Exemplar MCQs

Hint

Initial velocity \(=\mathrm{v}_{\mathrm{i}}=50 \mathrm{j}\)
Final velocity \(=\mathrm{v}_{\mathrm{f}}=-50 \mathrm{i}\)
Change in velocity \(=\Delta \mathrm{v}=\mathrm{v}_{\mathrm{f}}-\mathrm{v}_{\mathrm{i}}\)
Magnitude of change in velocity
\(|\Delta V|=\sqrt{50^{2}+50^{2}+2 \times 50 \times 50 \cos \left(90^{\circ}\right)}=70 \mathrm{~km} / h
\)
It is towards the southwest.

Hint

(c) Velocity is uniform in both cases; that is, acceleration is zero.
\(
\begin{aligned}
&x=v_{1} t_{1} \Rightarrow t_{1}=\frac{x}{v_{1}} \\
&x=v_{2} t_{2} \Rightarrow t_{2}=\frac{x}{v_{2}}
\end{aligned}
\)
Total displacement, \(x^{\prime}=2 x\)
Total time, \(t=t_{1}+t_{2}\)
\(
\begin{aligned}
&\therefore \text { Average velocity, } v=\frac{x^{\prime}}{t}=\frac{2 v_{1} v_{2}}{v_{1}+v_{2}} \\
&\Rightarrow \frac{2}{v}=\frac{1}{v_{1}}+\frac{1}{v_{2}}
\end{aligned}
\)

Hint

(d) Gravity is the only force acting on the stone when it is released. And, we know that gravity is always in the downward direction.

Hint

(c) \(
v_{\mathrm{A}}=v_{\mathrm{B}}
\)
Total energy of any particle \(=\frac{1}{2} m v^{2}+m g h\)
Both the particles were at the same height and thrown with equal initial velocities, so their initial total energies are equal. By the law of conservation of energy, their final energies are equal.
At the ground, they are at the same height. So, their P.E. are also equal; this implies that their K.E. should also be equal. In other words, their final velocities are equal.

Hint

(c) In projectile motion, velocity is perpendicular to acceleration only at the highest point. Here, velocity is along the horizontal direction and acceleration is along the vertically downward direction.

Hint

(c) Given, \(\mathrm{u}_{1}>\mathrm{u}_{2}\)
In both cases, the downward acceleration and the initial vertical velocities are the same, therefore, they will take same time to hit the ground.
As we know
\(\mathrm{h}=\mathrm{u}_{\mathrm{y}} \mathrm{t}+\frac{1}{2} \mathrm{gt}^{2}\) (Applying in vertical direction)
\(
\begin{aligned}
&\Rightarrow \mathrm{h}=0+\frac{1}{2} \mathrm{gt}^{2} \\
&\Rightarrow \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}
\end{aligned}
\)
time of flight does not depend on the horizontal velocity.
Hence both will reach simultaneously.

Hint

(d) Given that,Range \(\mathrm{R}=50 \mathrm{~m}\)
We know that For the first case
\(\mathrm{R}=\frac{\mathrm{u}^{2}\left(\sin 2 \times 15^{0}\right)}{\mathrm{g}}\)
\(50=\frac{u^{2}}{2 g}\)
\(\frac{\mathrm{u}^{2}}{\mathrm{~g}}=100 \ldots . .(\mathrm{i})\)
For the second case
\(\mathrm{R}=\frac{\mathrm{u}^{2}\left(\sin 2 \times 45^{0}\right)}{\mathrm{g}}\)
\(R=\frac{u^{2}}{g}\)
\(
\mathrm{R}=100 \mathrm{~m}
\)
Hence, the range is \(100 \mathrm{~m}\)

Hint

(c) the information is insufficient to decide the relation of \(R_{A}\) with \(R_{B}\).
Horizontal range for the projectile, \(R=\frac{u^{2} \sin (2 \theta)}{g}\); Information of the initial velocity is not given in the question.

Hint

For a minimum time, the swimmer should swim perpendicular to the direction of river flow. So, he should swim due north.

Hint

(b) Let, \(v\) be the velocity of mass in an upwards direction.
Therefore, component of velocity along the rope \(=\mathrm{v} \cos \theta\)
Now, \(\mathrm{v} \operatorname{Cos} \theta=\mathrm{u}\)
Therefore, \(\mathrm{v}=\frac{\mathrm{u}}{\operatorname{Cos} \theta}\)

Hint

(a), (d) In a trip of one hour of a tip of the minute hand, its initial and final positions are the same. Displacement is the distance between the initial and final position Thus, its displacement in this trip is zero. Option (a) is correct. But it has covered some path in this trip, thus distance traveled by the tip of the hand is not zero. Thus, the Average speed that is defined as the ratio of total distance travelled with the total time taken is not zero. Options (b) and (c) are incorrect. Also, the average velocity is defined as the ratio of total displacement with the total time taken, as displacement is zero, the average velocity of the tip also remains zero. Option (d) is correct.

Hint

(c) & (d) We can write the equation as,
\(
X=u(t-2)+\frac{1}{2}(2 a)(t-2)^{2}
\)
clearly, \(\mathrm{X}=0\) at \(\mathrm{t}=2\)
Also, acceleration \(=2\) a when we compare the given equation with
\(
X=u t+\frac{1}{2} a t^{2}
\)

Hint

Correct options are (a), (b) and, (c)
Distance is greater than or equal to displacement. Hence, the average speed is greater than or equal to the average velocity. So, (a) is correct.
In uniform circular motion,
\(\left|\frac{d \vec{v}}{d t}\right|=\omega^{2} R = 0\)
\(\frac{\mathrm{d}|\overrightarrow{\mathrm{v}}|}{\mathrm{dt}}=0\) as \(|\overrightarrow{\mathrm{v}}|\) is constant. So, (b) is correct.
In a uniform circular motion, the instantaneous velocity is never zero but the average velocity in the time period of rotation is zero. So, (c) is correct.
If the average velocity of a particle is zero, its net displacement is zero. This can happen in a straight line motion only in two ways:
(i) The particle is at rest.
(ii). The particle reverses its direction at least once in the time interval. Both scenarios demand that the instantaneous velocity is zero at least once in the time interval.
So, option (d) is wrong.

Hint

option (b) & (d) are correct
Speed is a scalar quantity while velocity is a vector quantity. An object can have a constant speed but velocity will change since velocity is a vector quantity. Velocity and acceleration are vector quantities that can be changed by changing direction only (keeping magnitude constant).

