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When a negatively charged rod is brought near the cap of a charged gold leaf electroscope which has positive charges, the leaf collapses.

Q = CV = 0.1 x 10 = 1C

A 300Ω resistor is connected to it in parallel Let R\(_t\) = total resistance in parallel \(\frac{1}{R_t}\) = \(\frac{1}{R_1}\) + \(\frac{1}{R_2}\) = \(\frac{1}{200}\) + \(\frac{1}{300}\) R\(_t\) = \(\frac{200 \times 300}{200 + 300}\) = \(\frac{6000}{500}\) = 120 \(\Omega\)

Solenoid method: This involves placing the magnet inside a solenoid (an electromagnet) and gradually reducing the current. It provides a controlled environment for effective demagnetization.

Mass = ZIT kg Mass = 3.3 x 10-7 x 4 x 2 x 60 x 60 kg Mass = 9.5 x 10-3kg

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

\(\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2}\) = \(\frac{1}{5} + \frac{1}{20}\) = \(\frac{4 + 1}{20} = \frac{5}{20}\) r = \(\frac{20}{5}\) = 4\(\Omega\) R = R1 + R2 = 5 + 4 = 9\(\Omega\) but V = IR I = \(\frac{V}{R} = \frac{9}{9}\) = 1A

e = \(\frac{RA}{L} = \frac{10 \times 0.005}{100} = 0.0005\)

\(\frac{ES}{EP} = \frac{Ns}{Np} = \frac{Es}{150} = \frac{40}{400}\) Es = \(\frac{150 \times 40}{400}\) = 15V

\(\frac{Mv^2}{r}\) = qvB r = \(\frac{MV^2}{qVB} = \frac{9.11 \times 10^{31} \times (2.00 \times 10^7)^2}{1.61 \times 10^{-19} \times 2.00 \times 10^7 \times 1.2 \times 10^{-3}}\) = 9.43 x 10\(^{-2}\)

Electricity I = 0.36A \(\frac{x}{cm}\) \(\frac{I}{A}\) \(\frac{X^{-1}}{cm^{-1}}\) 20.0 1.14 0.050 35.0 0.90 0.029 45.0 0.84 0.022 60.0 0.78 0.017 80.0 0.66 0.013 Slope (s) = \(\frac{\bigtriangleup {X}^{-1}}{\bigtriangleup {I}}\) = \(\frac{0.08-0.01}{1.68-0.644} = \frac{0.07}{1.036}\) From the graph, When X\(^{-1}\) = 0 \(I_0\) of \(I_1\) = 0.448A x. Evaluate \(\frac{I_{0}}{I}\) = \(\frac{0.448}{0.36}\) = 1.244 Precaution; The key was opened when readings were not taken. I ensured tight connections of the connecting wire I avoided parallax errors when reading the potentiometer. (bi) Define emf of a battery: The e.m.f of a cell is the potential difference across its terminals when it is in an open circuit. i.e. not Supplying current to an external circuit. ii. A cell x emf is 1.018V. Lengt\(h_1\) on the potentiometer x = 50.0cm Lengt\(h_2\) on the potentiometer, y,= 75.0cm A cell Y emf is? Solution By formula; \(\frac{E_{1}}{E_{2}}\) = \(\frac{L_{2}}{L_{1}}\) \(\frac{1.018}{E_{2}} = \frac{75.0}{50.0}\) 75.\(E_2\) = 1.018 x 50 \(E_2\) = \(\frac{1.018\times 50}{75}\) \(E_2\) = 0.612V. see graph above

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

r = \(\frac{E}{I} - R\) r = \(\frac{1.5}{0.3} - 1\) r = 4\(\Omega\)

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

I = \(\frac{P}{v}\) I = \(\frac{2000}{250}\) I = 8A