Physics 2021 Jamb Examination

Temperature gradient is defined as the change in temperature per unit distance (thickness) across a conductor. It represents how rapidly the temperature changes as you move through the material.
The formula for temperature gradient is:\[ \text{Temperature Gradient} = \frac{\theta_2 - \theta_1}{L} \]Given:
\( \theta_2 = 80 \, ^\circ\text{C} \)
\( \theta_1 = 20 \, ^\circ\text{C} \)
\( L = 0.02 \, \text{m} \)
Calculation:\[ \text{Temperature Gradient} = \frac{80 - 20}{0.02} \] \[ = \frac{60}{0.02} \] \[ = \frac{6000}{2} \] \[ = 3000 \, \text{Km}^{-1} \] In standard scientific notation, this is expressed as \( 3.0 \times 10^3 \, \text{Km}^{-1} \). Note that a change in temperature in degrees Celsius is equivalent to a change in Kelvin, hence the unit Km\(^{-1}\).

This problem is based on the principle of conservation of energy (Method of Mixtures), where the heat lost by the hot copper is equal to the heat gained by the melting ice.
Data provided:
Mass of copper (\( m_c \)) = 2 kg
Initial temperature of copper (boiling water) = 100ºC
Final temperature of copper (contact with ice) = 0ºC
Specific heat capacity of copper (\( c_c \)) = 400 Jkg\(^{-1}\)K\(^{-1}\)
Latent heat of fusion of ice (\( L_f \)) = 3.3 x 10\(^{5}\) Jkg\(^{-1}\)

Heat lost by copper (\( Q \)):\[ Q = m_c c_c \Delta\theta \] \[ Q = 2 \times 400 \times (100 - 0) \] \[ Q = 800 \times 100 = 80,000 \, \text{J} \]Heat gained by ice to melt (\( Q \)):\[ Q = m_i L_f \] \[ 80,000 = m_i \times (3.3 \times 10^5) \] \[ m_i = \frac{80,000}{330,000} \] \[ m_i = \frac{8}{33} \, \text{kg} \]Thus, 8/33 kg of ice will be converted to water.

The standard equation for a progressive wave travelling in the positive x-direction is:\[ y = A \sin(\omega t - kx) \]Where:
\( A \) is the amplitude
\( \omega \) is the angular velocity
\( t \) is time
\( k \) is the wave number (\( 2\pi/\lambda \))
\( x \) is distance

By comparing the given equation \( y = 0.25 \times 10^{-3} \sin(500t - 0.025x) \) with the standard form:\[ \omega = 500 \, \text{rads}^{-1} \]In scientific notation, 500 is written as \( 5.00 \times 10^2 \, \text{rads}^{-1} \). The angular velocity represents the rate of change of the phase of the wave.

The speed of sound in air is directly proportional to the square root of its absolute temperature (\( v \propto \sqrt{T} \)). When sound moves from a cold region to a hot region, its speed increases.
According to the wave equation:\[ v = f \lambda \]The frequency (\( f \)) of a wave is determined by the source and does not change when the wave moves from one medium to another. Therefore, if the velocity (\( v \)) increases as it enters the hot air, the wavelength (\( \lambda \)) must also increase to maintain the equality. Consequently, sound waves 'stretch out' in warmer air.

For a stretched string, the harmonics are whole-number multiples of the fundamental frequency (\( f_0, 2f_0, 3f_0, 4f_0, \dots \)).
Given fundamental frequency \( f_0 = 40 \, \text{Hz} \).
The frequencies produced are:
1. Fundamental (1st Harmonic): 40 Hz
2. 1st Overtone (2nd Harmonic): \( 2 \times 40 = 80 \, \text{Hz} \)
3. 2nd Overtone (3rd Harmonic): \( 3 \times 40 = 120 \, \text{Hz} \)
4. 3rd Overtone (4th Harmonic): \( 4 \times 40 = 160 \, \text{Hz} \)
5. 4th Overtone (5th Harmonic): \( 5 \times 40 = 200 \, \text{Hz} \)

The overtones between 40 Hz and 180 Hz are 80 Hz, 120 Hz, and 160 Hz. There are exactly 3 overtones in this range.

In a plane mirror, the object distance is always equal to the image distance.
Initially, the man is 4m from the mirror. If the mirror moves 1m towards him, his new distance from the mirror becomes:\[ 4\text{m} - 1\text{m} = 3\text{m} \]Since the image is formed at the same distance behind the mirror, the new image distance is also 3m.
The total distance between the man and his new image is the sum of the object distance and the image distance:\[ 3\text{m} + 3\text{m} = 6\text{m} \]

A hollow metal sphere polished on the inside reflects light from its inner surface, which characterizes a concave (converging) mirror.
The radius of curvature (\( R \)) of the mirror is half the diameter of the sphere:\[ R = \frac{20 \, \text{cm}}{2} = 10 \, \text{cm} \]The focal length (\( f \)) of a spherical mirror is half its radius of curvature:\[ f = \frac{R}{2} = \frac{10 \, \text{cm}}{2} = 5 \, \text{cm} \]Thus, it forms a concave mirror with a focal length of 5cm.

The refractive index (\( n \)) of a medium relative to air is given by the ratio of the speed of light in air (\( c \)) to the speed of light in the medium (\( v \)). It is also defined by Snell's Law as the ratio of the sine of the angle of incidence (\( i \)) to the sine of the angle of refraction (\( r \)). \[ n = \frac{c}{v} = \frac{\sin i}{\sin r} \] \[ \frac{3.0 \times 10^8}{2.0 \times 10^8} = \frac{\sin i}{\sin 30^\circ} \] \[ 1.5 = \frac{\sin i}{0.5} \] \[ \sin i = 1.5 \times 0.5 \] \[ \sin i = 0.75 \] The sine of the angle of incidence is 0.75.

A telescope is in 'normal adjustment' when it is set up such that the viewer's eye is completely relaxed (unaccommodated). This state is achieved when the final image is formed at infinity.
In this configuration, the principal focus of the objective lens coincides with the principal focus of the eyepiece lens. Consequently, parallel rays from a distant object enter the objective, form a real image at the common focus, and this image acts as an object for the eyepiece, which then produces a virtual image at infinity. This prevents eye strain during long periods of observation.

Inductive reactance (\( X_L \)) is the opposition that an inductor offers to alternating current. It is measured in Ohms (\(\Omega\)) and depends on both the inductance and the frequency of the supply.
The formula is:\[ X_L = 2\pi f L \]Given:
Inductance (\( L \)) = 2 H
Frequency (\( f \)) = \( \frac{50}{\pi} \, \text{Hz} \)
Calculation:\[ X_L = 2 \times \pi \times \left(\frac{50}{\pi}\right) \times 2 \] Notice that \( \pi \) cancels out: \[ X_L = 2 \times 50 \times 2 \] \[ X_L = 200 \, \Omega \] The reactance of the inductor in this circuit is 200 Ohms.