Hint

(a), (b) & (d) are correct

(a) Acceleration is given by
\(
\begin{aligned}
&-a=\frac{d v}{d t} \\
&-a<0 \\
&\Rightarrow \frac{d v}{d t}<0 \\
&\Rightarrow V_{\text {final }}<V_{\text {initial }}
\end{aligned}
\)
(b) If the position and velocity have opposite signs, the particle moves towards the origin.
(d) If the velocity is zero in a certain time interval, then the change in the velocity in that time interval will also be zero. As acceleration is rate of change of velocity, it will also be zero at an instant in that time interval.

Hint

(a) is not true. Consider the case of a particle dropped vertically downward, its velocity at t=0 is zero but acceleration is not zero.
(b) is true because we do not have information about the velocity in the next instants. If in the next instants the velocity still remains zero, the acceleration will be zero.
(c) is true because, if acceleration is zero that means there is no change in velocity. The velocity remains zero in this interval.
(d) is true because if the speed remains zero from t=0 to t=10 s, that means the particle is at rest in this interval, so the acceleration is also zero in this interval.

Hint

option (a) is correct. At any instant, the magnitude of the velocity is also its speed. So (a) is correct. (b) is incorrect because velocity is displacement in unit time while speed is the distance covered in unit time. (c) is incorrect because if speed is zero in a time interval, that means the particle has not moved in this interval and no distance traveled. So average speed has to be zero. (d) is incorrect because if the speed of the particle is never zero, means it has traveled some distance. so the average speed in the interval will not be zero.

Hint

(a) is true because in a velocity-time graph the slope (which shows the rate of change of velocity) denotes acceleration. In this graph the slope is constant, so acceleration is constant.
(b) is not true because at \(t=10\) s the velocity changes its sign from positive to negative, which means it has turned around.
(c) is not true because displacement in the positive direction of \(v\) is given by the area of the graph above the time axis and displacement in the negative direction of \(v\) is given by the area of the graph below the time axis. In this case, later is greater than the former not equal. So displacement is not zero.
(d) is true because in 0 to \(10 \mathrm{~s}\) speed changes from 10 to \(0 \mathrm{~m} / \mathrm{s}\) and in 10 to \(20 \mathrm{~s}\) speed changes from 0 to \(10 \mathrm{~m} / \mathrm{s}\). So average speed will be the same.

Hint

(a) The slope of the curve at any point gives the velocity of the particle at that point. In the given graph the curve has three troughs and three crests. At these points, the slope of the curve (tangent is horizontal to the time axis) is zero. So at six points velocity is zero, which means it has come to rest. (b) is not true because the slope of the curve is not maximum at \(t=6 \mathrm{~s}\). \((\mathrm{c})\) is not true because gradient of the slope is not positive always between \(t=0\) to \(t=6 \mathrm{~s}\). (d) is not true because the final displacement is not less than the initial.

Hint

(d) is correct answer. Using parallelogram law, Let,
\(\overrightarrow{\mathrm{S}_{1}}  \overrightarrow{\mathrm{S}_{2}}\) be the arms of the parallelogram,
Given,
\(
\overrightarrow{\mathrm{S}}_{1}=\overrightarrow{\mathrm{S}}_{2}
\)
the angle between these two vectors may lie between \(0^{\circ}\) to \(180^{\circ}\)
Let us find the resultant taking \(\theta=0^{\circ}\) and \(180^{\circ}\)
\(
\begin{aligned}
&\mathrm{R}=\sqrt{\mathrm{S}_{1}^{2}+\mathrm{S}_{2}^{2}-2 \cdot \mathrm{S}_{1} \mathrm{~S}_{2} \cos \theta} \\
&\therefore \mathrm{R}=\sqrt{5^{2}+5^{2}-2 \cdot(5)(5) \cos 0^{\circ}}=0
\end{aligned}
\)
and,
\(
R=\sqrt{5^{2}+5^{2}-2 .(5)(5) \cos 180^{\circ}}=10
\)
\(\therefore\) The acceleration of frame \(S_{2}\) with respect to \(S_{1}\) lies between 0 and \(10 \mathrm{~ms}^{-2}\)

Hint

(a) is the correct answer. Distance travelled is \(2 \mathrm{~km}\).
\(
\begin{aligned}
\text { Time taken } &=\frac{1 \mathrm{~km}}{6 \mathrm{~km} / \mathrm{hr}}+\frac{1 \mathrm{~km}}{8 \mathrm{~km} / \mathrm{hr}} \\
&=\left(\frac{1}{6}+\frac{1}{8}\right) \mathrm{hr}=\frac{7}{24} \mathrm{hr} . \\
\text { Average speed } &=\frac{2 \mathrm{~km} \times 24}{7 \mathrm{hr}}=\frac{48}{7} \mathrm{~km} / \mathrm{hr} \\
& \approx 7 \mathrm{~km} / \mathrm{hr} .
\end{aligned}
\)

Hint

(b) Total distance travelled in 50 minutes \(=800 \mathrm{ft}\). Average speed \(=\frac{800}{50} \mathrm{ft} / \mathrm{min}=16 \mathrm{ft} / \mathrm{min}\).
At \(12.00\) noon he is at the door and at \(12.50 \mathrm{pm}\) he is again at the same door.
The displacement during the \(50 \mathrm{~min}\) interval is zero.
Average velocity \(=\) zero.

Hint

(a) \(x=A t^{3}+B t^{2}+C t+D\)
or, \(v=\frac{d x}{d t}=3 A t^{2}+2 B t+C\).
Thus, at \(t=4 \mathrm{~s}\), the velocity
\(
\begin{aligned}
&=3\left(1 \mathrm{~m} / \mathrm{s}^{3}\right)\left(16 \mathrm{~s}^{2}\right)+2\left(4 \mathrm{~m} / \mathrm{s}^{2}\right)(4 \mathrm{~s})+(-2 \mathrm{~m} / \mathrm{s}) \\
&=(48+32-2) \mathrm{m} / \mathrm{s}=78 \mathrm{~m} / \mathrm{s} .
\end{aligned}
\)
(b) \(v=3 A t^{2}+2 B t+C\)
or, \(a=\frac{d v}{d t}=6 A t+2 B\).
At \(t=4 \mathrm{~s}, a=6\left(1 \mathrm{~m} / \mathrm{s}^{3}\right)(4 \mathrm{~s})+2\left(4 \mathrm{~m} / \mathrm{s}^{2}\right)=32 \mathrm{~m} / \mathrm{s}^{2}\).
(c) \(x=A t^{3}+B t^{2}+C t+D\).
Position at \(t=0\) is \(x=D=5 \mathrm{~m}\).
Position at \(t=4 \mathrm{~s}\) is
\(
\begin{aligned}
&\left(1 \mathrm{~m} / \mathrm{s}^{3}\right)\left(64 \mathrm{~s}^{3}\right)+\left(4 \mathrm{~m} / \mathrm{s}^{2}\right)\left(16 \mathrm{~s}^{2}\right)-(2 \mathrm{~m} / \mathrm{s})(4 \mathrm{~s})+5 \mathrm{~m} \\
&\quad=(64+64-8+5) \mathrm{m}=125 \mathrm{~m} .
\end{aligned}\)

