Question

Let P represent radiation pressure, c represent speed of light and l represent radiation energy striking a unit area per second, then p^x×l^y×c^z will be dimensionless for what value of x,y,z

Answer 1

dimension of P = [ML⁻¹T⁻²] [ ∵ pressure = force/area , Force = [MLT⁻²] ]
dimension of I = [ML²T⁻²]/[L²][T] = [MT⁻³] [ ∵ energy = [ML²T⁻²] ]
dimension of c = [LT⁻¹]

Now, P^x × I^y × c^z = dimensionless
[ML⁻¹T⁻²]^x × [MT⁻³]^y × [LT⁻¹]^z = [M⁰L⁰T⁰]
[M]^(x + y) [L]^(-x + z) [T]^(-2x +3y -z) = [M⁰L⁰T⁰]
Compare both sides,
(x + y) = 0
(-x + z) = 0
(-2x + 3y - z) = 0
After solving equations , we get x = 0, y = 0 and z = 0
Hence, x = 0, y = 0, z = 0

Answer 2

"The values of (x,y,z) is (0,0,0).

Solution:

We know that,

Radiation pressure$= \frac{Radiation force}{Area}$

$\Rightarrow p=\frac{\text {Mass} \times \text {Acceleration}}{\text {Area}}$

Thus, the dimensional formula for radiation pressure, $\mathrm{p}=\left[\mathrm{M}^{1} \mathrm{L}^{-1} \mathrm{T}^{-2}\right]$

We know that,

Speed of light $= \frac {Distance travelled by light}{Time taken}$

Thus, dimensional formula for speed, $c=\left[M^{0} L^{1} T^{-1}\right]$

We know that,

Radiation energy striking a unit area per second = $\frac { Mass \times Velocity^2}{2\times Area\times Time}$

Thus, dimensional formula for radiation energy striking a unit area per second, $\mathrm{I}=\left[\mathrm{M}^{1} \mathrm{L}^{0} \mathrm{T}^{-3}\right]$

From the question, we know that,

$p^{x} l^{y} c^{z}=\text { Dimensionless }=\left[M^{0} L^{0} T^{0}\right]$

$\left[M^{1} L^{-1} T^{-2}\right]^{x}\left[M^{1} L^{0} T^{-3}\right]^{y}\left[M^{0} L^{1} T^{-1}\right]^{z}=\left[M^{0} L^{0} T^{0}\right]$

$\left[M^{x+y} L^{-x+z} T^{-2 x-3 y-z}\right]=\left[M^{0} L^{0} T^{0}\right$]"