Numerical Problems Based on Derivation of Dimensional Formulae for Class 11 Physics

(i) Angle ($\theta$):

(ii) Angular velocity ($\omega$):

(iii) Angular acceleration ($\alpha$):

(iv) Torque ($\tau$):

(v) Angular momentum ($L$):

(vi) Moment of inertia ($I$):

(i) Impulse ($J$):

(ii) Power ($P$):

(iii) Surface energy:

(iv) Coefficient of viscosity ($\eta$):

(v) Bulk modulus ($B$):

(vi) Force constant ($k$):

First, find the dimensional formula for energy per unit volume ($u$):

Next, find the dimensional formula for pressure ($P$):

Since both quantities possess the identical dimensional formula $[\text{ML}^{-1}\text{T}^{-2}]$, it is proven that energy per unit volume is dimensionally equivalent to pressure.

The dimensional formula for angular momentum ($L = mvr$) is:

Using the energy relation $E = h\nu$, we can solve for Planck’s constant $h$:

Substituting the dimensions for Energy ($[\text{ML}^2\text{T}^{-2}]$) and frequency $\nu$ ($[\text{T}^{-1}]$):

Since $[L] = [h]$, angular momentum has the same physical units as Planck’s constant (e.g., $\text{kg m}^2\text{s}^{-1}$ or $\text{J s}$).

Density ($\rho$) is defined as mass per unit volume:

We must express mass ($m$) in terms of Force ($F$), Length ($L$), and Time ($T$). From Newton’s second law, $F = m \times a$:

Substitute this back into the density formula:

Let Force $F \propto \rho^a v^b \nu^c$. Writing the standard dimensions for each:

• $[F] = [\text{MLT}^{-2}]$
• $[\rho] = [\text{ML}^{-3}\text{T}^0]$
• $[v] = [\text{M}^0\text{LT}^{-1}]$
• $[\nu] = [\text{M}^0\text{L}^0\text{T}^{-1}]$

Substitute these into the proportionality:

Equating the powers of M, L, and T:

1. For M: $a = 1$
2. For L: $-3a + b = 1 \implies -3(1) + b = 1 \implies b = 4$
3. For T: $-b – c = -2 \implies -4 – c = -2 \implies c = -2$

Therefore, the dimensional formula for Force is:

Next, for Impulse ($J$), we know $J = \text{Force} \times \text{Time}$. Since Time $t = \frac{1}{\nu} = \nu^{-1}$: