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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