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JAMB UTME - Physics - 1993

Question 1 Report

A 40-W instrument has a resistance of \(90\ \Omega\). On what voltage should it be operated normally?

60V

150V

225V

3600V

Answer Details

To calculate the voltage required to operate the 40-W instrument with a resistance of 90 Ω, we can use Ohm's law, which states that the voltage (V) equals the product of the current (I) and the resistance (R): V = I × R. First, we need to calculate the current that flows through the instrument. We can use the formula for power (P) in terms of current and resistance: P = I^2 × R. Rearranging this equation, we get: I = sqrt(P / R). Substituting the values given in the problem, we get: I = sqrt(40 W / 90 Ω) ≈ 0.632 A Now we can use Ohm's law to find the voltage: V = I × R = 0.632 A × 90 Ω ≈ 56.9 V Therefore, the voltage required to operate the instrument normally is approximately 56.9 volts. (60V) is the closest to this value. Note that this is a simple example of how to apply Ohm's law to a practical problem. In real-life situations, the calculation may be more complicated, and other factors such as the temperature coefficient of resistance may need to be considered.

Read lesson note on Current Electricity (JAMB)

Current Electricity

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Question 2 Report

The external and internal diameter of a tube are measured as (32 ± 2)mm and (21 ± 1)mm respectively. Determine the percentage error in the thickness of the tube.

27%

14%

9%

3%

Answer Details

Actual thickness = 34 - 22 = 12

error thickness = 30 - 20 = 10

error = 12 - 10 = 2mm that is, (11 ± 1)mm

%error = $\frac{1}{11}$ x $\frac{100}{1}$

= 9%

Read lesson note on Measurements And Units (JAMB)

Measurements And Units

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Question 5 Report

Which of the following correctly describes the energy changes in the generation of light by a hydroelectric power station?

Electrical

\to

mechanical

\to

potential

potential

\to

mechanical

\to

electrical

\to

light

mechanical

\to

sound

\to

electrical

\to

light

kinetic

\to

mechanical

\to

electrical

\to

light

Answer Details

The correct option that describes the energy changes in the generation of light by a hydroelectric power station is: "Potential energy → mechanical energy → electrical energy → light." This is because a hydroelectric power station converts the potential energy of water stored at a certain height into mechanical energy by allowing the water to flow down through turbines. The turbines are connected to generators that convert the mechanical energy into electrical energy. Finally, the electrical energy is used to power light bulbs and other electrical devices, producing light.

Read lesson note on Current Electricity (JAMB)

Read lesson note on Sound Waves (WAEC)

Read lesson note on Magnetic Field (Part 2) (WAEC)

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

Current Electricity

Sound Waves

Magnetic Field (Part 2)

Work, Energy And Power

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Question 7 Report

200g of water at 90oC is mixed with 100g of water at 30oC. What is the final temperature?

50oC

60oC

70oC

80oC

Answer Details

Heat lost = heat gained

200 x c x (90 - $θ$ )

$θ$ = 70oC

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

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Question 8 Report

A beam of light is incident from air to water at an angle of \(30^\circ\text{C}\). Find the angle of refraction if the refractive index of water is \(\frac{4}{3}\)

15oC

18oC

22oC

24oC

Answer Details

\frac{4}{3}

=

\frac{s i n 30}{s i n r}

r = 22oC

View Answer

Question 9 Report

Natural radioactivity is the random spontaneous disintegration of the nuclei of heavy isotopes of certain elements with the emission of

α

, X, and

β

α

,

β

and

γ

-rays

X,

α

,

γ

-ray

α

,

β

and X-rays

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

Electric Field

Refraction Of Light Through Plane And Curved Surfaces

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Question 15 Report

Two forces of magnitudes 7N and 3N act at right angles to each other. The angle \( \theta \) between the resultant and the 7N force is given by

cos

θ

=

\frac{3}{7}

sin

θ

=

\frac{3}{7}

tan

θ

=

\frac{3}{7}

cot

θ

=

\frac{3}{7}

Answer Details

θ

= tan-1

\frac{3}{7}

View Answer

Question 23 Report

A parallel plate capacitor of area \(10\text{cm}^2\) in vacuum has a capacitance of \(10^{-2}\,\mu\text{F}\). What is the distance between the plates? \((e = 9 x 10^{-12}\text{fm})\)

