Physics M.C.Q (1 Marks)

$\overrightarrow{ F } \cdot \overrightarrow{ r }=6+2+2=10$

$\overrightarrow{ F } \times \overrightarrow{ r }=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 1 & -1 \\ 3 & 2 & -2\end{array}\right|$

$=\hat{ i }(0)-\hat{ j }(-1)+\hat{ k }(1)$

$=\hat{ j }+\hat{ k }$

$|\overrightarrow{ F } \times \overrightarrow{ r }|=\sqrt{2}$

$\begin{array}{l}
or\,\,\sqrt {{A^2} + {B^2} + 2AB\cos \theta } \\
\,\,\,\, = \sqrt {{A^2} + {B^2} - 2AB\cos \theta } \\
 \Rightarrow \,\,4\,AB\cos \theta  = 0\\
\,\,\,\,4\,AB \ne 0,\,\,\therefore \cos \theta  = 0\,or\,\theta  = {90^ \circ }
\end{array}$

Unit vector in the direction of $A$ will be $\hat A = \frac{{3\hat i + 4\hat j}}{5}$

So required vector =$25\left( {\frac{{3\hat i + 4\hat j}}{5}} \right)\,$ $ = 15\hat i + 20\hat j$

$|\overrightarrow A |\, = \sqrt {{{(2)}^2} + {{(4)}^2} + {{( - 5)}^2}} \, = \,\sqrt {45} $

 $\cos \alpha = \frac{2}{{\sqrt {45} }},\,\,\,\,\,\cos \beta = \frac{4}{{\sqrt {45} }},\,\,\,\,\cos \gamma = \frac{{ - 5}}{{\sqrt {45} }}$

So all the force will pass through one point and all forces will be balanced. i.e. their resultant will be zero.

$A=|\vec{A}|=\sqrt{4+9}=\sqrt{13}$

$\cos \beta=\frac{y}{A}=\frac{3}{\sqrt{13}}$

$\cos \beta=\frac{3}{\sqrt{13}}$

$\sin \beta=\frac{x}{A}=\frac{2}{\sqrt{13}}$

$\operatorname{Tan} \beta =\frac{x}{y}=\frac{2}{3}$

$=\tan ^{-1}\left(\frac{2}{3}\right)$

$⇒$ $|A| = \sqrt {{1^2} + {1^2}} = \sqrt 2 $

$\cos \alpha = \frac{{{A_x}}}{{|A|}} = \frac{1}{{\sqrt 2 }} = \cos 45^\circ $ 
$⇒$ $\alpha = 45^\circ $

$AC = \sqrt {{{(AB)}^2} + {{(BC)}^2}} = \sqrt {{{(400)}^2} + {{(300)}^2}} = 500m$

Distance $ = AB + BC = 400 + 300 = 700\,m$

$\overrightarrow R = \overrightarrow A + \overrightarrow B = 4\hat i + 3\hat j + 6\hat k - \hat i + 3\hat j - 8\hat k$

$\overrightarrow R = 3\hat i + 6\hat j - 2\hat k$

$\hat R = \frac{{\overrightarrow R }}{{|\vec R|}} = \frac{{3\hat i + 6\hat j - 2\hat k}}{{\sqrt {{3^2} + {6^2} + {{( - 2)}^2}} }} = \frac{{3\hat i + 6\hat j - 2\hat k}}{7}$

$ = 4\hat i - \hat j - 3\hat i + 2\hat j - \hat k$

$ = \hat i + \hat j - \hat k$

$\hat r = \frac{{\vec r}}{{|r|}} = \frac{{\hat i + \hat j - \hat k}}{{\sqrt {{1^2} + {1^2} + {{( - 1)}^2}} }} = \frac{{\hat i + \hat j - \hat k}}{{\sqrt 3 }}$

= $\frac{{\sqrt {36 + 64 + 100} }}{1}$$ = 10\sqrt 2 \;kg$

$=6\left(\frac{2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}}{3}\right) $

$=4 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}$

$ = \hat j - [(\hat i - 3\hat j + 2\hat k) + (3\hat i + 6\hat j - 7\hat k)]$

$ = - \,4\hat i - \,2\hat j + 5\hat k$

$ = \sqrt {{{10}^2} + {{12}^2} + {{14}^2}} $

$ = \sqrt {100 + 144 + 196} $

$ = \sqrt {440} = 20.97\approx 21 \,m$

at $t = 0$, ${\vec r_1} = 7\hat k$

at $t = 10\sec $, ${\vec r_2} = 300\hat i + 400\hat j + 7\hat k$,

$\overrightarrow {\Delta r} = {\vec r_2} - {\vec r_1} = 300\hat i + 400\hat j$

$|\overrightarrow {\Delta r} |\, = \,|{\vec r_2} - {\vec r_1}|\, = \sqrt {{{(300)}^2} + {{(400)}^2}} = 500m$

