Question

The velocity of the water wave (v) may depend on their wavelength $’\lambda’$, the density of water ‘$\rho$’, and acceleration due to gravity ‘$g$’. The method of dimensions gives the relation ${v^2} \propto \lambda g\rho$ between these quantities as
(A) ${v^2} \propto {\lambda g \rho}$
(B) ${v^2} \propto {\lambda ^{ - 1}}{g^{ - 1}}{\rho ^{ - 1}}$
(C) ${v^2} \propto g\lambda$
(D) ${v^2} \propto \lambda \rho$

Answer 1

To solve the questions we are using the following dimensional formulas, as shown below:
(i) $v = \left[ {L{T^{ - 1}}} \right]$
(ii) $\lambda = \left[ L \right]$
(iii) $\rho = \left[ {M{L^{ - 3}}} \right]$
(iv) $g = \left[ {L{T^{ - 2}}} \right]$

Complete answer:
Let us write the relation between the quantities below.
$v \propto {\lambda ^x}{\rho ^y}{g^z}$ (1)
Here, x,y, and z correspond to the powers of the $\lambda, \rho$ and $g$ respectively.
Let us use the dimensional formula for all the quantities. we get,
$\left[ {L{T^{ - 1}}} \right] \propto {\left[ L \right]^x}{\left[ {M{L^{ - 3}}} \right]^y}{\left[ {L{T^{ - 2}}} \right]^z}$
Let us simplify to find out the value of $x,y$ and $z$, we get,
$\Rightarrow \left[ {L{T^{ - 1}}} \right] \propto \left[ {{M^y}{L^{x - 3y + z}}{T^{ - 2z}}} \right]$
To simplify further, let us now equate the power of the same quantities. On equating the values of powers we get, as follows,
For mass(M):$y = 0$ (2)
Since there is no term of mass in the LHS thus the power corresponds to the mass thus is equal to the $y=0$
For length(L): $1=x-3y+z$ (3)
Since in LHS, the power of $L$ is equal to 1 and in RHS the power of $L$ is equal to the $x-3y+z$.
For time(T):$- 1 = - 2z \Rightarrow z = \dfrac{1}{2}$ (4)
Since in LHS the power of $T$ is equal to $-1$ and in RHS the power of $T$ is equal to the $-2z$.
From (2),(3), and (4) we get the values of $x,y$ and $z$.
Let us use the values of x and y from (2) and (4) in (3).
$1 = x - 3 \times 0 + \dfrac{1}{2} \Rightarrow x = \dfrac{1}{2}$
Now, let us substitute the values of x, y and z in equation (1), we get,
${v^2} \propto {\lambda ^{x}}{\rho ^{z}}{g^{y}} \propto {\lambda ^{\dfrac{1}{2}}}{\rho ^{0}}{g^{\dfrac{1}{2}}}$
As the above relation is not available in the given options
Hence, Let us square the equation.
On squaring the equation we get,
${v^2} \propto \lambda g$

$\therefore$ Option (C) that is ${v^2} \propto \lambda g$ is the correct option.

Additional information:
The dimensional formula is the representation of quantities in terms of mass, length, time, current, amount of substance, temperature, and luminous intensity.
Dimensional formulas are used to convert units of one system into units of other systems, to check the correctness of an equation, and to establish a relationship between various physical quantities.
By the dimensional method, the value of any dimensionless constant involved in the formula cannot be obtained.

Note:
If a physical quantity depends on more than the factors then using this method relation cannot be established.
The equations containing trigonometric terms, exponentials terms, and terms like a log cannot be analyzed.
If we know the factors on which a given physical quantity may depend then using dimensions we can find the formula relating the quantity with those factors.
If we know the dimensional formula of a physical quantity, we can find its unit.