Physics 2023 Jamb Examination

The relationship between the primary and secondary voltages and the number of turns in a transformer is governed by the transformer equation:\[\frac{V_p}{V_s} = \frac{N_p}{N_s}\]Given:
Primary voltage \(V_p = 2.2\,\text{kV} = 2200\,\text{V}\)
Secondary voltage \(V_s = 110\,\text{V}\)
Secondary turns \(N_s = 25\)

Substituting the values:\[\frac{2200}{110} = \frac{N_p}{25}\]\[20 = \frac{N_p}{25}\]\[N_p = 20 \times 25 = 500\]The primary winding has 500 turns.

$W = mg = 0.5 \times 10 = 5\,\text{N}$
Since it's light, neglect the weight of the metre rule.
The effective tension $T$ acting in the vertical direction $= T \sin 30^{\circ}$
From the second condition of equilibrium, sum of clockwise moments equals sum of anticlockwise moments.
Taking moment at $E$:\[ \Rightarrow T \sin 30^{\circ} \times 30 = 5 \times 50 \]\[ \Rightarrow T(0.5) \times 30 = 250 \]\[ \Rightarrow 15T = 250 \]\[ \therefore T = \frac{250}{15} = 16.67\,\text{N} \]

First, calculate the resistance of the series combination:\[R_{series} = 35\,\text{k}\Omega + 40\,\text{k}\Omega = 75\,\text{k}\Omega\]Next, we find the resistance \(R\) to be connected in parallel with this \(75\,\text{k}\Omega\) to get a total equivalent resistance (\(R_{eq}\)) of \(25\,\text{k}\Omega\):\[\frac{1}{R_{eq}} = \frac{1}{R_{series}} + \frac{1}{R}\]\[\frac{1}{25} = \frac{1}{75} + \frac{1}{R}\]\[\frac{1}{R} = \frac{1}{25} - \frac{1}{75}\]\[\frac{1}{R} = \frac{3 - 1}{75} = \frac{2}{75}\]\[R = \frac{75}{2} = 37.5\,\text{k}\Omega\]

First, find the mass per unit length (\(\mu\)) and the velocity (\(v\)) of the wave on the string:
Length \(L = 0.5\,\text{m}\), Mass \(m = 0.01\,\text{kg}\)\[\mu = \frac{m}{L} = \frac{0.01}{0.5} = 0.02\,\text{kg/m}\]\[v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{800}{0.02}} = \sqrt{40000} = 200\,\text{ms}^{-1}\]The fundamental frequency (\(f_1\)) is:\[f_1 = \frac{v}{2L} = \frac{200}{2 \times 0.5} = 200\,\text{Hz}\]A piano wire (fixed at both ends) has harmonics \(f_n = n \cdot f_1\). The third overtone is the 4th harmonic (\(n=4\)):\[f_4 = 4 \times 200 = 800\,\text{Hz}\]

Using the mirror formula:\[\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\]For a convex mirror, the focal length is negative. So, \(f = -15\,\text{cm}\) and \(u = 35\,\text{cm}\).\[\frac{1}{-15} = \frac{1}{35} + \frac{1}{v}\]\[\frac{1}{v} = -\frac{1}{15} - \frac{1}{35}\]Finding a common denominator (105):\[\frac{1}{v} = \frac{-7 - 3}{105} = -\frac{10}{105}\]\[v = -\frac{105}{10} = -10.5\,\text{cm}\]The negative sign indicates the image is virtual and located 10.5 cm behind the mirror.

1. Find the mass of water that fills the bottle:\[\text{Mass of water} = 66\,\text{g} - 19\,\text{g} = 47\,\text{g}\]2. Use the definition of relative density (R.D.):\[\text{R.D.} = \frac{\text{Mass of substance}}{\text{Mass of equal volume of water}}\]\[0.8 = \frac{\text{Mass of alcohol}}{47}\]\[\text{Mass of alcohol} = 0.8 \times 47 = 37.6\,\text{g}\]3. Calculate total mass of bottle filled with alcohol:\[\text{Total mass} = \text{Mass of empty bottle} + \text{Mass of alcohol}\]\[\text{Total mass} = 19\,\text{g} + 37.6\,\text{g} = 56.6\,\text{g}\]

The work done against gravity is equal to the increase in gravitational potential energy of the object:\[W = mgh\][Image showing work done lifting an object against gravity]
Given:
\(m = 3.0\,\text{kg}\)
\(g = 10\,\text{ms}^{-2}\) (approximate value standard for JAMB unless 9.8 is specified)
\(h = 240\,\text{m}\)
\[W = 3.0 \times 10 \times 240 = 7200\,\text{J}\]To convert to kilojoules (kJ):\[W = \frac{7200}{1000} = 7.2\,\text{kJ}\]

According to Snell's Law:\[n_1 \sin \theta_1 = n_2 \sin \theta_2\]Where:
\(n_1\) = refractive index of water = 1.33
\(\theta_1\) = angle of incidence in water = 40°
\(n_2\) = refractive index of glass = 1.5
\(\theta_2\) = angle of refraction in glass
\[1.33 \times \sin(40^{\circ}) = 1.5 \times \sin \theta_2\]\[1.33 \times 0.6428 = 1.5 \times \sin \theta_2\]\[0.8549 = 1.5 \times \sin \theta_2\]\[\sin \theta_2 = \frac{0.8549}{1.5} = 0.5699\]\[\theta_2 = \sin^{-1}(0.5699) \approx 34.75^{\circ}\]

Absolute pressure (\(P_{abs}\)) is the sum of atmospheric pressure (\(P_{atm}\)) and gauge pressure (\(P_g\)). In an open-tube manometer, if the mercury is higher in the open tube, the gas pressure is greater than atmospheric pressure.
\[P_{abs} = P_{atm} + h\rho g\]Given:
\(P_{atm} = 101,325\,\text{Pa}\)
\(h = 25\,\text{cm} = 0.25\,\text{m}\)
\(\rho = 13600\,\text{kg/m}^3\)
\(g = 9.8\,\text{m/s}^2\)
\[P_g = 0.25 \times 13600 \times 9.8 = 33,320\,\text{Pa}\]\[P_{abs} = 101,325 + 33,320 = 134,645\,\text{Pa}\]

Sensitivity in thermometry refers to the ability of the thermometer to detect small changes in temperature. A highly sensitive thermometer will show a significant change in its thermometric property (like the expansion of mercury or change in resistance) even for a very small change in temperature. Option B describes 'responsiveness' (how fast it reacts), and Option C describes the 'range' of the thermometer. Therefore, the most accurate definition of sensitivity is the smallest temperature change that can be measured.