Answer

Scalar: Volume, mass, speed, density, number of moles, angular frequency

Vector: Acceleration, velocity, displacement, angular velocity

A scalar quantity is specified by its magnitude only. It does not have any direction associated with it. Volume, mass, speed, density, number of moles, and angular frequency are some of the scalar physical quantities.

A vector quantity is specified by its magnitude as well as the direction associated with it. Acceleration, velocity, displacement, and angular velocity belong to this category.

Answer

Work and current are scalar quantities.

Work done is given by the dot product of force and displacement. Since the dot product of two quantities is always a scalar, work is a scalar physical quantity.

Current is described only by its magnitude. Its direction is not taken into account. Hence, it is a scalar quantity.

Answer

Impulse

Impulse is given by the product of force and time. Since force is a vector quantity, its product with time (a scalar quantity) gives a vector quantity.

Answer

(a) Meaningful

(b) Not Meaningful

(c) Meaningful

(d) Meaningful

(e) Meaningful

(f) Meaningful

Explanation:

(a) The addition of two scalar quantities is meaningful only if they both represent the same physical quantity.

(b) The addition of a vector quantity with a scalar quantity is not meaningful.

(c) A scalar can be multiplied with a vector. For example, force is multiplied with time to give impulse.

(d) A scalar, irrespective of the physical quantity it represents, can be multiplied with another scalar having the same or different dimensions.

(e) The addition of two vector quantities is meaningful only if they both represent the same physical quantity.

(f) A component of a vector can be added to the same vector as they both have the same dimensions.

Answer

(a) True

(b) False

(c) False

(d) True

(e) True

Explanation:

(a) The magnitude of a vector is a number. Hence, it is a scalar.

(b) Each component of a vector is also a vector.

(c) Total path length is a scalar quantity, whereas displacement is a vector quantity. Hence, the total path length is always greater than the magnitude of displacement. It becomes equal to the magnitude of displacement only when a particle is moving in a straight line.

(d) It is because of the fact that the total path length is always greater than or equal to the magnitude of displacement of a particle.

(e) Three vectors, which do not lie in a plane, cannot be represented by the sides of a triangle taken in the same order.

Answer

Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

Here, we can write:

$$ \begin{align*} & |\overrightarrow{{}OM}|=|\vec{a}| \tag{i} \end{align*} $$

$$ \begin{align*} & |\overrightarrow{{}MN}|=|\overrightarrow{{}OP}|=|\vec{b}| \tag{ii}\\ & |\overrightarrow{{}ON}|=|\vec{a}+\vec{b}| \tag{iii} \end{align*} $$

In a triangle, each side is smaller than the sum of the other two sides.

Therefore, in $\triangle OMN$, we have:

$ON<(OM+MN)$

$$|\vec{a}+\vec{b}|<|\vec{a}|+|\vec{b}| \tag{iv}$$

If the two vectors $\vec{a}$ and $\vec{b}$ act along a straight line in the same direction, then we can write:

$$ |\vec{a}+\vec{b}|=|\vec{a}|+|\vec{b}| \tag{iv}$$

Combining equations (iv) and (v), we get:

$|\vec{a}+\vec{b}| \leq|\vec{a}|+|\vec{b}|$

Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the adjacent sides of a parallelogram OMNP, as shown in the given figure.

Here, we have:

$$ \begin{align*} & |\overrightarrow{{}OM}|=|\vec{a}| \tag{i} \end{align*} $$

$$ \begin{align*} & |\overrightarrow{{}MN}|=|\overrightarrow{{}OP}|=|\vec{b}| \tag{ii} \\ & |\overrightarrow{{}ON}|=|\vec{a}+\vec{b}| \tag{iii} \end{align*} $$

In a triangle, each side is smaller than the sum of the other two sides.

