Physics 2015 Jamb Examination

To determine the time taken for an object to fall from a height under gravity, we use the second equation of motion for an object starting from rest (\(u = 0\)):\[ h = ut + \frac{1}{2}gt^2 \]Given:
Height \(h = 50 \; \text{m}\)
Acceleration due to gravity \(g = 10 \; \text{ms}^{-2}\)
Initial velocity \(u = 0\)
Substituting the values:\[ 50 = 0(t) + \frac{1}{2}(10)t^2 \]\[ 50 = 5t^2 \]\[ t^2 = \frac{50}{5} = 10 \]\[ t = \sqrt{10} \approx 3.16 \; \text{sec} \]Rounding to one decimal place, we get 3.2 seconds.

Heat capacity (C) is the amount of heat required to change the temperature of an entire object by 1 Kelvin. Specific heat capacity (c) is the heat required per unit mass to achieve the same temperature change. The relationship between the two is given by the formula:\[ C = mc \]Given:
Heat capacity \(C = 40 \; \text{J/K}\)
Specific heat capacity \(c = 380 \; \text{J/kg K}\)
Solving for mass (m):\[ m = \frac{C}{c} \]\[ m = \frac{40}{380} \]\[ m = \frac{4}{38} \approx 0.10526 \; \text{kg} \]Rounding to two decimal places, the mass is approximately 0.11kg.

Using the equation of motion for a free-falling body starting from rest:\[ h = \frac{1}{2}gt^2 \]Given:
\(h = 20 \; \text{m}\)
\(g = 10 \; \text{ms}^{-2}\)
Rearranging to solve for time (t):\[ 20 = \frac{1}{2} \times 10 \times t^2 \]\[ 20 = 5t^2 \]\[ t^2 = \frac{20}{5} = 4 \]\[ t = \sqrt{4} = 2 \; \text{sec} \]It takes exactly 2 seconds for the ball to reach the ground.

Velocity Ratio (V.R) is defined as the ratio of the distance moved by the effort to the distance moved by the load in the same time interval. If we let \(x\) be the distance moved by the effort and \(y\) be the distance moved by the load, the formula is:\[ V.R = \frac{\text{Distance moved by Effort (x)}}{\text{Distance moved by Load (y)}} \]Given \(V.R = 5\):\[ 5 = \frac{x}{y} \implies \frac{x}{y} = \frac{5}{1} \]Therefore, the ratio of x to y is 5:1.

Mechanical Advantage (M.A) is a measure of the force amplification achieved by using a machine. It is defined as the ratio of the Load (the output force) to the Effort (the input force). \[ M.A = \frac{\text{Load}}{\text{Effort}} \]Given \(M.A = 4\):\[ 4 = \frac{\text{Load}}{\text{Effort}} \implies \frac{\text{Load}}{\text{Effort}} = \frac{4}{1} \]This means for every 1 Newton of effort applied, the machine can lift a load of 4 Newtons. The ratio is 4:1.

While mercury and alcohol thermometers are common for general use, the constant-volume gas thermometer is considered the most accurate and is used as a standard for calibrating other thermometers. This is because the expansion of gases is much more uniform and predictable over a wide range of temperatures compared to the expansion of liquids. Gases also have low thermal capacity and high sensitivity to temperature changes, making them the superior choice for high-precision scientific measurements.

According to Galileo's findings and the laws of motion, in the absence of air resistance, all objects fall with the same constant acceleration due to gravity (\(g \approx 9.8 \text{ ms}^{-2}\)) regardless of their mass. Since both objects start from the same height with the same initial velocity (zero) and experience the same acceleration, they will cover the same distance in the same amount of time. Therefore, they will hit the ground simultaneously.

Angular velocity (\(\omega\)) is related to linear velocity (\(v\)) and the radius of the circular path (\(r\)) by the following formula:\[ v = \omega r \]Given:
Linear speed \(v = 2 \; \text{ms}^{-1}\)
Radius \(r = 0.7 \; \text{m}\)
Mass \(m = 3 \; \text{kg}\) (Note: mass is not required for calculating angular velocity from linear speed).
Rearranging to solve for \(\omega\):\[ \omega = \frac{v}{r} \]\[ \omega = \frac{2}{0.7} \]\[ \omega = \frac{20}{7} \approx 2.857 \; \text{rads}^{-1} \]Looking at the provided options, none exactly match 2.86. However, if the radius were 0.5m, the answer would be 4. If the speed were 0.7m/s, the answer would be 1. Based on the values provided, there may be a typo in the question or options.

Different optical instruments are designed for different purposes. A telescope is used to view distant objects like stars and planets. A sextant is used to measure the angle between two visible objects, primarily for navigation. A periscope is used to see over, around, or through an obstacle. A microscope, however, is specifically designed to magnify very small, nearby objects that are otherwise difficult or impossible to see with the naked eye.

In nuclear notation, a particle is represented by the symbol \(^{A}_{Z}X\), where \(A\) is the mass number and \(Z\) is the atomic number or charge. An electron (beta particle) has a negligible mass relative to protons and neutrons, so its mass number \(A\) is 0. It carries a single negative charge, so its \(Z\) value is -1. Therefore, the symbol \(^{0}_{-1}e\) or \(^{0}_{-1}X\) represents an electron. A proton would be \(^{1}_{1}p\) and a neutron would be \(^{1}_{0}n\).