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JAMB UTME - Physics - 1985

Tambaya 1 Rahoto

Which of the following statements are true about the spring balance and the chemical balance? i. both are used to measure the mass of an object. ii. either of them may be used to measure the weight of an object. iii. the spring balance works on the principle of Hooke's law while the chemical balance works on the principle of moments. iv. a change in gravity does the readings of a spring balance but not that of a chemical balance.

i, and iv

ii and iii

ii and iv

i, ii and iii

i and iii

Equilibrium Of Forces

Tambaya 6 Rahoto

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

Bayanin Amsa

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

Tambaya 7 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Current Electricity (JAMB)

Current Electricity

Tambaya 8 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Electrostatics (JAMB)

Electrostatics

Tambaya 11 Rahoto

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

Bayanin Amsa

Height =

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

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

= 6.67mm

Karanta bayanin darasi akan Measurements And Units (JAMB)

Measurements And Units

Tambaya 12 Rahoto

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

Bayanin Amsa

\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

Karanta bayanin darasi akan Quantity Of Heat (JAMB)

Quantity Of Heat

Tambaya 13 Rahoto

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

Tambaya 18 Rahoto

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

Bayanin Amsa

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 $Ω$

Karanta bayanin darasi akan Current Electricity (JAMB)

Karanta bayanin darasi akan Heat Energy (WAEC)

Current Electricity

Heat Energy

Tambaya 21 Rahoto

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

Bayanin Amsa

R =

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

R = 14.14N

R = F = ma

14.14 = 2 x a

a = 7.07ms-1

Karanta bayanin darasi akan Scalars And Vectors (WAEC)

Karanta bayanin darasi akan Equilibrium Of Forces (WAEC)

Scalars And Vectors

Equilibrium Of Forces

Tambaya 23 Rahoto

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

Bayanin Amsa

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

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

= 2

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Karanta bayanin darasi akan Pressure (JAMB)

Karanta bayanin darasi akan Measurements And Units (JAMB)

Temperature And Its Measurement

Measurements And Units

Tambaya 24 Rahoto

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

Bayanin Amsa

β

= A2 - A1

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

A2 = 1.498m2

Tambaya 25 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Capacitors (JAMB)

Karanta bayanin darasi akan Current Electricity (JAMB)

Current Electricity

Tambaya 27 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Quantity Of Heat (JAMB)

Quantity Of Heat

Tambaya 30 Rahoto

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

Bayanin Amsa

\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

Karanta bayanin darasi akan Position, Distance And Displacement (WAEC)

Position, Distance And Displacement

Tambaya 31 Rahoto

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

Bayanin Amsa

F = ma

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

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

F = 25 000N

Karanta bayanin darasi akan Current Electricity (JAMB)

Current Electricity

Tambaya 36 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Current Electricity (JAMB)

Current Electricity

Tambaya 37 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Motion (JAMB)

Tambaya 38 Rahoto

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

Bayanin Amsa

Echo time = 1

\div

2 = 0.5sec

depth = velocity x time

= 1500 x 0.5

= 750m = 0.75km

Karanta bayanin darasi akan Sound Waves (WAEC)

Sound Waves

Tambaya 39 Rahoto

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}

Bayanin Amsa

Height inside H2O =

\frac{2}{3}

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

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

Tambaya 42 Rahoto

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

Bayanin Amsa

F =

μ

R

= $μ$ mg

8 = $μ$ x 5 x 10

$μ$ = 0.16

Karanta bayanin darasi akan Friction (JAMB)

Tambaya 43 Rahoto

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

Bayanin Amsa

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

Karanta bayanin darasi akan Elasticity (JAMB)

Tambaya 46 Rahoto

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

Bayanin Amsa

Heat lost = heat gained

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

$θ$ = 56oC

Karanta bayanin darasi akan Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

Tambaya 48 Rahoto

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

Bayanin Amsa

slope = pressure

\div

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

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

= Nm = work

Karanta bayanin darasi akan Gas Laws (JAMB)

Tambaya 50 Rahoto

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

Bayanin Amsa

\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

Karanta bayanin darasi akan Refraction Of Light Through Plane And Curved Surfaces (JAMB)

Refraction Of Light Through Plane And Curved Surfaces

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