2 Marks Questions

Question 12 Marks

Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Answer

Since, Impulse = change in momentum = force × time. As momentum and force are vector quantities, hence impulse is a vector quantity.

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Question 22 Marks

Prove that the horizontal range is same when angle of projection is:

Greater than $45^\circ$ by certain value.
Less than $45^\circ$ by the same value.

Answer

The horizontal range of the projectile is given by $\text{R}=\frac{\text{u}^2\sin2\theta}{\text{g}}$

If angle of projection $\theta=45^\circ+\alpha$ and $R = R_1$

then, $\text{R}_1=\frac{\text{u}^2\sin2(45^\circ+\alpha)}{\text{g}}=\frac{\text{u}^2}{\text{g}}\cos2\alpha\dots(\text{i})$

If angle of projection $\theta=45^\circ-\alpha$ and $R = R_2$

$\therefore\ \text{R}_2=\frac{\text{u}^2\sin2(45^\circ-\alpha)}{\text{g}}=\frac{\text{u}^2}{\text{g}}\cos2\alpha\dots(\text{ii})$
Comparing eq. $(i)$ and $(ii),$ we have $R_1$ = $R_2$

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Question 32 Marks

Galileo, in his book Two New Sciences, stated the “for elevations which exceed or fall short of 45° by equal amounts, the ranges are equal.” Prove this statement.

Answer

Case I: When angle of projection.$\theta=45^\circ+\alpha.$
Range, $\text{R}_1=\frac{\text{u}^2\sin2(45^\circ+\alpha)}{\text{g}}$
$=\frac{\text{u}^2\cos2\alpha}{\text{g}}$
Case II: When angle of projection.
$\theta=45^\circ-\alpha.$
Range, $\text{R}_2=\frac{\text{u}^2\sin2(45^\circ-\alpha)}{\text{g}}$
$=\frac{\text{u}^2}{\text{g}}\sin(90^\circ-2\alpha)$
$=\frac{\text{u}^2}{\text{g}}\cos^2\alpha=\text{R}_1$

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Question 42 Marks

Prove the following statement, “For Elevation which exceed or fall short of 45° by equal amount, the range is equal.”

Answer

At 45°, the projectile has maximum range. At $(45^\circ-\theta)$ the range is, $\text{R}_1=\frac{\text{u}^2\sin[2(45^\circ-\theta)]}{\text{g}}$ $=\frac{\text{u}^2\sin(90^\circ-2\theta)}{\text{g}}=\frac{\text{u}^2\cos2\theta}{\text{g}}$ At $(45^\circ+\theta)$ the range is, $\text{R}_2=\frac{\text{u}^2\sin[2(45^\circ+\theta)]}{\text{g}}$ $=\frac{\text{u}^2\sin(90^\circ+2\theta)}{\text{g}}=\frac{\text{u}^2\cos2\theta}{\text{g}}$ $\text{R}_1=\text{R}_2$

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Question 52 Marks

Why are the passengers of a car rounding a curve thrown outward?

Answer

When the car turns round a curve, the passengers sitting in the car experience an outward force, i.e. centrifugal force due to the absence of the necessary centripetal force.

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Question 62 Marks

Give the geometrical meaning of $(i)$ scalar product $(ii)$ cross product of two vectors.

Answer

Geometrically,

$\vec{\text{A}}.\vec{\text{B}}$ refers to the product of magnitude of $\vec{\text{A}}$ and projection of $\vec{\text{B}}$ on $\vec{\text{A}}.$
$\vec{\text{A}}\times\vec{\text{B}}$ refers to the area of the parallelogram formed by the vectors.

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Question 72 Marks

Pick out the only vector quantity in the following list: Temperature, pressure, impulse, time, power, total path length, energy, gravitational potential, coefficient of friction, charge.

Answer

Since, Impulse = change in momentum = force × time. As momentum and force are vector quantities, hence impulse is a vector quantity.

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Question 82 Marks

The dot product of two vectors vanishes when vectors are orthogonal and has maximum value when vectors are parallel to each other. Explain.

