Loading....






Press & Hold to Drag Around
Click Here to Close

JAMB UTME - Physics - 1985

Question 1 Report

Two thermo flasks of volume \(V_x\) and \(V_y\) are filled with liquid water at an initial temperature of 0^oC. After sometime the temperature were found to be \(\theta_x\), \(\theta_y\), respectively. Given that \(\frac{V_x}{V_y}=2\) and \(\frac{\theta_x}{\theta_y}=\frac{1}{2}\) the ratio of the heat flow into the flask is

1

4

1

2

4

1

2

Answer Details

Heat = mc

θ

Hx = Dx x Vx x Cx x $θ$ x

Hy = Dy x Vy x Cy x $θ$ y

$\frac{H_{x}}{H_{y}}$ = $\frac{D_{x} \times V_{x} \times C_{x} \times θ_{x}}{D_{x} \times V_{x} \times C_{x} \times θ_{x}}$

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

= 1

Read lesson note on Quantity Of Heat (JAMB)

Quantity Of Heat

View Answer

Question 2 Report

If the fundamental frequency of a closed pipe organ on a day when the speed of sound is 340ms-1 is 170Hz, then the length of the pipe is

50cm

70cm

100cm

150cm

200cm

Answer Details

λ

=

\frac{v}{f}

= $\frac{340}{170}$

= 2m ; for closed pipe

L = $\frac{1}{4}$ $λ$

= $\frac{1}{4}$ x $\frac{2}{1}$

= 0.5m

= 50cm

Read lesson note on Current Electricity (JAMB)

Current Electricity

View Answer

Question 3 Report

A quantity of gas occupies a certain volume when the temperature is -73oC and the pressure is 1.5 atmospheres. If the pressure is increased to 4.5 atmospheres and the volume is halved at the same time, what will be the new temperature of the gas?

573oC

327oC

300oC

110oC

27oC

Answer Details

\frac{P_{1} V_{1}}{T_{1}}

=

\frac{P_{2} V_{2}}{T_{2}}

V1 = 2V2

$\frac{1.5 \times 2 V_{2}}{- 73 + 273}$ = $\frac{4.5 \times V_{2}}{T_{2}}$

T2 = 300K = 27oC

Read lesson note on Quantity Of Heat (JAMB)

Quantity Of Heat

View Answer

Question 4 Report

A block of mass 2.04kg resting on a smooth horizontal plane is acted upon simultaneously by two-forces, 10N due North and 10N due East. The magnitude of the acceleration produced by the force on the block is

0.10ms-2

7.07ms-2

10.00ms-2

14.10ms-2

20.00ms-2

Answer Details

R =

\sqrt{10^{2} + 10^{2}}

R = 14.14N

R = F = ma

14.14 = 2 x a

a = 7.07ms-1

Read lesson note on Scalars And Vectors (WAEC)

Read lesson note on Equilibrium Of Forces (WAEC)

Scalars And Vectors

Equilibrium Of Forces

View Answer

Question 5 Report

A wire P has half the diameter and half the length of a wire Q of similar material. The ratio of the resistance of P to that of Q is

8 : 1

4 : 1

2 : 1

1 : 1

1 : 4

Answer Details

Resistance

α

\frac{l e n g t h}{a r e a}

$\frac{R_{Q}}{R_{P}}$ = $\frac{L_{Q} A_{P}}{L_{P} A_{Q}}$

LQ = 2LP

DQ = 2DP

AP = $π$ $\frac{(2 D_{P})^{2}}{2}$

= $π$ DP

AQ = $π$ $\frac{(D_{P})^{2}}{2}$

= $\frac{1}{4}$ $π$ DP

therefore, $\frac{R_{Q}}{R_{P}}$ = $\frac{2 L_{P} \times π D_{P}}{L_{P} \times \frac{1}{4} π D_{P}}$

= 2 : 1

Read lesson note on Capacitors (JAMB)

Read lesson note on Current Electricity (JAMB)

Current Electricity

View Answer

Question 6 Report

If a car starts from rest and moves with a uniform acceleration of 10ms-2 for ten seconds, the distance it covers in the last one second of its motion is

95m

100m

500m

905m

1000m

Answer Details

Distance in 105, using, x = ut +

\frac{1}{2}

at2

x = (0 x 10) + ( $\frac{1}{2}$ x 10 x 10 x 10) = 500m

distance in 9s; x - (0 x 9) + ( $\frac{1}{2}$ x 10 x 9 x 9) = 405m

distance in last is = difference

= 500 - 405

= 95m

Read lesson note on Motion (JAMB)

