Physical World Units and Measurements - Result Question 28
30. P represents radiation pressure, $c$ represents speed of light and $S$ represents radiation energy striking unit area per sec. The non zero integers $x, y, z$ such that $P^{x} S^{y} c^{z}$ is dimensionless are (a) $x=1, y=1, z=1$
(b) $x=-1, y=1, z=1$
(c) $x=1, y=-1, z=1$
(d) $x=1, y=1, z=-1$
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Answer:
Correct Answer: 30. (c)
Solution:
- (c) Let the expression, $\alpha=P^{x} S^{y} c^{z} \quad$..(i) and given that dimension of $\alpha=[M^{0} L^{0} T^{0}] \ldots$ (ii) $=$ dimensionless
Dimension fo radiation pressure $P=\frac{\text{ Force }}{\text{ Area }}$
$ =\frac{[MLT^{-2}]}{[L^{2}]}=[ML^{-1} T^{-2}] $
Dimension of radiation energy/unit area unit time
$S=\frac{\text{ Energy }}{\text{ Area } \times \text{ Time }}=\frac{[ML^{2} T^{-2}]}{[L^{2}][T]}=[MT^{-3}]$
Dimension of speed of light, $c=[LT^{-1}]$
By equation (i) we get,
So, the dimension of $\alpha=[ML^{-1} T^{-2}]^{x}[MT^{-3}]^{y}$
$[LT^{-1}]^{z}$
According to equation (ii),
$\Rightarrow[M^{0} L^{0} T^{0}]=[ML^{-1} T^{-2}]^{x}[MT^{-3}]^{y}[LT^{-1}]^{z}$
$\Rightarrow[M^{0} L^{0} T^{0}]=[M^{x+y} L^{-x+z} T^{-2 x-3 y-z}]$
Applying the principle of homegenity of dimension we get, $x+y=0$ $-x+z=0$ $-2 x-3 y-z=0$
After solving above three equation we get, $x=1 ; y=-1 ; z=1$
Try out the given alternatives. When $x=1, y=-1, z=1$
$ \begin{aligned} P^{x} y _{c^{z}} & =P^{1} S^{-1} c^{1}=\frac{P c}{S} \\ & =\frac{[M L^{-1} T^{-2}][L T^{-1}]}{[M L^{2} T^{-2} / L^{2} T]}=[M^{0} L^{0} T^{0}] \end{aligned} $
