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JAMB UTME - Physics - 1997

Swali 1 Ripoti

What happens when a gas expands at constant temperature?

its pressure decreases

the total momentum of its molecules remains constant

its pressure decreases and the total momentum of its moleculees remains constant

its pressure decreses and the total kinetic energy of its molecule decreases

Maelezo ya Majibu

When a gas expands at constant temperature, its pressure decreases. This is due to the fact that the gas molecules now have more space to move around, and as a result, they collide with the walls of the container less frequently, causing the pressure to decrease. However, the total momentum of the gas molecules remains constant. This is because momentum is a conserved quantity, and the gas molecules only transfer their momentum to other molecules or to the walls of the container. Therefore, the total momentum of the gas remains the same, even though the individual molecules may have different speeds and directions. Option (C) is correct, as it correctly identifies both of these phenomena. Option (D) is incorrect, as the total kinetic energy of the gas molecules does not necessarily decrease when the gas expands at constant temperature. While individual molecules may lose kinetic energy due to collisions, others may gain kinetic energy due to the same collisions.

Soma maelezo ya somo kwenye Gas Laws (JAMB)

Tazama Jibu

Swali 16 Ripoti

In the diagram above, which of the angles \( \theta_1 \), \( \theta_2 \), \( \theta_3 \) and \( \theta_4 \) is the angle of deviation of the ray of light passing through the glass prism XYZ?

θ3

θ2

θ_{1}

θ4

Tazama Jibu

Swali 17 Ripoti

A body of mass 2kg moving vertically upwards has its velocity increased uniformly from 10ms-1 to 40ms-1 in 4s. Neglecting air resistance, calculate the upward vertical force acting on the body.[g = 10 ms-1]

15N

20N

35N

45N

Maelezo ya Majibu

F = mg + ma

F = 2 [10 + $\frac{(40 - 10)}{4}$ ]

F = 35 N

Soma maelezo ya somo kwenye Scalars And Vectors (WAEC)

Scalars And Vectors

Tazama Jibu

Swali 18 Ripoti

An electric water pump rate 1.5 kW, lifts 200kg of water through a vertical height of 6 meters in seconds. What is the efficiency of the pump? [g = 10ms-2, neglecting air resistance]

90.0%

85.0%

80.0%

65.0%

Maelezo ya Majibu

Efficiency = output x 100

= $\frac{\frac{200 \times 10 \times θ}{10}}{1500}$ x $\frac{100}{1}$

= 80%

Soma maelezo ya somo kwenye Electrical Energy And Power (JAMB)

Electrical Energy And Power

Tazama Jibu

Swali 21 Ripoti

The distance xm travelled by a particle in time t seconds is described by the equation x = 10 + 12t2, Find the average speed of the particle between the time interval t = 2s and t = 5s

60 ms-1

72 ms-1

84 ms-1

108 ms-1

Maelezo ya Majibu

speed =

\frac{d x}{d t}

$\frac{d x}{d t}$ = 10 + 12t2

$\frac{d x}{d t}$ = 24t

$\frac{d x}{d t}$ =24 x 3

$\frac{d x}{d t}$ = 72 ms-1

Soma maelezo ya somo kwenye Motion (WAEC)

Tazama Jibu

Swali 24 Ripoti

Two bodies X and Y are projected on the same horizontal plane, with the same initial speed but at angles 30 and 60 respectively to the horizontal. Neglecting air resistance, the ratio of the range of X to that of Y is

1:1

1:2

√3:1

√3

Maelezo ya Majibu

Range =

\frac{U^{2} s i n 2 θ}{g}

$\frac{R_{x}}{R_{y}}$ = $\frac{U_{x}^{2} s i n 60}{U_{y}^{2} s i n 120}$

sin60o = sin120o

$\frac{R_{x}}{R_{y}}$ = $\frac{1}{1}$

hence, = 1:1

Soma maelezo ya somo kwenye Motion (JAMB)

Tazama Jibu

Swali 29 Ripoti

The voltage of the domestic electric supply is represented by the equation \(V = 311 \sin 314.2t\). Determine the frequency of the a.c. supply \([\pi = 3.142]\)

