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JAMB UTME - Physics - 1997

Vraag 1 Verslag

A boy on looking into a mirror observed that his face appeared to have grown bigger. The boy must have been looking at a

convex mirror with his face at the focus

concave mirror with his face between the focus and the mirror

concave mirror with his face at the focus

convex mirror with his face between the focus and the mirror

Antwoorddetails

When an object is placed in front of a concave mirror, the image formed can either be real or virtual depending on the position of the object relative to the mirror. If the object is placed between the focus and the mirror, the image formed is virtual, erect and magnified. Therefore, the boy must have been looking at a concave mirror with his face between the focus and the mirror.

Lees lesnotitie op Electrostatics (JAMB)

Electrostatics

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Vraag 5 Verslag

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

Antwoorddetails

F =

\frac{1}{2 L}

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

F1L1 = F2L2

500 x 0.9 = 150 x L2

therefore, L2 = 3m

Lees lesnotitie op Motion (JAMB)

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Vraag 6 Verslag

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%

Antwoorddetails

Efficiency = output x 100

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

= 80%

Lees lesnotitie op Electrical Energy And Power (JAMB)

Electrical Energy And Power

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Vraag 7 Verslag

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

Antwoorddetails

\frac{θ - 0}{100 - 0}

=

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

$θ$ = 75.0oC

Lees lesnotitie op Measurements And Units (JAMB)

Measurements And Units

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Vraag 11 Verslag

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

Antwoorddetails

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

Lees lesnotitie op Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

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Vraag 15 Verslag

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

Antwoorddetails

F = mg + ma

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

F = 35 N

Lees lesnotitie op Scalars And Vectors (WAEC)

Scalars And Vectors

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Vraag 24 Verslag

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

Antwoorddetails

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

Bekijk antwoord

Vraag 32 Verslag

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

Antwoorddetails

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

Lees lesnotitie op Current Electricity (JAMB)

Current Electricity

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Vraag 36 Verslag

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

Antwoorddetails

g =

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

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

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

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

Lees lesnotitie op Gravitational Field (JAMB)

Gravitational Field

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Vraag 37 Verslag

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

Bekijk antwoord

Vraag 39 Verslag

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

Antwoorddetails

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

Lees lesnotitie op Scalars And Vectors (WAEC)

Scalars And Vectors

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Vraag 43 Verslag

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

Antwoorddetails

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

Lees lesnotitie op Motion (WAEC)

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Vraag 45 Verslag

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

Antwoorddetails

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

Lees lesnotitie op Motion (JAMB)

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