Question 21
Let the unknown distance for the 15N force be \(x\).
Anticlockwise moment = Clockwise moment\[ 10\text{N} \times 30\text{cm} = 15\text{N} \times x \]\[ 300 = 15x \]\[ x = \frac{300}{15} \]\[ x = 20\text{cm} \]Therefore, the 15N force must be placed 20cm from the pivot on the opposite side to achieve horizontal balance.
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Question 22
The vertical component of initial velocity (\(u_y\)) is given by:\[ u_y = u \sin \theta \]Using the equation of motion \(v = u + at\), where \(v = 0\), \(a = -g\), and \(u = u \sin \theta\):\[ 0 = u \sin \theta - gt \]\[ gt = u \sin \theta \]\[ t = \frac{u \sin \theta}{g} \]Given:
\(u = 300 \text{ ms}^{-1}\)
\(\theta = 30^{\circ}\)
\(g = 10 \text{ ms}^{-2}\)
Calculation:\[ t = \frac{300 \times \sin 30^{\circ}}{10} \]\[ t = \frac{300 \times 0.5}{10} \]\[ t = \frac{150}{10} \]\[ t = 15\text{s} \]
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Question 23
Initial velocity \(u = 20 \text{ ms}^{-1}\)
Final velocity \(v = 0 \text{ ms}^{-1}\) (since it comes to a complete halt)
Deceleration \(a = -2 \text{ ms}^{-2}\) (acceleration is negative for deceleration)
Substituting the values into the formula:\[ 0 = 20 + (-2)t \]\[ 2t = 20 \]\[ t = \frac{20}{2} \]\[ t = 10\text{s} \]It takes 10 seconds for the train to stop.
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Question 24

There is an inverse relationship between frequency and wavelength: waves with the highest energy have the highest frequency and the shortest wavelength. In the spectrum, Gamma rays occupy the high-frequency end, possessing more energy and shorter wavelengths than X-rays, ultraviolet, visible light, infrared, microwaves, and radio waves. Therefore, Gamma rays have the shortest wavelength among the listed options.
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Question 25
Given:
\(\lambda = 10^{-6} \text{ s}^{-1}\)
Calculation:\[ T_{1/2} = \frac{0.693}{10^{-6}} \]\[ T_{1/2} = 0.693 \times 10^6 \]\[ T_{1/2} = 6.93 \times 10^5 \text{s} \]
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Question 26
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Question 27
\(P = 7.6 \times 10^6 \text{ Nm}^{-2}\)
\(V = 10^{-3} \text{ m}^3\)
\(n = 6 \text{ moles}\)
\(R = 8.3 \text{ Jmol}^{-1}\text{K}^{-1}\)
Solving for temperature (T) in Kelvin:\[ T = \frac{PV}{nR} \]\[ T = \frac{(7.6 \times 10^6) \times 10^{-3}}{6 \times 8.3} \]\[ T = \frac{7600}{49.8} \]\[ T \approx 152.61 \text{K} \]Wait, the calculation gives roughly 153K. To convert to Celsius (\(^{\circ}\text{C}\)):\[ heta = T - 273 \]\[ heta = 152.6 - 273 = -120.4^{\circ}\text{C} \]Looking at the options provided, they appear to be in Celsius but positive. Re-evaluating the calculation: if the result 153 was intended to be the final answer in Kelvin, but the options are marked Celsius, there might be a data discrepancy in the question paper. However, strictly following the physics, 153K is the absolute temperature result.
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Question 28
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Question 29
At absolute zero, theoretically, all molecular motion ceases. If the molecules are not moving, they cannot collide with the container walls, and therefore the pressure of the gas drops to zero. This is reflected in Gay-Lussac's Law (\(P \propto T\)), where pressure is directly proportional to the absolute temperature.
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Question 30
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