Question Library

To calculate the length of the wire, we use the formula for resistance: \(R = \frac{\rho \cdot L}{A}\) Where: - \( R \) is the resistance (21 \Omega), - \( \rho \) is the resistivity (\( 1.1 \times 10^{-8} \, \Omega \, m \)), - \( L \) is the length of the wire, - \( A \) is the cross-sectional area (\( 2\pi \times 10^{-8} \, m^2 \)). Rearranging the formula to solve for length \( L \): \(L = \frac{R \cdot A}{\rho}\) Substituting the known values, we first calculate the area \( A \): \(A = 2\pi \times 10^{-8} \, m^2 = 2 \times \frac{22}{7} \times 10^{-8} \approx 8.0 \times 10^{-8} \, m^2\) Now substitute into the length formula: \(L = \frac{21 \, \Omega \cdot 8.0 \times 10^{-8} \, m^2}{1.1 \times 10^{-8} \, \Omega \, m}\) Calculating: \(L = \frac{21 \cdot 8.0}{1.1}\)m \(L = \frac{168}{1.1} \, m \approx 120\)m

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option A.

(a)G) Root-mean-square value of an alternating Current 1s the value of a.c. current that has the same heating effect as d.c. current. (ii) Impedance is the effective resistance of an electric circuit arising from the combined effects of ohmic resistance, capacitance and inductive reactance. (b) Current in circuit I = 60/120 = 0.5A Voltage across device V\(_R\) = 120v Voltage across capacitor = V\(_C\) = \sqrt[V\(^2\) - V\(_R\)\(^2\)] V\(_C\) = \sqrt[240\(^2\) - 120^\circ] V\(_C\) = 207.847v I\(_C\) = I = 0.5A X\(_C\) = V\(_C\) / I\(_C\) = \(\frac{207.846}{0.5) → 415.7\Omega Capicitance, C = \(\frac{X_c}{2\pif}\) = \(\frac{415.7}{2 * 3.142 * 50}\) = 1.323F The capacitance C of a capacitor is the ratio of the amount of charge q on its plates to the potential difference between them, i.e C = q/v (ii) I - charge on both capacitors are the same q1 = q2 Total capacitance, C = \(\frac{C_1 C_2}{C_1 + C_2}\) C = \(\frac{C_2}{1 + C_1/C_2}\) = \(\frac{C_2}{1 + V_1 + C_2}\) C = \(\frac{C_2 V_2}{V_1 + C_2}\) = \(\frac{C_2 V_2}{2}\) C = 1/2 = \(C_2 V_2\) II - voltage across V\(_1\) = \(\frac{C_2}{C_1 + C_2}\) * V V\(_1\) = \(\frac{1}{C_1_{C_2} + 1}\) * V but \(\frac{C_1}{C_2}\) = \(\frac{C_1}{C_2}\) = \(\frac{5}{2}\) V\(_1\) = \(\frac{1}{5}{2} + 1}\) * 2 = \(\frac{2}{7}\) * 2 = \(\frac{4}{7}\) V Voltage across C\(_2\) : V\(_2\) = 2 = \(\frac{4}{7}\) = \(\frac{10}{7}\) V

(a) Functions of parts labeled: A - Produces the alternating current (a.c.) B- Smoothens the rectified current. (b) Output signal: see the diagram above.

SEE GRAPH ABOVE (vii) Slope = \(\frac{\Deltad}{\DeltaI^{-1}}\) = \(\frac{d_2 - d_1}{I_2 - I^{-1}\) → \(\frac{80 - 30}{3.333 - 15}\} = \(\frac{50}{1.833}\) Slope = 27.2cm / A Precaution: We should avoid parallax error by placing both eyes at the measured level. We should always switch the fan off to ensure no external environment interferes with the result. Diameter d = 0.6cm = 0.006m : r = \(\frac{0.006}{2}\) = 0.003m Resistivity = 1.0 x 10\(^{-6}\) R = 4\Omega R = \(\frac{ρL}{A}|) → L = \(\frac{RA}{ρ}\) L = \(\frac{ 4 x 22 x 0.0000009}{7 x 1.0 x 10^{-6}}\) L = 6.289 x 2 x 9 x 10\(^{-6}\) x 10\(^{-6}\) L = 113.14 x 10\(^{-6 + 6}\) L = 113.14

No expanded working has been attached for this question yet. The verified answer is Option B.

When K is open. Total Resistance = 8 + 2 = 10\(\Omega\) 1 = \(\frac{V}{\Omega}\) = \(\frac{6v}{\Omega}\) = 0.6A Where K is closed, \(\frac{1}{8} + \frac{1}{8} = \frac{1}{\Omega}\) \(\frac{2}{8} = \frac{1}{\Omega}\) \(\Omega\) = \(\frac{8}{2}\) = 4\(\Omega\) \(\Omega\) = \(\frac{8}{2}\) = 4\(\Omega\) 4\(\Omega\) in series with 2\(\Omega\)= 6\(\Omega\) 1 = \(\frac{V}{\Omega}\) = \(\frac{6}{6}\) = 1A 1 - 0.6A = 0.4A increase

For transformation, Es = e.m.f. of secondary coil Ep = Np Np = e.m.f. of secondary coil Ns = number of turns in secondary coil Ns = number of turns in primary coil \(\frac{30}{300}\) = \(\frac{x}{100}\) = 300 X = 3000 X = 10 volts

No expanded working has been attached for this question yet. The verified answer is Option D.

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option B.

As the distance between the plates increases, the ability of the capacitor to store charge (capacitance) decreases. Since the battery is disconnected, the charge remains constant, but the voltage across the plates increases.

No expanded working has been attached for this question yet. The verified answer is Option A.

No expanded working has been attached for this question yet. The verified answer is Option A.