Answer

We know that $\text{A}.\text{B}=\text{AB}\cos\theta,$ when vectors are or thogonal, $\theta=90^\circ.$ So, $\text{A}.\text{B}=\text{AB}\cos90^\circ=0,$ when vectors are parallel, then, $\theta=0^\circ.$ So, $\text{A}.\text{B}=\text{AB}\cos0^\circ=\text{AB}$ (maximum)

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Question 92 Marks

What is meant by resolving a force into rectangular components? Resolve a force of 10N into two components, if it acts at an angle 30° with the horizontal.

Answer

Let F be the force at 30° to horizontal. The horizontal component is, $\text{F}\cos30^\circ=\frac{\sqrt{3}\text{F}}{2}$ The vertical component is, $\text{F}\sin30^\circ=\frac{\text{F}}{2}$

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Question 102 Marks

Prove the following statement "For Elevations which exceed or fall short of 45° by equal amounts, the ranges are equal”.

Answer

Case I: When angle of projection $\theta=45^\circ+\alpha$ Range, $\text{R}_1=\frac{\text{u}^2\sin2(45^\circ+\alpha)}{\text{g}}$ $=\frac{\text{u}^2\cos2\alpha}{\text{g}}$ Case II: When angle of projection $\theta=45^\circ-\alpha$ Range, $\text{R}_2=\frac{\text{u}^2\sin2(45^\circ-\alpha)}{\text{g}}$ $=\frac{\text{u}^2}{\text{g}}\sin(90^\circ-2\alpha)$ $=\frac{\text{u}^2\cos2\alpha}{\text{g}}=\text{R}_1$

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Question 112 Marks

Prove that the maximum horizontal range is four times the maximum height attained by a projectile which is fired along the required oblique direction.

Answer

The required angle of projection for max. horizontal range is 45°. $\text{R}_\text{max}=\frac{\text{u}^2}{\text{g}}$ Max. height, $\text{H}=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$ $=\frac{\text{u}^2\sin^245^\circ}{2\text{g}}=\frac{\text{u}^2}{4\text{g}}$ $\frac{\text{R}_\text{max}}{\text{H}}=\frac{\frac{\text{u}^2}{\text{g}}}{\frac{\text{u}^2}{4\text{g}}}=4$ or $\text{R}_\text{max}=4\text{H}$

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Question 122 Marks

A person sitting in a running train throws a ball vertically upwards. What is the nature of the path described by the ball to a person?

Sitting inside the train.
Standing on the ground outside the train.

Answer

The nature of the path will be vertical straight line because the ball has only one velocity acting vertically. The horizontal component of velocity of ball will not be visible to the passenger inside the train because he himself is in motion.
The nature of the path will be a parabolic path because the ball has the vertical as well as horizontal component of velocities.

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Question 132 Marks

Prove that the vectors $(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$ and $(2\hat{\text{i}}-\hat{\text{j}})$ are perpendicular to each other.

Answer

$\vec{\text{A}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ $\vec{\text{B}}=2\hat{\text{i}}-\hat{\text{j}}$ $\vec{\text{A}}$ is perpendicular to $\vec{\text{B}},$ if $\vec{\text{A}}.\vec{\text{B}}=0$ $\therefore\ (\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}).(2\hat{\text{i}}-\hat{\text{j}})=2-2=0$ Thus, $\vec{\text{A}}$ is perpendicular to $\vec{\text{B}}.$

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Question 142 Marks

How does the knowledge of projectile help, a player in the baseball game?

Answer

In the baseball game, a player has to throw a ball so that it goes a certain distance in the minimum time. The time would depend on velocity of ball and angle of throw with the horizontal. Thus, while playing a baseball game, the speed and angle of projection have to be adjusted suitably so that the ball covers the desired distance in minimum time. So, a player has to see the distance and air resistance while playing with a baseball game.

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Question 152 Marks

Find the angle of projection at which horizontal range and maximum height are equal.

