| Q.No |
Answer |
Q.No |
Answer |
Q.No |
Answer |
| 1 | (c) Angular Velocity (Axial vector) |
23 | $F_x = 50\text{ N}, F_y = 50\sqrt{3}\text{ N}$ |
45 | $\hat{i} - 4\hat{j} - 5\hat{k}$ |
| 2 | $13$ units |
24 | $8\hat{i} + 6\hat{j}$ |
46 | $\frac{4\hat{i} + 5\hat{j} + 7\hat{k}}{\sqrt{90}}$ |
| 3 | $\frac{3\hat{i} + 6\hat{j} - 2\hat{k}}{7}$ |
25 | $2/7, -3/7, 6/7$ |
47 | $\sqrt{107}$ sq. units |
| 4 | $\pm \sqrt{0.2}$ (or $\pm 0.447$) |
26 | Proof based on $\cos^2\alpha+\cos^2\beta+\dots = 1$ |
48 | $|\vec{A}||\vec{B}|$ or $-|\vec{A}||\vec{B}|$ |
| 5 | $x=2, y=-3, z=1$ |
27 | $Mg\sin\theta$ (parallel), $Mg\cos\theta$ (perp) |
49 | $16\hat{i} - 13\hat{j} - 7\hat{k}$ |
| 6 | $2\hat{i} - \hat{j} + 4\hat{k}$ |
28 | $45^\circ$ ($\cos\gamma = \sqrt{2}/2$) |
50 | $16.5$ sq. units |
| 7 | $3\hat{i} + 3\hat{j} + 3\hat{k}$, Mag = $3\sqrt{3}$ |
29 | $\frac{5}{\sqrt{2}} \left( \frac{\hat{i}+\hat{j}}{\sqrt{2}} \right)$ |
51 | $2(\vec{A} \times \vec{B})$ |
| 8 | Magnitude $0$, arbitrary direction |
30 | $100\text{ N}, 50\sqrt{3}\text{ N}$ |
52 | $p = 2$ |
| 9 | Yes, $\vec{B} = 2\vec{A}$ |
31 | $15\hat{i} + 20\hat{j}$ |
53 | $1$ (since $\hat{i}\cdot\hat{i} + \hat{j}\cdot\hat{j} - \hat{k}\cdot\hat{k} = 1$) |
| 10 | Translation (Force) vs Rotation (Torque) |
32 | Yes ($\cos^2 45^\circ + \cos^2 60^\circ + \cos^2 120^\circ = 1$) |
54 | Proof (Lagrange's Identity) |
| 11 | $12$ units and $5$ units |
33 | $-11$ |
55 | $-1.2\hat{i} - 0.2\hat{j} + 0.4\hat{k}$ |
| 12 | $120^\circ$ |
34 | $\cos^{-1}(0)$ i.e., $90^\circ$ |
- | - |
| 13 | $90^\circ$ |
35 | $m = -1$ |
- | - |
| 14 | $13\text{ N}, \tan^{-1}(12/5)$ with $5\text{N}$ force |
36 | $12\text{ J}$ |
- | - |
| 15 | Proof using Cosine Law |
37 | $A^2 - B^2$. If $0 \implies |\vec{A}| = |\vec{B}|$ |
- | - |
| 16 | $90^\circ$ (Pythagorean triplet) |
38 | $120^\circ$ |
- | - |
| 17 | $10\text{ km}$ West |
39 | $140\text{ W}$ |
- | - |
| 18 | $\theta = 60^\circ$ |
40 | $7/\sqrt{14}$ |
- | - |
| 19 | $\frac{1}{\sqrt{2}}\text{ m/s}^2$ North-West |
41 | Proof (Dot products = 0) |
- | - |
| 20 | (b) True (Polygon inequality) |
42 | $45^\circ$ (since $\tan\theta = 1$) |
- | - |
| 21 | $P\sqrt{3}$ at $30^\circ$ with $2P$ |
43 | Yes (Dot product of two sides = 0) |
- | - |
| 22 | Proof (Square the expression) |
44 | $20\text{ N m}^2\text{/C}$ |
- | - |