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JAMB UTME - Physics - 2018

Question 1 Report

The density of 400cm3 of palm oil was 0.9gcm-3 before frying. If the density of the oil was 0.6gcm-3 after frying, assuming no loss of oil due to spilling, its new volume was?

1360cm

^{3}

600cm

^{3}

240cm

^{3}

8000m

^{3}

Answer Details

The density of a substance is defined as its mass per unit volume. Therefore, the mass of the palm oil before frying was: Mass = Density x Volume = 0.9 g/cm³ x 400 cm³ = 360 g After frying, the mass of the palm oil remains the same, but its density changes to 0.6 g/cm³. Therefore, the new volume of the palm oil can be calculated by rearranging the density formula: Volume = Mass / Density = 360 g / 0.6 g/cm³ = 600 cm³ So the new volume of the palm oil after frying is 600 cm³. is the correct answer.

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Question 3 Report

An electric heating coil rated at 1KW is used to heat 2kg of water for 2 minutes. The initial water temperature is 30\(^{o}\)C. Taking the specific heat of the water as 4,000Jkg\(^{-1}\) and neglecting that of the container, the final water temperature is

15.0

^{o}

C

45.0

^{o}

C

52.5

^{o}

C

25.0

^{o}

C

Answer Details

To determine the final temperature of the water, we can use the formula: Q = mcΔT 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. We know that the power of the electric heating coil is 1KW, which means it transfers 1000 Joules of energy per second. In 2 minutes, or 120 seconds, it transfers 120,000 Joules of energy to the water. The mass of the water is given as 2kg and the specific heat capacity of water is 4000 J/kg°C. We can assume that the initial temperature of the water is 30°C. Using the formula, we can solve for the change in temperature: 120,000 J = (2 kg)(4000 J/kg°C)(ΔT) ΔT = 15°C Therefore, the final temperature of the water is 30°C + 15°C = 45°C. So, the final water temperature is 45.0oC.

Read lesson note on Quantity Of Heat (JAMB)

Quantity Of Heat

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Question 4 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 7 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 8 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 9 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 11 Report

The efficiency of energy conversion on the energy flow through a hydroelectric power is?

E1

E2

\frac{E 3}{E 2}

\frac{E 2}{E 1}

+ E2

\frac{E 2}{E 3}

+ E1

Answer Details

Efficiency =

\frac{useful energy output from machine}{energy input into machine}

=

\frac{E 3}{E 2}

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

Work, Energy And Power

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Question 20 Report

If the focal length of a camera is 20cm, the distance from the film at which the lens must be set to produce a sharp image of 100cm away is

17cm

20cm

25cm

100cm

Answer Details

F = 20cm
V = 100cm
U = ?

\frac{1}{U}

+

\frac{1}{V}

=

\frac{1}{F}

\frac{1}{20}

+

\frac{1}{100}

=

\frac{1}{F}

\frac{5 + 1}{100}

=

\frac{1}{F}

F =

\frac{100}{6}

= 16.7cm
= 17cm

Read lesson note on Vapours (JAMB)

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Question 22 Report

An object moves in a circular path of radius 0.5m with a speed of \(1ms^{-1}\). What is its angular velocity?

8rads

^{- 1}

4rads

^{- 1}

1rads

^{- 1}

2rads

^{- 1}

Answer Details

Angular velocity is a measure of how fast an object is rotating around a center point. It's usually measured in radians per second (rad/s). To calculate angular velocity, we use the formula: angular velocity = linear velocity / radius. In this case, the linear velocity is 1 m/s, and the radius is 0.5 m. So, the angular velocity would be: 1 m/s / 0.5 m = 2 rad/s Therefore, the answer is 2 rad/s or 2rads^-1

Read lesson note on Motion (JAMB)

Read lesson note on Speed And Velocity (WAEC)

Speed And Velocity

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