Question

The speed $\left( v \right)$ of ripples on surface of water depends upon the surface tension $\left( \sigma \right)$ density $\left( \rho \right)$ and wavelength $\left( \lambda \right)$ the speed $v$ is proportional to
$\left( A \right){\left( {\dfrac{\sigma }{{\lambda \rho }}} \right)^{\dfrac{1}{2}}}$
$\left( B \right){\left( {\dfrac{\rho }{{\lambda \sigma }}} \right)^{\dfrac{1}{2}}}$
$\left( C \right){\left( {\dfrac{\lambda }{{\sigma \rho }}} \right)^{\dfrac{1}{2}}}$
$\left( D \right){\left( {\rho g\sigma } \right)^{\dfrac{1}{2}}}$

Answer 1

First we need to know the fundamental dimension of physics to solve this problem. As speed is proportional to the tension, density and wavelength. First writes this statement mathematically by assuming some power in it. Then write each of the terms in its fundamental unit equating both side powers so we can get the solution of this problem.

Complete Step By Step Answer:
As per the problem statement:
The speed $\left( v \right)$ of ripples on surface of water depends upon the surface tension $\left( \sigma \right)$ density $\left( \rho \right)$ and wavelength $\left( \lambda \right)$ .
We need to calculate the speed $v$ is proportional to
Statement of the problem:
speed $\left( v \right)$ $\alpha$ surface tension $\left( \sigma \right)$ density $\left( \rho \right)$ wavelength $\left( \lambda \right)$ .
Mathematically writing the give statement we can get,
$v\,\alpha \,{\sigma ^a}{\rho ^b}{\lambda ^c}$
Equation both we get a proportionally constant,
$v\, = k\,{\sigma ^a}{\rho ^b}{\lambda ^c} \ldots \ldots \left( 1 \right)$
Proportionality constant = k
We know,
$v = \left[ {{M^0}{L^1}{T^{ - 1}}} \right]$
$\sigma = \left[ {{M^1}{L^0}{T^{ - 2}}} \right]$
$\rho = \left[ {{M^1}{L^{ - 3}}{T^0}} \right]$
$\lambda = \left[ {{M^0}{L^1}{T^0}} \right]$
Putting all the fundamental values in the equation $\left( 1 \right)$ respectively we will get,
$\left[ {{M^0}{L^1}{T^{ - 1}}} \right] = {\left[ {{M^1}{L^0}{T^{ - 2}}} \right]^a}{\left[ {{M^1}{L^{ - 3}}{T^0}} \right]^b}{\left[ {{M^0}{L^1}{T^0}} \right]^c}$
Simplifying the terms we will get,
$\left[ {{M^0}{L^1}{T^{ - 1}}} \right] = {\left[ {{M^1}{T^{ - 2}}} \right]^a}{\left[ {{M^1}{L^{ - 3}}} \right]^b}{\left[ {{L^1}} \right]^c}$
Taking out the brackets we will get,
$\left[ {{M^0}{L^1}{T^{ - 1}}} \right] = {M^a}{T^{ - 2a}}{M^b}{L^{ - 3b}}{L^c}$
$\left[ {{M^0}{L^1}{T^{ - 1}}} \right] = \left[ {{M^a}^{ + b}{L^{ - 3b + c}}{T^{ - 2a}}} \right]$
Comparing LHS and RHS we can write,
$a + b = 0 \ldots \ldots \left( 2 \right)$
$- 3b + c = 1 \ldots \ldots \left( 3 \right)$
$- 2a = - 1$
Further simplifying the above equation we will get,
$\Rightarrow a = \dfrac{1}{2} \ldots \ldots \left( 4 \right)$
Putting equation $\left( 4 \right)$ in equation $\left( 1 \right)$ we will get,
$a + b = 0$
$\Rightarrow \dfrac{1}{2} + b = 0$
$\Rightarrow b = - \dfrac{1}{2} \ldots \ldots \left( 5 \right)$
Putting equation $\left( 5 \right)$ in equation $\left( 3 \right)$ we will get,
$- 3b + c = 1$
$\Rightarrow - 3\left( {\dfrac{{ - 1}}{2}} \right) + c = 1$
Further solving the equation we will get,
$\Rightarrow c = 1 - \dfrac{3}{2}$
$\Rightarrow c = \dfrac{{ - 1}}{2} \ldots \ldots \left( 6 \right)$
Putting equation $\left( 4 \right),\left( 5 \right)$ and $\left( 6 \right)$ in equation $\left( 1 \right)$ we will get,
$v\, = k\,{\sigma ^a}{\rho ^b}{\lambda ^c}$
$\Rightarrow v\, = k\,{\sigma ^{\dfrac{1}{2}}}{\rho ^{ - \dfrac{1}{2}}}{\lambda ^{ - \dfrac{1}{2}}}$
$\Rightarrow v\, = k\,\dfrac{{{\sigma ^{\dfrac{1}{2}}}}}{{{\rho ^{\dfrac{1}{2}}}{\lambda ^{\dfrac{1}{2}}}}}$
$\Rightarrow v\, = k\,\sqrt {\dfrac{\sigma }{{\rho \lambda }}}$
$v\,\alpha \,\sqrt {\dfrac{\sigma }{{\rho \lambda }}}$
Therefore the correct option is $\left( A \right)$ .

Note :
Before comparing RHS and LHS first see that both the sides are arranged in the same order first mass then length then time then only equate both the LHS and RHS power or else we will make a mistake. Write proper fundamental function before solving and recheck it.