Question

Mercury of density 13.6×10^3 kg/m^3 is contained in a utube with its arms vertical. Neglecting damping find the time period of oscillation of the mercury if the total length of the utube occupied by the mercury is 0.3m . If the area of cross section of the tube is 2×10^-4 m^2 , find the energy of the motion when the amplitude is 0.05m.

Answer 1

HERE YOYR ANSWER

where , P is the pressure in tube column,

ρ is the density of liquid,

g is the acceleration due to gravity,

h is the height of Mercury level.

now,Let difference in the level of Mercury in two arms in indicated by gauge pressure is ∆h

here, P = 1.03 × 10^5 N/m²

ρ = 13.6 × 10³ Kg/m³ and g = 9.8 m/s²

so,∆h = 1.03 × 10^5/13.6 × 10^3 × 9.8

= 103/13.6 × 9.8 m

= 0.772 m = 77.2 cm

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Comments Report

prmkulk1978 Ace

Given :

density of mercury=13.6 x 10³kg/m3

pressure=P=1.03 × 10⁵ pascals

Formula to be used :P=ρgh

wher ρ=density of liquid

g-acceleration due to gravity

h is height of liquid column

Difference of mercury in both arms:

P=ρgΔh

Δh=P/ρ g

=1.03x10⁵ pa/13.6 x 1000 x9.8

=0.772m

=77.2cm