2003 - JAMB Physics Past Questions and Answers - page 5
41
When a nucleus is formed by bringing protons and neutrons together, the actual mass of the formed nucleus is less than the sum of the masses of the energy equivalent of this mass difference is the
A
lost energy
B
work function
C
binding energy
D
stability energy
correct option: c
Users' Answers & Comments42
A bead traveling on a straight wire is brought to rest at 0.2m by friction. If the mass of the bead is 0.01kg and the coefficient of friction between the bead and the wire is 0.1 determine the work done by the friction
A
\( 2 \times 10^{-4}J \)
B
\( 2 \times 10^{-3}J \)
C
\( 2 \times 10^{1}J \)
D
\( 2 \times 10^{2}J \)
correct option: b
Work done by friction = Friction force x Displacement from the relation, f = uR
where u = coefficient of friction; R= Normal reaction
F = u mg (R =mg)
= 0.1 x 0.01 x 10
work done = F x Displacement
= 0.1 x 0.01 x 10 x 0.2
= 0.002 or \( 2 \times 10^{-3}J \)
Users' Answers & Commentswhere u = coefficient of friction; R= Normal reaction
F = u mg (R =mg)
= 0.1 x 0.01 x 10
work done = F x Displacement
= 0.1 x 0.01 x 10 x 0.2
= 0.002 or \( 2 \times 10^{-3}J \)
43
A ray of light is incident on an equilateral triangular glass prism of refractive index 3/2, Calculate the angle through which the ray is minimally deviated in the prism
A
30.0\(^{\circ} \)
B
37.2\(^{\circ} \)
C
42.0\(^{\circ} \)
D
48.6\(^{\circ} \)
correct option: b
For the equilateral glass prism A = 600
\( \frac{3}{2} \text{Sin} \frac{\left( \frac{Dm + 60}{2}\right)}{\left(\frac{60}{2}\right)}= \frac{\left( \text{ Sin }\frac{Dm + 60}{2} \right)}{0.5} \\
\implies Dm = 37.2^{\circ} \)
Users' Answers & Comments\( \frac{3}{2} \text{Sin} \frac{\left( \frac{Dm + 60}{2}\right)}{\left(\frac{60}{2}\right)}= \frac{\left( \text{ Sin }\frac{Dm + 60}{2} \right)}{0.5} \\
\implies Dm = 37.2^{\circ} \)
44
In the calibration of an ammeter using faraday's laws of electrolysis, the ammeter reading is kept constant at 1.2A. If 0.990g of copper is deposited in 40 minutes , the correction to be applied to the ammeter is
A
0.03A
B
0.04A
C
0.05A
D
0.06A
correct option: c
From faraday's law of electrolysis
M = I t Z
0.990 = I x (40 x 60) x 3.3 x 10-4
\( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \\ \text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A \)
Users' Answers & CommentsM = I t Z
0.990 = I x (40 x 60) x 3.3 x 10-4
\( \implies \frac{0.990}{2400 \times 3.3 \times 10^{-4}} = 1.25A \\ \text{Thus the correction to be made } = 1.25 - 1.20 = 0.05A \)
45
An electron makes a transition from a certain energy level Ek to the ground state E0. If the frequency of emission is 8.0 x 1014Hz The energy emitted is
A
8.25 x 10-19J
B
5.28 x 10-19J
C
5.28 x 1019J
D
8.25 x 1019J
correct option: b
Atomic emission and absorption of energy is usually in packets referred to as photon or quantum of energy of magnitude E =hf, where f is the frequency and h is the plancks constant
Energy emitted = hf
= 6.6 x 10-34 x 8.0 x 10-14
= 5.28 x 10-19J
Users' Answers & CommentsEnergy emitted = hf
= 6.6 x 10-34 x 8.0 x 10-14
= 5.28 x 10-19J
46
A force of 100N was used to kick a football of mass 0.8kg. Find the velocity with which the ball moves if it takes 0.8s to be kicked
A
\(32ms^{-1} \)
B
\(50ms^{-1} \)
C
\(64ms^{-1} \)
D
\(100ms^{-1} \)
correct option: d
From Newton's second law:
\( F = \frac{mV - mµ}{t} \)
Let the final velocity = V; and since the football is initially at rest, its initial vel µ = 0, M = 0.8kg
\( t = 0.8s \\
\implies 100 = \frac{0.8V - 0.8 \times 0}{0.8} \\
\text{Therefore } V = \frac{100 \times 0.8}{0.8} = 100ms^{-1} \)
Users' Answers & Comments\( F = \frac{mV - mµ}{t} \)
Let the final velocity = V; and since the football is initially at rest, its initial vel µ = 0, M = 0.8kg
\( t = 0.8s \\
\implies 100 = \frac{0.8V - 0.8 \times 0}{0.8} \\
\text{Therefore } V = \frac{100 \times 0.8}{0.8} = 100ms^{-1} \)
47
Given that young's modulus for aluminium is 7.0 x 1010Nm-2 and density is 2.7 x 103kgm-3 find the speed of the sound produced if a solid bar is struck at one end with a hammer
A
5.1 x 103ms-1
B
4.2 x 103ms-1
C
3.6 x 103ms-1
D
2.8 x 103ms-1
correct option: a
The speed of sound in a material is given as:
\( V = \sqrt{\frac{E}{P}}, Where\\
E = \text{Young's modulus} \\ P = \text{Density }\\
\text{Therefore } V = \sqrt{\frac{7.0 \times 10^{10}}{2.7 \times 10^3}} = 5091.75\\
= 5.1 \times 10^3MS^{-1} \)
Users' Answers & Comments\( V = \sqrt{\frac{E}{P}}, Where\\
E = \text{Young's modulus} \\ P = \text{Density }\\
\text{Therefore } V = \sqrt{\frac{7.0 \times 10^{10}}{2.7 \times 10^3}} = 5091.75\\
= 5.1 \times 10^3MS^{-1} \)
48
If the uncertainty in the measurement of the position of a particle is 5.0 x 10-10m, the uncertainty in the momentum of the particle is
A
1.32 x 10-44Ns
B
3.30 x 10-44Ns
C
1.32 x 10-24Ns
D
3.30 x 10-24Ns
correct option: c
From the uncertainty principle the product of the position and momentum of a sub-atomic particle is equal or greater than planck's constant.
i.e \( \Delta x. \Delta p \geq h. \\ \text{Thus} 5 \times 10^{-10} \times \Delta p = 6.6 \times 10^{-34} \\
\Delta p = \frac{6.6 \times 10^{-34}}{5 \times 10^{-10}} = 1.32 \times 10^{-24}Ns\\
\text{Therefore Uncertainty in the momentum } = 1.32 \times 10^{-24}Ns \)
Users' Answers & Commentsi.e \( \Delta x. \Delta p \geq h. \\ \text{Thus} 5 \times 10^{-10} \times \Delta p = 6.6 \times 10^{-34} \\
\Delta p = \frac{6.6 \times 10^{-34}}{5 \times 10^{-10}} = 1.32 \times 10^{-24}Ns\\
\text{Therefore Uncertainty in the momentum } = 1.32 \times 10^{-24}Ns \)
49
In a fission process, the decrease in mass is 0.01%. How much energy could be obtained from the fission of 0.1g of the material
A
\( 9.0 \times 10^{9}J \)
B
\( 9.0 \times 10^{10}J \)
C
\( 6.3 \times 10^{11}J \)
D
\( 9.0 \times 10^{11}J \)
correct option: a
In general, the energy released during nuclear fission is given by the Einstein's mass-energy equation, given by
\( E = \Delta MC^2 \)
Where \( \Delta m\) is the mass defect, and C is the speed of light.
Therefore
\( \Delta m = 0.01% \text{of} 1.0g = \frac{0.01}{100}
\times 1.0g \\
= 1.0 \times 10^{-4} \\ = 1.0 \times 10^{-7}kg\\
\text{Energy Released } = \Delta MC^2 \\
= 1.0 \times 10^{-7} \times (3.0 \times 10^8)^2\\ = 1.0 \times 10^{-2} \times 9.0 \times 10^{16} \\
= 9.0 \times 10^9 J \)
Users' Answers & Comments\( E = \Delta MC^2 \)
Where \( \Delta m\) is the mass defect, and C is the speed of light.
Therefore
\( \Delta m = 0.01% \text{of} 1.0g = \frac{0.01}{100}
\times 1.0g \\
= 1.0 \times 10^{-4} \\ = 1.0 \times 10^{-7}kg\\
\text{Energy Released } = \Delta MC^2 \\
= 1.0 \times 10^{-7} \times (3.0 \times 10^8)^2\\ = 1.0 \times 10^{-2} \times 9.0 \times 10^{16} \\
= 9.0 \times 10^9 J \)