Therefore, in $\triangle OMN$, we have:

$ \begin{aligned} & ON+MN>OM \\ & ON+OM>MN \\ & |\overrightarrow{{}ON}|>|\overrightarrow{{}OM}-\overrightarrow{{}OP}| \quad(\because OP=MN) \\ & |\vec{a}+\vec{b}|>|| \vec{a}|-| \vec{b}|| _{\ldots(i v)} \end{aligned} $

If the two vectors $\vec{a}$ and $\vec{b}$ act along a straight line in the same direction, then we can write:

$|\vec{a}+\vec{b}|=|| \vec{a}|-| \vec{b}|| _{\ldots(v)}$

Combining equations (iv) and (v), we get:

$ |\vec{a}+\vec{b}| \geq|| \vec{a}|-| \vec{b}|| $

Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

Here we have:

$$ \begin{align*} & |\overrightarrow{{}OR}|=|\overrightarrow{{}PS}|=|\vec{b}| \tag{i} \end{align*} $$

$$ \begin{align*} & |\overrightarrow{{}OP}|=|\vec{a}| \tag{ii} \end{align*} $$

In a triangle, each side is smaller than the sum of the other two sides. Therefore, in $\triangle OPS$, we have:

$$ \begin{align*} & OS<OP+PS \end{align*} $$

$$ \begin{align*} & |\vec{a}-\vec{b}|<|\vec{a}|+|-\vec{b}| \\ & |\vec{a}-\vec{b}|<|\vec{a}|+|\vec{b}| \tag{iii} \end{align*} $$

If the two vectors act in a straight line but in opposite directions, then we can write:

$$ \begin{equation*} |\vec{a}-\vec{b}|=|\vec{a}|+|\vec{b}| \tag{iv} \end{equation*} $$

Combining equations (iii) and (iv), we get:

$ |\vec{a}-\vec{b}| \leq|\vec{a}|+|\vec{b}| $

Let two vectors $\vec{a}$ and $\vec{b}$ be represented by the adjacent sides of a parallelogram PORS, as shown in the given figure.

The following relations can be written for the given parallelogram.

$$ \begin{align*} & OS+PS>OP \tag{i}\\ & OS>OP-PS \tag{ii}\\ & |\vec{a}-\vec{b}|>|\vec{a}|-|\vec{b}| \tag{iii} \end{align*} $$

The quantity on the LHS is always positive and that on the RHS can be positive or negative. To make both quantities positive, we take modulus on both sides as:

$$ \begin{align*} & || \vec{a}-\vec{b}||>|| \vec{a}|-| \vec{b}|| \\ & |\vec{a}-\vec{b}|>|| \vec{a}|-| \vec{b}|| \tag{iv} \end{align*} $$

If the two vectors act in a straight line but in the opposite directions, then we can write:

$$ \begin{equation*} |\vec{a}-\vec{b}|=| \vec{a}|-| \vec{b}|| \tag{v} \end{equation*} $$

Combining equations (iv) and (v), we get:

$|\vec{a}-\vec{b}| \geq|| \vec{a}|-| \vec{b}||$

Answer

(a) Incorrect

In order to make $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{0}$, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.

(b) Correct

$\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{0}$

$\mathbf{a}+\mathbf{c}=-(\mathbf{b}+\mathbf{d})$

Taking modulus on both the sides, we get:

$|\mathbf{a}+\mathbf{c}|=|-(\mathbf{b}+\mathbf{d})|=|\mathbf{b}+\mathbf{d}|$

Hence, the magnitude of $(\mathbf{a}+\mathbf{c})$ is the same as the magnitude of $(\mathbf{b}+\mathbf{d})$.

(c) Correct

$\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{0}$

$\mathbf{a}=(\mathbf{b}+\mathbf{c}+\mathbf{d})$

Taking modulus both sides, we get:

$|\mathbf{a}|=|\mathbf{b}+\mathbf{c}+\mathbf{d}|$

$|\mathbf{a}| \leq|\mathbf{a}|+|\mathbf{b}|+|\mathbf{c}| \ldots(i)$

Equation (i) shows that the magnitude of a is equal to or less than the sum of the magnitudes of $\mathbf{b}, \mathbf{c}$, and $\mathbf{d}$.

