1. **Problem Statement:** Derive the dimensions of the following physical quantities: i) Angular Momentum ii) Specific Heat Capacity 2. **Angular Momentum Dimension Derivation:** Angular momentum $L$ is defined as the product of moment of inertia $I$ and angular velocity $\omega$: $$L = I \times \omega$$ - Moment of inertia $I$ has dimensions of mass times length squared: $$[I] = M L^2$$ - Angular velocity $\omega$ is the rate of change of angle with respect to time, so its dimension is: $$[\omega] = T^{-1}$$ Therefore, the dimension of angular momentum is: $$[L] = [I][\omega] = M L^2 T^{-1}$$ 3. **Specific Heat Capacity Dimension Derivation:** Specific heat capacity $c$ is defined as the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree: $$c = \frac{Q}{m \Delta T}$$ - Heat energy $Q$ has the dimension of energy: $$[Q] = M L^2 T^{-2}$$ - Mass $m$ has dimension: $$[m] = M$$ - Temperature difference $\Delta T$ has dimension: $$[\Delta T] = \Theta$$ Thus, the dimension of specific heat capacity is: $$[c] = \frac{M L^2 T^{-2}}{M \Theta} = L^2 T^{-2} \Theta^{-1}$$ --- 4. **Problem Statement:** Solve the differential equation: $$(x^2 + y^2) \frac{dy}{dx} = xy$$ 5. **Solution Steps:** Rewrite the equation: $$\frac{dy}{dx} = \frac{xy}{x^2 + y^2}$$ Divide numerator and denominator by $x^2$ (assuming $x \neq 0$): $$\frac{dy}{dx} = \frac{y/x}{1 + (y/x)^2}$$ Let $v = \frac{y}{x}$, so $y = vx$. Then: $$\frac{dy}{dx} = v + x \frac{dv}{dx}$$ Substitute into the differential equation: $$v + x \frac{dv}{dx} = \frac{v}{1 + v^2}$$ Rearranged: $$x \frac{dv}{dx} = \frac{v}{1 + v^2} - v = v \left(\frac{1}{1 + v^2} - 1\right) = v \left(\frac{1 - (1 + v^2)}{1 + v^2}\right) = -\frac{v^3}{1 + v^2}$$ Divide both sides by $x$: $$\frac{dv}{dx} = -\frac{v^3}{x(1 + v^2)}$$ Separate variables: $$\frac{1 + v^2}{v^3} dv = -\frac{1}{x} dx$$ Rewrite left side: $$\left(\frac{1}{v^3} + \frac{v^2}{v^3}\right) dv = \left(\frac{1}{v^3} + \frac{1}{v}\right) dv$$ Integrate both sides: $$\int \left( v^{-3} + v^{-1} \right) dv = -\int \frac{1}{x} dx$$ Calculate integrals: $$\int v^{-3} dv = -\frac{1}{2 v^2}, \quad \int v^{-1} dv = \ln |v|$$ So: $$-\frac{1}{2 v^2} + \ln |v| = -\ln |x| + C$$ Rewrite: $$\ln |v| + \ln |x| = \frac{1}{2 v^2} + C'$$ Combine logarithms: $$\ln |v x| = \frac{1}{2 v^2} + C'$$ Recall $v = \frac{y}{x}$, so $v x = y$: $$\ln |y| = \frac{1}{2 (y/x)^2} + C' = \frac{x^2}{2 y^2} + C'$$ This implicit solution relates $x$ and $y$: $$\ln |y| - \frac{x^2}{2 y^2} = C''$$ where $C''$ is an arbitrary constant. **Final answers:** - Dimension of angular momentum: $M L^2 T^{-1}$ - Dimension of specific heat capacity: $L^2 T^{-2} \Theta^{-1}$ - Solution to differential equation: implicit form $$\ln |y| - \frac{x^2}{2 y^2} = C$$