Question

The velocity of sound in air$\left( V \right)$, pressure$(P)$ and density of air$\left( d \right)$ are related as $V \propto {p^x}{d^y}$. The values of $x$ and $y$ respectively are.
A) $1,\dfrac{1}{2}$
B) $- \dfrac{1}{2}, - \dfrac{1}{2}$.
C) $\dfrac{1}{2},\dfrac{1}{2}$.
D) $\dfrac{1}{2}, - \dfrac{1}{2}$.

Answer 1

The dimensional analysis can help us in solving this problem and finding the correct option to this problem, the dimensional analysis tells us the relationship between different physical quantities. We can replace the dimensional formula of volume, pressure and density into the given relation and on comparing we can find out the value of x and y.

Step by step solution:
Step 1.
The dimensional formula for velocity is $V = {M^0}L{T^{ - 1}}$ the dimensional formula for pressure is $P = M{L^{ - 1}}{T^{ - 2}}$ and dimensional formula for density is$d = M{L^{ - 3}}$.
Step 2.
The given relation $V \propto {p^x}{d^y}$ can be rewritten as $V = k \cdot {p^x}{d^y}$ where $p$ is pressure $d$ is density and $k$ is constant.
Step 3.
Using the new relation$V = k \cdot {p^x}{d^y}$. Put the dimensional formula for each of the physical quantities in the new relation.
$V = k \cdot {p^x}{d^y}$
As $k$ is constant therefore and does not have any dimension therefore we can drop it.
The modified relation for the calculation would be$V = {p^x}{d^y}$. Let us put the dimensional formula for each of the physical quantities.
$\Rightarrow V = {p^x}{d^y}$
$\Rightarrow {M^0}L{T^{ - 1}} = {\left( {M{L^{ - 1}}{T^{ - 2}}} \right)^x} \cdot {\left( {M{L^{ - 3}}} \right)^y}$
$\Rightarrow {M^0}L{T^{ - 1}} = {\left( M \right)^{x + y}} \cdot {\left( L \right)^{ - x - 3y}} \cdot {\left( T \right)^{ - 2x}}$………eq.(1)
Step 4.
Comparing the powers in equation (1) we get
$x + y = 0$, $- x - 3y = 1$ and $- 2x = - 1$.
We have got three equations that let us find out the values for $x$ and$y$.
$\Rightarrow - 2x = - 1$
$\Rightarrow x = \dfrac{1}{2}$
So the value of $x$ is $x = \dfrac{1}{2}$.
Now put $x = \dfrac{1}{2}$ in $- x - 3y = 1$
$\Rightarrow - x - 3y = 1$
$\Rightarrow x + 3y = - 1$
$\Rightarrow \dfrac{1}{2} + 3y = - 1$
$\Rightarrow 3y = - 1 - \dfrac{1}{2}$
$\Rightarrow 3y = - \dfrac{3}{2}$
$\Rightarrow y = - \dfrac{1}{2}$
So the value of $y$ is$y = - \dfrac{1}{2}$.
Hence the value of $x$ is $x = \dfrac{1}{2}$ and the value of $y$ is$y = - \dfrac{1}{2}$.

So the correct option for this problem is option D.

Note: Students should remember the dimensional formula for most of the physical quantities as sometimes it is required while solving the problem. It is advised to observe the units of any physical quantity in case you cannot remember the dimensional formula of any physical quantity.