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JAMB UTME - Physics - 2018

Question 1 Report

The photo cell works on the principle of the

Voltaic cell

Emission of electron by incident radiation

Emission of protons by incident radiation

Photographic plate

Answer Details

The photocell works on the principle of the emission of electrons by incident radiation. In simple terms, a photocell is a device that converts light energy into electrical energy. It does this by using a material (such as silicon) that releases electrons when it is exposed to light. These electrons can then be collected and used to produce a current, which can be used to power an electrical device. The more light that hits the photocell, the more electrons are released and the greater the electrical current.

Read lesson note on Electrostatics (JAMB)

Electrostatics

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Question 5 Report

An object weighs 30N in air and 21N in water. The weight of the object when completely immersed in a liquid of relative density 1.4 is

25.2N

17.4N

12.6N

9.0N

Answer Details

Weight of water displaced = upthrust = 30 - 21 = 9N
Mass of water displaced = $\frac{9}{10}$ = 0.9kg
Volume of object = 9 × 10 $^{- 4}$ m $^{3}$
= (9 × 10 $^{- 4}$ ) (1.4 ×103)

= 1.26kg = 12N
30 - 12.6 = 17.4N

Read lesson note on Gravitational Field (JAMB)

Gravitational Field

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Question 6 Report

The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the efficiency of the machine is 40, then the value of W is?

28.0

48.0N

233.0N

50.0N

Answer Details

Let the total number of pulleys used in both the blocks be

n

.
In a block-and-tackle pulley system, the velocity ratio is equal to n.
Efficiency =

\frac{MA}{VR} \times 100 %

MA = \frac{L}{E}, VR = n

Efficiency =

\frac{L}{E} \times \frac{1}{n} \times 100 %

E = \frac{L}{Eff. \times n} \times 100 %

E = \frac{120 N}{40 % \times 6} \times 100 %

E = 50 N

Read lesson note on Simple Machines (JAMB)

Simple Machines

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Question 7 Report

A body of mass 100g moving with a velocity if 10ms-1 collides with a wall. If after collision it moves with a velocity of 2.0ms−1 in the opposite direction, calculate the change in momentum.

0.8Ns

1.2Ns

12.0Ns

80.0Ns

Answer Details

Momentum = Mass x Velocity
Change in Momentum = Mass x (change in velocity)
Change in Momentum = Mass x (v - u)
Mass = 100g = 0.1kg
U = 2ms

^{- 1}

V = 10ms

^{- 1}

Change in momentum = m (v − u)
= 0.1 (10 − 2)
= 0.1 (8)
= 0.8Ns

Read lesson note on Motion (JAMB)

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Question 18 Report

To keep a vehicle moving at a constant speed V requires power P from the engine. The force provided by the engine is

P

V

(

\frac{1}{2}

)V

PV

P

V²

Answer Details

The force provided by the engine to keep a vehicle moving at a constant speed is proportional to the power (P) required from the engine. This power is proportional to the product of the speed (V) and force (P), so the relationship can be expressed as P = kV, where k is a proportionality constant.

Read lesson note on Work, Energy And Power (WAEC)

Work, Energy And Power

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Question 35 Report

Water of mass 150g at \(60^\circ\)c is added to 300g of water at \(20^\circ\)c and the mixture is well stirred. Calculate the temperature of the mixture.(neglect heat losses to the surroundings)

33

^{o}

c

40

^{o}

c

25

^{o}

c

10

^{o}

c

Answer Details

To solve this problem, we can use the principle of conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted from one form to another. In this case, the energy transferred is in the form of heat. We can use the formula: Q = m*c*(ΔT) where Q is the heat transferred, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. First, we can calculate the heat transferred from the hot water to the cold water: Q1 = 150g * 4.18 J/(g°C) * (60°C - T) Q1 = 627 * (60 - T) where T is the temperature of the mixture. Next, we can calculate the heat transferred from the cold water to reach the final temperature of the mixture: Q2 = 300g * 4.18 J/(g°C) * (T - 20°C) Q2 = 1254 * (T - 20) Since the heat transferred between the two water samples must be equal, we can set Q1 equal to Q2 and solve for T: 627 * (60 - T) = 1254 * (T - 20) 37620 - 627T = 1254T - 25080 1881T = 62760 T = 33.4°C Therefore, the temperature of the mixture is approximately 33°C. Answer: 33°C

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

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Question 38 Report

A beam of light is incident from air to water at an angle of 300. Find the angle of refraction if the refractive index of water is 4/3

15

^{o}

18

^{o}

22

^{o}

24

^{o}

Answer Details

X =

\frac{4}{3}

r = ?
Shell?s law:. 7 =

\frac{S i n 2^{0}}{S i n r^{0}}

\frac{V}{g}

=

\frac{S i n 30^{0}}{S i n r^{0}}

Sin

r^{0}

=

\frac{3 S i n 30^{0}}{4}

Sin

r^{0}

= 0.375
R

^{o}

= Sin-1 (0.375)
R

^{o}

= 22.02

^{o}

R

^{o}

= 22

^{o}

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

Read lesson note on Refraction Of Light Through Plane And Curved Surfaces (JAMB)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Refraction Of Light Through Plane And Curved Surfaces

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