2023 - JAMB Physics Past Questions and Answers - page 2

11

 

An open-tube mercury manometer is used to measure the pressure in a gas tank. When the atmospheric pressure is 101,325 \(P_a\), what is the absolute pressure in \(P_a\) in the tank if the height of the mercury in the open tube is 25 cm higher

A

108,986 Pa

B

165,238 Pa

C

122,364 Pa

D

134,645 Pa

correct option: d

The absolute pressure (\(P_{\text{abs}}\)) in the tank can be calculated using the equation \(P_{\text{abs}} = P_{\text{atm}} + \rho gh\), where:

- \(P_{\text{atm}}\) is the atmospheric pressure (given as \(101,325 \, \text{Pa}\)),
- \(\rho\) is the density of mercury (\(13,600 \, \text{kg/m}^3\)),
- \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)),
- \(h\) is the height difference of the mercury column (\(0.25 \, \text{m}\)).

Substituting the values:

\[
P_{\text{abs}} = 101,325 + 13,600 \times 9.8 \times 0.25 \\
P_{\text{abs}} = 101,325 + 33,320 \\
P_{\text{abs}} = 134,645 \, \text{Pa}
\]

Therefore, the correct answer is: \(134,645 \, \text{Pa}\).

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12

 

What is the name of the model of the atom that describes electrons as orbiting the nucleus in specific energy levels?

A

Bohr model

B

Dalton model

C

Rutherford model

D

Thomson model

correct option: a

The Bohr model of the atom, proposed by Niels Bohr in 1913, describes electrons as orbiting the nucleus in specific energy levels or shells. According to this model, electrons occupy quantized orbits, and each orbit is associated with a specific energy level. Electrons can jump from one energy level to another by absorbing or emitting energy in discrete packets or quanta. The Bohr model was an improvement over the earlier Rutherford model and provided a better explanation for the spectral lines of hydrogen.

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13

 

A missile is launched with a speed of 75 ms-1 at an angle of 22° above the surface of a warship. Find the horizontal range achieved by the missile. Ignore the effects of air resistance.

[Take g = 10 ms-1]

A

195

B

271 m

C

391 m

D

136 m

correct option: c

The horizontal range (R) of a projectile launched at an angle can be calculated using the formula:

\[ R = \frac{v^2 \sin(2\theta)}{g} \]

where:
- \( v \) is the initial speed of the projectile,
- \( \theta \) is the launch angle,
- \( g \) is the acceleration due to gravity.

In this case, \( v = 75 \, \text{m/s} \), \( \theta = 22^\circ \), and \( g = 10 \, \text{m/s}^2 \).

\[ R = \frac{(75)^2 \sin(2 \times 22^\circ)}{10} \]

Calculating this expression yields approximately 391 m.

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14

 

Which of the following liquids has the highest surface tension?

A

Soapy water

B

Water

C

Mercury

D

Oil

correct option: c

Surface tension is a measure of the force that tends to pull adjacent parts of a liquid's surface together, minimizing the surface area. Mercury has a relatively high surface tension compared to the other liquids listed, making it the correct answer in this case.

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15

 

A parallel plate capacitor separated by an air gap is made of \(0.8m^2\) tin plates and 20 mm apart. It is connected to 120 V battery. What is the charge on each plate?

Take \(ε_o\) = \(8.85×10^-12 Fm^-1\)

A

3.54nC

B

42.5nC

C

35.4nC

D

4.25nC

correct option: b

The charge (\(Q\)) on each plate of the capacitor can be calculated using the formula \(Q = CV\), where:

- \(C\) is the capacitance,
- \(V\) is the voltage.

The capacitance (\(C\)) for a parallel plate capacitor with air as the dielectric is given by \(C = \frac{\varepsilon_0 A}{d}\), where:

- \(\varepsilon_0\) is the vacuum permittivity (\(8.85 \times 10^{-12} \, \text{F/m}\)),
- \(A\) is the area of the plates (\(0.8 \, \text{m}^2\)),
- \(d\) is the separation between the plates (\(0.02 \, \text{m}\)).

Let's substitute the values:

\[
C = \frac{(8.85 \times 10^{-12} \, \text{F/m})(0.8 \, \text{m}^2)}{0.02 \, \text{m}} = 3.54 \times 10^{-10} \, \text{F}
\]

Now, calculate the charge (\(Q\)):

\[
Q = CV = (3.54 \times 10^{-10} \, \text{F})(120 \, \text{V}) = 4.25 \times 10^{-8} \, \text{C}
\]

Convert the charge to nanocoulombs:

\[
4.25 \times 10^{-8} \, \text{C} = 42.5 \, \text{nC}
\]

Therefore, the correct answer is: \(42.5 \, \text{nC}\).

