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JAMB UTME - Physics - 2019

Question 1 Report

A man on a bench will exert the greatest pressure on the bench when he

lies flat on his back

lies flat on his belly

stands on both feet

stands on the toes of one foot

Answer Details

The man on the bench will exert the greatest pressure when he stands on the toes of one foot. This is because when he stands on one foot, all his weight is concentrated on a smaller surface area of the bench, resulting in more pressure. The pressure he exerts is calculated by dividing his weight by the surface area in contact with the bench. When he stands on one foot, the surface area is smaller, which means the pressure exerted is greater. In comparison, when he lies flat on his back or belly, or when he stands on both feet, his weight is distributed over a larger surface area, resulting in less pressure.

Read lesson note on Pressure (JAMB)

Read lesson note on Measurements And Units (JAMB)

Measurements And Units

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Question 7 Report

An a.c of 1A at a frequency of 800 cycles per second flows through a coil, the inductance of which is 2.5mH and the resistance of which is 5Ω. What is the power absorbed in the Coil?

5W

6W

7W

8W

Answer Details

I = 1A, F = 800 cycles/s = 800Hz
R = 5Ω, L = 2.5mH
P = I

^{2}

R = I

^{2}

× 5 = 5W

View Answer

Question 8 Report

A body moves in SHM between two point 20m on the straight line Joining the points. If the angular speed of the body is 5 rad/s. Calculate its speed when it is 6m from the center of the motion.

20 m/s

30 m/s

40 m/s

50 m/s

Answer Details

From two parts 20m apart
a = 10m, x = 6m, A = 5
V = ω

\sqrt{A^{2} - X^{2}}

= 5

\sqrt{10^{2} - 6^{2}}

= 40m/s

Read lesson note on Simple Harmonic Motion (WAEC)

Simple Harmonic Motion

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Question 10 Report

The resultant capacitance in the figure above is

0.4μf

0.8μf

1.8μf

4.5μf

Answer Details

For the parallel arrangement = 2 + 4 = 6μf

For

the

series

arrangement

=

1C

_{T}

=

12

+

13

+

16

+

14

1C

_{T}

=

1512

C

_{T}

=

1215

=

0.8μf

Read lesson note on Capacitors (JAMB)

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Question 13 Report

The lead-acid accumulator consists of

lead as the positive electrode

lead acid as the negative electrode

hydrochloric acid as the electrolyte

tetraoxosulphate (vi) acid as the electrolyte

Answer Details

- the positive pole is lead peroxide (PbO

_{2}

)
- the negative pole is head
- the electrolyte is H

_{2}

SO

_{4}

Read lesson note on Electric Cells (JAMB)

Electric Cells

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Question 16 Report

The value of T in the figure above is

30N

10.0N

20N

40N

Answer Details

Tsin30 + Tsin30 =40

2Tsin30 = 40

Tsin30 = 40/2 = 20

T( $\frac{1}{2}$ ) = 20

T = 20 x 2 = 40N

Read lesson note on Scalars And Vectors (JAMB)

Read lesson note on Motion (JAMB)

Scalars And Vectors

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Question 20 Report

A rectangular solid black has length 10cm, breadth 5cm and height 2cm. If it lies on a horizontal surface, and has density \(100\text{kg/m}^3\), calculate the pressure it exerts on the surface.

200 N/m

^{2}

100 N/m

^{2}

400 N/m

^{2}

Answer Details

To calculate the pressure that the rectangular solid exerts on the surface, we need to use the formula for pressure: Pressure = Force / Area In this case, the force is the weight of the rectangular solid, which we can calculate using the formula: Weight = Mass x Gravity The mass of the rectangular solid can be calculated using its density and volume: Mass = Density x Volume The volume of the rectangular solid is simply its length x breadth x height: Volume = Length x Breadth x Height = 10 cm x 5 cm x 2 cm = 100 cm3 We need to convert this volume to cubic meters to use the density given in kg/m3: Volume = 100 cm3 = 0.0001 m3 Now we can calculate the mass: Mass = Density x Volume = 100 kg/m3 x 0.0001 m3 = 0.01 kg The gravity is the acceleration due to gravity, which we can assume to be 9.81 m/s2. Therefore, the weight is: Weight = Mass x Gravity = 0.01 kg x 9.81 m/s2 = 0.0981 N Now we can use this weight to calculate the pressure on the surface. The surface area in contact with the rectangular solid is simply its length x breadth: Area = Length x Breadth = 10 cm x 5 cm = 50 cm2 We need to convert this area to square meters: Area = 50 cm2 = 0.005 m2 Therefore, the pressure is: Pressure = Force / Area = 0.0981 N / 0.005 m2 = 19.62 N/m2 We can convert this to units of N/cm2 or N/mm2 if desired. This is equivalent to: Pressure = 0.1962 N/cm2 = 0.0001962 N/mm2 So the pressure that the rectangular solid exerts on the surface is 19.62 N/m2, which is approximately 20 N/m2. Therefore, the answer is 200 N/m2.