Thus, the displacement during 0 to \(4 \mathrm{~s}\) is \(125 \mathrm{~m}-5 \mathrm{~m}=120 \mathrm{~m}\).
Average velocity \(=\frac{120 \mathrm{~m}}{4 \mathrm{~s}}=30 \mathrm{~m} / \mathrm{s}\).
(d) \(v=3 A t^{2}+2 B t+C\).
Velocity at \(t=0\) is \(C=-2 \mathrm{~m} / \mathrm{s}\).
Velocity at \(t=4 \mathrm{~s}\) is \(=78 \mathrm{~m} / \mathrm{s}\).
Average acceleration \(=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}=20 \mathrm{~m} / \mathrm{s}^{2}\).

Hint

At \(t=0\), the particle is at rest, say at the origin. After that, the velocity is positive, so that the particle moves in the positive \(x\) direction. Its speed increases till 1 second when it starts decreasing. The particle continues to move further in the positive \(x\) direction. At \(t=2 \mathrm{~s}\), its velocity is reduced to zero, it has moved through a maximum positive \(x\) distance. Then it changes its direction, velocity being negative, but increasing in magnitude. At \(t=3 \mathrm{~s}\) velocity is maximum in the negative \(x\) direction and then the magnitude starts decreasing. It comes to rest at \(t=4 \mathrm{~s}\).
(a) Distance during 0 to \(2 \mathrm{~s}=\) Area of \(O A B\)
\(
=\frac{1}{2} \times 2 \mathrm{~s} \times 10 \mathrm{~m} / \mathrm{s}=10 \mathrm{~m} .
\)
(b) Distance during 2 to \(4 \mathrm{~s}=\) Area of \(B C D=10 \mathrm{~m}\). The particle has moved in negative\( x\) direction during this period.
(c) The distance travelled during 0 to \(4 \mathrm{~s}=10 \mathrm{~m}+10 \mathrm{~m}\) \(=20 \mathrm{~m}\).
(d) displacement during 0 to \(4 \mathrm{~s}=10 \mathrm{~m}+(-10 \mathrm{~m})=0\).
(e) at \(t=1 / 2 \mathrm{~s}\) acceleration \(=\) slope of line \(O A=10 \mathrm{~m} / \mathrm{s}^{2}\).
(f) at \(t=2 \mathrm{~s}\) acceleration \(=\) slope of line \(A B C=-10 \mathrm{~m} / \mathrm{s}^{2}\).

Hint

Correct option is (c)

(a) Velocity at time \(t\) second is
\(100 \mathrm{~m} / \mathrm{s}=a .(t)\) second
and velocity at time \((t+1)\) second is
\(150 \mathrm{~m} / \mathrm{s}=a .(t+1)\) second.
Solving above two equations,  \(a=50 \mathrm{~m} / \mathrm{s}^{2}\)
(b) Consider the interval \(t\) second to \((t+1)\) second, time elapsed \(=1 \mathrm{~s}\)
initial velocity \(=100 \mathrm{~m} / \mathrm{s}\)
final velocity \(=150 \mathrm{~m} / \mathrm{s}\).
Thus, \((150 \mathrm{~m} / \mathrm{s})^{2}=(100 \mathrm{~m} / \mathrm{s})^{2}+2\left(50 \mathrm{~m} / \mathrm{s}^{2}\right) x\)
or, \(x=125 \mathrm{~m}\).

Hint

(d) is the correct option. Consider the accelerated \(24 \mathrm{~cm}\) motion of the stone.
Initial velocity \(=0\)
Final velocity \(=2 \cdot 2 \mathrm{~m} / \mathrm{s}\)
Distance travelled \(=24 \mathrm{~cm}=0.24 \mathrm{~m}\)
Using \(v^{2}=u^{2}+2 a x\),
\(
a=\frac{4 \cdot 84 \mathrm{~m}^{2} / \mathrm{s}^{2}}{2 \times 0.24 \mathrm{~m}}=10 \cdot 1 \mathrm{~m} / \mathrm{s}^{2} \text {. }
\)

Hint

Option (a) is the correct answer.

Suppose the pickpocket is caught at a time \(t\) after the motorcycle starts. The distance travelled by motorcycle during this interval is \(
s=\frac{1}{2} a t^{2} \ldots (i) \)
During this interval, the jeep travels a distance
\(
s+d=v t \ldots (ii)
\)
By (i) and (ii),
\(
\frac{1}{2} a t^{2}-v t+d=0
\)
or,
\(
t=\frac{v \pm \sqrt{v^{2}-2 a d}}{a}
\)
The pickpocket will be caught if \(t\) is real and positive. This will be possible if
\(
v^{2} \geq 2 a d \quad \text { or, } v \geq \sqrt{2 a d} .
\)

Hint

(c) is the right choice. At \(t=0\), the stone was going up with a velocity of \(5.0 \mathrm{~m} / \mathrm{s}\). After that, it moved as a freely falling particle with downward acceleration \(g\). Take vertically upward as the positive \(X\)-axis. If it reaches the ground at time \(t\)
\(
x=-50 \mathrm{~m}, \quad u=5 \mathrm{~m} / \mathrm{s}, \quad a=-10 \mathrm{~m} / \mathrm{s}^{2} .
\)
We have \(\quad x=u t+\frac{1}{2} a t^{2}\)
or, \(\quad-50 \mathrm{~m}=(5 \mathrm{~m} / \mathrm{s}) . t+\frac{1}{2} \times\left(-10 \mathrm{~m} / \mathrm{s}^{2}\right) t^{2}\)
or, \(\quad t=\frac{1 \pm \sqrt{41}}{2} \mathrm{~s} .\)
or, \(\quad t=-2.7 \mathrm{~s} \quad\) or, \(3.7 \mathrm{~s} .\)
Negative \(t\) has no significance in this problem. The stone reaches the ground at \(t=3.7 \mathrm{~s}\). During this time the balloon has moved uniformly up. The distance covered by it is
\(
5 \mathrm{~m} / \mathrm{s} \times 3.7 \mathrm{~s}=18.5 \mathrm{~m} \text {. }
\)
Hence, the height of the balloon when the stone reaches the ground is \(50 \mathrm{~m}+18.5 \mathrm{~m}=68.5 \mathrm{~m}\).