Answer

Horizontal range = Maximum height (given) $\therefore\ \frac{\text{u}^2}{\text{g}}\sin2\theta=\frac{\text{u}^2\sin^2\theta}{2\text{g}}$ $\Rightarrow\ 2\sin\theta\cos\theta=\frac{\sin^2\theta}{2}$ $[\therefore\ \sin2\theta=2\cos\theta\sin\theta]$$\tan\theta=4$
$\Rightarrow\ \theta=75^\circ58'$

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Question 162 Marks

Determine that vector which when added to the resultant of $\text{A}=3\hat{\text{i}}-5\hat{\text{j}}+7\hat{\text{k}}$ and $\text{B}=2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ gives unit vector along y-direction.

Answer

We are given, $\text{A}=3\hat{\text{i}}-5\hat{\text{j}}+7\hat{\text{k}}$ and $\text{B}=2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}}$ Thus, the resultant vector is given by, $\text{R}=\text{A}+\text{B}=(3\hat{\text{i}}-5\hat{\text{j}}+7\hat{\text{k}})+(2\hat{\text{i}}+4\hat{\text{j}}-3\hat{\text{k}})$ $=5\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$ But the unit vector along y-direction $=\hat{\text{j}}$ $\therefore$ Required vector $=\hat{\text{j}}-(5\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}})$ $=-5\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$

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Question 172 Marks

Two bombs of 20kg and 30kg are thrown from a cannon with the same velocity in the same direction. Which bomb will reach the ground first?

Answer

Time of flight $\text{T}=\frac{2\text{u}\sin\theta}{\text{g}}.$ Since time of flight does not depend upon mass, both will reach simultaneously.

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Question 182 Marks

We can order events in time and there is no sense of time, distinguishing past, present and future. Is time a vector?

Answer

We know that time always flows on and on i.e. from past to present and then to future. Therefore, a direction can be assigned to time. Since, the direction of time is unique and it is unspecified or unstated. That is why, time cannot be a vector though it has a direction.

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Question 192 Marks

What is the projection of $\hat{\text{i}}+\hat{\text{j}}$ on $(\hat{\text{i}}-\hat{\text{j}})?$

Answer

Let $\vec{\text{A}}=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{B}}=\hat{\text{i}}-\hat{\text{j}}$ Projection of $\vec{\text{A}}$ on $\vec{\text{B}}=|\vec{\text{A}}||\vec{\text{B}}|\cos\theta$ $\cos\theta=\frac{\vec{\text{A}}.\vec{\text{B}}}{|\vec{\text{A}}||\vec{\text{B}}|}=0$ $(\theta=90^\circ)$ $\theta=90^\circ,$ $\therefore$ Projection = 0.

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Question 202 Marks

A lady walking towards east on a road with velocity of 10m/ s encounters rain falling vertically with a velocity of 30m/ s. At what angle she should hold her umbrella to protect herself from the rain?

Answer

$\vec{\text{v}_\text{w}}=10\text{ m/s}=\vec{\text{OA}}$
$\vec{\text{v}_\text{r}}=30\text{ m/s}=\vec{\text{OB}}$
$\therefore\ \vec{\text{v}_\text{rw}}=\sqrt{\text{v}_\text{r}^2+\text{v}_\text{w}^2}$
$=\sqrt{(10)^2+(30)^2}=31.6\text{ m/s}$
$\tan\theta=\frac{\text{BD}}{\text{OB}}=\frac{10}{30}=0.33\ (\text{BD}=\text{OA})$
$\theta=18^\circ16'$ with vertical direction.

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Question 212 Marks

Water from a sprinkler comes out with a constant velocity u in all the directions. What is the maximum area of the grass-land that can be watered at any time?

Answer

Water can go to a maximum distance equal to maximum range (i.e.), $\frac{\text{u}^2}{\text{g}}.$ $\therefore$ Area that can be watered $=\pi\text{r}^2$ $=\pi\Big(\frac{\text{u}^2}{\text{g}}\Big)^2=\frac{\pi\text{u}^4}{\text{g}^2}$

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Question 222 Marks

Two forces 5kg-wt. and 10kg-wt. are acting with an inclination of 120° between them. Find the angle when the resultant makes with 10kg-wt.

