Q. 195.0( 1 Vote )

One end of a U-tu

Answer :

Now here the liquid on side of U-tube rises whereas on the other side of the U-tube containing mercury it has depressed by same amount due to the suction pump, the weight of the excess mercury will exert force in downward direction in downward direction to restore to original equilibrium position, let the area of cross section be A and ride in height of mercury be X


The situation has been shown in the figure



so total excess height of mercury on side of tube is 2x


now we know


Volume = Area × length


Here surface area of cross section of tube is A, excess height of mercury on side of tube is 2x so the excess volume of mercury on one side of the U-Tube is


V = 2xA


Now we know mass is given as


m = V × ρ


where m is the mass of the body, V is its volume and ρ is the density


so excess mass of mercury on side of tube is


m = 2Axρ


where A is the area of cross section of cork, ρ is density of mercury


we know weight of a body is given as


W = mg


Here m is the mass of the body and g is acceleration due to gravity, so weight of the excess mercury on one side of tube is


W = 2Axρg


Now this excess weight of mercury one side will exert force and tend to bring mercury in the tube to its original position hence the mercury in tube will start oscillating and height will decrease and increase periodically


now total mass of the mercury in U –Tube will be


M = Volume × Density


Let the total length of U – Tube be l and area of cross section is A we know


Volume = Area × length


so total Volume of liquid is


Vm = Al


So total mass of mercury is


M = Vmρ = Alρ


Where ρ is the density of mercury


We know force on a Body is


F = ma


Where m is the mass of Body and a is the acceleration of the Body, so acceleration of the body can be written as


a = F/m


here mass of the total mercury in the tube is


M = Alρ


where A is the area of cross section of cork, ρ is its density and l is the length of U - tube


so net acceleration of mercury in U-Tube is


a = -W/M


(negative sign because acceleration of mercury at each point inside tube is opposite to the direction of displacement)


i.e. a = -(2Axρg)/Alρ


= -(2g/l)x


we know condition for simple harmonic motion is that acceleration is proportional to distance from mean position and is directed toward mean position


and we have relation between acceleration and displacement as


a = -ω2x


where a is the acceleration of Body undergoing Simple Harmonic Motion with angular frequency ω, x is displacement from equilibrium or mean position


here in this case the acceleration of mercury is always directed towards mean position and also acceleration is given as


a = -(2g/l)x


here acceleration due to gravity g, length of tube l are constant so we can say acceleration is proportional to displacement from mean position, i.e. mercury in tube is undergoing simple harmonic motion, so comparing with the equation for acceleration of simple harmonic motion we get


ω2 = 2g/l


i.e. the angular frequency is



We know relation between time period T and angular frequency ω T = 2π/ω


So we have the time period of the oscillation as



So time period of the Simple harmonic motion mercury in tube is



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