Units and Dimensions

Ans. (2) $[ML^{-3}] = [MLT^{-2}]^a [LT^{-1}]^b [T]^c$ $= [M^a L^{a + b} T^{-2a - b + c}]$ $a = 1, a + b = -3,$ $\Rightarrow \quad b = -4,$ also $-2a - b + c = 0$ $c = -2$

Ans. (1) $v = \lambda^{a} g^{b} \rho^{c}$ Using dimension formula $\Rightarrow [M^{0}L^{1}T^{-1}] = [L^{1}]^{a}[L^{1}T^{-2}]^{b}[M^{1}L^{-3}]^{c}$ $\Rightarrow [M^{0}L^{1}T^{-1}] = [M^{c} L^{a+b-3c} T^{-2b}]$ $\therefore c = 0, a + b - 3c = 1, -2b = -1 \Rightarrow b = \frac{1}{2}$ Now $a + b - 3c = 1$ $\Rightarrow a + \frac{1}{2} - 0 = 1$ $\Rightarrow a = \frac{1}{2}$ $\therefore a = \frac{1}{2}, b = \frac{1}{2}, c = 0$

Ans. (4)
$\left| \left\langle \vec {v} \right\rangle \right| = \frac {\left| \vec {r} _ {f} - \vec {r} _ {i} \right|}{\Delta t} = \frac {2 R \cos \left[ \frac {\pi - \theta}{2} \right]}{\frac {2 \pi R}{3 v}} = 3 \cos 3 0 ^ {0}$
$. 5 \sqrt {3} m / s$
Correct option is (4)

Ans. (1)
Let instantaneous velocity be v. time,
$t = \frac {\text { Arc Length }}{v} = \frac {2 \pi \frac {R}{4}}{v} = \frac {\pi R}{2 v}$
Average velocity,
$\langle v \rangle = \frac {A B}{t} = \frac {R \sqrt {2} (2 v)}{\pi R} = \frac {2 \sqrt {2} v}{\pi}$

Ans. (A) $Y = c^{\alpha} h^{\beta} G^{\gamma}$ $ML^{-1} T^{-2} = (LT^{-1})^{\alpha} (ML^{2} T^{-1})^{\beta} (M^{-1} L^{3} T^{-2})^{\gamma}$ $1 = \beta - \gamma$ ...(1) $-1 = \alpha + 2\beta + 3\gamma$ ...(2) $-2 = -\alpha - \beta - 2\gamma$ ...(3) $-3 = \beta + \gamma$ $1 = \beta - \gamma$ $-2 = 2\beta$ $\Rightarrow \beta = -1, \gamma = -2$ $-1 = \alpha - 2 - 6$ $\therefore \alpha = 7$

Ans. (4)
$S = \frac {Q}{m \Delta T} = \frac {J}{K g ^ {\circ} C}$
$L = \frac {Q}{m} = \frac {J}{K g}$

Ans. (4) $m \propto t^{a}v^{b}\ell^{c}$ $m \propto [T]^{a}[LT^{-1}]^{b}[ML^{2}T^{-1}]^{c}$ $M^{1}L^{0}T^{0} = M^{c}L^{b + 2c}T^{a - b - c}$ Comparing powers $c = 1, b = -2, a = -1$ $m \propto t^{-1}v^{-2}\ell^{1}$

Ans. (2) $[A] = [MLT^{-2}]$ $[B] = [L^{-1}]$ $[D] = [T^{-1}]$ $\left[\frac{AD}{B}\right] = \frac{[MLT^{-2}][T^{-1}]}{[L^{-1}]}$ $\left[\frac{AD}{B}\right] = [ML^2 T^{-3}]$

Ans. (4)
$E = M L ^ {2} T ^ {- 2}$
$L = M L ^ {2} T ^ {- 1}$
m = M
$G = M ^ {- 1} L ^ {+ 3} T ^ {- 2}$
$P = \frac {E L ^ {2}}{M ^ {5} G ^ {2}}$
$[ P ] = \frac {(M L ^ {2} T ^ {- 2}) (M ^ {2} L ^ {4} T ^ {- 2})}{M ^ {5} (M ^ {- 2} L ^ {6} T ^ {- 4})} = M ^ {0} L ^ {0} T ^ {0}$

