Solveeit Logo
Login

Question

Question: The velocity of ripples on the water surface depends upon the wavelength \(\lambda \), density of wa...

The velocity of ripples on the water surface depends upon the wavelength λ\lambda , density of water d, and acceleration due to gravity g. Which of the following relations is correct among these quantities?
A.V2∝gλ B.V2∝λg C.V2∝λgd D.V2∝gλd  A.{V^2} \propto g\lambda \\\ B.{V^2} \propto \dfrac{\lambda }{g} \\\ C.{V^2} \propto \dfrac{\lambda }{{gd}} \\\ D.{V^2} \propto g\lambda d \\\

Explanation

Solution

Even though you may apply fundamental concepts to solve this problem, the quickest and the easiest method to solve this problem is using dimensional analysis. Dimensional analysis is based on the principle that the dimensions on the left-hand side of the formula must be equal to the right-hand side of the equals sign.

Complete step-by-step answer:
Step 1: Write down the quantities in the basic dimension form.
There are 4 quantities here – velocity, acceleration due to gravity, wavelength, and density.
We can write all these quantities in forms of basic quantities such as mass M, length L and time T.
Velocity – Measured in meter/second. Dimension = LT−1=M0L1T−1L{T^{ - 1}} = {M^0}{L^1}{T^{ - 1}}
Acceleration due to gravity – Measured in ms−2m{s^{ - 2}}. Dimension = LT−2=M0L1T−2L{T^{ - 2}} = {M^0}{L^1}{T^{ - 2}}
Wavelength – Measured in meters Dimension = L=M0L1T0L = {M^0}{L^1}{T^0}
Density – Measured in kg/m3kg/{m^3} Dimension = ML−3=M1L−3T0M{L^{ - 3}} = {M^1}{L^{ - 3}}{T^0}
Step 2: Write down the main equation.
The main equation will be used to solve the dimensions. Here, the velocity will be expressed in proportional to powers of acceleration, wavelength and density as shown:
V∝gaλbdcV \propto {g^a}{\lambda ^b}{d^c} where a, b and c are exponents.
Step 3: Solve for the exponents a, b, and c.
By substituting for the dimensions of these quantities, we can solve for the exponents a, b, and c.
V∝gaλbdc M0L1T−1∝[M0L1T−2]a[M0L1T0]b[M1L−3T0]c  V \propto {g^a}{\lambda ^b}{d^c} \\\ {M^0}{L^1}{T^{ - 1}} \propto {\left[ {{M^0}{L^1}{T^{ - 2}}} \right]^a}{\left[ {{M^0}{L^1}{T^0}} \right]^b}{\left[ {{M^1}{L^{ - 3}}{T^0}} \right]^c} \\\
We will rearrange the equation in such a way that we can equate the M, L and T terms, we can find the exponents a, b and c
M0L1T−1∝[M0L1T−2]a[M0L1T0]b[M1L−3T0]c M0L1T−1∝McLa+b−3cT−2b  {M^0}{L^1}{T^{ - 1}} \propto {\left[ {{M^0}{L^1}{T^{ - 2}}} \right]^a}{\left[ {{M^0}{L^1}{T^0}} \right]^b}{\left[ {{M^1}{L^{ - 3}}{T^0}} \right]^c} \\\ {M^0}{L^1}{T^{ - 1}} \propto {M^c}{L^{a + b - 3c}}{T^{ - 2b}} \\\
Equating, we have –
c=0 a+b−3c=1 \-2b=−1  c = 0 \\\ a + b - 3c = 1 \\\ \- 2b = - 1 \\\
From the equations,
b=12 c=0 a=1−3c−b=12  b = \dfrac{1}{2} \\\ c = 0 \\\ a = 1 - 3c - b = \dfrac{1}{2} \\\
Step 4: Obtain the expression by substituting exponents a, b, and c
V∝gaλbdc V∝g12λ12d0 V∝gλ V2∝gλ  V \propto {g^a}{\lambda ^b}{d^c} \\\ V \propto {g^{\dfrac{1}{2}}}{\lambda ^{\dfrac{1}{2}}}{d^0} \\\ V \propto \sqrt {g\lambda } \\\ {V^2} \propto g\lambda \\\

Hence, the correct option is Option A.

Note: Students can go wrong and confuse in Step 3 while equating the exponential terms a, b, and c along with the constants. I request you to cross-check this particular step for a minimum of two times before going to the next step because this step is the make-or-break step. A small computation error in this step can cause your entire answer to go wrong, from full marks to negative marking for wrong answer.