Hence, the magnitude of vector $a$ can never be greater than the sum of the magnitudes of $\mathbf{b}, \mathbf{c}$, and $\mathbf{d}$.

(d) Correct

For $\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0$

$\mathbf{a}+(\mathbf{ b}+\mathbf{c})+\mathbf{d}=0$

The resultant sum of the three vectors $\mathbf{a},(\mathbf{b}+\mathbf{c})$, and $\mathbf{d}$ can be zero only if $(\mathbf{b}+\mathbf{c})$ lie in a plane containing a and $\mathbf{d}$, assuming that these three vectors are represented by the three sides of a triangle.

If $\mathbf{a}$ and $\mathbf{d}$ are collinear, then it implies that the vector $(\mathbf{b}+\mathbf{c})$ is in the line of $\mathbf{a}$ and $\mathbf{d}$. This implication holds only then the vector sum of all the vectors will be zero.

Answer

Displacement is given by the minimum distance between the initial and final positions of a particle. In the given case, all the girls start from point $P$ and reach point $Q$. The magnitudes of their displacements will be equal to the diameter of the ground.

Radius of the ground $=200 m$

Diameter of the ground $=2 \times 200=400 m$

Hence, the magnitude of the displacement for each girl is $400 m$. This is equal to the actual length of the path skated by girl B.

Answer

Displacement is given by the minimum distance between the initial and final positions of a body. In the given case, the cyclist comes to the starting point after cycling for 10 minutes. Hence, his net displacement is zero.

Average velocity is given by the relation:

Average velocity $=\frac{\text{ Net displacement }}{\text{ Total time }}$

Since the net displacement of the cyclist is zero, his average velocity will also be zero.

Average speed of the cyclist is given by the relation:

Average speed $=\frac{\text{ Total path length }}{\text{ Total time }}$

Total path length $=OP+PQ+QO$

$=1+\frac{1}{4}(2 \pi \times 1)+1$

$=2+\frac{1}{2} \pi=3.570 km$

Time taken $=10 \min =\frac{10}{60}=\frac{1}{6} h$

$\therefore$ Average speed

$ =\frac{3.570}{\frac{1}{6}}=21.42 km / h $

Answer

The path followed by the motorist is a regular hexagon with side $500 m$, as shown in the given figure

Let the motorist start from point $P$.

The motorist takes the third turn at $S$.

$\therefore$ Magnitude of displacement $=PS=PV+VS=500+500=1000 m$

Total path length $=PQ+QR+RS=500+500+500=1500 m$

The motorist takes the sixth turn at point $P$, which is the starting point.

$\therefore$ Magnitude of displacement $=0$

Total path length $=PQ+QR+RS+ST+TU+UP$

$=500+500+500+500+500+500=3000 m$

The motorist takes the eight turn at point $R$

$\therefore$ Magnitude of displacement $=PR$

$ \begin{aligned} & =\sqrt{PQ^{2}+QR^{2}+2(PQ) \cdot(QR) \cos 60^{\circ}} \\ & =\sqrt{500^{2}+500^{2}+(2 \times 500 \times 500 \times \cos 60^{\circ})} \\ & =\sqrt{250000+250000+(500000 \times \frac{1}{2})} \\ & =866.03 m \\ & \beta=\tan ^{-1}(\frac{500 \sin 60^{\circ}}{500+500 \cos 60^{\circ}})=30^{\circ} \end{aligned} $

Therefore, the magnitude of displacement is $866.03 m$ at an angle of $30^{\circ}$ with PR.