Hint

(c) is the correct choice

(a) Take the origin at the point where the ball is kicked, vertically upward as the \(Y\)-axis and the horizontal in the plane of motion as the \(X\)-axis. The initial velocity has the components
\(u_{x}=(20 \mathrm{~m} / \mathrm{s}) \cos 45^{\circ}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}\)
and \(\quad u_{y}=(20 \mathrm{~m} / \mathrm{s}) \sin 45^{\circ}=10 \sqrt{2} \mathrm{~m} / \mathrm{s}\).
When the ball reaches the ground, \(y=0\).
Using \(y=u_{y} t-\frac{1}{2} g t^{2}\),
\(0=(10 \sqrt{2} \mathrm{~m} / \mathrm{s}) t-\frac{1}{2} \times\left(10 \mathrm{~m} / \mathrm{s}^{2}\right) \times t^{2}\)
or, \(\quad t=2 \sqrt{2} \mathrm{~s}=2.8 \mathrm{~s}.\)
Thus, it takes \(2.8 \mathrm{~s}\) for the football to fall on the ground.
(b) At the highest point \(v_{y}=0\). Using the equation
\(
\begin{aligned}
v_{y}^{2} &=u_{y}^{2}-2 g y, \\
0 &=(10 \sqrt{2} \mathrm{~m} / \mathrm{s})^{2}-2 \times\left(10 \mathrm{~m} / \mathrm{s}^{2}\right) H \\
\text { or, } \quad H &=10 \mathrm{~m} .
\end{aligned}
\)
Thus, the maximum height reached is \(10 \mathrm{~m}\).
(c) The horizontal distance travelled before falling to the ground is \(x=u_{x} t\)
\(
=(10 \sqrt{2} \mathrm{~m} / \mathrm{s})(2 \sqrt{2} \mathrm{~s})=40 \mathrm{~m} .
\)

Hint

(b) is the right answer. The velocity of the food packet at the time of release is \(u\) and is horizontal. The vertical velocity at the time of release is zero.
Vertical motion: If \(t\) be the time taken by the packet to reach the victim, we have for vertical motion,
\(
H=\frac{1}{2} g t^{2} \quad \text { or, } \quad t=\sqrt{\frac{2 H}{g}} \ldots \quad (i).
\)
Horizontal motion: If \(D\) be the horizontal distance travelled by the packet, we have \(D=u t\). Putting \(t\) from (i),
\(
D=u \sqrt{\frac{2 H}{g}} .
\)
The distance between the victim and the packet at the time of release is
\(
\sqrt{D^{2}+H^{2}}=\sqrt{\frac{2 u^{2} H}{g}+H^{2}}
\)

Hint

(d) is the correct answer. Take \(X, Y\)-axes as shown in the figure below. Suppose that the particle strikes the plane at a point \(P\) with coordinates \((x, y)\). Consider the motion between \(A\) and \(P\).
Figure the figure Motion in \(x\)-direction :
Initial velocity \(=u\)
Acceleration \(=0\)
\(
x=u t .\dots (i)
\)
Motion in \(y\)-direction :
Initial velocity \(=0\)
Acceleration \(=g\)
\(
y=\frac{1}{2} g t^{2} \dots (ii)
\)

Eliminating \(t\) from (i) and (ii)
\(
y=\frac{1}{2} g \frac{x^{2}}{u^{2}} .
\)
Also \(\quad y=x \tan \theta\).
Thus, \(\frac{g x^{2}}{2 u^{2}}=x \tan \theta\) giving \(x=0\), or, \(\frac{2 u^{2} \tan \theta}{g}\).
Clearly the point \(P\) corresponds to \(x=\frac{2 u^{2} \tan \theta}{g}\),
then \(y=x \tan \theta=\frac{2 u^{2} \tan ^{2} \theta}{g}\).
The distance \(A P=l=\sqrt{x^{2}+y^{2}}\)
\(
\begin{aligned}
&=\frac{2 u^{2}}{g} \tan \theta \sqrt{1+\tan ^{2} \theta} \\
&=\frac{2 u^{2}}{g} \tan \theta \sec \theta .
\end{aligned}
\)

Hint

(a) is the correct choice. Solution : Let the speed be \(v\) when it makes an angle \(\alpha\) with the horizontal. As the horizontal component of velocity remains constant,
\(
\begin{aligned}
v \cos \alpha &=u \cos \theta \\
\text { or, } \quad v &=u \cos \theta \sec \alpha .
\end{aligned}
\)

Hint

The situation is shown in the figure below. The object starts falling from the point \(B\). Draw a vertical line \(B C\) through \(B\). Suppose the bullet reaches the line \(B C\) at a point \(D\) and it takes a time \(t\) in doing so. Consider the vertical motion of the bullet. The initial vertical velocity \(=0\). The distance travelled vertically \(=B D=\frac{1}{2} g t^{2}\). In time \(t\) the object also travels a distance \(\frac{1}{2} g t^{2}=B D\). Hence at time \(t\), the object will also be at the same point \(D\). Thus, the bullet hits the object at point \(D\).

