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JAMB UTME - Physics - 1995

Question 1 Report

Which of the following is required to convert a milliammeter to an ammeter?

A high resistance in parallel

A high resistance in series

A low resistance in parallel

A low resistance in series

Answer Details

To convert a milliammeter (which measures milliamperes) to an ammeter (which measures amperes), a low resistance in parallel is required. This is because an ammeter should ideally have zero resistance, and adding a low resistance in parallel with a milliammeter will effectively reduce its resistance, allowing it to measure higher current values without affecting the circuit being measured.

Read lesson note on Electrostatics (JAMB)

Electrostatics

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Question 2 Report

A vehicle of mass m is driven by an engine of power P from rest. Find the minimum time it will take to acquire a speed

\frac{m v^{2}}{t}

mv²

2t

mv

t

mv

2t

Work, Energy And Power

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Question 11 Report

A well-lagged bar of length 100cm has its ends maintained at 100oC and 40oC respectively. What is the temperature at a point 60cm from the hotter end?

58oC

62oC

64oC

76oC

Answer Details

100oC ....100cm

60oC ....60cm

40oC .... 0cm

$\frac{θ - 40}{100 - 40}$ = $\frac{60 - 0}{100 - 0}$

$θ$ = 76oC

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

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Question 13 Report

A temperature scale has a lower fixed point of 40mm and a upper fixed point of 200mm. What is the reading on this scale when a thermometer reads 60oC?

33.3mm

36.0mm

96.0mm

136.0mm

Answer Details

200mm ..... 100oC

Xmm ..... 60oC

400mm ..... 0oC

$\frac{x - 40}{200 - 40}$ = $\frac{60 - 0}{100 - 0}$

X = 136mm

Read lesson note on Temperature And Its Measurement (JAMB)

Temperature And Its Measurement

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Question 17 Report

What is the acceleration due to gravity 'g' on the moon, if g is 10-2 on the earth?

0. 10 ms-2

0.74ms-2

1.6ms-2

10.00ms-2

Answer Details

s = 10-2 on earth

g = 6 ms-2 on earth

Earth on moon = $\frac{10}{6}$

= 1.6ms-2

Read lesson note on Gravitational Field (JAMB)

Gravitational Field

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Question 18 Report

Which of the following is the dimension of pressure?

ML-1T-2

MLT-2

ML2T-3

ML-3

Answer Details

Pressure = heg

= m x $\frac{k g}{m^{3}}$ x $\frac{m}{5^{2}}$

= ML-1T-2

Read lesson note on Measurements And Units (JAMB)

Measurements And Units

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Question 26 Report

A rectangular metal block of volume 10-6m3 at 273k is heated to 573k. If its coefficient of linear expansion is 1.2 x 10-5k-1, the percentage change of it volume is

1.5%

1.1%

0.4%

0.1%

Answer Details

Y =

\frac{Δ V}{V \times Δ θ}

3 x 10-5 x 1.2 = $\frac{Δ V}{10^{- 6} \times (573 - 273)}$

$Δ$ V = 1080 x 10-11m3

% $Δ$ V = $\frac{1080 \times 10}{10^{- 6}}$ x 100

= 1.08%

$\approx$ = 1.1%

Read lesson note on Elasticity (JAMB)

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Question 28 Report

An alternating source is connected in series to a capacitor of capacity reactance \(10^3\sqrt{3}\) and a resistor of resistance \(10^3\) Ohms. The Impedance of the circuit is...

2000 \sqrt{3}

Ohms

2000 Ohms

1000 \sqrt{3}

Ohms

1000 Ohms

Simple A.C Circuits

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Question 33 Report

An electric current of 2 amperes flows in a heating coil of resistance 50 ohms for 3 minutes 20 seconds. Determine the heat produced

0.5kJ

8.0kJ

20.0kJ

40.0kJ

Answer Details

Heat energy = Elect. energy

heat energy= I2RT

= 2 X 2 X 50 X 200

= 40000J

to convert to kJ, divide J by 1000

hence, $\frac{40000}{1000}$ = 40kJ

Read lesson note on Current Electricity (JAMB)

Current Electricity

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Question 37 Report

If a resistance is halved in value and the potential difference across it is tripled, then the ratio of the new current to the old one is

1 : 6

1 : 3

2 : 1

6 : 1

Answer Details

\frac{V_{1}}{V_{2}}

=

\frac{I_{1} R_{1}}{I_{2} R_{2}}

$\frac{I_{1}}{I_{2}}$ V2 = 3V1 and R1 = 2R2

$\frac{V_{1}}{3 V_{1}}$ = $\frac{I_{1} \times 2 R_{2}}{I_{2} \times R_{2}}$

$\frac{I_{1}}{I_{2}}$ = $\frac{6}{1}$

Read lesson note on Current Electricity (JAMB)

Read lesson note on Simple A.C Circuits (JAMB)

Current Electricity

Simple A.C Circuits

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Question 41 Report

A hydraulic press has a large circular piston of radius 0.8m and a circular plunger of radius 0.2m. A force of 500N is exerted by the plunger. Find the force exerted on the piston

8000N

4000N

2000N

31N

Answer Details

Pressure on piston = pressure on the plunger

$\frac{520}{π \times 0.8 \times 0.8}$ = $\frac{F}{π \times 0.2 \times 0.2}$

F = 31N

Read lesson note on Elasticity (JAMB)

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