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JAMB UTME - Physics - 2019

Question 1 Report

A thermocouple thermometer is connected to a millivoltmeter which can read up to 10mV. When one junction is in ice at 0°C and the other is steam at 100°C, the millivoltmeter reads 4mV. What is the maximum temperature which this arrangement can measure

100

^{o}

C

248

^{o}

C

250

^{o}

C

350

^{o}

C

Answer Details

The maximum temperature which this arrangement can measure is 250°C. A thermocouple thermometer works by using the thermoelectric effect, which is the phenomenon that occurs when two dissimilar metals are joined together to form a loop and a temperature difference is established between the two junctions. This temperature difference generates a small electrical voltage, which can be measured using a millivoltmeter. The voltage generated is proportional to the temperature difference between the two junctions. In the case of the thermocouple thermometer described, one junction is in ice at 0°C and the other is steam at 100°C, and the millivoltmeter reads 4mV. This means that the voltage generated by the thermocouple is 4 millivolts, which corresponds to a temperature difference of 100°C. However, the millivoltmeter can only read up to 10mV, so the maximum temperature difference it can measure is 10mV / 4mV/°C = 250°C. This means that the maximum temperature which this arrangement can measure is 250°C.

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

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Question 3 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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Question 6 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 7 Report

A body moves in SHM between two point 20m on the straight line Joining the points. If the angular speed of the body is 5 rad/s. Calculate its speed when it is 6m from the center of the motion.

20 m/s

30 m/s

40 m/s

50 m/s

Answer Details

From two parts 20m apart
a = 10m, x = 6m, A = 5
V = ω

\sqrt{A^{2} - X^{2}}

= 5

\sqrt{10^{2} - 6^{2}}

= 40m/s

Read lesson note on Simple Harmonic Motion (WAEC)

Simple Harmonic Motion

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Question 17 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 19 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

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Question 22 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 28 Report

A car moving at 20m/s with its horn blowing (f = 1200Hz) is chasing another car going at 15m/s. What is the apparent frequency of the horn as heard by the driver being chased?

1.22KHz

2.40KHz

8.12KHz

1.22Hz

Answer Details

f

_{1}

=

f(v - v

_{o}

)v - v

_{s}

=

1200(340 - 15)340 - 20

=

1.22KHz

Read lesson note on Position, Distance And Displacement (WAEC)

Position, Distance And Displacement

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Question 38 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)

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Question 40 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

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