Hint

(a) The situation is shown in the figure below. The \(X\)-axis is chosen along the river flow and the origin at the starting position of the man. The direction of the velocity of man with respect to the ground is along the \(Y\)-axis (perpendicular to the river). We have to find the direction of the velocity of the man with respect to water.
Let \(\vec{v}_{r, g}=\) velocity of the river with respect to the ground \(=2 \mathrm{~km} / \mathrm{h}\) along the \(X\)-axis

\(\vec{v}_{m, r}=\) velocity of the man with respect to the river \(=3 \mathrm{~km} / \mathrm{h}\) making an angle \(\theta\) with the \(Y\)-axis and \(\vec{v}_{m, g}=\) velocity of the man with respect to the ground along the \(Y\)-axis. We have
\(
\vec{v}_{m, g}=\vec{v}_{m, r}+\vec{v}_{r, g} . \dots (i)
\)
Taking components alongthe \(X\)-axis
\(
0=-(3 \mathrm{~km} / \mathrm{h}) \sin \theta+2 \mathrm{~km} / \mathrm{h}
\)
or, \(\quad \sin \theta=\frac{2}{3}\).
(b) Taking components in equation (i) along the \(Y\)-axis,
\(
v_{m, g}=(3 \mathrm{~km} / \mathrm{h}) \cos \theta+0
\)
or,
\(
v_{m, g}=\sqrt{5} \mathrm{~km} / \mathrm{h} \text {. }
\)
\(
\text { Time }=\frac{\text { Displacement in } y \text { direction }}{\text { Velocity in } y \text { direction }}
\)
\(
=\frac{0.5 \mathrm{~km}}{\sqrt{5} \mathrm{~km} / \mathrm{h}}=\frac{\sqrt{5}}{10} \mathrm{~h}
\)

Hint

The correct option is (a). When the man is at rest with respect to the ground, the rain comes to him at an angle \(30^{\circ}\) with vertical. This is the direction of the velocity of raindrops with respect to the ground. The situation when the man runs is shown in figure (b) below.
Here \(\vec{v}_{r, g}=\) velocity of the rain with respect to the ground

\(\vec{v}_{m, g}=\) velocity of the man with respect to the ground

and, \(\vec{v}_{r, m}=\) velocity of the rain with respect to the man.

We have, \(\quad \vec{v}_{r, g}=\vec{v}_{r, m}+\vec{v}_{m, g} \dots (i)\)
Taking horizontal components, equation (i) gives

(a)
\(v_{r, g} \sin 30^{\circ}=v_{m, g}=10 \mathrm{~km} / \mathrm{h}\)
or, \(v_{r, g}=\frac{10 \mathrm{~km} / \mathrm{h}}{\sin 30^{\circ}}=20 \mathrm{~km} / \mathrm{h}\), Taking vertical components, equation (i) gives

(b) \(
\text { or, } \begin{aligned}
v_{r, g} \cos 30^{\circ} &=v_{r, m} \\
v_{r, m} &=(20 \mathrm{~km} / \mathrm{h}) \frac{\sqrt{3}}{2} \\
&=10 \sqrt{3} \mathrm{~km} / \mathrm{h} .
\end{aligned}
\)

Hint

A vector is changed whenever the direction or the magnitude of the drawn line is altered to any extent.

Hence, a vector definitely changes if it is rotated through an arbitrary angle or cross multiplied by a unit vector, as the direction of the vectors changes in both cases.

Likewise, the vector changes due to the multiplication by an arbitrary scalar, due to the change of its magnitude.

However, the vector remains unchanged only if it is slid parallel to itself, as no change occurs in its direction or magnitude in this case.

Hint

1,2 and 1 may represent the magnitudes of three vectors adding to zero. For example one of the vector of length 1 should make an angle of \(135^{\circ}\) with \(\mathrm{x}\) axis and the other vector of length 1 makes an angle of \(225^{\circ}\) with \(\mathrm{x}\) axis. The third vector of length 2 should lie along \(x\) axis.

Hint

The resultant of two vectors is closer to the vector with the greater magnitude.
Thus, \(\alpha<\beta\) if \(A>B\)

From the given data, the total angle between Vector A and Vector \(B=\alpha+\beta\)
Now,
\(
\begin{aligned}
& A=R \cos \alpha \\
& B=R \cos \beta \\
& \text { If } \alpha<\beta
\end{aligned}
\)
Then,
\(
\begin{aligned}
& \cos \alpha>\cos \beta \\
& =R \cos \alpha>R \cos \beta
\end{aligned}
\)
Hence,
\(
\mathrm{A}>\mathrm{B}
\)
thus, it can be said
\(
\mathrm{A}>\mathrm{B} \text { and } \alpha<\beta
\)

Hint

All the given options are incorrect. The component of a vector may be less than, greater than or equal to its magnitude, depending upon the vector and its components.

Explanation of Correct Option:
In both figures, we can see a vector A and its component From Figure (i), it is clear that the magnitude of the component of A is less than the magnitude of the vector \(A\).

Consider fig.(ii), When the component is taken in the same direction as that of the vector, the magnitude of the component is maximum, i.e., the magnitude of the component is equal to the magnitude of the vector.
Since the magnitude of the component can be less than or equal to the magnitude of the vector, Option \(D\) is correct.

Hint

Here \(z\)-axis is vertically upward means normal to plane of paper as shown in figure.
Thus, \(\overrightarrow{\mathrm{A}}=\mathrm{Ak}\) and \(\overrightarrow{\mathrm{B}}=\mathrm{B} \hat{\mathrm{j}}\)
So, \(\vec{A} \times \vec{B}=A \hat{k} \times B \hat{j}=-A B \hat{i}\)
Thus, it is along negative \(\mathrm{x}\)-axis means along west.

Hint

Area of a circle, \(\mathrm{A}=\pi r^2 \mathrm{On}\) putting the values, we get:
\(
\begin{aligned}
& A=\frac{22}{7} \times 2.12 \times 2.12 \\
& \Rightarrow A=14.1 \mathrm{~cm}^2
\end{aligned}
\)
The rules to determine the number of significant digits says that in the multiplication of two or more numbers, the number of significant digits in the answer should be equal to that of the number with the minimum number of significant digits. Here, 2.12 \(\mathrm{cm}\) has a minimum of three significant digits. So, the answer must be written in three significant digits.

Hint

(a) the value of a scalar
(c) a vector
(d) the magnitude of a vector
The value of a scalar, a vector, and the magnitude of a vector do not depend on a given set of coordinate axes with different orientations. However, the components of a vector depend on the orientation of the axes.

Hint

Statements (a), (c) and (d) are incorrect.
Given: \(\vec{C}=\vec{A}+\vec{B}\)
Here, the magnitude of the resultant vector may or may not be equal to or less than the magnitudes of \(\vec{A}\) and \(\vec{B}\) or the sum of the magnitudes of both the vectors if the two vectors are in opposite directions.

