2023 - JAMB Physics Past Questions and Answers - page 4

Classical mechanics

Classical mechanics is not an example of elementary modern physics. It is a branch of physics that deals with the motion of objects at everyday speeds and scales, which was well-developed before the advent of modern physics.

Quantum mechanics, special relativity, and nuclear physics are examples of elementary modern physics, as they deal with phenomena at the atomic and subatomic scales and have significantly contributed to our understanding of the fundamental principles governing the universe.

To calculate relative humidity, you can use the formula:

\[ \text{Relative Humidity (\%)} = \left( \frac{\text{Partial Pressure of Water Vapor}}{\text{Saturation Vapor Pressure at the Given Temperature}} \right) \times 100 \]

The saturation vapour pressure is the maximum pressure that water vapor can exert at a given temperature. It depends on the temperature and is a function of the water vapor pressure at 100% humidity.

The saturation vapour pressure at 40°C can be obtained from tables or calculations. For this example, let's assume it's \(50 \, \text{mmHg}\).

Now, substitute the given values into the formula:

\[ \text{Relative Humidity (\%)} = \left( \frac{38.8}{50} \right) \times 100 \]

\[ \text{Relative Humidity (\%)} \approx 77.6 \]

So, the closest option is: 70

Please note that the saturation vapour pressure at 40°C may vary slightly depending on the source used, and it's advisable to use accurate data for precise calculations.

To determine the location of the image formed by a convex mirror, you can use the mirror formula:

\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]

where:
- \( f \) is the focal length of the mirror,
- \( d_o \) is the object distance (distance of the object from the mirror),
- \( d_i \) is the image distance (distance of the image from the mirror).

Given that \( f = 15 \) cm and \( d_o = -35 \) cm (negative because it's in front of the mirror for convex mirrors), you can rearrange the formula to solve for \( d_i \):

\[ \frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} \]

\[ \frac{1}{d_i} = \frac{1}{15} - \frac{1}{-35} \]

\[ \frac{1}{d_i} = \frac{1}{15} + \frac{1}{35} \]

\[ \frac{1}{d_i} = \frac{7}{105} + \frac{3}{105} \]

\[ \frac{1}{d_i} = \frac{10}{105} \]

\[ d_i = \frac{105}{10} \]

\[ d_i = 10.5 \, \text{cm} \]

The negative sign indicates that the image is formed on the same side as the object, which is expected for convex mirrors.

So, the correct option is 10.5 cm behind the mirror

Given the forces acting on the mass (\(m\)):

1. The weight of the mass (\(mg\)) acts vertically downward.
2. The tension in \(T_1\) acts at an angle of 38° to the horizontal.
3. The tension in \(T_2\) acts horizontally.

The vertical component of \(T_1\) (\(T_{1v}\)) is given by \(T_1 \cdot \sin(38^\circ)\), and the horizontal component of \(T_1\) (\(T_{1h}\)) is given by \(T_1 \cdot \cos(38^\circ)\).

The net force in the vertical direction is equal to the weight of the mass (\(mg\)):

\[ T_{1v} = mg \]

The net force in the horizontal direction is zero (assuming no acceleration in the horizontal direction):

\[ T_{1h} = T_2 \]

Now, set up the equations:

1. In the vertical direction:
\[ T_1 \cdot \sin(38^\circ) = mg \]

2. In the horizontal direction:
\[ T_1 \cdot \cos(38^\circ) = T_2 \]

Given that \(m = 220 \, \text{kg}\) and \(g = 9.8 \, \text{m/s}^2\), you can solve these equations to find \(T_1\) and \(T_2\).

Let's solve these equations:

1. \[ T_1 = \frac{mg}{\sin(38^\circ)} \]
2. \[ T_2 = T_1 \cdot \cos(38^\circ) \]

Substitute the values and calculate:

\[ T_1 \approx \frac{(220 \, \text{kg} \cdot 9.8 \, \text{m/s}^2)}{\sin(38^\circ)} \]

\[ T_2 = T_1 \cdot \cos(38^\circ) \]

Now, let's calculate these values:

\[ T_1 \approx 3502 \, \text{N} \]
\[ T_2 \approx 2760 \, \text{N} \]

So, the correct option is: \(T_1 \approx 3502 \, \text{N}, \quad T_2 \approx 2760 \, \text{N}\)

Instrument resolution is not inherently a limitation of experimental measurements; instead, it is a characteristic of the measuring instrument. Resolution refers to the smallest change in the measured quantity that the instrument can detect or display. While it influences the precision of the measurements, it is not a limitation in the same sense as errors or uncertainties.