Ans. (1)
Density = $[F^a L^b T^c]$
$[ML^{-3}] = [M^a L^a T^{-2a} L^b T^c]$
$[M^{1}L^{-3}] = [M^a L^{a+b} T^{-2a+c}]$
a = 1 ; a + b = -3 ; -2a + c = 0
1 + b = -3 ; c = 2a
b = -4 ; c = 2
So, density = $[F^{1}L^{-4}T^{2}]$.

Ans. (2)
$[M] = [F]^a [T]^b [V]^c$
$[M^1] = [M^{1}L^{1}T^{-2}]^a [T^1]^b [L^{1}T^{-1}]^c$
a = 1, b = 1, c = -1 $\therefore$ [M] = [FTV^{-1}]

Ans. (2)
We know, $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF} + \overrightarrow{OG} + \overrightarrow{OH} = \vec{0}$ By triangle law of vector addition, we can write $\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB}; \overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC}$ $\overrightarrow{AD} = \overrightarrow{AO} + \overrightarrow{OD}; \overrightarrow{AE} = \overrightarrow{AO} + \overrightarrow{OE}$ $\overrightarrow{AF} = \overrightarrow{AO} + \overrightarrow{OF}; \overrightarrow{AG} = \overrightarrow{AO} + \overrightarrow{OG}$ $\overrightarrow{AH} = \overrightarrow{AO} + \overrightarrow{OH}$ Now $\overrightarrow{AB} + \overrightarrow{AC} + \overrightarrow{AD} + \overrightarrow{AE} + \overrightarrow{AF} + \overrightarrow{AG} + \overrightarrow{AH}$ $= (7\overrightarrow{AO}) + \overrightarrow{OB} + \overrightarrow{OC} + \overrightarrow{OD} + \overrightarrow{OE} + \overrightarrow{OF} + \overrightarrow{OG} + \overrightarrow{OH}$ $= (7\overrightarrow{AO}) + (\vec{0} - \overrightarrow{OA})$ $= (7\overrightarrow{AO}) + \overrightarrow{AO}$ $= 8\overrightarrow{AO} = 8(2\hat{i} + 3\hat{j} - 4\hat{k})$ $= 16\hat{i} + 24\hat{j} - 32\hat{k}$

Ans. (4)
$\vec {A}. \vec {B} = | \vec {A} \times \vec {B} |$
$A B \cos \theta = A B \sin \theta \Rightarrow \theta = 4 5 ^ {\circ}$
$\begin{array}{c} \left| \vec {A} - \vec {B} \right| = \sqrt {A ^ {2} + B ^ {2} - 2 A B \cos 4 5 ^ {\circ}} \\ = \sqrt {A ^ {2} + B ^ {2} - \sqrt {2} A B} \end{array}$

Ans. (4)
$(A \cos \theta) \hat {B} = A \left(\frac {\overrightarrow {A . B}}{A B}\right) \hat {B} = \frac {\overrightarrow {A . B}}{B} \hat {B}$
$= \frac {2}{\sqrt {2}} \biggl (\frac {\hat {\imath} + \hat {j}}{\sqrt {2}} \biggr) = \hat {\imath} + \hat {j}$

Ans. (1)
$\overrightarrow {A D} = 2 \overrightarrow {A O}\tag{... (i}$
A: $\overrightarrow{AB} +\overrightarrow{AC} +\overrightarrow{AD} = (\overrightarrow{AO} +\overrightarrow{OB}) + (\overrightarrow{AO} +\overrightarrow{OC}) + 2\overrightarrow{AO}$
$\overrightarrow {A B} + \overrightarrow {A C} + \overrightarrow {A D} = 4 \overrightarrow {A O} + \overrightarrow {O B} + \overrightarrow {O C}\tag{... (ii}$
R: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD} = \overrightarrow{AD}$
$\overrightarrow {A B} + \overrightarrow {B C} + \overrightarrow {C D} + \overrightarrow {A D} = 2 \overrightarrow {A D} = 4 \overrightarrow {A O}\tag{... (iii}$
So, A is correct but R is not correct.

