2001 - JAMB Physics Past Questions and Answers - page 2
11
The driving mirror of a car has a radius of curvature of 1m. A vehicle behind the car is 4m from the mirror. Find the image distance behind the mirror.
A
8/7
B
4/9
C
9/2
D
4/7
correct option: b
A driving mirror is always a convex mirror which is always of negative focal length and always forms a virtual image
Thus if r = 1m = 100cm
f = 1/2 = -50
∴ if U = 4cm = 400cm
f = -50cm
∴1/u + 1/v = 1/f
1/400 + 1/v = -1/50
∴ -1/v = 1/400 + 1/50
-1/v = (1 + 8)/400 = 9/400
∴ v = -400/9
v = 4/9
Users' Answers & CommentsThus if r = 1m = 100cm
f = 1/2 = -50
∴ if U = 4cm = 400cm
f = -50cm
∴1/u + 1/v = 1/f
1/400 + 1/v = -1/50
∴ -1/v = 1/400 + 1/50
-1/v = (1 + 8)/400 = 9/400
∴ v = -400/9
v = 4/9
12
The pressure of a given mass of a gas changes from 200Nm-2 to 120Nm-2 while the temperature drops from 127°C to -73°C. The ratio of the final volume to the initial volume is
A
2 : 5
B
5 : 4
C
5 : 2
D
4 : 5
correct option: b
P1 = 300Nm-2; P2 = 120Nm-2
T1 = 127 + 273; T2 = -73 + 273
= 400K; = 200K
Let the initial volume V1; and the final volume = V2
∴ P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
∴ V2/V1 = P1T2/P2T1
∴ V2/V1 = (300 x 200)/(120 x 400) = 5/4
∴ V2 : V1 = 5 : 4
Users' Answers & CommentsT1 = 127 + 273; T2 = -73 + 273
= 400K; = 200K
Let the initial volume V1; and the final volume = V2
∴ P1V1/T1 = P2V2/T2
P1V1T2 = P2V2T1
∴ V2/V1 = P1T2/P2T1
∴ V2/V1 = (300 x 200)/(120 x 400) = 5/4
∴ V2 : V1 = 5 : 4
13
The thermometric substance of an absolute thermometer is
A
alcohol
B
mercury
C
helium
D
platinum
correct option: b
Users' Answers & Comments14
A plane sound wave of frequency 85.5Hz and velocity 342ms-1 is reflected from a vertical wall. At what distance from the wall does the wave have antinode?
A
2m
B
4m
C
1m
D
3m
correct option: c
If F = 85.5Hz, V = 342ms-1,
Then wavelength = V/F = 342/85.5 = 4m
Then if wavelength = 4m
=> wavelength/2 = 4/2 = 2m = NN,
Thus NA = wavelength/4 = 4/4 = 1m.
Users' Answers & CommentsThen wavelength = V/F = 342/85.5 = 4m
Then if wavelength = 4m
=> wavelength/2 = 4/2 = 2m = NN,
Thus NA = wavelength/4 = 4/4 = 1m.
15
A string is fastened tightly between two walls 24cm apart. The wavelength of the second overtone is
A
24cm
B
16cm
C
12cm
D
8cm
correct option: b
For a string fixed at both ends and plucked at the middle, the fundamental frequency is given
by f0 = v/λ, where v = velocity of sound, λ = wave length.
At its fundamental, note, the the length of the string is given l = λ/2, => λ0 = 2l, where λ0 = wave length fundamental note. for the first over tone, λ1 = l.
For the second over tone, λ2 = 21/3 = (2 x 24)/3 = 16cm.
i.e. λ2 = 16cm.
Users' Answers & Commentsby f0 = v/λ, where v = velocity of sound, λ = wave length.
At its fundamental, note, the the length of the string is given l = λ/2, => λ0 = 2l, where λ0 = wave length fundamental note. for the first over tone, λ1 = l.
For the second over tone, λ2 = 21/3 = (2 x 24)/3 = 16cm.
i.e. λ2 = 16cm.
16
A gas with initial volume of 2 X 10-6m3 is allowed to expand to six times its initial value at constant pressure of 2 x 105Nm-2. The work done is
A
2.0J
B
4. 0 J
C
12.0 J
D
2.4 J
correct option: d
P = F/A => F = PA = PV/L
W = FD = (PV/L) x L = PV
= 2 x 105
(2 x 10-6 x 6)
= 2 x 105 x 1.2 x 10-5
= 2.4 J
Users' Answers & CommentsW = FD = (PV/L) x L = PV
= 2 x 105
(2 x 10-6 x 6)
= 2 x 105 x 1.2 x 10-5
= 2.4 J
17
A cell of internal resistance r supplies current to a 6.0Ω resistor and its efficiency is 75%. Find the value of r.
A
4.5Ω
B
1.0Ω
C
8.0Ω
D
2.0Ω
correct option: d
Efficiency = power out put/power input * 100/1
= power out put/ power + that due * 100/1
to internal resistance
=> 75/1 = 12R *100/1
12(R+r)
i.e 75/100 = 6/6+r => 75 (6+r) = 100/*6
i.e 450 + 75r = 600
75r =600 - 450
i.e r = 150/75 = 2Ω
i.e r = 2Ω
Users' Answers & Comments= power out put/ power + that due * 100/1
to internal resistance
=> 75/1 = 12R *100/1
12(R+r)
i.e 75/100 = 6/6+r => 75 (6+r) = 100/*6
i.e 450 + 75r = 600
75r =600 - 450
i.e r = 150/75 = 2Ω
i.e r = 2Ω
18
A resistance R is connected across the terminal of an electric cell of internal resistance 2Ω and the voltage was reduced to 3/5 of its nominal value.
The value of R is
The value of R is
A
3 Ω
B
2 Ω
C
1 Ω
D
6 Ω
correct option: a
1 = V/R = E/R+r , if r = 2Ω
V = 3/5 E
=> 3E/5R = E/R+2
i.e 3E(R+2) = 5RE
3RE + 6E = 5RE
i.e 3R + 6 = 5R
6 = 5R - 3R
=> 2r = 6 i.e R = 6/2 = 3Ω
Users' Answers & CommentsV = 3/5 E
=> 3E/5R = E/R+2
i.e 3E(R+2) = 5RE
3RE + 6E = 5RE
i.e 3R + 6 = 5R
6 = 5R - 3R
=> 2r = 6 i.e R = 6/2 = 3Ω
19
A student is at a height 4m above the ground during a thunderstorm. Give that the potential difference between the thundercloud and the ground is 107v, the electric field created by the storm is
A
2.0 * 106 NC -1
B
2. 5 * 106 NC -1
C
1.0 * 107 NC -1
D
4.0 * 107 NC -1
20
A working electric motor takes a current of 1.5 A when the p.d across it is 250V. If its efficiency is 80%, the power output is
A
300.0W
B
469.0W
C
133.0W
D
4.8W
correct option: a
Efficiency = power output/power Input
=> 80/100 = power output/1.5 * 250
i.e power output = 1.5*250 *80/100 =300w
Users' Answers & Comments=> 80/100 = power output/1.5 * 250
i.e power output = 1.5*250 *80/100 =300w