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JAMB UTME - Physics - 1998

Question 1 Report

In free fall, a body of mass 1kg drops from a height of 125m from rest in 5s. How long will it take another body of mass 2kg to fall from rest from the same height? [g = 10ms-1·]

5s

10s

12s

15s

Answer Details

The time taken for an object to fall from a given height in free fall depends only on the height and the acceleration due to gravity. It is independent of the mass of the object. The formula for the time taken for an object to fall from rest from a height h in free fall is given by: t = √(2h/g) where h is the height and g is the acceleration due to gravity. In this question, we are given that a body of mass 1 kg falls from a height of 125 m in 5 s. Using the formula above, we can find the value of g: t = √(2h/g) 5 s = √(2 x 125 m / g) 25 s^2 = 250 m / g g = 250 m / (25 s^2) g = 10 m/s^2 We are then asked to find how long it will take for a body of mass 2 kg to fall from rest from the same height. Using the same formula as before, but with the new value of g, we get: t = √(2h/g) t = √(2 x 125 m / 10 m/s^2) t = √(25) t = 5 s Therefore, it will take the body of mass 2 kg the same amount of time as the body of mass 1 kg to fall from the same height, which is 5 s. The correct option is (a): 5s.

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Question 2 Report

A parallel plate capacitor has a common plate area of \(5 \times 10^{-8}\,\mathrm{m}^2\) and plate separation of \(2 \times 10^3\,\mathrm{m}\), Assuming free space, What is the capacitance? [\(\epsilon_o = 9.0 \times 10^{-12}\,\mathrm{C}^2\mathrm{N}^{-1}\mathrm{m}^{-2}\)]

2.25 x 10-17F

4.50 x 10-17F

2.25 x 10-16F

4.50 x 10-16F

Answer Details

C =

\frac{ε_{o} A}{d}

= $\frac{5 \times 10^{- 8} \times 10^{- 12} \times 9}{2 \times 10^{- 3}}$

= 22.5 x 10-17F

Read lesson note on Capacitors (JAMB)

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Question 9 Report

The force of repulsion between two point positive charges \(5\,\mu\text{C}\) and \(8\,\mu\text{C}\) separated at a distance of \(0.02\) map art is. \(\left[\dfrac{1}{4\pi\varepsilon_o}=9 \times 10^9\text{ Nm}^2\text{C}^{-2}\right]\)

1.8 x 10-10N

9.0 x 10-8N

9.0 x 102N

4.5 x 103N

Answer Details

F =

\frac{1}{4 π ε_{o}}

x

\frac{q_{1} q_{2}}{r^{2}}

F = $\frac{5 \times 10 \times 8 \times 10}{0.002 \times 0.002}$ x 9 x 10

F = 900N

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Properties Of Waves: Reflection, Refraction, Diffraction

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Question 11 Report

The linear expansivity of brass is 2 x 10-5C-1. If the volume of a piece of brass is 15.00cm3 at 00C, What is the volume at 100oC?

16.03cm3

16.00cm3

15.09cm3

15cm3

Answer Details

linear expansivity =

\frac{V_{2} - V_{1}}{V_{1} \times Δ θ}

3 x 2 x 10-5 = $\frac{V_{2} - 15}{15 \times (100 - 0)}$

V2 = 15.09cm3

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Temperature And Its Measurement

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Question 19 Report

An object is placed in front of two plane mirrors inclined at an angle of \(0^\circ\). If the total number of images formed is 7, Find the value of \(\theta\)

30o

45o

51o

90o

Answer Details

Number of image =

\frac{360}{θ}

-1

7 = $\frac{360}{θ}$ -1

thus, $θ$ = 45o

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Electrostatics

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Question 20 Report

A platinum resistance thermometer records \(3.0\,\Omega\) at \(0^\circ\mathrm{C}\) and \(8.0\,\Omega\) at \(100^\circ\mathrm{C}\). If it records \(6.0\,\Omega\) in a certain environment, The temperature of the medium is

80oC

60oC

50oC

30oC

Answer Details

8

Ω

.....100oC

6 $Ω$ ..... $θ$ oC

3 $Ω$ .....0oC

$\frac{6 - 3}{8 - 3}$ = $\frac{θ - 0}{100 - 0}$

therefore, $θ$ = 60oC

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Temperature And Its Measurement

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Question 28 Report

How many grams of water at 17oC must be added to 42g of ice at 0oC to melt the ice completely? [specific latent heat of fusion of ice = 3.4 x 105Jkg-1, specific heat capacity of water = 4200Jkg-1k-1]

200g

300g

320g

400g

Answer Details

Heat lost = Heat gained

mc $θ$ = mL

m x 4200 x 17 = 42 x 3.4 x 105

therefore, m = 200g

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Quantity Of Heat

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Question 31 Report

The velocity of a sound wave at 27oC is 360ms-1. Its velocity at 127oC is

120√3 ms-1

240 ms-1

240√3 ms-1

720√3 ms-1

Answer Details

Velocity

α \sqrt{t e m p e r a t u r e}

$\frac{V_{1}}{V_{2}} = \sqrt{\frac{T_{1}}{T_{2}}}$

therefore, $\frac{360}{V_{2}} = \sqrt{\frac{27 + 273}{127 + 273}}$

$V_{2} = 240 \sqrt{3}$ m/s

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Current Electricity

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Question 39 Report

A 500W heater is used to heat 0.6kg of water from \(25^\circ\mathrm{C}\) to \(100^\circ\mathrm{C}\) in \(t_2\) seconds. If another 1000W heater is used to heat 0.2kg of water from \(10^\circ\mathrm{C}\) to \(100^\circ\mathrm{C}\) in t, seconds, find \(\frac{t_1}{t_2}\)

50

5

5/4

1/5

Answer Details

Heat energy = Electrical energy

mc $θ$ = power x time

0.6 x c x (100 - 25) = 500 x t1

and, 0.2 x c x (100 - 10)= 100 x t2
$\frac{t_{1}}{t_{2}}$ = 5

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Heat Energy

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Question 44 Report

A solid of weight 0.600N is totally immersed in oil and water respectively. If the upthrust of oil is 0.210N and the relative density of oil is 0.875. Find the upthrust in water.

0.600N

0.360N

0.240N

0.180N

Answer Details

Rd =

\frac{upthrust in oil}{upthrust in H_{2} O}

thus; 0.875 = $\frac{0.21}{upthrust in H_{2} O}$

upthrust in H2O = 0.24N

Read lesson note on Gravitational Field (JAMB)

Gravitational Field

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