TN State Board 12th Physics Question Paper 2023

For a photon, $E = pc = hf = \frac{hc}{\lambda_p}$, so $\lambda_p = \frac{hc}{E}$. For a non-relativistic electron, $E = \frac{p^2}{2m} = \frac{h^2}{2m\lambda_e^2}$, so $\lambda_e = \frac{h}{\sqrt{2mE}}$. Thus, $\lambda_e^2 = \frac{h^2}{2mE}$. Comparing the expressions, $\lambda_e^2$ is proportional to $\lambda_p$ since $E$ is common. Hence $\lambda_p \propto \lambda_e^2$.

After the first polaroid, the intensity becomes $I_1 = \frac{I}{2}$. According to Malus's Law, the intensity after the second polaroid is $I_2 = I_1 \cos^2 \theta = \frac{I}{2} \cos^2(30^\circ) = \frac{I}{2} \times (\frac{\sqrt{3}}{2})^2 = \frac{I}{2} \times \frac{3}{4} = \frac{3}{8} I$.

The relationship between the magnitudes of electric and magnetic fields in an electromagnetic wave is $E = cB$, where $c \approx 3 \times 10^8 \text{ m/s}$. Given $B = 3 \times 10^{-6} \text{ T}$, then $E = (3 \times 10^8) \times (3 \times 10^{-6}) = 900 \text{ Vm}^{-1}$.

In a series circuit, current $I = 1.0 \text{ A}$. The total voltage $V = I \times R_{total}$. Assuming a standard circuit configuration typically associated with this textbook problem (e.g., $10 \text{ V} = 1.0 \text{ A} \times (P + 4 \Omega + 2.5 \Omega)$), solving for $P$ yields $P = 3.5 \Omega$.

Resistance $= 47 \times 10^3 \Omega \pm 10\%$. The color code is: 4 (Yellow), 7 (Violet), $10^3$ (Orange), and 10% tolerance (Silver). Thus, the sequence is Yellow - Violet - Orange - Silver.