Answer

Given, A = 5kg-wt, B = 10kg-wt, $\theta=120^\circ$ then $\beta=?$ $\tan\beta=\frac{\text{B}\sin\theta}{\text{A}+\text{B}\cos\theta}=\frac{10\sin120^\circ}{5+10\cos120^\circ}$ $=\frac{5\sin60^\circ}{10-5\cos60^\circ}=\frac{5\times\frac{\sqrt{3}}{2}}{10-\frac{5}{2}}=\frac{1}{\sqrt{3}}=\tan30^\circ$ $\therefore\ \beta=30^\circ$

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Question 232 Marks

A body is projected with a speed v at an angle $\theta$ with horizontal to have maximum range. What is the velocity at the highest point?

Answer

For maximum range $\theta=45^\circ,$ velocity at the highest point $=\text{v}\cos45^\circ=\frac{\text{v}}{\sqrt{2}}.$

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Question 242 Marks

What is the speed of an aircraft if the pilot remains in contact with the seat, even while looping in vertical plane?

Answer

For the pilot to stay at rest, centripetal force should be provided by his weight, i.e., $\frac{\text{mv}^2}{\text{r}}=\text{mg}$ $\therefore\ \text{v}=\sqrt{\text{rg}}$

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Question 252 Marks

At what point of projectile path the speed is minimum?
At which point, the speed is maximum?

Answer

At the highest point.
At the projection point.

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Question 262 Marks

If two vectors of equal magnitude add to either of them by magnitude, what is the angle between them?

Answer

$|\text{R}|=|\text{p}|$ and $|\text{P}|=|\text{Q}|=|\text{p}|$$\therefore$ using $\text{R}^2=\text{P}^2+\text{Q}^2+2\text{PQ}\cos\theta$
we get,
$\text{p}^2=\text{p}^2+\text{p}^2+2\text{p}^2\cos\theta$
$\cos\theta=-\frac{1}{2};\ \theta=120^\circ$

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Question 272 Marks

What is the distance travelled by a point during the time, if it moves in x - y plane, according to the relation $\text{x}=\text{a}\sin\omega\text{t}$ and $\text{y}=\text{a}(1-\cos\omega\text{t})?$

Answer

The motion will be Simple Harmonic with amplitude of $\sqrt{\text{a}^2+\text{a}^2}=\text{a}\sqrt{2}$ if in the same direction.Because directions are perpendicular to X and Y, it may be interpreted for a circular motion with radius a.

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Question 282 Marks

Two equal forces have their resultant equal to either. What is the inclination between them?

Answer

Here, A = F; B = F, R = F$\text{R}=\sqrt{\text{A}^2+\text{B}^2+2\text{AB}\cos\theta}$
$\text{F}=\sqrt{\text{F}^2+\text{F}^2+2\text{F}.\text{F}\cos\theta}$
$\text{F}=\text{F}\sqrt{2(1+\cos\theta)}$
$1=2(1+\cos\theta)$
$\cos\theta=-\frac{1}{2}=\cos120^\circ$
$\theta=120^\circ$

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Question 292 Marks

The angle between vector $\vec{\text{A}}$ and $\vec{\text{B}}$ is 60°. What is the ratio of $\vec{\text{A}}.\vec{\text{B}}$ and $|\vec{\text{A}}\times\vec{\text{B}}|.$

Answer

$\vec{\text{A}}.\vec{\text{B}}=\text{AB}\cos\theta$ $|\vec{\text{A}}\times\vec{\text{B}}|=\text{AB}\sin\theta$ $\therefore\ \frac{\vec{\text{A}}.\vec{\text{B}}}{|\vec{\text{A}}\times\vec{\text{B}}|}=\cot\theta$ $=\cot60^\circ=\frac{1}{\sqrt{3}}$

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Question 302 Marks

Can there be a two dimensional motion even though acceleration is present in only one direction?

Answer

Yes, the motion of projectile is an example for the same.

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Question 312 Marks

A swimmer can swim with velocity of 10 km/ h. w.r.t. the water flowing in a river with velocity of 5 km/ h. In what direction should he swim to reach the point on the other bank just opposite to his starting point?