Total path length $=$ Circumference of the hexagon $+PQ+QR$

$=6 \times 500+500+500=4000 m$

The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table

Answer

Total distance travelled $=23 km$

Total time taken $=28 \min =\frac{28}{60} h$

$\therefore$ Average speed of the taxi $=\frac{\text{ Total distance travelled }}{\text{ Total time taken }}$

$=\frac{23}{(\frac{28}{60})}=49.29 km / h$

Distance between the hotel and the station $=10 km=$ Displacement of the car

$=\frac{10}{28}=21.43 km / h$

$\therefore$ Average velocity

60

Therefore, the two physical quantities (averge speed and average velocity) are not equal.

Answer

Speed of the ball, $u=40 m / s$

Maximum height, $h=25 m$

In projectile motion, the maximum height reached by a body projected at an angle $\theta$, is given by the relation:

$h=\frac{u^{2} \sin ^{2} \theta}{2 g}$

$25=\frac{(40)^{2} \sin ^{2} \theta}{2 \times 9.8}$

$\sin ^{2} \theta=0.30625$ $\sin \theta=0.5534$

$\therefore \theta=\sin ^{-1}(0.5534)=33.60^{\circ}$

Horizontal range, $R=\frac{u^{2} \sin 2 \theta}{g}$

$=\frac{(40)^{2} \times \sin 2 \times 33.60}{9.8}$

$=\frac{1600 \times \sin 67.2}{9.8}$

$=\frac{1600 \times 0.922}{9.8}=150.53 m$

Answer

Maximum horizontal distance, $R=100 m$

The cricketer will only be able to throw the ball to the maximum horizontal distance when the angle of projection is $45^{\circ}$, i.e., $\theta=45^{\circ}$.

The horizontal range for a projection velocity $v$, is given by the relation:

$$ \begin{align*} & R=\frac{u^{2} \sin 2 \theta}{g} \\ & 100=\frac{u^{2}}{g} \sin 90^{\circ} \\ & \frac{u^{2}}{g}=100 \tag{i} \end{align*} $$

The ball will achieve the maximum height when it is thrown vertically upward. For such motion, the final velocity $v$ is zero at the maximum height $H$.

Acceleration, $a=-g$

Using the third equation of motion:

$$ \begin{aligned} & v^{2}-u^{2}=-2 g H \\ & H=\frac{1}{2} \times \frac{u^{2}}{g}=\frac{1}{2} \times 100=50 m \end{aligned} $$

Answer

Length of the string, $l=80 cm=0.8 m$

Number of revolutions $=14$

Time taken $=25 s$

Frequency, $\quad v=\frac{\text{ Number of revolutions }}{\text{ Time taken }}=\frac{14}{25} Hz$

Angular frequency, $\omega=2 \pi \nu$

$=2 \times \frac{22}{7} \times \frac{14}{25}=\frac{88}{25} rad s^{-1}$

Centripetal acceleration, $a_c=\omega^{2} r$

$=(\frac{88}{25})^{2} \times 0.8$

$=9.91 m / s^{2}$

The direction of centripetal acceleration is always directed along the string, toward the centre, at all points.

Answer

Radius of the loop, $r=1 km=1000 m$

Speed of the aircraft, $v=900 km / h=900 \times \frac{5}{18}=250 m / s$

Centripetal acceleration, $a_c=\frac{v^{2}}{r}$

$ =\frac{(250)^{2}}{1000}=62.5 m / s^{2} $

Acceleration due to gravity, $g=9.8 m / s^{2}$

$ \frac{a_c}{g}=\frac{62.5}{9.8}=6.38 $

$ a_c=6.38 g $

Answer

(a) False

The net acceleration of a particle in circular motion is not always directed along the radius of the circle toward the centre. It happens only in the case of uniform circular motion.

(b) True

At a point on a circular path, a particle appears to move tangentially to the circular path. Hence, the velocity vector of the particle is always along the tangent at a point.

(c) True

In uniform circular motion (UCM), the direction of the acceleration vector points toward the centre of the circle. However, it constantly changes with time. The average of these vectors over one cycle is a null vector.

