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JAMB UTME - Physics - 2014

Question 1 Report

The maximum power transfer occur in a cell when the external resistance is

Twice the internal resistance of the cell

The same as the internal resistance of the cell

Greater than the internal resistance of the cell

Less than the internal resistance of the cell

Answer Details

When we talk about a cell, it has some internal resistance, which is the resistance that opposes the flow of current within the cell. The external resistance is the resistance outside the cell that allows the flow of current. The power transfer from a cell to the external circuit is at its maximum when the external resistance is equal to the internal resistance of the cell. This is because the maximum power transfer theorem states that the maximum power will be transferred from a source to a load when the resistance of the load is equal to the internal resistance of the source. When the external resistance is equal to the internal resistance, the maximum current flows through the circuit, resulting in maximum power transfer. If the external resistance is less than the internal resistance, then the current would be too high, leading to internal heating and reduced power transfer. If the external resistance is greater than the internal resistance, then the current would be too low, leading to reduced power transfer. Therefore, the correct answer to the question is: "The same as the internal resistance of the cell."

Read lesson note on Current Electricity (JAMB)

Read lesson note on Electric Cells (JAMB)

Current Electricity

Electric Cells

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Question 2 Report

An object is placed 10m from a pinhole camera of length 25cm. Calculate the linear magnification.

2.5x102

2.5x10-2

2.5x10-1

2.5x101

Answer Details

Linear Magnification = \frac{Length of Camera}{Distance of the Object}

$Linear Magnification = \frac{25}{1000}$

$Linear Magnification = 2.5 \times 10^{- 2}$

Read lesson note on Electrostatics (JAMB)

Electrostatics

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Question 8 Report

A wheel and an axle is used to raise a load whose weight is 800N when an effort of 250N is applied. If the radii of the wheel and axle are 800mm and 200mm respectively, the efficiency of the machine is

90%

80%

85%

87%

Answer Details

VR = 800/200 = 4

E f f i c i e n c y = \frac{M A}{V R} \times \frac{100}{1}

$E f f i c i e n c y = \frac{L}{E} \times \frac{1}{V R} \times \frac{100}{1}$

$E f f i c i e n c y = \frac{800}{250} \times \frac{1}{4} \times \frac{100}{1}$

= 78.74%
= 80%

Read lesson note on Simple Machines (JAMB)

Simple Machines

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Question 18 Report

A car accelerates uniformly from rest at 3ms-2. its velocity after traveling a distance of 24m is

12ms-1

144ms-1

72ms-1

36ms-1

Answer Details

V2 = U2 x 2ax
V2 = 0 x 2(3)(24)
V2 = 144
V =

\sqrt{144}

V = 12ms-1

Read lesson note on Motion (JAMB)

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Question 21 Report

A machine whose efficiency is 75% is used to lift a load of 1000N. Calculate the effort put in to the machine if it has a Velocity ratio of 4.

343.32N

233.33N

333.33N

334.33N

Answer Details

\frac{M A}{V R} \times \frac{100}{1} = e f f i c i e n c y

$\frac{L}{E} \frac{1}{V R} \times \frac{100}{1} = 75$

$\frac{1000}{E} \frac{1}{4} \times \frac{100}{1} = 75$

$E = \frac{250 \times 100}{75}$

E = 333.3N

Read lesson note on Simple Machines (JAMB)

Simple Machines

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Question 29 Report

A force of 500N1 is applied to a steel wire of cross-sectional area 0.2m2, The tensile stress is

2.5x104Nm-2

1.0x102Nm-2

1.0x103Nm-2

2.5x103Nm-2

Answer Details

$S t r e s s = \frac{F o r c e}{A r e a}$
$S t r e s s = \frac{500}{0.2} = 2500$
= 2.5 x 103Nm-3

Read lesson note on Elasticity (JAMB)

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Question 41 Report

The focal length of a concave mirror is 2.0cm. If an object is placed 8.0cm from it, the image is at

2.7m

2.0m

2.3m

2.5m

Answer Details

\frac{1}{U} + \frac{1}{V} = \frac{1}{f}

$\frac{1}{8} + \frac{1}{V} = \frac{1}{2}$

$V = \frac{8}{3}$

$V = 2.7 c m$

Read lesson note on Structure Of The Nucleus (WAEC)

Structure Of The Nucleus

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