“Iron rusts from disuse, stagnant water
loses its purity and in cold weather becomes frozen; so does inaction sap the
vigors of the mind.”
– Leonardo da Vinci
If you have a concrete idea about manipulation of vectors you will be at an advantage in problem solving in many situations. Today we shall discuss a few multiple choice practice questions on vectors.
(1) In the following
figure two vectors P and Q are shown. One of the vectors shown in
options (a), (b), (c), (d) and (e) represents their resultant. Pick out the
correct one.
The correct option
is (b).
You can verify this
from the following figure in which a parallelogram is constructed with the
vectors P and Q as adjacent sides. The diagonal of the parallelogram indeed
represents the resultant R of the
vectors P and Q.
(2) A small toy car
is in uniform circular motion with speed 2 ms–1 along horizontal ground. When the acceleration
of the car is directed towards east its velocity is directed towards south. If
the magnitude of its acceleration is 4
ms–2, the circular path followed
by the car is
(a) clockwise with radius 1 m
(b) anticlockwise with radius 1 m
(c) clockwise with radius 2 m
(d) anticlockwise with radius 2 m
(e) clockwise with radius 4 m
The path of the car
is shown in the adjoining figure in which the normal acceleration (centripetal
acceleration) vector is indicated eastwards and the velocity vector is
indicated southwards. Evidently the path is anticlockwise. The normal
acceleration ‘a’ is given by
a = v2/r where ‘v’ is
the speed and ‘r’ is the radius of the circular path.
Therefore, r = v2/a
= 22/4 = 1 m.
The correct option
is (b).
(3) A motor boat can
cross a 60 m wide river in a minimum time of 10 s when the water is still. What
will be the minimum time required by the boat to cross the river when the water
in the river flows steadily at a speed of 1 ms–1?
(a) 9.86 s
(b) 10.14 s
(c) 8.57 s
(d) 12 s
(e) 10 s
The minimum time required will be 10 s itself. In
still water the boat will have to move in a direction at right angles to the
bank in order to reach the other bank in minimum
time. When the river is flowing, the boat will reach the opposite bank in
the same minimum time if the engine drives the boat at right angles to the
bank. In this case the velocity component of the boat at right angles to the river
bank will be unaffected by the flow of the river. But, because of the flow of
the river the boat will be carried downstream through 10 m by the time it reaches
the other bank.
The following questions
are meant for AP Physics C aspirants:
(4) A and B shown in the figure are electric field vectors. The component of
vector A along the direction of
vector B is
(a) (A × B)/ | A |
(b) (A × B)/ | B |
(c) (A . B)/ | A |
(d) (A . B)/ | B |
(e) | (A × B) |
If the angle
between A and B is θ,
the component of A along the direction of B is A cosθ. We have
A . B = AB cosθ where A
and B are the magnitudes of A and B respectively.
Therefore, A cosθ = (A . B)/B = (A . B)/ | B |
(5) Specific charge
of proton is 9.6×107 C/kg. A proton
having velocity (4 i + 6 j) m/s enters a magnetic field of flux density (j + 2 k) tesla where i, j and k are unit vectors along the x, y and z directions respectively.
The acceleration produced in the proton in m/s2 is
(a) 3.84×108
(3i – 2j + k)
(b) 3.84×108
(6i – 4j + 2k)
(c) 9.6×107(6i – 4j)
(d) 1.92×108
(3i – 2j + k)
(e) 1.92×108
(3i + k)
The magnetic force F on the proton is
given by
F
= e(v×B) where e
is the charge on
the proton, v is its velocity and B is the magnetic
flux density.
The acceleration a of the proton is given by
a = e(v×B)/m where m is the mass of the proton.
Or, a = (e/m) (v×B)
Substituting for the specific charge (e/m), v
and B, we have
a = 9.6×107(4 i
+ 6 j) × (j + 2 k)
Or, a = 9.6×107(4
k – 8 j + 12 i)
This gives a = 3.84×108 (3i – 2j + k), as given in option (a).
[Remember that i × j =
k, i × k
= – j, j
× j = 0, j × k = i]
You will find a
couple of multiple choice questions (with solution) on vectors here.
No comments:
Post a Comment