2007 - JAMB Physics Past Questions and Answers - page 1
1
The unit of moment of a couple can be expressed in?
A
N m-1
B
N m-2
C
N m
D
N m2
2
A gramophone record takes 5s to reach its constant angular velocity of 4\(\pi\) rads-1 from rest. Find its constant angular acceleration.
A
0.4\(\pi\) rads-2
B
0.8\(\pi\) rads-2
C
1.3\(\pi\) rads-2
D
20.0\(\pi\) rads-2
correct option: b
From the radiation:
V = +at, and if it starts from N & T, U = 0
And for angular motion, W = O + at
4R = aχ5
∴ a = (4a)/5 = 0. 8\(\pi\) rads-2
Users' Answers & CommentsV = +at, and if it starts from N & T, U = 0
And for angular motion, W = O + at
4R = aχ5
∴ a = (4a)/5 = 0. 8\(\pi\) rads-2
3
A car of mass 1500 kg goes round a circular curve of radius 50m at a speed of 40 ms-1. The magnitude of the centripetal force on the car is ?
A
4.8 x 104 N
B
4.8 x 103 N
C
1.2 x 103 N
D
1.2 x 102 N
correct option: a
Centipetal force of the car = ma = (MV)2/r
= (1500 x 40 x40)/50
= 48000 = 4.8 x 104 N
Users' Answers & Comments= (1500 x 40 x40)/50
= 48000 = 4.8 x 104 N
4
Two spheres of masses 5.0 kg and 10.0 kg are 0.3 m apart. Calculate the force of attraction between them?
[G = 6.67 X 10-11 Nm2 Kg2]
[G = 6.67 X 10-11 Nm2 Kg2]
A
3.50 x 10-10N
B
3.71 x 10-8 N
C
3.57 x 10-2N
D
4.00 x 102N
correct option: b
The force of attraction between the two spheres is given by F (GM1 M2)/r2 = (6.67X 10-11 X5X10)/(0.3)2
= 3.71 X 10-8N.
Users' Answers & Comments= 3.71 X 10-8N.
5
A machine has a velocity ratio of 4. if it requires 800 N to overcome a load of 1600 N,what is the efficiency of the machine?
A
60%
B
50%
C
40%
D
2%
correct option: b
Efficiency = (M.A)/V.R X (100)/1
But M.A = (Load)/Effort = (1600)/800 = 2
Efficiency = (2)/4 x (100)/1 = 50%
Users' Answers & CommentsBut M.A = (Load)/Effort = (1600)/800 = 2
Efficiency = (2)/4 x (100)/1 = 50%
6
A reservoir 500 m deep is filled with a fluid of density 850 kgm-3. If the atmospheric pressure is 1.05 x 105 Nm-2, the pressure at the bottom of the reservoir is.
[g = 10ms-2]
[g = 10ms-2]
A
4.72 x 106 Nm-2
B
4.36 x 106 Nm-2
C
4.28 x 106 Nm-2
D
4.25 x 106 Nm-2
correct option: d
In general, pressure varies with depth in the relation p = lgh
= 850 x 10 x 500
= 4250000
= 4.25 x 106 NM-2
Users' Answers & Comments= 850 x 10 x 500
= 4250000
= 4.25 x 106 NM-2
7
A piece of iron weights 250 N in air and 200 N in a liquid of density 1000 kgm-3. the volume of the iron is?
A
5.0 x 10-2 ms-2
B
4.5 x 10--2 ms-3
C
2.5 x 10-2 ms-8
D
2.0 x 10-2 ms-3
correct option: c
With in the air = 250N
wt. in the liquid = 200N
loss in wt. in the liquid = 250 - 200 = 50N
∴ Wt. the liquid of equal vol. of the iron = 50N
∴If the density of the liquid is 1000 lgm-3
then the volume of the liquid = (N)/D = V = (25kg)/1000 =0.025m3 = 2.5 x 10-2 m8.
Users' Answers & Commentswt. in the liquid = 200N
loss in wt. in the liquid = 250 - 200 = 50N
∴ Wt. the liquid of equal vol. of the iron = 50N
∴If the density of the liquid is 1000 lgm-3
then the volume of the liquid = (N)/D = V = (25kg)/1000 =0.025m3 = 2.5 x 10-2 m8.
8
A clinical thermometer is different from other mercury in glass thermometers owing to
A
the constriction on its stem
B
the grade of mercury used in it
C
its long stem
D
its wide range of temperatures
correct option: a
Users' Answers & Comments9
I. Use a liquid with a high melting point.
II. use a liquid of high volume expansivity.
III. use a capillary tube of large diameter.
which of the above best describes how the sensitivity of a liquid-in-glass thermometer can be enhanced?
II. use a liquid of high volume expansivity.
III. use a capillary tube of large diameter.
which of the above best describes how the sensitivity of a liquid-in-glass thermometer can be enhanced?
A
I only
B
II. only
C
I and II only
D
II and III only
10
2 kg of water is heated with a heating coil which draws 3.5 A from a 200 V mains for 2 minutes. what is the increase in temperature of the water?
[specific heat capacity of water = 4200 jkg -1k-1]
[specific heat capacity of water = 4200 jkg -1k-1]
A
30°
B
25°
C
15°
D
10°
correct option: d
Electricity heat energy supplies by the heater = lvt = 3.5 x 200 x (2 x 60) = 84000j.
and this heat is supplied to the 2 kg of the water MC ∆ t = 84000
2 x 4200 x ∆ t = 84000
∴ ∆t = (84000)/8400 = 10°c
∴ increase in temp. of water = 10°c
Users' Answers & Commentsand this heat is supplied to the 2 kg of the water MC ∆ t = 84000
2 x 4200 x ∆ t = 84000
∴ ∆t = (84000)/8400 = 10°c
∴ increase in temp. of water = 10°c