(a) Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of the earth due to the earth rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre? What is it at latitude 0°? How does these accelerations compare with $g = 9.8 \text{ m/s}^2$?
(b) Earth also moves in circular orbit around the sun once every year with an orbital radius of $1.5 \times 10^{11}$ m. What is the acceleration of the earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with $g = 9.8 \text{ m/s}^2$?
$\begin{array}{ll} \text{(a)} & \text{Radius of the earth } (R) = 6400 \text{ km} = 6.4 \times 10^6 \text{ m} \\ & \text{Time period } (T) = 1 \text{ day} = 24 \times 60 \times 60 \text{ s} = 86400 \text{ s} \\ & \\ & \text{Centripetal acceleration } (a_c) = \omega^2 R = R \left(\frac{2\pi}{T}\right)^2 = \frac{4\, \pi^2\, R}{T^2} \\ &= \frac{4 \times (22/7)^2 \times 6.4 \times 10^6}{(24 \times 60 \times 60)^2} \\ &= \frac{4 \times 484 \times 64 \times 10^6}{49 \times (24 \times 3600)^2} \\ &= 0.034 \text{ m/s}^2 \\ & \\ & \text{At equator, latitude } \theta = 0^\circ \\ & \therefore \frac{a_c}{g} = \frac{0.034}{9.8} = \frac{1}{288} \\ \end{array}$
(b) Orbital radius of the earth around the sun $(R) = 1.5 \times 10^{11}\ \text{m}$
Time period = 1 yr = 365 day
$\qquad \qquad \qquad \quad = 365 \times 24 \times 60 \times 60\ \text{s} = 3.15 \times 10^{7}\ \text{s}$
Centripetal acceleration $(a_c) = R\omega^2 = \frac{4\pi^2 R}{T^2}$
$\qquad \ \qquad \qquad \qquad \ = \frac{4 \times (22/7)^2 \times 1.5 \times 10^{11}}{(3.15 \times 10^{7})^2} $
$\qquad \qquad \qquad \quad \ \ = 5.97 \times 10^{-3}\ \text{m/s}^2$
$\therefore \quad \frac{a_c}{g} = \frac{5.97 \times 10^{-3}}{9.8} = \frac{1}{1642}$
Given below in Column I are the relations between vectors a, b and c and in Column II are the orientations of a, b and c in the XY-plane. Match the relation in Column I to correct orientations in Column II.
| Column I | Column II |
|---|---|
| (a) $ a + b = c $ | (i)  |
| (b) $ a - c = b $ | (ii) |
| (c) $ b - a = c $ | (iii)  |
| (d) $ a + b + c = 0 $ | (iv)  |
Consider the adjacent diagram in which vectors A and B are corrected by head and tail. Resultant vector C = A + B
(a) from (iv) it is clear that c = a + b
(b) from (iii) c + b = a → a - c = b
(c) from (i) b = a + c → b - a = c
(d) from (ii) -c = a + b → a + b + c = 0
If $|A|=2$ and $|B|=4$, then match the relation in Column I with the angle $\theta$ between $A$ and $B$ in Column II.
| Column I | Column II |
|---|---|
| (a) $\mathbf{A \cdot B} = 0$ | (i) $\theta = 0^\circ$ |
| (b) $\mathbf{A \cdot B} = +8$ | (ii) $\theta = 90^\circ$ |
| (c) $\mathbf{A \cdot B} = 4$ | (iii) $\theta = 180^\circ$ |
| (d) $\mathbf{A \cdot B} = -8$ | (iv) $\theta = 60^\circ$ |
Given $|A| = 2$ and $|B| = 4$:
(a) $A \cdot B = AB \cos \theta = 0$
$\Rightarrow 2 \times 4 \cos \theta = 0$
$\Rightarrow \cos \theta = 0 = \cos 90^\circ$
$\Rightarrow \theta = 90^\circ$
Thus, option (a) matches with option (ii).
(b) $A \cdot B = AB \cos \theta = 8$
$\Rightarrow 2 \times 4 \cos \theta = 8$
$\Rightarrow \cos \theta = 1 = \cos 0^\circ$
$\Rightarrow \theta = 0^\circ$
Thus, option (b) matches with option (i).
