Loading....






Press & Hold to Drag Around
Click Here to Close

JAMB UTME - Physics - 2019

Question 1 Report

The pin-hole camera produces a less sharply defined image when the

pin-hole is larger

illumination is less

screen is further from the pin-hole

object is further from the pin-hole

Answer Details

The pin-hole camera produces a less sharply defined image when the pin-hole is larger. A pin-hole camera works by allowing light to pass through a small hole (the pin-hole) and project an inverted image of the outside world onto a screen or surface located behind the hole. The smaller the pin-hole, the sharper the resulting image, as light passing through a smaller hole produces less diffraction or spreading out of the light. When the pin-hole is larger, more light enters the camera, but the light rays also become more scattered, resulting in a less well-defined image. This is because the larger opening allows more light rays to enter at different angles, creating a wider range of paths that the light can take as it travels through the camera and onto the screen. As a result, the image is less clear and less defined, with less sharp edges and more blurring. is the correct answer because it correctly identifies the effect of a larger pin-hole on the image produced by the pin-hole camera. less illumination, would actually produce a dimmer image, but it would not affect the sharpness or definition of the image. the distance of the screen from the pin-hole, and the distance of the object from the pin-hole, would affect the size of the image and the scale of the objects, but they would not affect the sharpness or definition of the image.

Read lesson note on Light Energy (JAMB)

Read lesson note on Structure Of The Nucleus (WAEC)

Light Energy

Structure Of The Nucleus

View Answer

Question 11 Report

A well 1km deep is filled with a liquid of density \(950\text{kg/m}^3\) and \(g = 10\text{m/s}^2\), the pressure at the bottom of the well is

9.6 × 10

^{6}

N/m

^{2}

9.5 x 10

^{5}

N/m

^{2}

9.4 x 10

^{6}

N/m

^{2}

9.5 x 10

^{6}

N/m

^{2}

Answer Details

P = Pa + ρgh = (1.00 × 10

^{5}

) + (950 × 10 × 1000)
P = 10

^{5}

+ (95 × 10

^{5}

) = 10

^{5}

(1 + 95) = 96 × 10

^{5}

P = 9.6 × 10

^{6}

N/m

^{2}

Read lesson note on Pressure (JAMB)

Read lesson note on Motion (WAEC)

View Answer

Question 14 Report

The distance between an object and its real image in a convex lens is 40cm. If the magnification of the image is 3, calculate the focal length of the lens

6.5 cm

7.5 cm

8.5 cm

4.5 cm

Answer Details

u + v = 40

$\frac{v}{u}$ = 3

v = 3u

u + 3u = 40

4u = 40

u = 10cm

v = 3u = 30cm

f = $\frac{u v}{u + v} = \frac{10 (30)}{10 + 30} = \frac{300}{40}$

= 7.5 cm

Read lesson note on Gravitational Field (WAEC)

Gravitational Field

View Answer

Question 15 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)

View Answer

Question 17 Report

Which of the following equations is the correct definition of the reactance of an indicator L?

Reactance = (Amplitude of voltage)

\div

(Amplitude of current)

Reactance = (Amplitude of voltage) x (Amplitude of current)

Reactance = (Amplitude of current)

^{2}

\div

(Amplitude of voltage)

Reactance = (Amplitude of current)

^{2}

\div

(Amplitude of voltage)

\frac{1}{2}

Answer Details

The correct definition of the reactance of an inductor L is: Reactance = (Amplitude of voltage) ÷ (Amplitude of current) The reactance of an inductor is a measure of the opposition offered by the inductor to the flow of alternating current (AC). It is denoted by the symbol Xl and is measured in ohms. When AC flows through an inductor, a magnetic field is generated around the inductor, which opposes any changes in the current flowing through it. This opposition to the flow of current is called reactance. The reactance of an inductor depends on its inductance, frequency of the AC signal, and the amplitude of the AC signal. However, the reactance of an inductor is directly proportional to the frequency of the AC signal and the inductance of the inductor. The reactance of an inductor is also affected by the amplitude of the AC signal, but this effect is not as significant as the other two factors. is the correct definition of the reactance of an inductor, as it expresses the ratio of the amplitude of voltage to the amplitude of current, which is a common way to define reactance. is incorrect, as it represents the power delivered by the AC signal, not the reactance. and are also incorrect, as they involve squaring either the amplitude of current or the amplitude of voltage, which is not a valid method of calculating reactance. Therefore, the correct option is.

Read lesson note on Electric Field (WAEC)

Electric Field

View Answer

Question 23 Report

A siren having a ring of 200 hole makes 132 rev/min. A jet of air is directed on the set of holes. Calculate the frequency and wavelength in air of the note produced (take v = 350m/s)

0.875m

0.625m

0.335m

0.125m

Answer Details

n = 200, S = 132 rev/min, v = 350m/s $^{2}$

f

=

ns

=

200

×

132

revmin

×

1min60s

=

440Hz

λ

=

vf

=

350440

=

0.875m

Read lesson note on Quantity Of Heat (JAMB)

Quantity Of Heat

View Answer

Question 24 Report

Radio waves belongs to the class of ware whose velocity is about

340m/s

3 × 10

^{6}

m/s

3 × 10

^{8}

m/s

3 × 10

^{- 8}

m/s

Answer Details

Radio waves belong to the class of waves whose velocity is approximately 3 x 10^8 m/s. This velocity is commonly denoted as the speed of light, which is the speed at which all electromagnetic waves, including radio waves, travel in a vacuum. This constant velocity is one of the fundamental principles of physics and is important in understanding the behavior and properties of light and other electromagnetic waves. The speed of light is incredibly fast, and it's difficult for us to imagine just how fast it is. To put it into perspective, light can travel around the Earth's equator almost 7.5 times in just one second. This high speed is essential for radio communication, as it enables radio waves to travel long distances in a short amount of time, allowing us to communicate with people and devices far away from us.

Read lesson note on Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

View Answer

Question 25 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

View Answer

Question 28 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

View Answer

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40