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JAMB UTME - Physics - 2017

Question 1 Report

A string of length 5cm is extended by 0.04m when a load of 0.8kg is suspended at the end. How far will it extend if a force of 16N is applied? [\(g = 10ms^{-2}\)]

0.04m

0.12m

0.01m

0.08m

Answer Details

The extension of a string when a load is suspended from it is given by Hooke's Law which states that the extension is proportional to the force applied. The proportionality constant is called the spring constant, represented by k. Mathematically, we can express this relationship as: F = kx where F is the force applied, x is the extension, and k is the spring constant. To solve this problem, we need to find the spring constant, k, of the string. We can use the information given in the problem to calculate k as follows: k = F/x where F is the force applied and x is the extension. We are given that the string of length 5cm (which is 0.05m) is extended by 0.04m when a load of 0.8kg is suspended at the end. We can convert the mass to force using the formula F = mg, where g is the acceleration due to gravity (g = 10ms^-2). F = 0.8kg × 10ms^-2 = 8N Using the formula for the spring constant, we get: k = F/x = 8N / 0.04m = 200N/m Now, we can use this spring constant to find the extension when a force of 16N is applied. Again, we can use the formula F = kx and rearrange it to solve for x: x = F/k Plugging in the values, we get: x = 16N / 200N/m = 0.08m Therefore, the string will extend by 0.08m when a force of 16N is applied. The answer is (D) 0.08m.

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Question 3 Report

What is the speed of a body vibrating at 50 cyclic per second

100 ?rads

^{- 1}

200 ?rads

^{- 1}

50 ?rads

^{- 1}

400 ?rads

^{- 1}

Answer Details

1 cycle = 3600 = 2? rad

? 50 cycle = 2 ? rad × 50 = 100 ?rad

? = angular speed = 50 cycles per seconds = 100 ? rads $_{- 1}$

Read lesson note on Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

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Question 7 Report

Which of the following representations is correct for an atom X with 28 electrons and 30 neutrons?

5828X

3028X

2830X

5830X

Answer Details

An atom can be represented thus,

$_{A}^{Z} X$ where Z = mass number

A = atomic number

A = number of proton = number of electrons in a free state = 28

Z = number of protons + number of neutrons = 28 + 30 = 58

$_{28}^{58} X$ is the representation

Read lesson note on Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

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Question 8 Report

Three 3ω resistance are connected in parallel what is the equivalent resistance?

1ω

9ω

3ω

0.33ω

Answer Details

Resistor connected in a parallel have an equivalence given by

\frac{1}{Reff}

=

\frac{1}{3}

+

\frac{1}{3}

+

\frac{1}{3}

= $\frac{1 + 1 + 1}{3}$ = $\frac{3}{3}$ = 1

Reff = 1ω

Read lesson note on Scalars And Vectors (JAMB)

Read lesson note on Equilibrium Of Forces (WAEC)

Scalars And Vectors

Equilibrium Of Forces

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Question 13 Report

A cell of internal resistance 2 π supplies current to a 6 π resistor. The efficiency of the cell is

12.0%

25.0%

33.3%

75.0%

Answer Details

Internal resistance determine the maximum current that can be supplied

Efficiency = $\frac{r}{R}$ × 100

= $\frca 26$ × 100

= 33.3%

Read lesson note on Current Electricity (JAMB)

Read lesson note on Electric Cells (JAMB)

Current Electricity

Electric Cells

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Question 15 Report

From the diagram above, calculate the total current in the circuit

5.0A

3.7A

4.5A

4.0A

Answer Details

V = IR from ohms law

I = $\frac{V}{R}$

Since the resistance are in parallel

$\frac{1}{R e f f}$ = $\frac{1}{8}$ + $\frac{1}{12}$ + $\frac{1}{6}$

= $\frac{2 + 3 + 6}{36}$ = $\frac{11}{36}$

Reff = $\frac{36}{11}$ = 3.27

I = I1 + I2 + I3

= $\frac{V}{R e f f}$ = $\frac{12}{3.27}$

= 3.669A

I = 3.7A

Read lesson note on Current Electricity (JAMB)

Current Electricity

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Question 17 Report

From the diagram above, calculate the energy stored in the capacitor

4.0 x 10

^{- 2}

J

4.0 x 10

^{- 4}

J

8.0 x 10

^{- 4}

J

8.0 x 10

^{- 2}

J

Answer Details

c = 8µF, v = 10v

Energy stored = $\frac{1}{2}$ cv2

= $\frac{1}{2}$ × (8 × 10− 6) × 102

= 4 × 10 $^{- 4}$

Read lesson note on Capacitors (JAMB)

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Question 24 Report

The velocity ratio of an inclined plane where angle of inclination in θ is

Sinθ

Cosθ

Tanθ

\frac{1}{\sin θ}

Answer Details

Consider an inclined plane shown below, as the effort moves along OB, the load moves is lifted up through a vertical height AB