50.0Hz

100.0Hz

311.0Hz

314.2Hz

Maelezo ya Majibu

V = 311 sin314.2t compare with,

V = V sin2 $π$ Ft

2 $μ$ F = 314.2

F = $\frac{314.2}{2 \times 3.142}$

F = 50Hz

Tazama Jibu

Swali 30 Ripoti

Three electric cells each of e.m.f 1.5V and internal resistance of 1.0 \(\Omega\) are connected in parallel across an external resistance of \(\frac{2}{3}\Omega\). Calculate the value of the current in the resistor

0.5 A

0.9A

1.5A

4.5A

Maelezo ya Majibu

E.m.f = 1.5

r = $\frac{1 \times 1 \times 1}{(1 \times 1) + (1 \times 1) + (1 \times 1)}$

r = $\frac{1}{3}$ $Ω$

I = $\frac{E . m . f}{r + R}$

I = $\frac{1.5}{\frac{1}{3} + \frac{2}{3}}$

I = 1.5A

Soma maelezo ya somo kwenye Current Electricity (JAMB)

Current Electricity

Tazama Jibu

Swali 31 Ripoti

The length of mercury thread when it is at 0^oC, 100^oC and at an unknown temperature \( \theta \) is 25mm, 225mm and 175mm respectively. The value of \( \theta \) is

85.0oC

80.0oC

75.0oC

70.0oC

Maelezo ya Majibu

\frac{θ - 0}{100 - 0}

=

\frac{175 - 25}{225 - 25}

$θ$ = 75.0oC

Soma maelezo ya somo kwenye Measurements And Units (JAMB)

Measurements And Units

Tazama Jibu

Swali 33 Ripoti

An object of mass 100g projected vertically upwards from the ground level has a velocity of 20ms-1 at a height of 10m. calculate its initial kinetic energy at the ground level.[g = 10ms-2, neglect air resistance]

10 J

20 J

30 J

50 J

Maelezo ya Majibu

Using, v2 = u2 - 2gx

(20)2 = u21/2 x 10 x 10

u2 = 600(m/s)2

initial K.E = - mu2

= $\frac{1}{2}$ x 600

= 30 J

Soma maelezo ya somo kwenye Scalars And Vectors (WAEC)

Scalars And Vectors

Tazama Jibu

Swali 34 Ripoti

Under constant tension and constant mass per unit length, the note produced by a plucked string is 500Hz when the length of the string is 0.90m. At what length is the frequency 150Hz?

3m

4m

5m

6m

Maelezo ya Majibu

F =

\frac{1}{2 L}

\sqrt{\frac{T}{m}}

F1L1 = F2L2

500 x 0.9 = 150 x L2

therefore, L2 = 3m

Soma maelezo ya somo kwenye Motion (JAMB)

Tazama Jibu

Swali 37 Ripoti

In a meter bridge experiment, two resistor \(2\Omega\) and \(3\Omega\) occupy the left and right gaps respectively. Find the balance point from the left side of the bridge

20cm

40cm

60cm

80cm

Maelezo ya Majibu

Let the point be x

$\frac{2}{x}$ = $\frac{3}{100 - x}$

making x subject of formula,

2(100 - x) = 3x

200 - 2x = 3x

200 = 3x + 2x

200 = 5x

divide both side by 5

x = 40cm

Soma maelezo ya somo kwenye Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

Tazama Jibu

Swali 42 Ripoti

A planet has mass m1 and is at a distance r, from the sun. A second planet has mass m2 = 10m1 and at a distance of r2 = 2r1 from the sun. Determine the ratio of the gravitational force experienced by the planets.

1 : 5

2 : 5

3 : 5

4 : 5

Maelezo ya Majibu

g =

\frac{G M}{R^{2}}

$\frac{G_{1} M_{1}}{r_{1}^{2}}$ = $\frac{G_{2} M_{2}}{r_{2}^{2}}$

$\frac{G_{1} M_{1}}{r_{1}^{2}}$ = $\frac{G_{2} \times 10 M_{1}}{(2 r_{1})^{2}}$

$\frac{G_{1}}{G_{2}}$ = $\frac{2}{5}$

Soma maelezo ya somo kwenye Gravitational Field (JAMB)

Gravitational Field

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