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

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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.

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\(V_8 \frac{hf - w_o}{e} = \frac{5.0 \times 1.60 \times 10^{-19} - 3.0 \times 1.6 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.0V\)

E = DMC2 = 14,1141 x 1.67 x 10-27 x 9 x 1016 = 2.332 x 10-9J

\(U^{238 - 4}_{92 - 2}\)

Tables of value/observation S/N L 9cm) V (volts) L\(^{-1}\)(cm\(^{-1}\)) V\(^{-1}\)(volts\(^{-1}\)) L\(^{-1}\)(cm\(^{-1}\))x10\(^{-3}\) 1 15.0 0.60 0.067 1.667 67.0 2 25.0 0.80 0.040 1.250 40.0 3 35.0 1.10 0.029 0.909 29.0 4 45.0 1.30 0.022 0.909 22.0 5 55.5 1.60 0.018 0.625 18.0 6 65.0 1.80 0.015 0.556 15.0 (6) The graph of V\(^{-1}\)(volts\(^{-1}\)) vs \(\bigtriangleup\) L\(^{-1}\) (cm\(^{-1}\) Scale: V\(^{-1}\)(volts\(^{-1}\)) axis: 2cm = 0.15units \(\bigtriangleup\) L\(^{-1}\) (cm\(^{-1}\)) axis: 2cm = 7units (7) Slope/gradient = \(\frac{\bigtriangleup {V}^{-1}(volts ^{-1})}{\bigtriangleup {L}^{-1}(cm^{-1})}\) = \(\frac{1.575-0.405}{63-4.2} = \frac{1.17}{58.8}\) = 0.0199cm/volts (8) K = \(\frac{1}{s} = \frac{1}{0.0199}\) = 50.25 volts/cm precautions are: I would ensure that the key is opened when the circuit is not in use to avoid the cell being used up. I would ensure that the electrical wires are tightly connected I would ensure that I take note of the zero error in the pointer instruments to avoid systematic error I would ensure that I take pointer readings within the vertical eye level to avoid parallax I used short connecting wires to ensure that the required energy to dive the electrical charges was not used up in overcoming the resistance of the wire. (b)i. The factors affecting the resistance of a wire are: - length of wire (R&L) - cross-sectional area of the (R \(\propto\)) - Nature of material - Temperature of the wire ii. R = 100cm x 0.02\(\Omega\)cm = 2\(\Omega\), v =1.5v t = 1min = 60s H =p x t = \(\frac{V^{2}t}{R} = \frac{(1.5)^{2}}{2}\) x 60 = 67.5J OR I = \(\frac{V}{R} = \frac{1.5}{2}\) = 0.75A H = P x t = I\(^{2}\)Rt = (0.7)\(^{2}\) x 60 x 2 = \(\frac{9}{16}\)x120 = 67.5J

In general, the velocity of sound in air varies directly as the square root of temperature measured in kelvin. That V \( \propto \sqrt{T} \implies V^2 \propto T.\) \(\text{Therefore} \frac{V^2_1}{T_1} = \frac{V^2_2}{T_2}\) \(\text{Thus Let } V_1\) = 4m/s T\(_1\) = 10K \(\text{Therefore } V_2 = 2V_1\) = 8m/s \(\implies \frac{4^2}{10} = \frac{8^2}{T_2}\) T\(_2 = \frac{64 \times 10}{16}\) = 40K T\(_2 = 4T_1 \) Thus, when the velocity of sound in air is doubled, its absolute temperature will be quadrupled.

Let the masses of the two objects be M1 and M2 and their distance apart = r Therefore the force acting F1 = \( \frac{GM1M2}{r^2}\) for a new distance of r/2, we have a force of F2 = \( \frac{GM2M2}{r/2}\) \(\text{Thus } F1 \times r^2 = GM_1M_1 \text{and}\) F2 \(\times (r/2)^2 = GM_1M_1\) \(\text{Since } GM1M1 = GM_1M_1\) \(\implies F_1 \times r^2 = F2 \times (r/2)^2\) \(\text{Therefore } 200 \times r^2 = F2 \times r^2/4\) F2 = \(\frac{4 \times 200 \times r^2}{r^2} = 800N\)

l =l1 + l2 = 3 + 6 = 9H

For Hooke's law F = Ke \( \implies \frac{F}{e} = K\) \(\frac{f_1}{e_1} = \frac{f_2}{e_2}\) \(\text{Let the original length} = t_0 \) \(\text{therefore}\) e\(_1\) = (36 - t_0)\)cm ; e\(_2 = (46 - t_0)\)cm \(f_1\) = 40N, \(f_2\) = 60N \(\frac{40}{36 - t_0} = \frac{60}{45 - t_0}\) \(\implies 40(45 - t_0) = 60(36 - t_0)\) \(\text{therefore } 1800 - 40t_0\) = 2160 - 60\(t_0\) 60\(t_0\) - 40\(t_0\) = 2160 - 1800 20\(t_0\) = 360 \(t_0\) = 18cm