Answer

$\vec{v}(t)=(3.0 \hat{\mathbf{i}}-4.0 t \hat{\mathbf{j}}) ; \vec{a}=-4.0 \hat{\mathbf{j}}$

The position of the particle is given by:

$\vec{r}=3.0 t \hat{\mathbf{i}}-2.0 t^{2} \hat{\mathbf{j}}+4.0 \hat{\mathbf{k}}$

Velocity $\vec{v}$, of the particle is given as:

$\vec{v}=\frac{d \vec{r}}{d t}=\frac{d}{d t}(3.0 t \hat{\mathbf{i}}-2.0 t^{2} \hat{\mathbf{j}}+4.0 \hat{\mathbf{k}})$

$\therefore \vec{v}=3.0 \hat{\mathbf{i}}-4.0 t \hat{\mathbf{j}}$

Acceleration $\vec{a}$, of the particle is given as:

$\vec{a}=\frac{d \vec{v}}{d t}=\frac{d}{d t}(3.0 \hat{\mathbf{i}}-4.0 t \hat{\mathbf{j}})$

$\therefore \vec{a}=-4.0 \hat{\mathbf{j}}$

$8.54 m / s, 69.45^{\circ}$ below the $x$-axis

We have velocity vector, $\vec{v}=3.0 \hat{\mathbf{i}}-4.0 t \hat{\mathbf{j}}$

At $t=2.0 s$ :

$\vec{v}=3.0 \hat{\mathbf{i}}-8.0 \hat{\mathbf{j}}$

The magnitude of velocity is given by:

$|\vec{v}|=\sqrt{3^{2}+(-8)^{2}}=\sqrt{73}=8.54 m / s$

Direction, $\theta=\tan ^{-1}(\frac{v_y}{v_x})$

$=\tan ^{-1}(\frac{-8}{3})=-\tan ^{-1}(2.667)$

$=-69.45^{\circ}$

The negative sign indicates that the direction of velocity is below the $x$-axis.

Answer

Velocity of the particle, $\vec{v}=10.0 \hat{\mathbf{j}} m / s$

Acceleration of the particle $\vec{a}=(8.0 \hat{\mathbf{i}}+2.0 \hat{\mathbf{j}})$

Also,

$ \vec{a}=\frac{d \vec{v}}{d t}=8.0 \hat{\mathbf{i}}+2.0 \hat{\mathbf{j}} $

But,

$d \vec{v}=(8.0 \hat{\mathbf{i}}+2.0 \hat{\mathbf{j}}) d t$

Integrating both sides:

$\vec{v}(t)=8.0 t \hat{\mathbf{i}}+2.0 t \hat{\mathbf{j}}+\vec{u}$

Where,

$\vec{u}=$ Velocity vector of the particle at $t=0$

$\vec{v}=$ Velocity vector of the particle at time $t$

But, $\vec{v}=\frac{d \vec{r}}{d t}$

$d \vec{r}=\vec{v} d t=(8.0 t \hat{\mathbf{i}}+2.0 t \hat{\mathbf{j}}+\vec{u}) d t$

Integrating the equations with the conditions: at $t=0 ; r=0$ and at $t=t ; r=r$

$ \begin{aligned} \vec{r} & =\vec{u} t+\frac{1}{2} 8.0 t^{2} \hat{\mathbf{i}}+\frac{1}{2} \times 2.0 t^{2} \hat{\mathbf{j}} \\ & =\vec{u} t+4.0 t^{2} \hat{\mathbf{i}}+t^{2} \hat{\mathbf{j}} \\ & =(10.0 \hat{\mathbf{j}}) t+4.0 t^{2} \hat{\mathbf{i}}+t^{2} \hat{\mathbf{j}} \\ x & \hat{\mathbf{i}}+y \hat{\mathbf{j}}=4.0 t^{2} \hat{\mathbf{i}}+(10 t+t^{2}) \hat{\mathbf{j}} \end{aligned} $

