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JAMB UTME - Physics - 2018

Question 1 Report

An object is placed 20cm from a concave mirror of focal length 10cm. The linear magnification of the image produced is?

0

1

2

1

2

Answer Details

The linear magnification of an image is given by the formula: magnification = height of image / height of object = -v/u where v is the image distance, u is the object distance, and the negative sign indicates that the image is inverted. In this problem, the object is placed 20cm from a concave mirror of focal length 10cm. Since the object is placed beyond the focal point, the image will be real and inverted. Using the mirror formula 1/f = 1/v + 1/u, we can find the image distance v: 1/10 = 1/v + 1/20 Solving for v, we get: v = -20 cm Now, we can use the magnification formula to find the linear magnification: magnification = -v/u = -(-20)/20 = 1 Therefore, the linear magnification of the image produced is 1, which means the image is the same size as the object and is also inverted. The answer is: 1.

Read lesson note on Electrostatics (JAMB)

Electrostatics

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Question 5 Report

Which of the following is the dimension of pressure

ML

^{- 1}

T

^{- 2}

MLT

^{- 2}

ML

^{2}

T

^{3}

ML

^{- 3}

Answer Details

The dimension of pressure is ML^-1T^-2 Pressure is defined as the force per unit area. This means that pressure is dependent on the force applied and the area over which it is applied. The unit of force is measured in Newtons (N), and the unit of area is measured in square meters (m²). Therefore, the unit of pressure is N/m², which is also known as Pascals (Pa). To determine the dimension of pressure, we need to break down the units into their fundamental dimensions of mass (M), length (L), and time (T). Force is measured in N, which is kg m/s². Area is measured in m², which is L². Therefore, the dimension of pressure can be calculated as (kg m/s²)/(L²), which simplifies to ML^-1T^-2.

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Question 13 Report

An object moves in a circular path of radius 0.5m with a speed of \(1ms^{-1}\). What is its angular velocity?

8rads

^{- 1}

4rads

^{- 1}

1rads

^{- 1}

2rads

^{- 1}

Answer Details

Angular velocity is a measure of how fast an object is rotating around a center point. It's usually measured in radians per second (rad/s). To calculate angular velocity, we use the formula: angular velocity = linear velocity / radius. In this case, the linear velocity is 1 m/s, and the radius is 0.5 m. So, the angular velocity would be: 1 m/s / 0.5 m = 2 rad/s Therefore, the answer is 2 rad/s or 2rads^-1

Read lesson note on Motion (JAMB)

Read lesson note on Speed And Velocity (WAEC)

Speed And Velocity

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Question 15 Report

For what values of F and ? will the forces shown in the diagram below be in equilibrium.

60N and 60

^{o}

60N and 30

^{o}

20?3 N and 60

^{o}

20?3N and 30

^{o}

Answer Details

Resolve vertically, 40cos? = 20, ? = 60

^{o}

Resolve horizontally, f = 40sin? = 40sin60

^{o}

= 40(

\frac{\sqrt{3}}{2}

)
= 20?3 N

Read lesson note on Equilibrium Of Forces (WAEC)

Equilibrium Of Forces

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Question 16 Report

The diagram shown represents a block-and-tackle pulley system on which an effort of W Newtons supports a load of 120.0N. If the efficiency of the machine is 40, then the value of W is?

28.0

48.0N

233.0N

50.0N

Answer Details

Let the total number of pulleys used in both the blocks be

n

.
In a block-and-tackle pulley system, the velocity ratio is equal to n.
Efficiency =

\frac{MA}{VR} \times 100 %

MA = \frac{L}{E}, VR = n

Efficiency =

\frac{L}{E} \times \frac{1}{n} \times 100 %

E = \frac{L}{Eff. \times n} \times 100 %

E = \frac{120 N}{40 % \times 6} \times 100 %

E = 50 N

Read lesson note on Simple Machines (JAMB)

Simple Machines

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Question 23 Report

The efficiency of energy conversion on the energy flow through a hydroelectric power is?

E1

E2

\frac{E 3}{E 2}

\frac{E 2}{E 1}

+ E2

\frac{E 2}{E 3}

+ E1

Answer Details

Efficiency =

\frac{useful energy output from machine}{energy input into machine}

=

\frac{E 3}{E 2}

Read lesson note on Work, Energy And Power (WAEC)

Work, Energy And Power

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Question 30 Report

If the focal length of a camera is 20cm, the distance from the film at which the lens must be set to produce a sharp image of 100cm away is

17cm

20cm

25cm

100cm

Answer Details

F = 20cm
V = 100cm
U = ?

\frac{1}{U}

+

\frac{1}{V}

=

\frac{1}{F}

\frac{1}{20}

+

\frac{1}{100}

=

\frac{1}{F}

\frac{5 + 1}{100}

=

\frac{1}{F}

F =

\frac{100}{6}

= 16.7cm
= 17cm

Read lesson note on Vapours (JAMB)

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Question 33 Report

In the diagram given the hanging mass m2 is adjusted until m1 is on the verge of sliding. The coefficient of static

m₁

m₂

m₁	g
m₂

m₂

m₁

m₂	g
m₁

Answer Details

I think the correct option is C (

\frac{m_{2}}{m_{1}}

). The coefficient of friction is a ratio of two forces and hence g will cancel out.

Read lesson note on Liquids At Rest (JAMB)

Liquids At Rest

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