$\overrightarrow R = \overrightarrow A + \overrightarrow B = 4\hat i - 3\hat j + 8\hat i + 8\hat j$$ = 12\hat i + 5\hat j$

$\hat R = \frac{{\overrightarrow R }}{{|R|}} = \frac{{12\hat i + 5\hat j}}{{\sqrt {{{(12)}^2} + {{(5)}^2}} }}$$ = \frac{{12\hat i + 5\hat j}}{{13}}$

For $7\, N$ both the vectors should be antiparallel i.e. angle between them should be $180^o$

For $13\, N$ both the vectors should be perpendicular to each other i.e. angle between them should be $90^o$

$|\overrightarrow A + \overrightarrow B |\, = \,\sqrt {{{(10)}^2} + {{(5)}^2}} $ $ = \,5\sqrt 5 $

$\tan \theta = \frac{5}{{10}} = \frac{1}{2}$ $ \Rightarrow $$\theta = {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$

$\Delta \vec v  = \vec v _2 - \vec v _1 = - 20(\hat i + \hat j)$

$|\Delta \vec v |\, = \,20\sqrt 2 $ and direction

$\theta = {\tan ^{ - 1}}(1) = 45^\circ $ i.e. $S - W$

$ = 2\hat i + \hat j - 6\hat j + 2\hat k + 18\hat i - 6\hat k$

=$20\hat i - 5\hat j - 4\hat k$

and ${\overrightarrow P _2} = m\,v\,\sin \theta \,\hat i + \,m\,v\,\cos \theta \,\hat j$

So change in momentum

$\overrightarrow {\Delta P} = {\overrightarrow P _2} - {\overrightarrow P _1} = 2\,m\,v\,\cos \theta \,\hat j,$$|\Delta \overrightarrow P |\, = 2\,m\,v\,\cos \theta $

By substituting, $A = F,\;B = F$ and $R = F$ we get

$\cos \theta = -\frac{1}{2}\therefore \theta = 120^\circ $

$| A + B |=\sqrt{| A |^2+| B |^2+2 AB \cos \theta}$

Now in the problem we've,

$| A + B |=| A |+| B |$

Squaring on both sides,

$\begin{array}{l}| A + B |^2=(| A |+| B |)^2 \\| A |^2+| B |^2+2| A || B | \cos \theta=| A |^2+| B |^2+2| A || B | \\\Rightarrow \cos \theta=1\end{array}$

assuming neither of the vectors are $zero$ vectors.

$|\overrightarrow{\mathrm{C}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}+2 \mathrm{AB} \cos 120^{\circ}}$

$|\overrightarrow{\mathrm{C}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-\mathrm{AB}} \quad\left[\text { Ascos } 120^{\circ}=-\frac{1}{2}\right]$

similarly, $|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|=\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}-2 \mathrm{AB} \cos 120^{\circ}}$

$=\sqrt{A^{2}+B^{2}+A B}$

$|\overrightarrow{\mathrm{A}}-\overrightarrow{\mathrm{B}}|>\mathrm{C}$

The angle between $A$ and $B$ is $\frac{\pi }{2}$

$|\vec A| = \sqrt {{3^2} + {{( - 2)}^2} + {1^2}} = \sqrt {9 + 4 + 1} = \sqrt {14} $

$|\vec B| = \sqrt {{1^2} + {{( - 3)}^2} + {5^2}} = \sqrt {1 + 9 + 25} = \sqrt {35} $

$|\vec A| = \sqrt {{2^2} + {1^2} + {{( - 4)}^2}} = \sqrt {4 + 1 + 16} = \sqrt {21} $

As $B = \sqrt {{A^2} + {C^2}} $ therefore $ABC$ will be right angled triangle.

The value of $C$ lies between $A - B$ and $A + B$

$|\vec C|\; < \;|\vec A|\;\;{\rm{or}}\;\;|\vec C|\; < \;|\vec B|$

$\cos \theta = \frac{Q}{R} = \frac{{12}}{{13}}$

$\theta = {\cos ^{ - 1}}\left( {\frac{{12}}{{13}}} \right)$

 $\tan 90^\circ = \frac{{B\sin \theta }}{{A + B\cos \theta }} \Rightarrow A + B\cos \theta = 0$

$\cos \theta = - \frac{A}{B}$

Hence, from $(i)$ $\frac{{{B^2}}}{4} = {A^2} + {B^2} - 2{A^2} \Rightarrow A = \sqrt 3 \frac{B}{2}$