So, the correct answer is Instrument resolution

Given:
- Weight of the box (\(W\)) = 400 N
- Applied force (\(P\)) = 100 N
- Angle (\(\theta\)) = 30°

To find the coefficient of kinetic friction (\(\mu_k\)), the forces acting along the horizontal direction are the frictional force (\(F_r\)) and the horizontal component of the applied force (\(P \cos 30^\circ\)).

The equation for equilibrium in the horizontal direction is:
\[ P \cos 30^\circ - F_r = ma \]

Since the box is moving at a constant speed, its acceleration is zero:
\[ P \cos 30^\circ - F_r = 0 \]

Now, we know that \(F_r = \mu R\), where \(R\) is the normal reaction. The normal reaction (\(R\)) is found by considering forces in the vertical direction:
\[ R - P \sin 30^\circ - W = 0 \]

Solving for \(R\):
\[ R = P \sin 30^\circ + W \]

Now, substitute \(R\) back into the equation for \(F_r\):
\[ P \cos 30^\circ - \mu R = 0 \]

\[ P \cos 30^\circ = \mu R \]

\[ \mu = \frac{P \cos 30^\circ}{R} \]

Substitute the given values:
\[ \mu = \frac{100 \cos 30^\circ}{450} \]

Simplify:
\[ \mu = \frac{100 \cdot \frac{\sqrt{3}}{2}}{450} \]

\[ \mu \approx 0.19 \]

Therefore, the correct option is 0.19

n=1.458, c=\(3.00 ×10^8 ms^-1\) ,λo = 589nm;  f=?

Speed of light in a medium (v)=\(\frac{c}{n}\)  where n is the refractive index of the medium

⇒λn=\(\frac{589}{1.458}\) = 404nm

  v=fλ
  
⇒f=\(\frac{v}{λ}\) 

=\(\frac{2.06×10^8}{404×10^-9}\) (\1nano=10^-9\)

∴f=\(5.09×10^14\) Hz

Given:
- Length of the wire (\(L\)) = 50 cm = 0.5 m
- Total mass of the wire (\(m\)) = 10 g = 0.01 kg
- Tension in the wire (\(T\)) = 800 N

First, calculate the linear mass density (\(\mu\)):

\[ \mu = \frac{m}{L} = \frac{0.01 \, \text{kg}}{0.5 \, \text{m}} = 0.02 \, \text{kg/m} \]

Now, calculate the fundamental frequency (\(f_0\)) using the formula:

\[ f_0 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]

\[ f_0 = \frac{1}{2 \times 0.5} \sqrt{\frac{800}{0.02}} \]

\[ f_0 = \frac{1}{1} \sqrt{40000} \]

\[ f_0 = 200 \, \text{Hz} \]

Now, for the third overtone:
\[ f_3 = 4 \times f_0 \]

\[ f_3 = 4 \times 200 = 800 \, \text{Hz} \]

Therefore, the correct option is 800 Hz

(i) It is an intensive property: False
   - Heat capacity is an extensive property, meaning it depends on the amount of substance. The heat capacity of an object is directly proportional to its mass.

(ii) Its SI unit is \(J \cdot K^{-1}\): True
   - The SI unit of heat capacity is indeed joules per kelvin (\(J \cdot K^{-1}\)).

(iii) It is an extensive property: True
   - As mentioned earlier, heat capacity is an extensive property because it depends on the amount (mass) of the substance.

(iv) Its SI unit is \(J \cdot kg^{-1}\): False
   - The correct SI unit for heat capacity is \(J \cdot K^{-1}\) or \(J/^\circ C\). The unit \(J \cdot kg^{-1}\) is specific heat capacity, which is an intensive property.

Therefore, the statements (i) and (iv) are not true about the heat capacity of a substance.

(i) The identification of gases: True
   - Electrical conduction in gases can be used for the identification of gases in a process called gas discharge tubes. Different gases emit characteristic colors when subjected to electrical discharges, allowing for identification.

(ii) Lighting/fluorescent tubes: True
   - Fluorescent tubes and some types of lighting tubes operate by using electrical conduction in gases. The electrical discharge in these tubes excites the gas, producing ultraviolet radiation, which then causes phosphor coating to emit visible light.

(iii) Advertising industry/Neon signs: Not mentioned in the provided options.
   - Neon signs in the advertising industry also use electrical conduction in gases. The electrical discharge in neon gas produces the characteristic glow in neon signs.

(iv) Cathode ray oscilloscope/T.V. tubes: True
   - Cathode ray oscilloscopes and TV tubes use electron beams in a vacuum rather than electrical conduction through gases.

Therefore, the correct statements are (i), (ii), and (iv).