Numerical Problems Based on Deriving Relationship between Physical Quantities for Class 11 Physics

(Note: The original prompt had a typo “$\lambda \propto \frac{\lambda}{mv}$”, corrected here to $\frac{h}{mv}$).

Let the relation be $\lambda \propto m^a v^b h^c$, which can be written as $\lambda = K m^a v^b h^c$, where $K$ is a dimensionless constant.

Writing the dimensions of each quantity:

• $[\lambda] = [\text{L}]$
• $[m] = [\text{M}]$
• $[v] = [\text{LT}^{-1}]$
• $[h] = [\text{ML}^2\text{T}^{-1}]$

Substitute these into the equation:

Applying the Principle of Homogeneity, equate the powers of M, L, and T:

1. For M: $a + c = 0 \implies a = -c$
2. For T: $-b – c = 0 \implies b = -c$
3. For L: $b + 2c = 1$

Substitute $b = -c$ into equation (3):

Since $c = 1$, we get $a = -1$ and $b = -1$.

Substituting the values of $a, b, c$ back into the original relation:

Therefore, $\lambda \propto \frac{h}{mv}$.

Let the force depend on mass $m$, velocity $v$, and radius $r$ as follows: $F = K m^a v^b r^c$.

Writing the dimensions of each physical quantity:

Equating the powers of M, L, and T:

1. For M: $a = 1$
2. For T: $-b = -2 \implies b = 2$
3. For L: $b + c = 1 \implies 2 + c = 1 \implies c = -1$

Substituting $a, b, c$ into the formula:

Since it is given that $K = 1$, the final expression is:

Let $v = K m^a r^b g^c$.

Writing the dimensions of the respective quantities:

Equating the powers of M, L, and T:

1. For M: $a = 0$ (This shows velocity is independent of mass $m$).
2. For T: $-2c = -1 \implies c = \frac{1}{2}$
3. For L: $b + c = 1 \implies b + \frac{1}{2} = 1 \implies b = \frac{1}{2}$

Substituting the values of $a, b,$ and $c$ into the assumed relation:

Let the viscous force $F$ be expressed as $F = K \eta^a r^b v^c$.

Writing dimensions on both sides:

Equating the corresponding powers of M, L, and T:

1. For M: $a = 1$
2. For T: $-a – c = -2 \implies -1 – c = -2 \implies c = 1$
3. For L: $-a + b + c = 1 \implies -1 + b + 1 = 1 \implies b = 1$

Substituting $a=1, b=1, c=1$ back into the equation:

Given $K = 6\pi$, we get Stokes’ Law:

Let the velocity $v$ depend on height $h$ and gravity $g$ as follows: $v = K h^a g^b$.

Writing the dimensional formulas on both sides:

Equating the powers of L and T:

1. For T: $-2b = -1 \implies b = \frac{1}{2}$
2. For L: $a + b = 1 \implies a + \frac{1}{2} = 1 \implies a = \frac{1}{2}$

Substituting $a$ and $b$ back into the original relation:

Hence proved.