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JAMB UTME - Physics - 1997

Pregunta 1 Informe

An arrow of mass 0.1kg moving with a horizontal velocity of 15ms-1 is shot into a wooden block of mass 0.4kg lying at rest on a smooth horizontal surface. Their common velocity after impact is

15.0 ms-1

7.5 ms-1

3.8 ms-1

3.0 ms-1

Detalles de la respuesta

We can solve this problem using the law of conservation of momentum, which states that the total momentum of a closed system remains constant before and after an interaction. In this case, the arrow and the wooden block form a closed system. Let the velocity of the arrow after impact be v. We can set up the following equation: (mass of arrow * initial velocity of arrow) = (total mass * final velocity) 0.1 kg * 15 m/s = (0.1 kg + 0.4 kg) * v 1.5 kg m/s = 0.5 kg * v v = 3 m/s Therefore, the common velocity of the arrow and the wooden block after impact is 3 m/s. None of the answer options provided matches this result.

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Pregunta 3 Informe

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

Detalles de la respuesta

g =

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

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

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

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

Lee la nota de la lección en Gravitational Field (JAMB)

Gravitational Field

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Pregunta 5 Informe

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

Detalles de la respuesta

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

Lee la nota de la lección en Motion (JAMB)

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Pregunta 7 Informe

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%

Detalles de la respuesta

Efficiency = output x 100

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

= 80%

Lee la nota de la lección en Electrical Energy And Power (JAMB)

Electrical Energy And Power

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Pregunta 13 Informe

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

Detalles de la respuesta

\frac{θ - 0}{100 - 0}

=

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

$θ$ = 75.0oC

Lee la nota de la lección en Measurements And Units (JAMB)

Measurements And Units

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Pregunta 14 Informe

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

Detalles de la respuesta

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

Lee la nota de la lección en Current Electricity (JAMB)

Current Electricity

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Pregunta 17 Informe

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

Detalles de la respuesta

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

Lee la nota de la lección en Motion (WAEC)

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Pregunta 19 Informe

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

Detalles de la respuesta

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

Lee la nota de la lección en Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

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Pregunta 35 Informe

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

Detalles de la respuesta

F = mg + ma

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

F = 35 N

Lee la nota de la lección en Scalars And Vectors (WAEC)

Scalars And Vectors

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Pregunta 37 Informe

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

Detalles de la respuesta

F =

\frac{1}{2 L}

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

F1L1 = F2L2

500 x 0.9 = 150 x L2

therefore, L2 = 3m

Lee la nota de la lección en Motion (JAMB)

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Pregunta 40 Informe

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

Detalles de la respuesta

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

Lee la nota de la lección en Scalars And Vectors (WAEC)

Scalars And Vectors

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Pregunta 42 Informe

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

Detalles de la respuesta

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

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Pregunta 43 Informe

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

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