$⇒ \cos \theta = - \frac{A}{B} = - \frac{{\sqrt 3 }}{2}$

$⇒ \theta = 150^\circ $

Required vector $\vec R$ = $ - 2\hat i + \hat j - \hat k$

$⇒$ $P + Q\cos \theta = 0$

$\cos \theta = \frac{{ - P}}{Q}$ 

$⇒$ $\theta = {\cos ^{ - 1}}\left( {\frac{{ - P}}{Q}} \right)$

By solving we get $P = 2\;{\rm{and }}Q = 1$

$\therefore \frac{P}{Q} = 2$ 

$⇒$ $P = 2Q$

and $\quad F 2=2 F$

Resultant force $=F_{\text {net }}=2 F$

$F_{net}$ $=\sqrt{F_{1}^{2}+F_{2}^{2}+2 F_{1} F_{2}} \cos \theta$

$2 F=\sqrt{F^{2}+2 F^{2}+2 \times F \times(2 F) \cos \theta}$

$4 F^{2}=5 F^{2}+4 F^{2} \cos \theta$

$4 \cos \theta=-1$

$\therefore \cos \theta=\frac{-1}{4}$

and $B=C-A$

Since, $C \perp A$, therefore

$B^{2}=C^{2}+A^{2}$

Also, $|C|=|A|,$ therefore

$B^{2}=2 A^{2} \Rightarrow B=\sqrt{2} A$

$\mathrm{Now}, A^{2}+B^{2}+2 A B \cos \theta=C^{2}=A^{2}$

$\therefore \cos \theta=\frac{1}{\sqrt{2}}$

This gives $\theta=\frac{3 \pi}{4}$ rad

$= 2 \times v \times \sin 90^\circ $

$ = 2 \times 100 = 200\;km/hr$

New position $=7 \hat i+0 \hat k $

Displacement $=$ (New position) $-$ (old position) $=(7 \hat i+0 \hat j)-(25 \hat i-6 \hat j)=-18 \hat i+6 \hat j $

Required displacement $=-18 \hat i + 6 hat j $

$=\frac{60}{\sqrt{2}}+20=(30 \sqrt{2}+20) k m$

$y=\sqrt{2}$

$y=75+60 \sin 45^{\circ}$

$=75+\frac{60}{\sqrt{2}}=(75+30 \sqrt{2}) k m$

$S=\sqrt{X^{2}+Y^{2}}=\sqrt{(20+30 \sqrt{2})^{2}+(75+30 \sqrt{2})^{2}}$

$S=\sqrt{(62.4)^{2}+(117.4)^{2}}$

$=132 \mathrm{km}$

$\theta = \frac{{360}}{N}$$ = \frac{{360}}{3} = 120^\circ $

If these three vectors are represented by three sides of triangle then they form equilateral triangle

If $A = B = P$ and $\theta = 120^\circ $ then $R = P$

magnitude of $\overrightarrow R = \,|\overrightarrow R |\, = \sqrt {49 + 225} $$ = \sqrt {274} $

$\overrightarrow A $ and $\overrightarrow B $ are parallel to each other

$\frac{{{a_1}}}{{{b_1}}} = \frac{{{a_2}}}{{{b_2}}} = \frac{{{a_3}}}{{{b_3}}}$ i.e. $\frac{2}{{ - 4}} = \frac{3}{{ - 6}} = \frac{{ - 1}}{\lambda }$$ \Rightarrow \,\,\lambda = 2$.

$⇒$  $|{\vec V_{{\rm{net}}}}|\; = \;|{\vec V'_{{\rm{net}}}}|$

So ${V_1}$ and ${V_2}$ will be mutually perpendicular.

$\overrightarrow {BC} = - 30\sqrt 2 \;cos{45^o}\overrightarrow i $$ - 30\sqrt 2 \;\sin {\rm{4}}{{\rm{5}}^{\rm{o}}}\overrightarrow j $ $ = - 30\;\overrightarrow i - 30\;\overrightarrow j $

Net displacement, $\overrightarrow {OC} = \overrightarrow {OA} + \overrightarrow {AB} + \overrightarrow {BC} $$ = - 10\;\overrightarrow i + 0\;\overrightarrow j $
$|\overrightarrow {OC} |\; = 10\;m.$

$⇒$ $\theta = 120^\circ $

$⇒$ ${\left( {\frac{F}{3}} \right)^2}$$ = {F^2} + {F^2} + 2{F^2}\cos \theta $

 $⇒$ $\cos \theta = \left( { - \frac{{17}}{{18}}} \right)$

${(2R)^2} = {(6P)^2} + {(2P)^2} + 2 \times 6P \times 2P \times \cos \theta $…(ii)

by solving (i) and (ii), $\cos \theta = - 1/2$

$⇒$ $\theta = 120^\circ $