Physics 2012 Jamb Examination

To eliminate parallax error when reading a metre rule, the observer’s eye must be directly above the measurement mark. Viewing from an angle causes displacement between the object and its image, leading to incorrect readings. Therefore, the correct method is to position the eye vertically downwards on the markings, ensuring alignment with the scale and avoiding any apparent shift due to angular viewing.

Since the load moves at uniform speed, the horizontal component of the tension equals the frictional force.\[ F_{f} = T \cos \theta \]\[ F_{f} = 1500 \cos 45^{\circ} \]\[ F_{f} = 1500 \times \dfrac{1}{\sqrt{2}} = 1060.66 \; \text{N} \]\( \therefore F_{f} = 1061 \; \text{N (approximately)} \). The frictional force opposes motion and balances the horizontal pull component at constant speed.

Using the kinematic relation:\[ v^{2} = u^{2} + 2as \]where \(v = 0\) (train stops), \(u = 30\, \text{m/s}\), and \(a = -0.1\, \text{m/s}^2\).\[ 0 = (30)^2 - 2(0.1)s \]\[ s = \dfrac{900}{0.2} = 4500\, \text{m} \]\( \therefore \text{The total distance covered is } 4500\, \text{m.} \)

From the equation of motion:\[ s = ut + \dfrac{1}{2} a t^{2} \]\(u = 0\), \(a = 30\, \text{m/s}^2\), and \(t = 20\, \text{s}\).\[ s = 0 + \dfrac{1}{2} (30)(20)^2 = 15 \times 400 = 6000\, \text{m} = 6\, \text{km} \]\( \therefore \text{The car covers } 6\, \text{km.} \)

According to Newton’s law of universal gravitation, the force between two masses is given by\[ F = \dfrac{Gm_{1}m_{2}}{r^{2}} \]\(F\) is inversely proportional to the square of the distance \(r\). As the rocket moves away from the Earth, \(r\) increases, causing \(F\) to decrease. Conversely, if \(r\) reduces, \(F\) increases. Hence, the force increases as \(r\) reduces.

For a freely suspended object acting as a simple pendulum, a restoring force acts towards the equilibrium position. The acceleration is directly proportional to the displacement from the equilibrium and directed towards it, satisfying the condition for simple harmonic motion. Hence, the system oscillates about its equilibrium position due to this restoring acceleration.

The relationship between linear velocity \(v\), angular velocity \(\omega\), and radius \(r\) is given by\[ v = \omega r \]\[ \omega = \dfrac{v}{r} = \dfrac{1}{0.5} = 2\, \text{rad/s} \]\( \therefore \text{The angular velocity is } 2\, \text{rad/s.} \)

At the point of sliding, the component of weight down the plane equals the limiting friction.\[ mg \sin \theta = \mu mg \cos \theta \]\[ \mu = \tan \theta = \tan 30^{\circ} = \dfrac{1}{\sqrt{3}} = 0.577 \approx 0.6 \]\( \therefore \text{The coefficient of static friction is } 0.6. \)

Useful work output is given by:\[ W_{out} = \text{Load} \times \text{Distance} = 250 \times 30 = 7500\, \text{J} \]\(W_{in} = W_{out} + \text{Work against friction} = 7500 + 1500 = 9000\, \text{J}\)\[ \text{Efficiency} = \dfrac{W_{out}}{W_{in}} \times 100 = \dfrac{7500}{9000} \times 100 = 83.3\% \]\( \therefore \text{The efficiency of the system is } 83.3\%. \)

From Hooke’s law, \(F = kx\). Here, \(F = mg = 1 \times 10 = 10\, \text{N}\), and \(x = 1.2\, \text{cm} = 0.012\, \text{m}\).\[ k = \dfrac{F}{x} = \dfrac{10}{0.012} = 833.33\, \text{N/m} \]\( \therefore \text{The force constant is } 833\, \text{N/m.} \)