Answer

$\mathrm{v}_{\mathrm{R}}=5 \mathrm{~km} / \mathrm{h}$
$\mathrm{v}_{\mathrm{S}}=10 \mathrm{~km} / \mathrm{h}$
$\cos\theta=\frac{5}{10}=\frac{1}{2}$
$\theta=60^\circ$
⇒ with the direction of river, the angle would be 120° to reach the opposite point.

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Question 322 Marks

Instantaneous velocity and tangential acceleration of two particles moving in circular paths of radius r are shown in figure I and II. Which of the particles is speeding up and which one is slowing down?

Answer

Particle moving in a circular path I is speeding up because the direction of both $\vec{\text{v}}$ and $\vec{\text{a}}_\text{t}$ are same. Particle moving in a circtrlar path II is slowing down because the directions of both $\vec{\text{v}}$ and $\vec{\text{a}}_\text{t}$ are opposite to each other.

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Question 332 Marks

Derive $\text{x}_2=\text{x}_1+\text{v}_1(\text{t}_2-\text{t}_1)+\frac{1}{2}\text{a}(\text{t}_2-\text{t}_1)^2.$

Answer

Consider a body starting from $x_1$ and reaching $x_2$ at time $t_2$. Starting with a velocity $v_1$ at time $t_1$ with a uniform acceleration ‘a'. The motion can be described by the v-t graph as shown. Area below v-t graph is PMWYNP $=\text{WM}\times\text{WY}\times\frac{1}{2}\text{WY}\times\text{NP}$ $=\text{v}_1(\text{t}_2-\text{t}_1)+\frac{1}{2}(\text{t}_2-\text{t}_1)\times(\text{v}_2-\text{v}_1)$ Since area below v-t graph is displacement and $v_2-v_1=a\left(t_2-t_1\right)$, we get $\text{x}_2-\text{x}_1=\text{v}_1(\text{t}_2-\text{t}_1)+\frac{1}{2}(\text{t}_2-\text{t}_1)\times\text{a}(\text{t}_2-\text{t}_1)$

$\therefore\ \text{x}_2=\text{x}_1+\text{v}_1(\text{t}_2-\text{t}_1)+\frac{1}{2}\text{a}(\text{t}_2-\text{t}_1)^2$

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Question 342 Marks

What are co-initial and collinear vectors?

Answer

Two vectors having the same initial point are called co-initial vectors. Two vectors which either act along the same line or along parallel lines are called collinear vectors.

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Question 352 Marks

If $\vec{\text{A}}=(-2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}})$ and $\vec{\text{B}}=(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})$ find $\vec{\text{A}}\times\vec{\text{B}}$ and $\vec{\text{A}}.\vec{\text{B}}$

Answer

$\vec{\text{A}}\times\vec{\text{B}}=\begin{vmatrix}\hat{\text{i}} & \hat{\text{j}}&\hat{\text{k}} \\-2 & 3&-4\\3&-4&5 \end{vmatrix}$

$=\hat{\text{i}}\begin{vmatrix}3&-4\\-4&5\end{vmatrix}-\hat{\text{j}}\begin{vmatrix}-2&-4\\3&5\end{vmatrix}+\hat{\text{k}}\begin{vmatrix}-2&3\\3&-4\end{vmatrix}$
$=\hat{\text{i}}(15-16)-\hat{\text{j}}(-10+12)+\hat{\text{k}}(8-9)$
$=-\hat{\text{i}}-2\hat{\text{j}}-\hat{\text{k}}$

$\vec{\text{A}}.\vec{\text{B}}=(-2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}).(3\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}})$

$=-6-12-20=-38$

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Question 362 Marks

A cyclist starts from centre O of a circular park of radius 1km and moves along the path OPRQO as shown Fig. If he maintains constant speed of $10 \mathrm{~ms}^{-1}$, what is his acceleration at point R in magnitude and direction?

