 kg m^-3 . Are the two densities of the same order of magnitude ? If so, why ?

A human heart discharges 75 cc of blood the through the artcries at each beat against an average pressure of 10 cm ofmercutry.Assuming that the pulse frequency is 72 per minute,the rate of working of heart is watt is (density of mercury = 13.6 g cm^-3 and g = 9.8 m s^(-2) ):

A plane is in level flight at constant speed and each of its two 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 )

It is known that density rho of air decreases with height y (in metres) as : rho = rho_(0) e^(-y//y_0) where rho_0 = 1.25 kg m^-3 is the density at sea level and y_0 is a constant. This density variation is called the law of atmospheres. Obtain this law assuming the temperature of atomoshpere remains constant (isothermal conditions). Also assume that the value of g remains constant.

A gas in equilibrium has uniform density and pressure throughout its volume. This is strictly true only if there are no external influences. A gas column under gravity, for example, does not have uniform density (and pressure). As you might expect, its density decreases with height. The precise dependence is given by the so-called 'law of atmospheres' n_2 = n_1 exp[-(mg)(H_2-H_1)/(kT)] where n_2,n_1 refer to number density at heights h_2 and h_1 respectively.Use this relation to derive the equation for sedimentation equilibrium of a suspension in a liquid column: n_2= n_1 exp[(-mgN_a)/(RT) (1-(rho)/(rho))(h_2-h_1)] where rho is the density of the suspended particle, and rho' that of surrounding medium. [N_A is Avogadro's number, and R the universal gas constant].