Question

If P represents radiation pressure, c represents speed of light and Q represents radiation energy striking a unit area per second, then non-zero integers x, y and z, such that ${P^x}{Q^y}{c^z}$ is dimensionless, may be
(A) $x = 1,{\text{ }}y = 1,{\text{ }}z = 1$
(B) $x = 1,{\text{ }}y = - 1,{\text{ }}z = 1$
(C) $x = - 1,{\text{ }}y = 1,{\text{ }}z = 1$
(D) $x = 0,{\text{ }}y = 0,{\text{ }}z = 0$

Answer 1

It is given that the product of P,Q and c are dimensionless so consider a function whose dimensions are zero so that the bases are the same and now by equating the powers find the values of x, y and z.

Complete step-by-step solution
According to the question,
P is the radiation pressure
$P = \dfrac{{Force}}{{Area}}$
Dimensional formula: $\left[ P \right] = \left[ {{M^1}{L^{ - 1}}{T^{ - 2}}} \right]$
Q is the radiation energy
$Q = \dfrac{{Energy}}{{Area \times Time}}$
Dimensional formula: $\left[ Q \right] = \left[ {{M^1}{L^0}{T^{ - 3}}} \right]$
c is the speed of light
Dimensional formula: $\left[ c \right] = \left[ {{M^0}{L^1}{T^{ - 1}}} \right]$
Let T be a function of product of P, Q and c and k is a dimensionless constant
$T = {M^0}{L^0}{T^0} \\\ T = {P^x}{Q^y}{c^z} \\\ {M^0}{L^0}{T^0} = {({M^1}{L^{ - 1}}{T^{ - 2}})^x}{({M^1}{L^0}{T^{ - 3}})^y}{({M^0}{L^1}{T^{ - 1}})^z} \\\ {M^0}{L^0}{T^0} = {M^{x + y}}{L^{ - x + z}}{T^{ - 2x - 3y - z}} \\\$
Now equate the corresponding powers,
$x + y = 0 \\\ \- x + z = 0 \\\ \- 2x - 3y - z = 0 \\\$
On solving these equations simultaneously, we get the values of x, y and z.
$x = 1,{\text{ }}y = - 1,{\text{ }}z = 1$
$\Rightarrow {P^1}{Q^{ - 1}}{c^1}$

Hence the correct option is B.

Note: Dimensional analysis can be used for:
(1) Finding units of physical quantity in a given system.
(2) Finding dimensions of physical constants or coefficients.
(3) Converting physical quantities to other systems.
(4) Checking the correctness of a relation.