View Answer

Question 7 Report

Two liquids, P at a temperature of 20oC and Q at a temperature of 80oC have specific heat capacities of 1.0Jkg-1oC-1 and 1.5J kg-1oC-1respectively. if equal masses of P and Q are mixed in a lagged calorimeter, then the equilibrium temperature is

44oC

50oC

56oC

60oC

70oC

Answer Details

Heat lost = heat gained

m x (80 - $θ$ ) x 1.5 = m x ( $θ$ - 20) x 1

$θ$ = 56oC

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

View Answer

Question 14 Report

In order to find the depth of the sea, a ship sends out a sound wave and receives an echo after one second. If the velocity of sound in water is 1500ms-1, what is the depth of the sea?

0.75km

1.50km

2.20km

3.00km

3.75km

Answer Details

Echo time = 1

\div

2 = 0.5sec

depth = velocity x time

= 1500 x 0.5

= 750m = 0.75km

Read lesson note on Sound Waves (WAEC)

Sound Waves

View Answer

Question 15 Report

Two divers G and H are at depths 20m and 40m respectively below the water surface in a lake. The pressure on G is P₁ while the pressure on H is P₂. If the atmospheric pressure is equivalent to 10m of water, then the value of \( \frac{p_2}{p_1} \) is

0.50

0.60

1.67

2.00

3.0

Answer Details

pressure at H = 40m ; pressure at G, p1 = 20m

$\frac{p_{1}}{p_{1}}$ = $\frac{40}{20}$

= 2

Read lesson note on Temperature And Its Measurement (JAMB)

Read lesson note on Pressure (JAMB)

Read lesson note on Measurements And Units (JAMB)

Temperature And Its Measurement

Measurements And Units

View Answer

Question 16 Report

The areas of the effort and load pistons of a hydraulic press are 0.5m2 and 5m2 respectively. If a force F1 of 100N is applied on the effort piston, the force F2 on the load is

10N

100N

500N

1000N

5000N

Answer Details

\frac{l o a d}{a r e a}

=

\frac{e f f o r t}{a r e a}

$\frac{l o a d}{5}$ = $\frac{100}{0.5}$

load = 1000N

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

Position, Distance And Displacement

View Answer

Question 18 Report

Two rays of light from a point below the surface of water are equally inclined to each other at \(60^\circ\) in water. What is the angle between the rays when they emerge into air? (Take the refractive index of water to be \(\frac{4}{3}\))

41.8o

44.1o

60.0o

83.6o

120.0o

Answer Details

\frac{4}{3}

=

\frac{s i n r_{1}}{s i n 30}

and

\frac{4}{3}

=

\frac{s i n r_{2}}{s i n 30}

r1 = 41.8o and r2 = 41.8o

angle between them = r1 + r2

= 83.6o

Read lesson note on Refraction Of Light Through Plane And Curved Surfaces (JAMB)

Refraction Of Light Through Plane And Curved Surfaces

View Answer

Question 21 Report

An electric kettle, connected to a 240V mains, produces 6.0 x 105J of heat energy to boil a quantity of water in 5 minutes. Find the resistance of the kettle.

14.4

Ω

28.8

Ω

144

Ω

288

Ω

2880

Ω

Answer Details

Heat energy = Electrical energy

6 x 105 = $\frac{V^{2}}{R}$ t

R = $\frac{240 \times 240 \times 5 \times 60}{6 \times 105}$

= 28.8 $Ω$

Read lesson note on Current Electricity (JAMB)

Read lesson note on Heat Energy (WAEC)

Current Electricity

Heat Energy

View Answer

Question 25 Report

A uniform cylindrical block of wood floats in water with one-third of its height above the water level, in a liquid of relative density 0.8, what fraction of its height will be above the liquid level?

1

6

1

5

1

3

4

5

\frac{5}{6}

Answer Details

Height inside H2O =

\frac{2}{3}

R.d = $\frac{height inside H_{2} O}{height inside liquid}$

0.8 = $\frac{\frac{2}{3}}{X}$

View Answer

Question 26 Report

A thin aluminum plate has a surface area of 1500m at 20oC. What will be its surface area when it is cooled to -20oC? (Take the linear expansivity of aluminum to be 2.5 x 10-5k-1).

1.503m2

1.500m2

1.498m2

1.497m2

1.490m2

Answer Details

β

= A2 - A1

2 X 10-5 x 2.5 = $\frac{A_{2} \times 1.5}{1.5 \times (- 20 - 20)}$

A2 = 1.498m2

View Answer

Question 27 Report

A 0.05kg bullet travelling at 500ms-1 horizontally strikes a thick vertical wall. It stops after penetrating through the wall horizontal distance of 0.25m. What is the magnitude of the average force the wall exerts on the bullet?

