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Table of values/Observation S/N m(g) x(cm) (cm) 1 50.00 49.75 0.0202 2 70.00 39.65 0.0252 3 90.00 35.10 0.0285 4 110.00 30.75 0.0325 5 130.00 27.40 0.0365 6 150.00 25.00 0.0400 Slope (s) = \(\frac{\bigtriangleup {m(g)}}{\bigtriangleup {\frac {1}{x}(cm^{-1})}}\) = \(\frac{130-10}{0.04-0.0145}\) \(\frac{140}{0.0255}\) = 4705.88(g cm) Deduction: When \(\frac{1}{x}\) = 0, m = \(\frac{1}{x}\)66.00g Precaution: - l avoided draught by switching off the fans. - I ensure that the suspended mass did not touch the table. - I avoided error due to parallax when reading the metre rule - Zero error of the metre rule was noted. (b)i. From the graph, when m = 0, \(\frac{1}{x}\) = 0.0125 \(\therefore\) \(\frac{1}{x} = \frac{1}{0.0125}\) = 80 cm (1) Sum of upward forces = sum of downward forces. (2) Algebraic sum of all the moments about the pivot is Zero. ii. when m = 100g. X = 0.0305 \(\therefore\) \(\frac{1}{x} = \frac{1}{0.0305}\) = 32.97cm

Table of values/observation S/N R\(\Omega\) L(cm) R\(^{-1}\)(\(\Omega^{-1}\)) L\(^{-1}\)(cm\(^{-1}\)) 1 30.00 88.80 0.033 0.01126 2 20.00 84.60 0.050 0.01182 3 10.00 79.80 0.100 0.01253 4 5.00 71.50 0.200 0.01399 5 3.00 69.50 0.333 0.01439 6 1.00 42.40 1.000 0.02358 Slope (s) = \(\frac{\bigtriangleup {R^{-1}(\Omega)}}{\bigtriangleup {L^{-1}(cm^{-1})}} = \frac{1.08-(-0.96)}{0.002175-0.00175}\) = \(\frac{1.08-(-0.96)}{0.002175-0.00175} = \frac{2.04}{0.00425}\) = 4800cm\(\Omega ^{-1}\) I = -1.14(\(\Omega ^{-1}\) (i) K I\(^{-1}\) \(\frac{1}{I} = \frac{-1}{1.14}\) = 0.877\(\Omega\) (ii) C = \(\frac{L_{o}}{S} = \frac{78.50cm}{4800cm \Omega ^{-1}}\) = 0.01635\(\Omega\) Precautions: - The key was opened when readings were not taken. - I ensured tight connections. - I avoided parallax error when reading potentiometre. (b)i. When R = 15\(\Omega\), R\(^{-1}\) = \(\frac{1}{R} = \frac{1}{15}\) = 0.0666\(\Omega ^{-1}\) From the graph, L\(^{-1}\) = 0.001175, \(\frac{1}{L}\) 0.01175, L = \(\frac{1}{0.01175}\) = 85.11cm I = 0.5+y\(^{-1}\), -1.14 = 0.5 + y\(^{-1}\), y\(^{-1}\) = 1.14 - 0.5 = -0.64, \(\frac{1}{y}\) = -0.64 \(\therefore\) y = \(\frac{-1}{0.64}\) = -1.6 (iii) e.m.f. is the p.d between the terminals of a cell when an open circuit or the work done in moving unit charge round a complete circuit.

V=rw, 8=2w, w=8/2= 4rad/s

\(v^2 = u^2 +2aS = 0 +2\times(25-5)\) \(v^2 =0+ 20\times 20 = 400\); v =\(\sqrt{400}= 20(ms^{-2}\)

M =0.1kg; F=\(m\left(\frac{v-u}{t}\right) = 0.1\left(\frac{15-(-10)}{0.01}\right)\) F=250N

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

To find the weight of the missile at an altitude using the inverse-square law of gravitation, we can use the formula for gravitational force: \(F = \frac{G \cdot m_1 \cdot m_2}{r^2}\) where: - \( F \) is the gravitational force (weight), - \( G \) is the gravitational constant, - \( m_1 \) is the mass of the Earth, - \( m_2 \) is the mass of the missile, - \( r \) is the distance from the center of the Earth to the missile. Step 1: Determine the radius at altitude 1. The radius of the Earth is \( R = 6.4 \times 10^6 \, \text{m} \). 2. The altitude \( h = 6.4 \times 10^6 \, \text{m} \). Thus, the distance from the center of the Earth to the missile at that altitude is: \(r = R + h = 6.4 \times 10^6 \, \text{m} + 6.4 \times 10^6 \, \text{m} = 12.8 \times 10^6 \, \text{m}\) Step 2: Calculate the weight at the altitude The weight of the missile at the surface is given as \( W_0 = 400 \, \text{N} \). The weight at altitude can be calculated using the ratio of the squares of the distances: \(W = W_0 \cdot \left( \frac{R}{r} \right)^2\) Substituting the values: \(W = 400 \cdot \left( \frac{6.4 \times 10^6}{12.8 \times 10^6} \right)^2\) Calculating this: \(W = 400 \cdot \left( \frac{1}{2} \right)^2 = 400 \cdot \frac{1}{4} = 100 \, \text{N}\) The weight of the missile at an altitude of \( 6.4 \times 10^6 \, \text{m} \) is \( 100 \, \text{N} \).

Upthrust = Volume of fluid displaced =vpg =0.2 x 0.2 x 0.2 x0.8 x 10= 0.064N

H=\(mc\Delta\theta \Rightarrow 60=0.03\times400\times\Delta\theta\\60= 12\Delta\theta therefore \Delta\theta=60/12 =5K\)

\(v_2=\frac{v_1p_1}{p_2}=\frac{20\times700}{750}=18.66=18.7cm^3\)

V=f\(\lambda, f =\frac{v}{\lambda}= \frac{3\times10^8}{150}\Rightarrow 2\times10^6\)Hz

\(\frac{I_s}{I_p}=\frac{n_p}{n_s}, I_s= \frac{I_pn_p}{n_s}=\frac{2.5\times50}{250}\Rightarrow0.5A\)

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 B.

Given: Peak current, \( I_{\text{peak}} = 5 \, \text{A} \), Resistance, \( R = 10 \, \Omega \) Calculate the rms current: The rms current is given by: \(I_{\text{rms}} = \frac{I_{\text{peak}}}{\sqrt{2}} = \frac{5}{\sqrt{2}} \approx 3.54 \, \text{A}\) Calculate the power dissipated in the resistor: Using the rms value, the power can be calculated as: \(P = I_{\text{rms}}^2 \cdot R\) Substituting the values: \(P = \left(\frac{5}{\sqrt{2}}\right)^2 \cdot 10\) Calculating: \(P = \frac{25}{2} \cdot 10 = 125 \, \text{W}\).