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JAMB UTME - Physics - 1991

Question 1 Rapport

The mass of a stone is 15.0g when completely immersed in water and 10.0g when completely immersed in a liquid of relative density 2.0. The mass of the stone in air is

5.0g

12.0g

20.0g

25.0g

Détails de la réponse

The buoyant force on an object is equal to the weight of the fluid displaced by the object. When the stone is immersed in water, the buoyant force is equal to the weight of the water that is displaced. Similarly, when the stone is immersed in the liquid of relative density 2.0, the buoyant force is equal to the weight of the liquid that is displaced. From the given information, we can calculate the volume of the stone. Let the volume of the stone be V. When completely immersed in water, the buoyant force on the stone is equal to the weight of the water displaced, which is equal to the weight of the stone in air minus the weight of the stone in water. Therefore, we have: Buoyant force = Weight of stone in air - Weight of stone in water V x density of water x g = (10.0g - 15.0g) x g V = 5.0g / (density of water x g) Similarly, when completely immersed in the liquid of relative density 2.0, we have: V x density of liquid x g = (10.0g - weight of liquid displaced) x g V x 2.0 x g = (10.0g - 15.0g) x g V = 5.0g / (2.0 x density of water x g) Since the volume of the stone is the same in both cases, we can set the two expressions for V equal to each other and solve for the density of water: 5.0g / (density of water x g) = 5.0g / (2.0 x density of water x g) density of water = 2.0 Now we can use the buoyant force equation to find the weight of the stone in air: Buoyant force = Weight of stone in air - Weight of stone in water V x density of water x g = (Weight of stone in air - 15.0g) x g Weight of stone in air = V x density of water x g + 15.0g Weight of stone in air = 5.0g + 15.0g Weight of stone in air = 20.0g Therefore, the mass of the stone in air is 20.0g. Answer: 20.0g.

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Question 3 Rapport

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

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Question 14 Rapport

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

Détails de la réponse

P.E =

\frac{1}{2}

Fe

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

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

= 1.9J

Lisez la note de cours sur Elasticity (JAMB)

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Question 18 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Simple Machines (JAMB)

Simple Machines

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Question 20 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Motion (JAMB)

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Question 21 Rapport

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

Détails de la réponse

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.

Lisez la note de cours sur Refraction Of Light Through Plane And Curved Surfaces (JAMB)

Lisez la note de cours sur Electrostatics (JAMB)

Refraction Of Light Through Plane And Curved Surfaces

Electrostatics

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Question 24 Rapport

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

Détails de la réponse

F =

\frac{u v}{u + v}

F = -ve

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

v = -3cm

Lisez la note de cours sur Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

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Question 29 Rapport

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

Détails de la réponse

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.

Lisez la note de cours sur Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

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Question 30 Rapport

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

Détails de la réponse

\frac{4}{3}

=

\frac{12}{Apparent depth}

Apparent depth = 9cm

Height by which coin is raised = 12 - 9

= 3cm

Lisez la note de cours sur Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

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Question 31 Rapport

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

Détails de la réponse

Heat by water = mc

θ

Heat by steam = mc $θ$ + mL

Difference = mL = 4 x 2,260,000

= 9,040,000J

Lisez la note de cours sur Heat Energy (WAEC)

Heat Energy

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Question 39 Rapport

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

Détails de la réponse

let the length be Lcm

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

L = 23cm

Lisez la note de cours sur Elasticity (JAMB)

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Question 41 Rapport

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

Détails de la réponse

Difference in pressure = 75 - 63

= 12cmHg

pressure = hge

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

= 1200m

Lisez la note de cours sur Pressure (JAMB)

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Question 42 Rapport

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

Détails de la réponse

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

Lisez la note de cours sur Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

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Question 47 Rapport

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

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