V.R = $\frac{distance moved by effort}{distance moved by load}$ = $\frac{O B}{A B}$

But Sinθ = $\frac{o p p}{h y p}$ = $\frac{A B}{O B}$

therefore; = $\frac{O B}{A B}$ = $\frac{1}{\sin θ}$

Read lesson note on Measurements And Units (JAMB)

Measurements And Units

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Question 26 Report

Which is the incorrect formula for a body accelerating uniformly?

a = v2 − u

\frac{2}{2}

v2 = u2 + 2as

s =

\frac{1}{2}

ut + at2

v2 − u2= 2as

Answer Details

s =

\frac{1}{2}

ut + at2 is a wrong equation, it should be s = ut +

\frac{1}{2}

at2

Read lesson note on Rectilinear Acceleration (WAEC)

Rectilinear Acceleration

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Question 27 Report

The net capacitance in the circuit above is

80µF

6.0µF

4.0µF

2.0µF

Answer Details

For capacitance i n parallel, 2µF and 2µF are in parallel,
their equivalence is 2µF and 2µF = 4µF

The 4µF generated is now in series with the remaining 4µF.

The net capacitance for series connection is

$\frac{1}{C}$ = $\frac{1}{4}$ + $\frac{1}{4}$ = 1 + $\frac{1}{4}$ = $\frac{2}{4}$

C = $\frac{4}{2}$

= 2µF

Read lesson note on Capacitors (JAMB)

Read lesson note on Measurements And Units (JAMB)

Measurements And Units

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Question 28 Report

When two mirrors are placed at an angle of 900 to each other, how many images will be formed when an object is placed in front of the mirrors

5

4

3

2

Answer Details

When two mirror are inclined at an angle θ, the number of images formed, n =

\frac{360}{θ}

− 1

∴ since $θ$ = 90o

∴ $\frac{360}{θ}$ − 1

= 4 − 1

= 3

Read lesson note on Properties Of Waves: Reflection, Refraction, Diffraction (WAEC)

Read lesson note on Electrostatics (JAMB)

Properties Of Waves: Reflection, Refraction, Diffraction

Electrostatics

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Question 29 Report

Calculate the electric field intensity between two plates of potential difference 6.5V when separated by a distance of 35cm.

18.57NC

^{- 1}

53.06NC

^{- 1}

2.28NC

^{- 1}

0.80NC

^{- 1}

Answer Details

The electric field intensity E between two parallel plates of potential difference V separated by a distance d is given by E = V/d. Substituting the given values, we have E = 6.5V/0.35m = 18.57 NC^−1. Therefore, the electric field intensity between the two plates is 18.57 NC^−1. Answer: 18.57NC^−1

Read lesson note on Electric Field (WAEC)

Electric Field

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Question 33 Report

Calculate the angle of minimum deviation of a 60o prism of a refractive index [sin-10.75 = 49o]

38.00o

19.47o

16.25o

49o

Answer Details

For a refractive index () =

\frac{\sin \frac{1}{2} (A + D)}{\sin \frac{1}{2} A}

D = angle of minimum deviation

A = refractive angle of the prism

1.5 = $\frac{\sin \frac{1}{2} (60 + D)}{\sin \frac{1}{2} \times 60}$

1.5 = $\frac{\sin \frac{1}{2} (60 + D)}{\sin 30}$

Sin $\frac{1}{2}$ (60 + D) = 1.5 * Sin 30

Sin $\frac{1}{2}$ (60 + D) = 0.75

$\frac{1}{2}$ (60 + D) = Sin-1 (0.75)

but Sin-1 (0.75) = 49o

$\frac{1}{2}$ (60 + D) = 49

60 + D = 2 * 49 = 98o

D = 98o - 60o

D = 38o

Read lesson note on Properties Of Waves: Reflection, Refraction, Diffraction (WAEC)

Properties Of Waves: Reflection, Refraction, Diffraction

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Question 38 Report

Calculate the specific latent heat of vaporization of steam of 1.13 x 106J pf heat energy is required to convert 15kg of it to water.

7.53 x 105 Jkg

^{- 1}

7.53 x 10-2Jkg

^{- 1}

7.53 x 104Jkg

^{- 1}

7.53 x 10-3Jkg

^{- 1}

Answer Details

The specific latent heat of vaporization of steam is the amount of heat energy required to convert one kilogram of water into steam at a constant temperature. The given problem states that 1.13 x 106 J of heat energy is required to convert 15 kg of steam to water. We can use the formula: heat energy = mass x specific latent heat where the mass is 15 kg and the heat energy is 1.13 x 106 J. Rearranging the formula, we get: specific latent heat = heat energy / mass Substituting the values, we get: specific latent heat = 1.13 x 106 J / 15 kg = 7.53 x 104 J/kg Therefore, the specific latent heat of vaporization of steam is 7.53 x 104 J/kg. The correct option is (3): 7.53 x 104 J/kg-1.

Read lesson note on Quantity Of Heat (JAMB)

Quantity Of Heat

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