Physics 2024 Jamb Examination

Weight is proportional to gravitational acceleration. Since the moon’s gravity is one-sixth of the earth’s, the weight of the bowling ball on the moon is also one-sixth of its earth weight.

Inserting a dielectric increases the permittivity between the plates, which increases the capacitance of the capacitor.

Using the heat conduction formula, the thermal conductivity is calculated as approximately \( 4.9 \times 10^{-2} \\text{Wm}^{-1}\\text{K}^{-1} \).

Frequency of a stretched string is proportional to the square root of the tension. Quadrupling the tension doubles the frequency, giving 100Hz.

Model rockets consist of a nose cone, body tube, fins, and recovery devices. Hence, recovery devices are part of a model rocket.

Efficiency (\( \eta \)) is the ratio of Mechanical Advantage (M.A.) to Velocity Ratio (V.R.), expressed as a percentage:\[ \eta = \frac{\text{M.A.}}{\text{V.R.}} \times 100 \]Given:
\( \eta = 60\% \)
\( \text{V.R.} = 2 \)
\( \text{Load} = 300\,\text{N} \)

First, find M.A.:\[ 0.60 = \frac{\text{M.A.}}{2} \implies \text{M.A.} = 1.2 \]Now, use \( \text{M.A.} = \frac{\text{Load}}{\text{Effort}} \):\[ 1.2 = \frac{300}{\text{Effort}} \implies \text{Effort} = \frac{300}{1.2} = 250\,\text{N} \]

According to the kinetic theory of gases, the average translational kinetic energy (\( K.E._{avg} \)) of a molecule is directly proportional to the absolute temperature (T) of the gas:\[ K.E._{avg} = \frac{3}{2} kT \]
Where \( k \) is the Boltzmann constant. It does not depend on the pressure, volume, or the nature of the gas, only on its temperature.

In a closed pipe (closed at one end), only odd harmonics are present (\( f_1, 3f_1, 5f_1, 7f_1, 9f_1, \dots \)).
1st overtone = 3rd harmonic (\( 3f_1 \))
2nd overtone = 5th harmonic (\( 5f_1 \))
3rd overtone = 7th harmonic (\( 7f_1 \))
4th overtone = 9th harmonic (\( 9f_1 \))

Given \( 9f_1 = 900\,\text{Hz} \):\[ f_1 = \frac{900}{9} = 100\,\text{Hz} \]

The efficiency (\( \eta \)) of a cell is the ratio of the potential difference (V) across the external resistor to the total electromotive force (E). This simplifies to the ratio of the external resistance (X) to the total resistance (X + r):\[ \eta = \frac{X}{X + r} \times 100 \]Given:
\( \eta = 75\% = 0.75 \)
\( r = 2\,\Omega \)
\[ 0.75 = \frac{X}{X + 2} \]\[ 0.75(X + 2) = X \]\[ 0.75X + 1.5 = X \]\[ 1.5 = 0.25X \implies X = \frac{1.5}{0.25} = 6\,\Omega \]

A potentiometer is more accurate for measuring e.m.f. because it works on a null-deflection principle. Unlike a standard voltmeter, which must draw some current from the source to provide a reading (thus causing a voltage drop across the source's internal resistance), a potentiometer measures voltage when no current flows through the circuit being measured. This allows it to measure the true e.m.f.