Since the motion of the particle is confined to the $x-y$ plane, on equating the coefficients of $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$, we get:

$x=4 t^{2}$

$t=(\frac{x}{4})^{\frac{1}{2}}$

And $y=10 t+t^{2}$

When $x=16 m$ :

$t=(\frac{16}{4})^{\frac{1}{2}}=2 s$

$\therefore y=10 \times 2+(2)^{2}=24 m$

Velocity of the particle is given by:

$ \begin{aligned} & \vec{v}(t)=8.0 t \hat{\mathbf{i}}+2.0 t \hat{\mathbf{j}}+\vec{u} \\ & \text{ at } t=2 s \\ & \vec{v}(t)=8.0 \times 2 \hat{\mathbf{i}}+2.0 \times 2 \hat{\mathbf{j}}+10 \hat{\mathbf{j}} \\ & \quad=16 \hat{\mathbf{i}}+14 \hat{\mathbf{j}} \end{aligned} $

$\therefore$ Speed of the particle:

$ \begin{aligned} |\vec{v}| & =\sqrt{(16)^{2}+(14)^{2}} \\ & =\sqrt{256+196}=\sqrt{452} \\ & =21.26 m / s \end{aligned} $

Answer

Consider a vector $\vec{P}$, given as:

$ \begin{aligned} & \stackrel{harpoonup}{P}=\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ & P_x \hat{\mathbf{i}}+P_y \hat{\mathbf{j}}=\hat{\mathbf{i}}+\hat{\mathbf{j}} \end{aligned} $

On comparing the components on both sides, we get:

$$ \begin{align*} & P_x=P_y=1 \\ & \overrightarrow{{}|P|}=\sqrt{P_x^{2}+P_y^{2}}=\sqrt{1^{2}+1^{2}}=\sqrt{2} \tag{i} \end{align*} $$

Hence, the magnitude of the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ is $\sqrt{2}$.

Let $\theta$ be the angle made by the vector $\vec{P}$, with the $x$-axis, as shown in the following figure.

$$\therefore \tan \theta=(\frac{P_y}{P_x})$$

$$\theta=\tan ^{-1}(\frac{1}{1})=45^{\circ} \tag{ii}$$

Hence, the vector $\hat{\mathbf{i}}+\hat{\mathbf{j}}$ makes an angle of $45^{\circ}$ with the $x$-axis.

Let $\vec{Q}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$

$Q_x \hat{\mathbf{i}}-Q_y \hat{\mathbf{j}}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$

$Q_x=Q_y=1$

$|\vec{Q}|=\sqrt{Q_x^{2}+Q_y^{2}}=\sqrt{2} \quad \quad \quad \quad \quad (iii)$

Hence, the magnitude of the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ is $\sqrt{2}$.

Let $\theta$ be the angle made by the vector $\vec{Q}$, with the $x$-axis, as shown in the following figure.

$\therefore \tan \theta=(\frac{Q_y}{Q_x})$

$\theta=-\tan ^{-1}(-\frac{1}{1})=-45^{\circ}\quad \quad \quad \quad \quad (iv)$

Hence, the vector $\hat{\mathbf{i}}-\hat{\mathbf{j}}$ makes an angle of $-45^{\circ}$ with the $x$-axis.

It is given that:

$ \begin{aligned} & \vec{A}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \\ & A_x \hat{\mathbf{i}}+A_y \hat{\mathbf{j}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \end{aligned} $

On comparing the coefficients of $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$, we have:

$ \begin{aligned} & A_x=2 \text{ and } A_y=3 \\ & |\vec{A}|=\sqrt{2^{2}+3^{2}}=\sqrt{13} \end{aligned} $

Let $ \vec{A} _x$ make an angle $\theta _{\text{with the }} x$-axis, as shown in the following figure.