Hint

Here, we have three vector \(A, B\) and \(C\).
\(
\begin{aligned}
& |\vec{A}+\vec{B}|^2=|\vec{A}|^2+|\vec{B}|^2+2 \vec{A} \cdot \vec{B} \ldots(i) \\
& |\vec{A}-\vec{B}|^2=|\vec{A}|^2+|\vec{B}|^2-2 \vec{A} \cdot \vec{B} \ldots(i)
\end{aligned}
\)
Subtracting (i) from (ii), we get:
\(
|\vec{A}+\vec{B}|^2-|\vec{A}-\vec{B}|^2=4 \vec{A} \cdot \vec{B}
\)
Using the resultant property \(\vec{C}=\vec{A}+\vec{B}\), we get:
\(
\begin{aligned}
& |\vec{C}|^2-|\vec{A}-\vec{B}|^2=4 \vec{A} \cdot \vec{B} \\
& \Rightarrow|\vec{C}|^2=|\vec{A}-\vec{B}|^2+4 \vec{A} \cdot \vec{B} \\
& \Rightarrow|\vec{C}|^2=|\vec{A}-\vec{B}|^2+4|\vec{A}||\vec{B}| \cos 120^{\circ}
\end{aligned}
\)
Since cosine is negative in the second quadrant, \(\mathrm{C}\) must be less than \(|A-B|\).

Hint

(a) is equal to the sum of the \(x\)-components of the vectors
(b) may be smaller than the sum of the magnitudes of the vectors
(d) may be equal to the sum of the magnitudes of the vectors.
The \(x\)-component of the resultant of several vectors cannot be greater than the sum of the magnitudes of the vectors.

Hint

(b) equal to \(A B\)
(c) less than \(A B\)
(d) equal to zero.
The magnitude of the vector product of two vectors \(|\vec{A}|\) and \(|\vec{B}|\) may be less than or equal to \(A B\), or equal to zero, but cannot be greater than \(A B\).

Hint

The angle between \(\vec{A}\) and \(\vec{B}=110^{\circ}-20^{\circ}=90^{\circ}\)
\(
|\vec{A}|=3 \text { and }|\vec{B}|=4 \mathrm{~m}
\)
Resultant \(R=\sqrt{A^2+B^2+2 A B \cos \theta}=5 \mathrm{~m}\)
Let \(\beta\) be the angle between \(\vec{R}\) and \(\vec{A}\)
\(
\beta=\tan ^{-1}\left(\frac{4 \sin 90^{\circ}}{3+4 \cos 90^{\circ}}\right)=\tan ^{-1}(4 / 3)=53^{\circ}
\)
\(\therefore\) Resultant vector makes angle \(\left(53^{\circ}+20^{\circ}\right)=73^{\circ}\) with x-axis.

Hint

Angle between \(\vec{A}\) and \(\vec{B}\) is \(\theta=60^{\circ}-30^{\circ}=30^{\circ}\)
\(|\vec{A}|\) and \(|\vec{B}|=10\) unit
\(
R=\sqrt{10^2+10^2+2 \cdot 10 \cdot 10 \cdot \cos 30^{\circ}}=19 \cdot 3 =20 \cos 15^{\circ}
\)
\(\beta\) be the angle between \(\vec{R}\) and \(\vec{A}\)
\(
\beta=\tan ^{-1}\left(\frac{10 \sin 30^{\circ}}{10+10 \cos 30^{\circ}}\right)=\tan ^{-1}\left(\frac{1}{2+\sqrt{3}}\right)=\tan ^{-1}(0.26795)=15^{\circ}
\)
Resultant makes \(15^{\circ}+30^{\circ}=45^{\circ}\) angle with \(x\)-axis

Hint

y component of \(\vec{a}=100 \sin 45^{\circ}=\frac{100}{\sqrt{2}}\)
y component of \(\overrightarrow{\mathrm{b}}=100 \sin 135^{\circ}=\frac{100}{\sqrt{2}}\)
\(\mathrm{y}\) component of \(\overrightarrow{\mathrm{c}}=100 \sin 315^{\circ}=\frac{-100}{\sqrt{2}}\)
Resultant of y component \(=\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}\) units
\(x\) component of \(\vec{a}=100 \cos 45^{\circ}=\frac{100}{\sqrt{2}}\)
\(x\) component of \(\vec{b}=100 \cos 45^{\circ}=\frac{-100}{\sqrt{2}}\)
\(\mathrm{x}\) component of \(\overrightarrow{\mathrm{c}}=100 \cos 45^{\circ}=\frac{100}{\sqrt{2}}\)
Resultant of \(\mathrm{x}\) component \(=\frac{100}{\sqrt{2}}-\frac{100}{\sqrt{2}}+\frac{100}{\sqrt{2}}=\frac{100}{\sqrt{2}}\) units
Total resultant of \(x\) and \(y\) component \(=\left(\frac{100}{\sqrt{2}}\right)^2+\left(\frac{100}{\sqrt{2}}\right)^2=100\)
Now, \(\tan \mathrm{D}=\mathrm{y}\)-component \(/ \mathrm{x}-\) component \(=1\)
\(
\mathrm{D}=\tan ^{-1}(1)=45^{\circ}
\)
So, the resultant is 100 unit and \(45^{\circ}\) with \(\mathrm{x}\)-axis

Hint

\(
\begin{aligned}
& \text { Given that, } \vec{A}=\hat{i}+\hat{j} \\
& \text { and } \vec{B}=\hat{i}-\hat{j} \\
& \because \vec{A} \cdot \vec{B}=0
\end{aligned}
\)
Since the dot product of the two vectors is zero, it means that the vectors are perpendicular to each other.
Hence, Angle between vectors, \(\vec{A}\) and \(\vec{B}\) is \(90^{\circ}\)

Hint

Clearly from the diagram \(\vec{u}=a \hat{i}+b \hat{j}\).
As \(u\) is in the first quadrant, hence both of its components \(a\) and \(b\) will be positive, and as \(v\) is in the fourth quadrant. For \(\vec{v}=p \hat{i}+q \hat{j}\), as it is in positive \(x\)-direction and located downward hence \(x\)-component \(p\) will be positive and \(y\)-component \(q\) will be negative.

