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JAMB UTME - Physics - 2019

Question 1 Report

The mass of water vapour in a given volume of air is 0.05g at 20°C, while the mass of water vapour required to saturate it at the same temperature is 0.15g. Calculate the relative humidity of the air.

3.33%

5.55%

33.33%

55.55%

Answer Details

Relative humidity is a measure of how much water vapor the air is holding compared to the maximum amount it could hold at a given temperature. It is expressed as a percentage. To calculate the relative humidity of the air in this problem, we need to use the formula: Relative humidity = (mass of water vapor in air / mass of water vapor required for saturation) x 100% We are given that the mass of water vapor in the air is 0.05g and the mass of water vapor required for saturation at the same temperature is 0.15g. Plugging these values into the formula, we get: Relative humidity = (0.05 / 0.15) x 100% = 33.33% Therefore, the relative humidity of the air is 33.33%. So the answer is 33.33%.

Read lesson note on Quantity Of Heat (JAMB)

Quantity Of Heat

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Question 12 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 13 Report

The value of T in the figure above is

30N

10.0N

20N

40N

Answer Details

Tsin30 + Tsin30 =40

2Tsin30 = 40

Tsin30 = 40/2 = 20

T( $\frac{1}{2}$ ) = 20

T = 20 x 2 = 40N

Read lesson note on Scalars And Vectors (JAMB)

Read lesson note on Motion (JAMB)

Scalars And Vectors

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Question 21 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 23 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 39 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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