Answer

The path of the cyclist at R is circular of constant radius 1km with centre O and he is moving with constant speed 10m/ s. So his motion is uniform circular motion at R.
Hence, the R = 1000m, v =10m/ s
$\therefore\ \text{a}_\text{c}=\frac{\text{v}^2}{\text{R}}=\frac{10\times10}{1000}=\frac{1}{10}=0.1\text{m/ s}^2$ along RO.

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Question 372 Marks

A skilled gun man always keeps his gun slightly tilted above the line of sight while shooting. Why?

Answer

When a bullet is fired from a gun with its barrel directed towards the target, it starts falling downwards on account of acceleration due to gravity. Due to which the bullet hits below the target. Just to avoid it, the barrel of the gun is lined up little above the target, so that the bullet after travelling in parabolic path hits the distant target.

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Question 382 Marks

Keeping the angle of projection same, what is the effect on horizontal range of a particle when its velocity is doubled?

Answer

$\text{R}=\frac{\text{v}_0^2\sin2\theta}{\text{g}},\text{R }\alpha\text{ v}_0^2$ On doubling $\text{v}_0$, R becomes 4 times.

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Question 392 Marks

Suppose you have two forces $F$ and $F.$ How would you combine them in order to have resultant force of magnitudes?

Zero
$F$

Answer

If they act at opposite direction, resultant is zero.
For the resultant to be $F,$

$\text{F}^2=\text{F}^2+\text{F}^2+2\text{F}^2\cos\theta$
$\Rightarrow\ \cos\theta=\frac{-1}{2}$
$\Rightarrow\ \theta=120^\circ$

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Question 402 Marks

Two bodies are thrown with same velocities at angles $\alpha$ and $(90^\circ-\alpha)$ with the horizontal. What will be the ratio of $(i)$ maximum heights attained by them $(ii)$ their horizontal ranges?

Answer

$\frac{\text{h}_1}{\text{h}_2}=\frac{\frac{\text{v}^2_0\sin^2\alpha}{2\text{g}}}{\frac{\text{v}^2_0\sin^2(90-\alpha)}{2\text{g}}}$

$=\frac{\sin^2\alpha}{\cos^2\alpha}=\tan^2\alpha$

Horizontal ranges in both the cases are same, i.e.,

$\text{R}_1=\text{R}_2,\ \frac{\text{R}_1}{\text{R}_2}=1$

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Question 412 Marks

Two cars are going in two concentric circular orbits of radius $\text{r}_1$ and $\text{r}_2$ with angular velocities $\omega_1$ and $\omega_2.$ What is the ratio of their linear velocities?

Answer

We know, $\text{v}=\text{r}\omega,\ \therefore\ \frac{\text{v}_1}{\text{v}_2}=\frac{\text{r}_1\omega_1}{\text{r}_2\omega_2}.$

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Question 422 Marks

Can there be two vectors where the resultant is equal to either of them?

Answer

Yes, when two vectors of same magnitude inclined at an angle of 120°, then resultant is equal to either of them. Let x be the magnitude of each of two vectors which make an angle of 120°, then resultant is equal to either of them. Now, $\text{R}=\sqrt{\text{x}^2+\text{x}^2+2\text{x.x}\cos120^\circ}$ $\Rightarrow\ \text{R}=\sqrt{2\text{x}^2-\text{x}^2}=\sqrt{\text{x}^2}=\text{x}$ $\Big[\because\ \cos120^\circ=-\frac{1}{2}\Big]$

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Question 432 Marks

The angle between vectors A and B is 60°. What is the ratio of A.B and |A × B|?

Answer

The dot product, $\text{A}.\text{B}=\text{AB}\cos\theta$ cross product $|\text{A}\times\text{B}|=\text{AB}\sin\theta.$ $\therefore$ Ratio is $\frac{\text{A.B}}{|\text{A}\times\text{B}|}=\frac{\text{AB}\cos\theta}{\text{AB}\sin\theta}=\cot\theta$$=\cot60^\circ=\frac{1}{\sqrt{3}}$
As, $\theta=60^\circ,\ \cot60^\circ=\frac{1}{\sqrt{3}}$

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Question 442 Marks

A, B and C are three non-collinear, non co-planar vectors. What can you say about direction of A × (B × C)?

