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JAMB UTME - Physics - 1992

Question 1 Report

The speed of light in air is 3 x 108ms-1. What is the speed in glass with a refractive index of 1.50?

1.5 x 108ms-1

3.0 x 108ms-1

2.0 x 108ms-1

6.0 x 108ms-1

Answer Details

The speed of light changes as it passes through different media with different refractive indices. The refractive index of a medium is the ratio of the speed of light in a vacuum to the speed of light in that medium. The formula for finding the speed of light in a medium is: Speed of light in a medium = (Speed of light in a vacuum) / (Refractive index of the medium) In this question, the speed of light in air is given as 3 x 10^8 ms^-1, and the refractive index of the glass is given as 1.50. So, we can use the formula to find the speed of light in glass: Speed of light in glass = (Speed of light in air) / (Refractive index of glass) = (3 x 10^8) / 1.50 = 2 x 10^8 ms^-1 Therefore, the answer is option C, 2.0 x 10^8 ms^-1.

Read lesson note on Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum (WAEC)

Electromagnetic Waves: Types Of Radiation In Electromagnetic Spectrum

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Question 6 Report

A concave mirror has a radius of curvature of 36cm. At what distance from the mirror should an object be placed to give a real image three times the size of the object?

12cm

24cm

48cm

108cm

Answer Details

F =

\frac{u v}{u + v}

v = 3u

18 = $\frac{3 u \times u}{3 u + u}$

u = 24cm

Read lesson note on Reflection Of Light At Plane And Curved Surfaces (JAMB)

Read lesson note on Properties Of Waves: Reflection, Refraction, Diffraction (WAEC)

Reflection Of Light At Plane And Curved Surfaces

Properties Of Waves: Reflection, Refraction, Diffraction

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Question 18 Report

A platinum resistance thermometer wire has a resistance of 5 ohms at 0oC and 5.5 ohms at 100oC. Calculate the temperature of the wire when the resistance is 5.2 ohms

80oC

60oC

40oC

10oC

Answer Details

100oC .... 5.5

$θ$ oC .... 5.3

0oC .... 5.0

$\frac{θ - 0}{100 - 0}$ = $\frac{5.5 - 5 .0}{5.5 - 5.0}$

$θ$ = 40oC

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

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Question 29 Report

The equation of a wave is y = 0.005 sin[x(0.5X - 200t)] where x and y are in meters and t is in seconds. What is the velocity of the wave?

4000ms-1

400ms-1

250ms-1

40ms-1

Answer Details

y = 0.005 sin[x(0.5X - 200t)]

compare to Y = A sin(kx - wt)

$λ$ = $\frac{2 π}{k}$ and F = $\frac{w}{2 π}$

V= $λ$ F

= $\frac{2 π}{0.5}$ x $\frac{200}{2 π}$

= 400ms-1

Read lesson note on Types Of Waves (WAEC)

Types Of Waves

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Question 30 Report

Which of the following apparatus will require the smallest fuse rating for its protection?

60W, 240V

60W, 40V

40W, 12V

40W, 5V

Answer Details

I =

\frac{P}{V}

=

\frac{60}{240}

= 0.25A

I = $\frac{P}{V}$ = $\frac{60}{40}$ = 1.5A

I = $\frac{P}{V}$ = $\frac{40}{12}$ = 3.3A

I = $\frac{P}{V}$ = $\frac{40}{5}$ = 8A

Read lesson note on Current Electricity (JAMB)

Read lesson note on Electrical Energy And Power (JAMB)

Current Electricity

Electrical Energy And Power

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Question 31 Report

A wire of length \(15\text{ m}\) made of a material of resistivity \(1.8 \times 10^{-6}\ \Omega\text{m}\) has a resistance of \(0.27\ \Omega\). Determine the area of the wire

1.5 x 10-4m2

1.0 x 10-4m2

2.7 x 10-5m2

7.3 x 10-6m2

Answer Details

Resistivity =

\frac{Resiatance x Area}{l e n g t h}

1.6 x 10-6 = $\frac{15 \times A r e a}{0.27}$

Area = 1.0 x 10-4m2

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Question 36 Report

A stone, Q is thrown with velocity U at an angle of 75o to the horizontal. Another stone R is thrown with the same velocity U but with an angle of 15o to the horizontal. The ranges covered by the stones will be

greater for Q

greater for R

the same for Q and R

greater for the heavier of the two stones

Answer Details

Range =

\frac{U^{2} s i n 2 θ}{g}

RangeQ = $\frac{U^{2} s i n 150}{g}$

RangeR = $\frac{U^{2} s i n 30}{g}$

Since, sin30o = sin150o

therefore, RQ = RR

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Question 38 Report

A machine requires 1000J of work to raise a load of 500N through a vertical distance of 1.5m. Calculate the efficiency of the

80%

75%

50%

33%

Answer Details

\frac{500 \times 1.5}{100.0}

x 100%

= 75%

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

Work, Energy And Power

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Question 40 Report

If a plastic sphere floats in water (density = 1000kgm3) with 0.5 of its volume submerged, and floats in oil with 0.4 of its volume submerged the density of the oil is

800kgm3

1200kgm3

1250kgm3

2000kgm3

Answer Details

R.d =

\frac{0.4}{0.5}

= 0.8

0.8 = $\frac{Density of oil}{Density of water}$

0.8 = $\frac{Density of oil}{1000}$

Density of oil = 800kgm3

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Question 43 Report

A 5-v battery is connected across the plates of a \(2.0\,\mu\text{F}\) uncharged parallel plate capacitor. The charge on the capacitor after a long time is

2.5

μ

C

10.0

μ

C

20.0

μ

C

50.0

μ

C

Answer Details

The charge stored in a capacitor is given by the formula Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor. Initially, the capacitor is uncharged, so Q = 0. When the 5-volt battery is connected to the plates of the capacitor, the voltage across the capacitor becomes 5 V. Therefore, the charge on the capacitor becomes: Q = CV = (2.0 μF)(5 V) = 10.0 μC The charge on the capacitor will not change after a long time because a capacitor is an ideal device that stores charge without dissipating it. Therefore, the correct answer is 10.0μC.

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