Hint

(b) Consider a vector \(\vec{R}\) in \(X\) – \(Y[latex] plane as shown in figure. If we draw orthogonal vectors [latex]\vec{R}_x\) and \(\vec{R}_y\) along \(x\) and \(y\) axes respectively, by law of vector addition, \(\vec{R}=\vec{R}_x+\vec{R}_y\)
The magnitude of component of \(r\) along \(X\)-axis
\(
\begin{aligned}
& r_x=|r| \cos \theta \\
& \left(r_x\right)_{\text {maximum }}=|r|(\cos \theta)_{\text {maximum }}
\end{aligned}
\)
\(
\begin{aligned}
r_x & =|r| \cos \theta \\
& =|r| \cos 0^{\circ}=|r| \quad\left(\because \cos \theta \text { is maximum if } \theta=0^{\circ}\right)
\end{aligned}
\)
As \(\theta=0^{\circ}\), \(r\) is along positive \(x\)-axis.

Hint

(c)

The horizontal range of the projectile fired at an angle is given by Range \(R=\frac{u^2 \sin 2 \theta}{g}\)
Where, \(\theta\) is the angle of projection, \(u\) is the initial velocity with which the projectile is fired, and \(g\) is the gravitational acceleration. According to the problem the angle of projection, \(\theta=15^{\circ}\) and horizontal range, \(R=50 \mathrm{~m}\) By substituting the values in the above relation,we get
\(
\begin{aligned}
& \Rightarrow \quad R=50 \mathrm{~m}=\frac{u^2 \sin \left(2 \times 15^{\circ}\right)}{g} \\
\Rightarrow & 50 \times g=u^2 \sin 30^{\circ}=u^2 \times \frac{1}{2} \\
\Rightarrow & 50 \times g \times 2=u^2 \\
\Rightarrow & u^2=50 \times 9.8 \times 2=100 \times 9.8=980 \\
\Rightarrow & u=\sqrt{980}=\sqrt{49 \times 20}=7 \times 2 \times \sqrt{5} \mathrm{~m} / \mathrm{s}=14 \times 223 \mathrm{~m} / \mathrm{s}=31.304 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
\(
\text { For } \theta=45^{\circ} ; \quad R=\frac{u^2 \sin 2 \times 45^{\circ}}{g}=\frac{u^2}{g} \quad\left(\because \sin 90^{\circ}=1\right)
\)
\(
\Rightarrow \quad R=\frac{(14 \sqrt{5})^2}{g}=\frac{14 \times 14 \times 5}{9.8}=100 \mathrm{~m}
\)

Hint

(b) We know that impulse \(J=F . \Delta t=\Delta p\), where \(F\) is force, At is time duration and Ap is change in momentum. As \(\Delta p\) is a vector quantity, hence impulse is also a vector quantity. Sometimes area can also be treated as vector direction of area vector is perpendicular to its plane.

Hint

(d) Speed (Instantaneous Speed): The magnitude of the velocity at any instant of time is known as Instantaneous Speed or simply speed at that instant of time. It is denoted by \(\mathrm{v}\).
Quantitatively: Speed = distance/ time
Mathematically, it is the time rate at which distance is being travelled by the particle.
– Speed is a scalar quantity. It can never be negative (as shown by the speedometer of our vehicle).
– Instantaneous speed is the speed of a particle at a particular instant of time.
Hence, Total distance travelled = Path length = (speed) \(x\) time taken Important point: We should be very careful with the fact that speed is related with total distance covered not with displacement.

Hint

(c) This motion is two-dimensional and given that instantaneous speed \(v_0\) is positive constant. Acceleration is defined as the rate of change of velocity (instantaneous speed), hence it will also be in the plane of motion.

Hint

(c) These types of problems can be solved by hit and trial method by picking up options one by one
Sum of vectors is given as \(\vec{A}+\vec{B}+\vec{C}=0\)
Hence, we can say that \(\vec{A}, \vec{B}\) and \(\vec{C}\) are in one plane and are represented by the three sides of a triangle taken in one order. Let us discuss all the options one by one.
(a) We can write
\(
\begin{array}{ll}
& \vec{B} \times(\vec{A}+\vec{B}+\vec{C})=\vec{B} \times 0=0 \\
\Rightarrow \quad & \vec{B} \times \vec{A}+\vec{B} \times \vec{B}+\vec{B} \times \vec{C}=0 \\
\Rightarrow \quad & \vec{B} \times \vec{A}+0 \times \vec{B} \times \vec{C}=0 \\
\Rightarrow \quad & \vec{B} \times \vec{A}=-\vec{B} \times \vec{C} \\
\Rightarrow \quad & \vec{A} \times \vec{B}=\vec{B} \times \vec{C} \\
\therefore \quad & (\vec{A} \times \vec{B}) \times \vec{C}=(\vec{B} \times \vec{C}) \times \vec{C}
\end{array}
\)
It cannot be zero.
If \(\vec{B} \| \vec{C}\), then \(\vec{B} \times \vec{C}=0\), then \((\vec{B} \times \vec{C}) \times \vec{C}=0\). Even if \(\vec{A} \| \vec{C}\) or \(\vec{A} \| \vec{B}\), then also the cross product in any manner will be zero. So, option (a) is not false.
(b) \((\vec{A} \times \vec{B}) \cdot \vec{C}=(\vec{B} \times \vec{A}) \cdot \vec{C}=0\) whatever be the positions of \(\vec{A}, \vec{B}\) and \(\vec{C} \cdot\) If \(\vec{B} \| \vec{C}\) then \(\vec{B} \times \vec{C}=0\), then \((\vec{B} \times \vec{C}) \times \vec{C}=0\). So, option (b) is not false.
(c) \((\vec{A} \times \vec{B})=\vec{X}=\vec{A} \vec{B}\) sin \(\theta\) and the direction of \(X\) is perpendicular to the plane containing \(\vec{A}\) and \(\vec{B} \cdot(\vec{A} \times \vec{B}) \times \vec{C}=X \times \vec{C}\). Its direction is in the plane of \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\). So, option (c) is false.
(d) If \(\vec{C}^2=\vec{A}^2+\vec{B}^2\), then angle between \(\vec{A}\) and \(\vec{B}\) is \(90^{\circ}\).
\(
\therefore \quad(\vec{A} \times \vec{B}) \cdot \vec{C}=\left(\vec{A} \vec{B} \sin 90^{\circ} X\right) \cdot \vec{C}=\vec{A} \vec{B}(X \cdot \vec{C})=A B C \cos 90^{\circ}=0
\)
Hence, option (d) is not false.