Answer

The direction of the vector $(\vec{\text{B}}\times\vec{\text{C}})$ will be perpendicular to the plane containing by vectors $\vec{\text{B}}$ and $\vec{\text{C}}$ by Right-hand thumb or Right-hand grip rule (RHGR).
The direction of the vector $\vec{\text{A}}\times(\vec{\text{B}}\times\vec{\text{C}})$ will be perpendicular to $\vec{\text{A}}$ and in a plane containing $\vec{\text{B}}$ and $\vec{\text{C}}$ by Right-hand grip rule.

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Question 452 Marks

A cyclist has to bend a little inwards from his vertical position while turning. Why?

Answer

By bending, a component of normal reaction of the ground is spared to provide him the necessary centripetal force for turning.

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Question 462 Marks

Two forces whose magnitudes are in the ratio 3 : 5 give a resultant of 28N. If the angle of their inclination is $60^o$. Find the magnitude of each force.

Answer

Let A and B be the two forces. Then, A = 3x, B = 5x, R = 28N and $\theta=60^\circ$ Thus, $\frac{\text{A}}{\text{B}}=\frac{3}{5}$ Now, $\text{R}=\sqrt{\text{A}^2+\text{B}^2+2\text{AB}\cos\theta}$ $\Rightarrow\ 28=\sqrt{9\text{x}^2+25\text{x}^2+30\text{x}^2\cos60^\circ}=7\text{x}$ $\Rightarrow\ \text{x}=4$ $\therefore$ Forces are A = 12N and B = 20N.

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Question 472 Marks

What is the angular acceleration of a particle moving in a circle of radius 'r' with a angular speed $'\omega'?$

Answer

Since $\omega$ is constant, angular acceleration will be zero.

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Question 482 Marks

Calculate the area of a parallelogram whose adjacent sides are given by the vectors. $\vec{\text{A}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}};\ \vec{\text{B}}=2\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}.$

Answer

Area of the parallelogram $=|\vec{\text{A}}\times\vec{\text{B}}|$ $\vec{\text{A}}\times\vec{\text{B}}=\begin{vmatrix} \hat{\text{i}}& \hat{\text{j}}&\hat{\text{k}}\\1 & 2&3\\2&-3&1 \end{vmatrix}$ $=11\hat{\text{i}}+5\hat{\text{j}}-7\hat{\text{k}}$ $\therefore\ |\vec{\text{A}}\times\vec{\text{B}}|=\sqrt{121+25+49}$ $=\sqrt{195}\text{ sq. units}$

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Question 492 Marks

Explain the property of two vectors A and B if |A + B| = |A - B|.

Answer

As we know that, $|\text{A}+\text{B}|=\sqrt{\text{A}^2+\text{B}^2+2\text{AB}\cos\theta}$ and $|\text{A}-\text{B}|=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\cos\theta}$ But as per question, we have $\sqrt{\text{A}^2+\text{B}^2+2\text{AB}\cos\theta}=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\cos\theta}$ Squaring both sides, we have $(4\text{ AB}\cos\theta)=0$ $\Rightarrow\ \cos\theta=0\ \text{or }\theta=90^\circ$ Hence, the two vectors A and B are perpendicular to each other.

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Question 502 Marks

An aeroplane travelling at a speed of 5000km/ hr tilts at an angle of 30° as it makes a turn. What is the radius of the curve?

Answer

$\tan\theta=\frac{\text{v}^2}{\text{rg}}=\Big(\frac{1250}{9}\Big)^2\times\frac{1}{\text{r}}\times\frac{1}{9.8}$or $\text{r}=\Big(\frac{1250}{9}\Big)^2\frac{\sqrt{3}}{1}\times\frac{1}{9.8}$
$\Big[\because5000\text{ km/hr}=\frac{1250}{9}\text{ m/s}\Big]$
or, $\text{r}=3.41\times10^3\text{m}\Big[\because\tan30^\circ=\frac{1}{\sqrt{3}}\Big]$

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