2000 - JAMB Physics Past Questions and Answers - page 4
31
A certain radioactive source emits radiations that were found to be deflected by both magnetic and electric fields. The radiations are
A
beta rays
B
gamma rays
C
x-rays
D
unltr-violet rays
correct option: a
Users' Answers & Comments32
A 2H inductor has negligible resistance and is connected to a 50/π Hz A.C. Supply. The reactance of the inductor is
A
200 Ω
B
50 Ω
C
100/π Ω
D
25/π Ω
correct option: a
Inductive reactance is given by
XL = WL = 2πfL
XL = 2 x π x (50/π) x 2
XL = 200Ω
Users' Answers & CommentsXL = WL = 2πfL
XL = 2 x π x (50/π) x 2
XL = 200Ω
33
A cell of internal resistance 1Ω supplies current to an external resistor of 3Ω. The efficiency of the cell is
A
25%
B
33%
C
50%
D
75%
correct option: d
Efficiency = Output/input x 100/1
Efficiency = 3/(3+1) x 100/1
= 3/4 x 100/1 = 75%
Users' Answers & CommentsEfficiency = 3/(3+1) x 100/1
= 3/4 x 100/1 = 75%
34
A conductor of length 2m carries a current of 0.8A while kept in a magnetic field of magnetic flux density 0.5T. The maximum force acting on it is
A
0.2N
B
0.8N
C
3.2N
D
8.0N
correct option: b
The force on the conductor is maximum when the conductor lies perpendicular to the field during which the maximum force is given as;
F = ILB = 0.8 x 2 x 0.5 = 0.8N
Users' Answers & CommentsF = ILB = 0.8 x 2 x 0.5 = 0.8N
35
Light of energy 5eV falls on a metal of work function 3eV and electrons are liberated. The stopping potential is
A
1.7V
B
2.0V
C
8.0V
D
15.0V
correct option: b
For a photoemitted electron,
eV = K.E.max, where V = stopping potential and K.E. = maximum kinetic energy of the photo emitted electron given by K.E. = Energy of incident rad - Work - function
= 5ev - 3ev = 2ev
∴ eV = 2eV
V = 2eV/e = 2V
∴ Stopping Potential = 2.0V
Users' Answers & CommentseV = K.E.max, where V = stopping potential and K.E. = maximum kinetic energy of the photo emitted electron given by K.E. = Energy of incident rad - Work - function
= 5ev - 3ev = 2ev
∴ eV = 2eV
V = 2eV/e = 2V
∴ Stopping Potential = 2.0V
36
The velocity y of a particle in a time t is given by the equation
y = 10 + 2t2
Find the instantaneous acceleration after 5 seconds
y = 10 + 2t2
Find the instantaneous acceleration after 5 seconds
A
60 ms-2
B
20 ms-2
C
15 ms-2
D
10 ms-2
correct option: b
The velocity V = 10 + 2t2
But acceleration, a = dv/dt
thus if V = 10 + 2t2
dv/dt = 4t
And t = 5s then a = 4x5 = 20m/s2
∴a = 20ms
Users' Answers & CommentsBut acceleration, a = dv/dt
thus if V = 10 + 2t2
dv/dt = 4t
And t = 5s then a = 4x5 = 20m/s2
∴a = 20ms
37
A simple pendulum has a period of 17.0s. When the length is shorten by 1.5m, its period is 8.5s. Calculate the original length of the pendulum
A
4.0 m
B
3.0 m
C
2.0 m
D
1.5 m
correct option: c
For a simple pendulum, the period T = 2π
√ 1/g
Since g, 2 and π are constant, => T ∝ √l
∴ T = K√l
Let the original length = l
thus for the length l, period T = 17s