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16

 

Rainbow formation is as a result of the combination of which of the following phenomena?
(i) Reflection
(ii) Dispersion
(iii) Total internal reflection
(iv) Refraction

A

(ii) and (iv) only

B

(i), (iii) and (iv) only

C

(i), (ii) and (iv) only

D

(ii), (iii) and (iv) only

correct option: d

The formation of a rainbow involves refraction, dispersion, and total internal reflection. Here's a more accurate summary:

When sunlight shines on a water droplet, the light undergoes refraction as it enters the droplet. The different colours of light are dispersed due to the varying wavelengths. Finally, total internal reflection occurs inside the droplet, leading to the formation of a rainbow.

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17

 

The surface temperature of a swimming pool on a warm day is 25oC and the temperature at the bottom is 15oC. If the swimming pool has a surface area of 620 m2 and a depth of 1.5m. Find the rate at which energy is transferred by conduction from the surface to the bottom of the swimming pool.

[Thermal conductivity of water (k) = 0.6071 Wm-1K-1]
 

A

2.5kw

B

250kw

C

300kw

D

3.0kw

correct option: a

The rate at which energy is transferred by conduction from the surface to the bottom of the swimming pool (\(\frac{q}{t}\)) can be calculated using the formula:

\[
\frac{q}{t} = K \cdot A \cdot \frac{\Delta \theta}{L}
\]

where:
- \(K\) is the thermal conductivity of water (\(0.6071 \, \text{W/m} \cdot \text{K}\)),
- \(A\) is the surface area of the pool (\(620 \, \text{m}^2\)),
- \(\Delta \theta\) is the temperature difference (\(25^\circ \text{C} - 15^\circ \text{C} = 10^\circ \text{C}\)),
- \(L\) is the depth of the pool (\(1.5 \, \text{m}\)).

Substitute the values:

\[
\frac{q}{t} = 0.6071 \times 620 \times \frac{10}{1.5} = 2509.35 \, \text{W}
\]

Convert the result to kilowatts (\(\text{kW}\)):

\[
\frac{2509.35 \, \text{W}}{1000} = 2.50935 \, \text{kW}
\]

Approximately, the rate of energy transfer is \(2.5 \, \text{kW}\), so the correct option is: \(2.5 \, \text{kW}\).

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18

 

A step-down transformer is used on a 2.2 kV line to deliver 110 V. How many turns are on the primary windings if the secondary has 25 turns?

A

20

B

250

C

200

D

500

correct option: d

The relationship between the number of turns on the primary (\(N_1\)) and secondary (\(N_2\)) windings of a transformer is given by the formula:

\[ \frac{{N_1}}{{N_2}} = \frac{{V_1}}{{V_2}} \]

where:
- \( N_1 \) is the number of turns on the primary winding,
- \( N_2 \) is the number of turns on the secondary winding,
- \( V_1 \) is the voltage on the primary side,
- \( V_2 \) is the voltage on the secondary side.

Given values:
\[ V_1 = 2.2 \, \text{kV} = 2200 \, \text{V} \]
\[ V_2 = 110 \, \text{V} \]
\[ N_2 = 25 \]

Now, rearrange the formula to solve for \( N_1 \):

\[ N_1 = N_2 \times \frac{{V_1}}{{V_2}} \]

Substitute the values:

\[ N_1 = 25 \times \frac{{2200}}{{110}} \]

\[ N_1 = 25 \times 20 \]

\[ N_1 = 500 \]

The number of turns on the primary winding is 500 when the secondary winding has 25 turns, and the transformer is used on a 2.2 kV line to deliver 110 V.

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19

 

Which of the following materials is a good insulator?

A

Silver

B

Water

C

Rubber

D

Copper

correct option: c

Insulators are materials that do not conduct electricity well. Rubber is known for its high electrical resistance, making it a good insulator. On the other hand, silver, copper, and water are conductors of electricity.

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20

The property of wave shown in the diagram above is?
 

A
Diffraction
B
Refraction
C
Polarisation
D
Interference
correct option: a

Diffraction is the bending of waves around the corners of an obstacle or through an aperture.

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