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JAMB UTME - Physics - 1991

Ibeere 1 Ìròyìn

A pilot records the atmospheric pressure outside its plane as 63cm of Hg while a ground observer records a reading of 75cm of Hg for the atmospheric pressure on the ground. Assuming that the density of the atmosphere is constant, calculate the height of the plane above the ground. (Relative density of Hg = 13.0 and that of air = 0.00013)

1200m

6300m

7500m

13800m

Awọn alaye Idahun

Difference in pressure = 75 - 63

= 12cmHg

pressure = hge

difference in height = $\frac{12}{10 \times \frac{0.00013}{13}}$

= 1200m

Ka ìwé ẹ̀kọ́ nípa Pressure (JAMB)

Ibeere 7 Ìròyìn

A trough 12.0cm deep is filled with water of refractive index \( \frac{4}{3} \). By how much would a coin at the bottom of the trough appear to be displaced when viewed vertically from above the water surface?

3.0cm

6.0cm

9.0cm

16.0cm

Awọn alaye Idahun

\frac{4}{3}

=

\frac{12}{Apparent depth}

Apparent depth = 9cm

Height by which coin is raised = 12 - 9

= 3cm

Ka ìwé ẹ̀kọ́ nípa Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Ibeere 12 Ìròyìn

A moving coil galvanometer of \(300\,\Omega\) resistance gives full scale deflection for 1.0mA. The resistance R, of the shunt that is required to convert the galvanometer into a 3.0A ammeter is

899.70

Ω

10.00

Ω

Ω

Ω

0.01

Ω

Electrostatics

Ibeere 13 Ìròyìn

When a plane mirror at which a ray is incident is rotated through an angle \(\theta\), the reflected ray will be rotated through

\frac{1}{2}

θ

θ

2

θ

3

θ

Awọn alaye Idahun

When a plane mirror is rotated through an angle θ, the reflected ray will be rotated through an angle of 2θ. To understand why, imagine a ray of light hitting a plane mirror at an incident angle θ. The angle of incidence is the angle between the incident ray and the perpendicular to the mirror surface at the point of incidence. According to the law of reflection, the angle of reflection is equal to the angle of incidence, and the reflected ray makes the same angle with the perpendicular as the incident ray. Therefore, if we rotate the mirror through an angle θ, the perpendicular to the mirror at the point of incidence will also be rotated through an angle θ. This means that the angle of incidence will also change by θ. Now, since the angle of reflection is equal to the angle of incidence, the angle of reflection will also change by θ. Therefore, the angle between the incident ray and the reflected ray will change by 2θ (θ for the change in angle of incidence and θ for the change in angle of reflection). Hence, the reflected ray will be rotated through an angle of 2θ.

Ka ìwé ẹ̀kọ́ nípa Refraction Of Light Through Plane And Curved Surfaces (JAMB)

Ka ìwé ẹ̀kọ́ nípa Electrostatics (JAMB)

Refraction Of Light Through Plane And Curved Surfaces

Electrostatics

Ibeere 16 Ìròyìn

A small metal ball thrown vertically upwards from the top of a tower with an initial velocity of 20ms-1. If the ball took a total of 6s to reach ground level, determine the height of the tower [g = 10ms-2]

60m

80m

100m

120m

Awọn alaye Idahun

Time from tower to top, t =

\frac{u - v}{g}

= $\frac{20 - 0}{10}$

= 2s

distance from tower to top

x = ut - $\frac{1}{2}$ gt2

= (20 x 2) - ( $\frac{1}{2}$ x 10 x 2 x 2)

= 20m

Time from the top to the ground, t = 6 - 2

= 4s

distance from top to ground,

x = ut - $\frac{1}{2}$ gt2

= (0 x 4) + ( $\frac{1}{2}$ x 10 x 4 x 4)

= 80cm

therefore, height of tower = 80 - 20

= 60m

Ka ìwé ẹ̀kọ́ nípa Motion (JAMB)

Ibeere 17 Ìròyìn

4g of a radioactive material of half-life 10 days is spilled on a laboratory floor. How long would it take to disintegrate 3.5g of the material?

