A U-tube of uniform bore of cross-sectional area 'A' has been set up vertically with open ends facing up. Now 'M' gram of a liquid of density 'd' is poured into it. The c

Question

A U-tube of uniform bore of cross-sectional area 'A' has been set up vertically with open ends facing up. Now 'M' gram of a liquid of density 'd' is poured into it. The column of liquid in this tube will oscillate with a period 'T', which is equal to [g = acceleration due to gravity]

TestHub · Accepted Answer

Let the total length of the liquid column be $L$ .

The mass of the liquid is given by $M = A L d$ , which gives $L = \frac{M}{A d}$ .

When the liquid is displaced by a small distance $x$ in one arm, the level in the other arm rises by $x$ . The difference in liquid levels is $2 x$ .

The restoring force acting on the liquid is the weight of the excess liquid column of height $2 x$ .

$F = - (A (2 x) d) g = - (2 A d g) x$

The acceleration of the liquid column is $a = \frac{F}{M} = - (\frac{2 A d g}{M}) x$ .

Comparing this with the standard equation of simple harmonic motion $a = - ω^{2} x$ , we get $ω^{2} = \frac{2 A d g}{M}$ .

The time period of oscillation is $T = \frac{2 π}{ω} = 2 π \frac{M}{2 A d g}$ .

Answer: $2 π \frac{M}{2 Adg}$

Physics - SHM Question with Solution | TestHub

Options:

Doubts & Discussion