Question

A U-tube of uniform cross-section contains mercury upto a height h in either limb. The mercury in one limb is depressed a little and then selected. Obtain an expression for the time period of oscillation assuming that $T$depends on $h,\rho ,g$.

Answer 1

As we know that mercury is dipped into the tube, it will show simple harmonic motion. Time period of that harmonic motion, we need to calculate. By using the dimensional analysis, we will try to find.

Complete step by step answer:
According to the dimensional analysis-
$T \propto {h^a}{\rho ^b}{g^c}$
Here T is the time period, $h,\rho ,g$are the height, density and gravity respectively.
If we remove the proportionality sign, one constant will be there, that is
$T = k{h^a}{\rho ^b}{g^c}$ --- (1)
Now, write the dimensional formula of L.H.S and R.H.S, we get-
$\left[ T \right] = {\left[ L \right]^a}{\left[ {M{L^{ - 3}}} \right]^b}{\left[ {L{T^{ - 2}}} \right]^c}$
$\left[ {{M^0}{L^0}{T^1}} \right] = {\left[ L \right]^{a - 3b + c}} + {\left[ M \right]^b} + {\left[ T \right]^{ - 3b - 2c}}$
Now compare the coefficients of this equation, we get-

$$a - 3b + c = 0 \\\ \Rightarrow b = 0 \\\ \Rightarrow - 3b - 2c = 1 \\\$$

These three equations, we got by comparison of the coefficients of M, L and T.
Now,
$b = 0,a = \dfrac{1}{2},c = - \dfrac{1}{2}$
Substitute all these values in equation, we get-
$T = k\dfrac{{\sqrt h }}{{\sqrt g }}$
$\therefore T = k\sqrt {\dfrac{h}{g}}$

So, this will be the expression for the time period of the oscillation which is depending on h and g.

Note: Basically, Dimensions of any physical quantity are the powers to which the fundamental units are raised to obtain the unit of that quantity. By using the dimensional analysis, we can easily determine the expression for the quantities. Also we can use dimensional analysis for the conversion of units, to check whether the equation is correct or not.