∴ 17 = K√1 .................. [1]
Again, for the new length [1 - 1.5]
the period T = 8.5s
∴ 8.5 = k
√ [1-1.5] ...............[2]
thus dividing eqn [1] by eqn[2] we have
17/8.5 = √(l/l-1.5)
∴2 = √(l/l-1.5)
=>4 = (l/l-1.5)
∴l = 4[l-1.5]
= 4l - 6.0
∴4l - l = 6.0
3l = 6.0
l = 2.0m
Users' Answers & Comments√ 1/g
Since g, 2 and π are constant, => T ∝ √l
∴ T = K√l
Let the original length = l
thus for the length l, period T = 17s
∴ 17 = K√1 .................. [1]
Again, for the new length [1 - 1.5]
the period T = 8.5s
∴ 8.5 = k
√ [1-1.5] ...............[2]
thus dividing eqn [1] by eqn[2] we have
17/8.5 = √(l/l-1.5)
∴2 = √(l/l-1.5)
=>4 = (l/l-1.5)
∴l = 4[l-1.5]
= 4l - 6.0
∴4l - l = 6.0
3l = 6.0
l = 2.0m
38
A sonometer wire is vibrating at frequency fo. If the tension in the wire is doubted while the length and the mass per unit length are kept constant, the new frequency of vibration is
A
fo/2
B
2fo
C
fo/√2
D
fo√2
correct option: d
For a vibrating sonometer wire, the frequency
F = l/2l √(T/m)
and if the length, l and mass per unit length, m, are constant, then Fo ∝ √T => FoK√T or Fo/√T = K
=> F1o/√F1 = F2o/√F2
∴ F1o/√T1 = F2o/√T2
=
∴F2 = Fo x √2 = Fox√2
Users' Answers & CommentsF = l/2l √(T/m)
and if the length, l and mass per unit length, m, are constant, then Fo ∝ √T => FoK√T or Fo/√T = K
=> F1o/√F1 = F2o/√F2
∴ F1o/√T1 = F2o/√T2
| thus F2 = | F3x√(2T1 |
| __√T1 |
=
| = | Fox√2 x √T1 |
| √T1 |
∴F2 = Fo x √2 = Fox√2
39
A travelling wave moving from left to right has an amplitude of 0.15m, a frequency of 550Hz and a wavelength of 0.01m. The equation describing the wave is
A
y = 0.15 sin 200π(x - 5.5t)
B
y = 0.15 sin π(0.01x - 5.5t)
C
y = 0.15 sin 5.5π(x - 200t)
D
y = 0.15 sin π(550x - 0.01t)
correct option: a
The general wave equation is given by Y = sin A 2π/λ [x - vt]
= Sin A [(2π X)/λ - (2π vt)/λ]
thus if A = 0.15;λ = 0.01m; f = 550Hz
then V = f x λ = 550 x 0.01 = 5.50m/s
∴ Y = 0.15 sin [(2 x λ x X)/0.01 - (2 x λ x 5.5t)/0.01]
= 0.15 sin[200λX - 200 x λ x 5.5t]
∴ Y = 0.15 sin 200λ[x - 5.5t]
Users' Answers & Comments= Sin A [(2π X)/λ - (2π vt)/λ]
thus if A = 0.15;λ = 0.01m; f = 550Hz
then V = f x λ = 550 x 0.01 = 5.50m/s
∴ Y = 0.15 sin [(2 x λ x X)/0.01 - (2 x λ x 5.5t)/0.01]
= 0.15 sin[200λX - 200 x λ x 5.5t]
∴ Y = 0.15 sin 200λ[x - 5.5t]
40
A transverse wave is applied to a string whose mass per unit length is 3 x 10-2kgm-1. If the string is under a tension 12N, the speed of the propagation of the wave is
A
5 ms-1
B
20 ms-1
C
30 ms-1
D
40 ms-1
correct option: b
For transverse wave on string
velocity V = √T/M
where T is the tension, and M is the mass per unit length
∴ for T = 12N and M = 3.0 x 10-2kgm-1
V=
√ 12 / (3.0x10-2
=
√ 12 / 0.03
=
√ 1200 / 3
=
√ 1200 / 3 = 20
∴V = 20m/s
Users' Answers & Commentsvelocity V = √T/M
where T is the tension, and M is the mass per unit length
∴ for T = 12N and M = 3.0 x 10-2kgm-1
V=
√ 12 / (3.0x10-2
=
√ 12 / 0.03
=
√ 1200 / 3
=
√ 1200 / 3 = 20
∴V = 20m/s