(c) $A \cdot B = AB \cos \theta = 4$
$\Rightarrow 2 \times 4 \cos \theta = 4$
$\Rightarrow \cos \theta = \frac{1}{2} = \cos 60^\circ$
$\Rightarrow \theta = 60^\circ$
Thus, option (c) matches with option (iv).
(d) $A \cdot B = AB \cos \theta = -8$
$\Rightarrow 2 \times 4 \cos \theta = -8$
$\Rightarrow \cos \theta = -1 = \cos 180^\circ$
$\Rightarrow \theta = 180^\circ$
Thus, option (d) matches with option (iii).
If $|A|=2$ and $|B|=4$, then match the relations in Column I with the angle $\theta$ between $A$ and $B$ in Column II
| Column I | Column II |
|---|---|
| (a) $\vert \mathbf{A \times B} \vert = 0$ | (i) $\theta = 30^\circ$ |
| (b) $\vert \mathbf{A \times B} \vert = 8$ | (ii) $\theta = 45^\circ$ |
| (c) $\vert \mathbf{A \times B} \vert = 4$ | (iii) $\theta = 90^\circ$ |
| (d) $\vert \mathbf{A \times B} \vert = 4\sqrt{2}$ | (iv) $\theta = 0^\circ$ |
Given $|A| = 2$ and $|B| = 4$
(a) $|A \times B| = AB \sin \theta = 0$
$ \Rightarrow 2 \times 4 \sin \theta = 0 $
$ \Rightarrow \sin \theta = 0 = \sin 0^\circ $
$ \Rightarrow \theta = 0^\circ $
$ \therefore \text{Option (a) matches with option (iv).} $
(b) $|A \times B| = AB \sin \theta = 8$
$ \Rightarrow 2 \times 4 \sin \theta = 8 $
$ \Rightarrow \sin \theta = 1 = \sin 90^\circ $
$ \Rightarrow \theta = 90^\circ $
$ \therefore \text{Option (b) matches with option (iii).} $
(c) $|A \times B| = AB \sin \theta = 4$
$ \Rightarrow 2 \times 4 \sin \theta = 4 $
$ \Rightarrow \sin \theta = \frac{1}{2} = \sin 30^\circ $
$ \Rightarrow \theta = 30^\circ $
$ \therefore \text{Option (c) matches with option (i).} $
(d) $|A \times B| = AB \sin \theta = 4 \sqrt{2}$
$ \Rightarrow 2 \times 4 \sin \theta = 4 \sqrt{2} $
$ \Rightarrow \sin \theta = \frac{1}{\sqrt{2}} = \sin 45^\circ $
$ \Rightarrow \theta = 45^\circ $
$ \therefore \text{Option (d) matches with option (ii).} $
A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800 m from the foot of the hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill? Take, $g = 10 \text{ m} / \text{s}^2$.
Given, speed of packets = 125 m/s
Height of the hill = 500 m.
To cross the hill, the vertical component of the velocity should be sufficient to cross such height.
$u_y \geq \sqrt{2gh}$
$\geq \sqrt{2 \times 10 \times 500}$
$\geq 100 \text{m/s}$
But
$u^2 = u_x^2 + u_y^2$
$\therefore$ Horizontal component of initial velocity,
$u_x = \sqrt{u^2 - u_y^2} = \sqrt{(125)^2 - (100)^2} = 75 \text{ m/s}$
Time taken to reach the top of the hill,
$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 500}{10}} = 10 \text{ s}$
Time taken to reach the ground from the top of the hill $t' = t = 10 \text{ s}$. Horizontal distance travelled in 10 s
$x = u_x \times t = 75 \times 10 = 750 \text{ m}$
$\therefore$ Distance through which canon has to be moved = 800 – 750 = 50 m
Speed with which canon can move = 2 m/s
$\therefore$ Time taken by canon = $\frac{50}{2}$ $\Rightarrow$ $t'' = 25 \text{ s}$
$\therefore$ Total time taken by a packet to reach on the ground = $t'' + t + t' = 25 + 10 + 10 = 45$ s