9 x 10-13m

9 x 10-7m

9 x 10-3m

9 x 107m

Answer Details

C =

\frac{ε_{o} A}{d}

10-2 x 10-6 = $\frac{9 \times 10^{- 12}}{d}$ x 10-4

= 9 x 10-7m

Read lesson note on Capacitors (JAMB)

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Question 24 Report

Which of the following expressions gives the linear magnification produced by a concave mirror of radius of curvature r, if U and V are the object and image distances respectively?

\frac{V}{r}

- 1

\frac{2 V}{r}

- 1

\frac{U}{r}

- 1

\frac{2 U}{r}

- 1

Answer Details

The expression that gives the linear magnification produced by a concave mirror of radius of curvature r, if U and V are the object and image distances respectively is "-V/U" or "V/U = -h'/h", where "h'" is the height of the image and "h" is the height of the object. This is a well-known formula in optics, which states that the magnification produced by a mirror is equal to the ratio of the height of the image to the height of the object. Moreover, the negative sign indicates that the image is inverted with respect to the object. None of the options provided match this formula exactly. Option "Vr" and "V/r-1" involve only the image distance and the radius of curvature, but not the object distance, which is necessary to calculate the magnification. Option "Ur" and "U/r-1" involve only the object distance and the radius of curvature, but not the image distance. Finally, option "2Vr" and "2U/r-1" involve a factor of 2 that is not present in the formula for magnification.

Read lesson note on Temperature And Its Measurement (JAMB)

Read lesson note on Gravitational Field (WAEC)

Temperature And Its Measurement

Gravitational Field

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Question 34 Report

Determine the distance between the consecutive antinodes XX if the wavelength is 66cm

15cm

30cm

60cm

120cm

Answer Details

Distance XX =

\frac{1}{2}

λ

= $\frac{1}{2}$ x $\frac{60}{1}$

= 30cm

Read lesson note on Propagation Of Sound Waves (JAMB)

Propagation Of Sound Waves

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Question 37 Report

A plane inclined at angle \( \theta \) has a velocity ratio of 10 : 1. The inclination of the plane to the horizontal is given by

tan

θ

=

\frac{1}{10}

cot

θ

=

\frac{1}{10}

cos

θ

=

\frac{1}{10}

sin

θ

=

\frac{1}{10}

Answer Details

V.R =

\frac{1}{s i n θ}

= 10

sin = $\frac{1}{10}$

Read lesson note on Scalars And Vectors (JAMB)

Scalars And Vectors

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Question 39 Report

The relationship between L and the wavelength I of the stationary wave is

2L =

λ

L =

λ

L =

\frac{λ}{3}

4L = 3

λ

Answer Details

L = -

λ

\Rightarrow

4L = 3

λ

Read lesson note on Structure Of The Atom (Nigeria Only) (WAEC)

Structure Of The Atom (Nigeria Only)

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Question 40 Report

An object of mass 400g and density 600kgm-3 is suspended with a sting so that half of it is immersed in paraffin of density 900kgm-3. The tension in the string is

1.0N

3.0N

4.0N

5.0N

Answer Details

volume of object =

\frac{400 \times 10^{- 2}}{600}

= 0.00067m3

volume of paraffin displaced = $\frac{1}{2}$ x 0.00067

= 0.00034m3

mass of paraffin = 900 x 34 x 10-5

= 0.3kg

weight of paraffin = 0.3 x 10

= 3N = upthrust = tension in the string

Read lesson note on Measurements And Units (JAMB)

Measurements And Units

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