$ \begin{aligned} & \therefore \tan \theta=(\frac{A_y}{A_x}) \\ & \theta=\tan ^{-1}(\frac{3}{2}) \\ & \quad=\tan ^{-1}(1.5)=56.31^{\circ} \end{aligned} $

Angle between the vectors $(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})$ and $(\hat{\mathbf{i}}+\hat{\mathbf{j}}), \theta^{\prime}=56.31-45=11.31^{\circ}$

Component of vector $\vec{A}$, along the direction of $\vec{P}$, making an angle $\theta^{\prime}$

$$ \begin{align*} & =(A \cos \theta) \hat{P}=(A \cos 11.31) \frac{(\hat{\mathbf{i}}+\hat{\mathbf{j}})}{\sqrt{2}} \\ & =\sqrt{13} \times \frac{0.9806}{\sqrt{2}}(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \\ & =2.5(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \\ & =\frac{25}{10} \times \sqrt{2} \\ & =\frac{5}{\sqrt{2}} \tag{v} \end{align*} $$

Let $\theta^{*}$ be the angle between the vectors $(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})$ and $(\hat{\mathbf{i}}-\hat{\mathbf{j}})$.

$ \theta^{\prime \prime}=45+56.31=101.31^{\circ} $

Component of vector $\vec{A}$, along the direction of $\vec{Q}$, making an angle $\theta^{\prime \prime}$

$$ \begin{align*} & =(A \cos \theta^{\prime \prime}) \vec{Q}=(A \cos \theta^{\prime \prime}) \frac{\hat{\mathbf{i}}-\hat{\mathbf{j}}}{\sqrt{2}} \\ & =\sqrt{13} \cos (901.31^{\circ}) \frac{(\hat{\mathbf{i}}-\hat{\mathbf{j}})}{\sqrt{2}} \\ & =-\sqrt{\frac{13}{2}} \sin 11.30^{\circ}(\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ & =-2.550 \times 0.1961(\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ & =-0.5(\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ & =-\frac{5}{10} \times \sqrt{2} \\ & =-\frac{1}{\sqrt{2}} \tag{vi} \end{align*} $$

Answer

(b) and (e)

(a) It is given that the motion of the particle is arbitrary. Therefore, the average velocity of the particle cannot be given by this equation.

(b) The arbitrary motion of the particle can be represented by this equation.

(c) The motion of the particle is arbitrary. The acceleration of the particle may also be non-uniform. Hence, this equation cannot represent the motion of the particle in space.

(d) The motion of the particle is arbitrary; acceleration of the particle may also be nonuniform. Hence, this equation cannot represent the motion of particle in space.

(e) The arbitrary motion of the particle can be represented by this equation.

Answer

(a) False

Despite being a scalar quantity, energy is not conserved in inelastic collisions.

(b) False

Despite being a scalar quantity, temperature can take negative values.

(c) False

Total path length is a scalar quantity. Yet it has the dimension of length.

(d) False

A scalar quantity such as gravitational potential can vary from one point to another in space.

(e) True

The value of a scalar does not vary for observers with different orientations of axes.

Answer

The positions of the observer and the aircraft are shown in the given figure.

Height of the aircraft from ground, $OR=3400 m$

Angle subtended between the positions, $\angle POQ=30^{\circ}$

Time $=10 s$

In $\triangle PRO$ :

$\tan 15^{\circ}=\frac{PR}{OR}$

$PR=OR \tan 15^{\circ}$

$=3400 \times \tan 15^{\circ}$

$\triangle PRO$ is similar to $\triangle RQO$.

$\therefore PR=RQ$

$PQ=PR+RQ$

$=2 PR=2 \times 3400 \tan 15^{\circ}$

$=6800 \times 0.268=1822.4 m$

$\therefore$ Speed of the aircraft $=\frac{1822.4}{10}=182.24 m / s$