1

\frac{1}{4}

days

8

\frac{1}{4}

days

30 days

80 days

Heat Energy

Ibeere 21 Ìròyìn

A man hears his echo from a nearby hill 2s after he shouted. If the frequency of his voice is 260Hz and the wavelength is 1.29m, how far away is the hill?

330.0m

335.4m

660.0m

670.8m

Awọn alaye Idahun

v =

λ

F

260 x 1.29 =335.4ms-1

Echo time = $\frac{2}{2}$ = 1sec.

distance = v x t

= 335.4 x 1 = 335.4m

Ka ìwé ẹ̀kọ́ nípa Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Ibeere 24 Ìròyìn

What is the difference in the amount of heat given out by 4kg of steam and 4kg of water when both are cooled from 100oC to 80oC? [the specific latent heat of steam is 2,260,000Jkg-1, specific heat capacity of water is 4200Jkg-1K-1]

4200J

2,260,OOOJ

9,040,000J

9,380,000J

Awọn alaye Idahun

Heat by water = mc

θ

Heat by steam = mc $θ$ + mL

Difference = mL = 4 x 2,260,000

= 9,040,000J

Ka ìwé ẹ̀kọ́ nípa Heat Energy (WAEC)

Heat Energy

Ibeere 29 Ìròyìn

The spiral spring of a spring balance is 25.0cm long when 5N hangs on it and 30cm long, when the weight is 10N. What is the length of the spring if the weight is 3N assuming Hooke's law is obeyed?

15.0cm

17.0cm

20.0cm

23.0cm

Awọn alaye Idahun

let the length be Lcm

$\frac{5 - 3}{25 - L}$ = $\frac{10 - 5}{30 - 25}$

L = 23cm

Ka ìwé ẹ̀kọ́ nípa Elasticity (JAMB)

Ibeere 31 Ìròyìn

A spring of force constant 1500Nm-1 is acted upon by a constant force of 75N. Calculate the potential energy stored in the string

1.9J

2.8J

3.45J

0.43J

Awọn alaye Idahun

P.E =

\frac{1}{2}

Fe

= $\frac{1}{2}$ $\frac{F}{K}$

= $\frac{1}{2}$ x $\frac{75 \times 75}{1500}$

= 1.9J

Ka ìwé ẹ̀kọ́ nípa Elasticity (JAMB)

Ibeere 42 Ìròyìn

A wheel and axle have radii 80cm and 10cm respectively. If the efficiency of the machine is 0.85, an applied force of 1200N to the wheel will raise a load of

8.0N

6.8N

8160N

9600N

Awọn alaye Idahun

Efficiency = m.A x

\frac{1}{V . R}

x 100%

85 = $\frac{l o a d}{1200}$ x $\frac{10}{80}$ x $\frac{100}{1}$

load = 8160N

Ka ìwé ẹ̀kọ́ nípa Simple Machines (JAMB)

Simple Machines

Ibeere 43 Ìròyìn

A body of weight W N rest on a smooth plane inclined at an angle \( \theta \)^o to the horizontal. What is the resolved part of the weight in Newtons along the plane?

θ

W cos

θ

W sec

θ

W tan

θ

Awọn alaye Idahun

When an object of weight W rests on an inclined plane, the weight force can be resolved into two components: one perpendicular to the plane (W cosθ) and the other parallel to the plane (W sinθ). The component of weight parallel to the plane is what we are interested in, as this is the force that will cause the object to slide down the plane if it is not held in place. Therefore, the resolved part of the weight in Newtons along the plane is W sinθ. This is because the weight force is proportional to the sine of the angle of inclination, meaning that the steeper the slope, the greater the force parallel to the plane. So, option A (W sinθ) is the correct answer.

Ka ìwé ẹ̀kọ́ nípa Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

Ibeere 46 Ìròyìn

Which of the following correctly, describes the image of an object, 4cm from a diverging lens of focal length 12cm?

The image is virtual, 3cm in front of the lens

The image is real, 6cm behind the lens

The image is virtual, 6cm in front of the lens

The image is real, 3cm in front the lens

Awọn alaye Idahun

F =

\frac{u v}{u + v}

F = -ve

-12 = $\frac{4 \times v}{4 + v}$

v = -3cm

Ka ìwé ẹ̀kọ́ nípa Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

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