25N

50N

250N

500N

25 000N

Answer Details

F = ma

= m $\frac{(v^{2} - u^{2})}{2 s}$

F = 0.05 $\frac{(0 - (500)^{2})}{2 \times 0.25}$

F = 25 000N

Read lesson note on Current Electricity (JAMB)

Current Electricity

View Answer

Question 28 Report

A force of 100N stretches an elastic string to a total length of 20cm. If an additional force of 100N stretches the string 5cm further, Find the natural length of the string

15cm

12cm

10cm

8cm

5cm

Answer Details

Using, f1 = f2

e1 = 20 - L ; e2 = 5cm

$\frac{f_{1}}{e_{1}}$ = $\frac{f_{2}}{e_{2}}$

$\frac{100}{20 - L}$ = $\frac{100}{5}$

hence, L = 15cm

Read lesson note on Elasticity (JAMB)

View Answer

Question 31 Report

Which of the following features is NOT a characteristic of natural radioactivity?

Radioactivity is a nuclear phenomenon

Radioactivity is exhibited only by elements of mass number greater than 206

The radioactivity of an element is affected by electric and magnetic fields in the surroundings

Radioactive substances emits three types of radiations, a-rays,

β

-rays and y-rays

The radioactive element is transformed into a new element

Properties Of Waves: Reflection, Refraction, Diffraction

View Answer

Question 37 Report

A simple pendulum with a period of 2.0s has its length doubled. Its new period is

1.00s

1.41s

0.3s

2.83s

4.00s

Answer Details

Using

\frac{T_{1}}{T_{2}}

=

\sqrt{\frac{L_{1}}{L_{2}}}

$\frac{2}{T_{2}}$ = $\sqrt{\frac{L_{1}}{2 L_{1}}}$

T2 = 2.83sec

View Answer

Question 41 Report

A metal cube of volume 103mm3 is lowered into a measuring cylinder containing water. If the internal cross-sectional area of the cylinder is 1.5 x 102mm2, by how much does the water level rise in the cylinder?

6.67 x 100mm

8.50 x 102mm

1.15 x 103mm

2.50 x 103mm

1.50 x 105mm

Answer Details

Height =

\frac{v o l u m e}{a r e a}

= Height = $\frac{10^{3}}{1.5 \times 10^{2}}$

= 6.67mm

Read lesson note on Measurements And Units (JAMB)

Measurements And Units

View Answer

Question 44 Report

In a gas experiment, the pressure of the gas is plotted against the reciprocal of the volume of the gas at a constant temperature. the unit of the slope of the resulting curve is

force

force/m

work

force/m3

energy/m3

Answer Details

slope = pressure

\div

\frac{1}{v o l u m e}

= $\frac{N}{m^{3}}$ x $\frac{m^{3}}{1}$

= Nm = work

Read lesson note on Gas Laws (JAMB)

View Answer

Question 45 Report

A transformer has a primary coil with 500 turns and a secondary coil with 2500 turns. When the voltage input to the primary coil is 120V, the output is

6000V

600V

240V

60V

24V

Answer Details

For transformers,

\frac{V_{p}}{V_{s}}

=

\frac{N_{p}}{N_{s}}

=

\frac{I_{p}}{I_{s}}

$\frac{500}{2500}$ = $\frac{120}{V_{s}}$

Vs = 600v

Read lesson note on Current Electricity (JAMB)

Current Electricity

View Answer

Question 47 Report

A metal block of mass 5kg lies on a rough horizontal platform. If a horizontal force of 8N applied to the block through its center of mass just slides the block on the platform, then the coefficient of limiting friction between the block and platform is

0.16

0.63

0.80

1.60

2.00

Answer Details

F =

μ

R

= $μ$ mg

8 = $μ$ x 5 x 10

$μ$ = 0.16

Read lesson note on Friction (JAMB)

View Answer

Question 49 Report

If the distance between two stationary charged particle is doubled, the magnitude of the electro-static force between them will be

doubled

halved

a quarter of its former value

unchanged

four times the original value

Answer Details

F

α

\frac{1}{d^{2}}

F x d2 = constant

F1 x (d1)2 = F2 x (d2)2

d2 = 2d1

F1 x (d1)2 = F2 x (2d1)2

F2 = $\frac{1}{4}$ F1

Read lesson note on Electrostatics (JAMB)

Electrostatics

View Answer

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

37

38

39

40

41

42

43

44

45

46

47

48

49

50