Hint

(b) According to the problem,
\(
|\vec{A}+\vec{B}|=|\vec{A}|
\)
By squaring both sides, we get
\(
\begin{aligned}
& |\vec{A}+\vec{B}|^2=|\vec{A}|^2 \\
\Rightarrow & |\vec{A}|^2+|\vec{B}|^2+2|\vec{A}||\vec{B}| \cos \theta=|\vec{A}|^2
\end{aligned}
\)
where \(\theta\) is the angle between \(\vec{A}\) and \(\vec{B}\).
\(
\begin{aligned}
& |\vec{B}|(|\vec{B}|+2|\vec{A}| \cos \theta)=0 \\
\Rightarrow & |\vec{B}|+2|\vec{A}| \cos \theta=0 \\
\Rightarrow & \cos \theta=-\frac{|\vec{B}|}{2|\vec{A}|} \dots(i)
\end{aligned}
\)
If \(A\) and \(B\) are antiparallel, then \(\theta=180^{\circ}\) Hence from Eq. (i),
\(
-1=-\frac{|\vec{B}|}{2|\vec{A}|} \Rightarrow|\vec{B}|=2|\vec{A}|
\)
Hence, correct answer will be \(\vec{A}\) and \(\vec{B}\) are antiparallel provided \(|\vec{B}|=2|\vec{A}|\).
It seems like option (a) is also correct but it is not for \(|\vec{A}+\vec{B}|=|\vec{A}|\), either \(\vec{B}=0\) or \(\vec{B}=-2 \vec{A}\), so this option will be false.

Hint

The answer is the option (a) and (b)
Explanation:
According to the formula. Max height of a projectile is
\(
H=\frac{u^2 \sin ^2 \theta}{2 g}
\)
Option a \(: H 1>H 2\)
\(
\begin{aligned}
& \sin ^2 \theta 1>\sin ^2 \theta 2 \\
& (\sin \theta 1+\sin \theta 2)(\sin \theta 1-\sin \theta 2)>0
\end{aligned}
\)
So, \((\sin \theta 1+\sin \theta 2)>0\) or \((\sin \theta 1-\sin \theta 2)>0\)
So, \(\theta 1>\theta 2\) and \(\theta\) lies between 0 and 90 degree i.e. acute
Option b :
\(
\frac{T 1}{T 2}=\frac{\sin \theta 1}{\sin \theta 2}
\)
\(T 1 \sin \theta 2=T 2 \sin \theta 1\)
Since \(\sin \theta 1>\sin \theta 2\)
\(
T 1>T 2
\)

Important points about time of flight: For complementary angles of projection \(\theta\) and \(90^{\circ}-\theta\)
(a) Ratio of time of flight \(=\frac{T_1}{T_2}=\frac{2 u \sin \theta / g}{2 u \sin \left(90^{\circ}-\theta\right) / g}=\tan \theta\)
\(
\Rightarrow \quad \frac{T_1}{T_2}=\tan \theta
\)
(b) Multiplication of time of flight \(=T_1 T_2=\frac{2 u \sin \theta}{g} \frac{2 u \cos \theta}{g}\)
\(
\Rightarrow \quad T_1 T_2=\frac{2 R}{g}
\)

Hint

Key concept: In such type of problems, we have to observe the nature of track that if there is a friction or not, as friction is not present in this track, the total energy of the particle will remain constant throughout the journey.
According to the problem, the path traversed by the particle on a frictionless track is parabolic, is given by the equation \(y=x^2\), thus total energy \((K E+P E)\) will be same throughout the journey.
Hence, total energy at \(A=\) total energy at \(P\)
At \(B\) the particle is having only \(K E\) but at \(P\) some \(K E\) is converted to \(P E\).
So, \((K E)_B>(K E)_P\)
Total energy at \(A=P E=\) Total energy at \(B=K E=\) Total energy at \(P=P E+K E\)
The potential energy at \(A\) is converted to \(K E\) and \(P E\) at \(P\), hence \((P E) P<(P E) A\)
Hence, (Height) \(P<\) (Height) A
As, Height of \(P<\) Height of \(A\)
Hence, path length \(A B>\) path length \(B P\)
Hence, time of travel from \(A\) to \(B \neq\) Time of travel from \(B\) to \(P\).

Hint

Option a:
The given relation is correct when the acceleration is uniform
Option c:
\(
\vec{r}=\frac{1}{2}\left(\vec{v}\left(t_2\right)-\vec{v}\left(t_1\right)\right) \div\left(t_2-t_1\right)
\)
This is the relationship given in the question, but it is not possible as the LHS and RHS dimensions \(\left[M^0 L^1 T^0\right]\) and \(\left[M^0 L^1 T^2\right]\) do not match and hence the relationship cannot be considered valid.

Hint

\((a, b, c)\) While a particle is in uniform circular motion. Then the following statements are true.
(i) speed will be always constant throughout.
(ii) velocity will be always tangential in the direction of motion at a particular point.
(iii) the centripetal acceleration \(a=v^2 / r\) and its direction will always towards centre of the circular trajectory.
(iv) angular momentum ( \(\mathrm{mvr}\) ) is constant in magnitude and direction. And its direction is perpendicular to the plane containing \(r\) and \(v\).
Important point: In uniform circular motion, the magnitude of linear velocity and centripetal acceleration is constant but the direction changes continuously.

Hint

(b, d) According to the problem, \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\)
\(
\begin{aligned}
& \Rightarrow \sqrt{|\vec{A}|^2+|\vec{B}|^2+2|\vec{A}||\vec{B}| \cos \theta}=\sqrt{|\vec{A}|^2+|\vec{B}|^2-2|A||B| \cos \theta} \\
& \Rightarrow|\vec{A}|^2+|\vec{B}|^2+2|\vec{A}||\vec{B}| \cos \theta=|\vec{A}|^2+|\vec{B}|^2-2|\vec{A}||\vec{B}| \cos \theta \\
& \Rightarrow 4|\vec{A}||\vec{B}| \cos \theta=0 \\
& \Rightarrow|\vec{A}||\vec{B}| \cos \theta=0 \\
& \Rightarrow|\vec{A}|=0 \text { or }|\vec{B}|=0 \text { or } \cos \theta=0
\end{aligned}
\)
i.e. \(\theta=90^{\circ}\)
When \(\theta=90^{\circ}\), we can say that \(\vec{A} \perp \vec{B}\).
Hence options (b) and (d) are correct.