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JAMB UTME - Physics - 2019

Question 1 Report

A train has an initial velocity of 44m/s and an acceleration of -4m/s\(^{2}\). Calculate its velocity after 10 seconds

10m/s

6m/s

8m/s

4m/s

Answer Details

The velocity of the train after 10 seconds can be calculated using the formula: v = u + at where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Substituting the given values, we get: v = 44 m/s + (-4 m/s^2) x 10 s v = 44 m/s - 40 m/s v = 4 m/s Therefore, the velocity of the train after 10 seconds is 4m/s. Answer option D is correct. Explanation: The train has an initial velocity of 44 m/s and an acceleration of -4 m/s^2. The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which means that the train is slowing down. After 10 seconds, the train's velocity decreases by 40 m/s (4 m/s^2 x 10 s) to reach a final velocity of 4 m/s.

Read lesson note on Motion (JAMB)

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Question 4 Report

When two objects A and B are supplied with the same quantity of heat, the temperature change in A is obtained to be twice that of B. The mass of P is half that of Q. The ratio of the specific heat capacity of A to B is

1 : 4

4 : 1

1 : 1

2 : 1

Answer Details

θA = 2θ $_{B}$ ,

m

_{A}

=

12

m

_{B}

H = MCθ
m $_{A}$ c $_{A}$ θ $_{A}$ = m $_{B}$ c $_{B}$ θ $_{B}$
( 1/2 m $_{B}$ )C $_{A}$ (2θ $_{B}$ ) = m $_{B}$ c $_{B}$ θ $_{B}$

C

_{A}

C

_{B}

=

11

⇒ 1 : 1

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

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Question 5 Report

In a slide wire bridge, the balance is obtained at a point 25cm from one end of wire 1m long. The resistance to be tested is connected to that end and a standard resistance of \(3.6\,\Omega\) is connected to the other end of the wire. Determine the value of the unknown resistance

3.2

Ω

1.4

Ω

3.21

Ω

1.2

Ω

Answer Details

$\frac{R}{3.6} = \frac{75}{25} = \frac{1}{3}$

3R = 3.6

R = 1.2 $Ω$

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Question 14 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

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Question 16 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 18 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

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Question 23 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 24 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 26 Report

Water and Kerosine are drawn respectively into the two limbs of a Hare's apparatus. The destiny of water is \(1.0\mathrm{g\,cm^{-3}}\) and the density of kerosine is \(0.80\mathrm{g\,cm^{-3}}\). If the height of the water column is 20.0cm, calculate the height of the kerosine column.

1.6cm

16.0cm

20.0cm

25.0cm

Answer Details

Devices with different liquids
d $_{1}$ h $_{1}$ = d $_{2}$ h $_{2}$
1 × 20 = 0.8 × h

h

=

200.8

=

25cm

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

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Question 30 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

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