2006 - JAMB Physics Past Questions and Answers - page 4
31
I. The number of neutrons released exceeds the number that caused the fission.
II. There is loss of mass.
III. Energy is produced.
IV. Smaller nuclei merge to form larger nuclei
which of the statement above are correct about nuclear fission?
II. There is loss of mass.
III. Energy is produced.
IV. Smaller nuclei merge to form larger nuclei
which of the statement above are correct about nuclear fission?
A
II,III and IV only
B
I, II and IV only
C
I,II and III only
D
I, III and IV only
correct option: c
Users' Answers & Comments32
A certain manufacturer wishes to make an n-type semiconductor. Which materials would he select for doping?
A
boron and antimony
B
boron and antimony
C
aluminium and indium
D
arsenic and antimony
correct option: d
An n- type semi conductor is usually formed by dopping the semi conductor(a four valent element)with a five valent
element.(arsenic and antimony)
Users' Answers & Commentselement.(arsenic and antimony)
33
A p-n junction diode is used as?
A
a rectifier in a d.c circuit
B
an amplifier in d.c circuit
C
an anplifier in an a.c circuit
D
a rectifier in an a.c circuit
correct option: d
Users' Answers & Comments34
An object placed 10cm from a concave mirror of focal length 5cm would have its image
A
at the centre of convature
B
at the radius of curvature
C
at infinity
D
10 cm from the focus
correct option: a
Users' Answers & Comments35
A submarine is observed to rise from a real depth off 80 m to 60 m in water. calculate the change in apparent depth.
[Refraction index of water (4)/3]
[Refraction index of water (4)/3]
A
45 m
B
80 m
C
15 m
D
60 m
correct option: c
Refraction index (n)= (Real depth)/App. depth
For 80 m Real depth: (4)/3 = (80)/App. depth
App. dept = (80 x 3)/4 = 60 m
For 60 m real depth; (4)/3 = (60)/ App. depth
App. depth = (60 x 3)/4 = 45 m
∴change in App. depth = 60 -45 = 15 m
Users' Answers & CommentsFor 80 m Real depth: (4)/3 = (80)/App. depth
App. dept = (80 x 3)/4 = 60 m
For 60 m real depth; (4)/3 = (60)/ App. depth
App. depth = (60 x 3)/4 = 45 m
∴change in App. depth = 60 -45 = 15 m
36
Non-luminous object can be seen because they
A
reflect light
B
are near
C
are polished
D
emit light
correct option: a
Users' Answers & Comments37
The wavelength of the first overtone of a note in a closed pipe of length 33 cm is?
A
33 cm
B
44 cm
C
22 cm
D
17 cm
correct option: b
In a closed tube, the fundamental note occurs when the length of the tube l = (λ)/4,
the first over tone occurs when the length of the tube l = (3)/4λ.
λ = (4)/3 l = (4)/3 x (33)/T = 44 cm
Users' Answers & Commentsthe first over tone occurs when the length of the tube l = (3)/4λ.
λ = (4)/3 l = (4)/3 x (33)/T = 44 cm
38
A person can focus objects when they lie beyond 75 cm from his eyes. The focal length of the lens required to reduce his least distance of distinct vision to 25 cm is?
A
37.50 cm
B
75.00 cm
C
18.75
D
25.00 cm
correct option: a
The least distance of distinct vision required = 25 cm.
∴ for the person to see an object clearly at 25 cm from the eye, the image must be formed at 75 cm on the same side of lens, at his near point.
∴ U = 25 cm; V = -75 cm (image should be virtual)
thus (1)/f = (1)/v +(1)/u
= (1)/75 + (1)/25
=(-1 + 3)/75 = (2)/75
∴f = (75)/2 = 37.50
Users' Answers & Comments∴ for the person to see an object clearly at 25 cm from the eye, the image must be formed at 75 cm on the same side of lens, at his near point.
∴ U = 25 cm; V = -75 cm (image should be virtual)
thus (1)/f = (1)/v +(1)/u
= (1)/75 + (1)/25
=(-1 + 3)/75 = (2)/75
∴f = (75)/2 = 37.50
39
If two charged plates are maintained at a potential difference of 3 kv, the work done in taking a charge of 600 µC across the field is?
A
0.8 J
B
1.8 J
C
9.0 J
D
18.0 J
correct option: b
Work done = product of the charge Q, and the p.d V.
(i)/e work done = Q.V = 600 x 10-6 x 3.0 x 10-3 = 1.8 J.
Users' Answers & Comments(i)/e work done = Q.V = 600 x 10-6 x 3.0 x 10-3 = 1.8 J.
40
A conductor has a diameter of 1.00 mm and length 2.00m. if the resistance of the material is 0.1Ω, its resistivity is?
A
2.55 x 105 Ωm
B
2.55 x 102 Ωm
C
3.93 x 10-6 Ωm
D
3.93 x 10-8 Ωm
correct option: d
R = (1)/a, P = (R.a)/1 = (0.1 x 3.14)/2 (10-3/2)3
∴ P = (0.1 x 3.14 x 10-6)/2 x 4
= 0.3925 x 10-7 OR -3.930 X 10-8 Ωm
Users' Answers & Comments∴ P = (0.1 x 3.14 x 10-6)/2 x 4
= 0.3925 x 10-7 OR -3.930 X 10-8 Ωm