Read lesson note on Pressure (JAMB)

Read lesson note on Measurements And Units (JAMB)

Measurements And Units

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Question 24 Report

A thermocouple thermometer is connected to a millivoltmeter which can read up to 10mV. When one junction is in ice at 0°C and the other is steam at 100°C, the millivoltmeter reads 4mV. What is the maximum temperature which this arrangement can measure

100

^{o}

C

248

^{o}

C

250

^{o}

C

350

^{o}

C

Answer Details

The maximum temperature which this arrangement can measure is 250°C. A thermocouple thermometer works by using the thermoelectric effect, which is the phenomenon that occurs when two dissimilar metals are joined together to form a loop and a temperature difference is established between the two junctions. This temperature difference generates a small electrical voltage, which can be measured using a millivoltmeter. The voltage generated is proportional to the temperature difference between the two junctions. In the case of the thermocouple thermometer described, one junction is in ice at 0°C and the other is steam at 100°C, and the millivoltmeter reads 4mV. This means that the voltage generated by the thermocouple is 4 millivolts, which corresponds to a temperature difference of 100°C. However, the millivoltmeter can only read up to 10mV, so the maximum temperature difference it can measure is 10mV / 4mV/°C = 250°C. This means that the maximum temperature which this arrangement can measure is 250°C.

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

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Question 28 Report

A supply of 400V is connected across capacitors of 3μf and 6μf in series. Calculate the charge

8 x 10

^{- 4}

C

4 x 10

^{- 2}

C

8 x 10

^{- 3}

C

4 x 10

^{- 8}

C

Answer Details

C

_{T}

=

C

_{1}

× C

_{2}

C

_{1}

+ C

_{2}

=	3 × 63 + 6

= $\frac{18}{9}$ = 2μf
Q = CV
⇒ 2 × 10 $^{- 6}$ × 400
⇒ 800 × 10 $^{- 6}$ C = 8 × 10 $^{- 4}$ C

Read lesson note on Capacitors (JAMB)

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Question 29 Report

A metal rod has a length of 100cm at \(200^\circ C\). At what temperature will its length be 99.4cm. If the linear expansivity of the material of the rod is \(2 \times 10^{-5}C^{-1}\)

200

^{o}

C

300

^{o}

C

100

^{o}

C

-100

^{o}

C

Answer Details

The linear expansivity of a material describes how its length changes with temperature. If the linear expansivity is given as 2 × 10^-5/°C, this means that for every 1°C change in temperature, the length of the material will change by 2 × 10^-5 times its original length. Given that the rod has a length of 100 cm at 200°C, we can use this information to find its length at a different temperature. If we let L be the length of the rod at temperature T, we can write the relationship as follows: L = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) To find the temperature at which the rod will have a length of 99.4 cm, we can set L equal to 99.4 cm and solve for T: 99.4 cm = 100 cm * (1 + 2 × 10^-5 * (T - 200°C)) 99.4 cm / 100 cm = 1 + 2 × 10^-5 * (T - 200°C) 0.994 = 1 + 2 × 10^-5 * (T - 200°C) -0.006 = 2 × 10^-5 * (T - 200°C) -0.006 / 2 × 10^-5 = T - 200°C -0.006 / (2 × 10^-5) = T - 200°C -0.006 / (2 × 10^-5) + 200°C = T So the temperature at which the rod will have a length of 99.4 cm is approximately equal to -0.006 / (2 × 10^-5) + 200°C, or -100°C. Therefore, the answer is -100°C.

Read lesson note on Thermal Expansion (JAMB)

Read lesson note on Temperature And Its Measurement (JAMB)

Thermal Expansion

Temperature And Its Measurement

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Question 31 Report

The diagram shows a uniform meter rule AB which balances horizontally at the 90cm mark when a mass of 0.2kg is suspended from B. Calculate the mass of the meter rule.

0.05kg

0.20kg

0.80kg

1.00kg

Answer Details

M $_{r}$ (90 - 50) = 0.2(100 - 90)
40M $_{r}$ = 0.2 × 10

M $_{r}$ = $\frac{2}{40}$ = 0.05kg

Read lesson note on Measurements And Units (JAMB)

Measurements And Units

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