Ans. (1)
Let magnitude be equal to $\lambda$ .
$\overrightarrow {O A} = \lambda [ \cos 3 0 ^ {\circ} \hat {\imath} + \sin 3 0 ^ {\circ} \hat {j} ] = \lambda \left[ \frac {\sqrt {3}}{2} \hat {\imath} + \frac {1}{2} \hat {j} \right]$
$\overrightarrow {O B} = \lambda [ \cos 6 0 ^ {\circ} \hat {\imath} - \sin 6 0 ^ {\circ} \hat {j} ] = \lambda \left[ \frac {1}{2} \hat {\imath} - \frac {\sqrt {3}}{2} \hat {j} \right]$
$\overrightarrow {O C} = \lambda [ \cos 4 5 ^ {\circ} (- \hat {\imath}) + \sin 4 5 ^ {\circ} \hat {j} ] = \lambda \left[ - \frac {1}{\sqrt {2}} \hat {\imath} + \frac {1}{\sqrt {2}} \hat {j} \right]$

$\therefore \overrightarrow {O A} + \overrightarrow {O B} - \overrightarrow {O C} = \lambda \left[ \left(\frac {\sqrt {3} + 1}{2} + \frac {1}{\sqrt {2}}\right) \hat {\imath} + \left(\frac {1 - \sqrt {3}}{2} - \frac {1}{\sqrt {2}}\right) \hat {\jmath} \right]$
∴ Angle with x-axis
$\tan^ {- 1} \left[ \frac {\frac {1}{2} - \frac {\sqrt {3}}{2} - \frac {1}{\sqrt {2}}}{\frac {\sqrt {3}}{2} + \frac {1}{2} + \frac {1}{\sqrt {2}}} \right] = \tan^ {- 1} \left[ \frac {1 - \sqrt {3} - \sqrt {2}}{\sqrt {3} + 1 + \sqrt {2}} \right]$

Ans. (3)
Angle between $\vec{A}$ and $\vec{B}$ , $\theta = 60^{\circ}$
Angle between $\vec{A}$ and $\vec{A} -\vec{B}$
$\tan \alpha = \frac{B \sin \theta}{A - B \cos \theta} = \frac{B \sqrt{\frac{3}{2}}}{A - B \times \frac{1}{2}}$


$\tan \alpha = \frac{\sqrt{3} B}{2 A - B}$

Ans. (3)
$[ h ] = M ^ {1} L ^ {2} T ^ {- 1}$
$[ C ] = L ^ {1} T ^ {- 1}$
$[ G ] = M ^ {- 1} L ^ {3} T ^ {- 2}$
$[ f ] = \sqrt {\frac {M ^ {1} L ^ {2} T ^ {- 1} \times L ^ {5} T ^ {- 5}}{M ^ {- 1} L ^ {3} T ^ {- 2}}} = M ^ {1} L ^ {2} T ^ {- 2}$

Ans. (4)
$S = \frac {P}{A} = \frac {M L ^ {2} T ^ {- 3}}{L ^ {2}} = M T ^ {- 3}$

Ans. (3)
$\left| \vec {A} \right| = \left| \vec {B} \right| = x\tag{... (i}$
$\left| \vec {A} + \vec {B} \right| = n \left| \vec {A} - \vec {B} \right|$
$A ^ {2} + B ^ {2} + 2 A B \cos \theta = n ^ {2} (A ^ {2} + B ^ {2} - 2 A B \cos \theta)\tag{... (ii}$
$2 x ^ {2} (1 + \cos \theta) = n ^ {2} \times 2 x ^ {2} (1 - \cos \theta)$
$1 + \cos \theta = n ^ {2} - n ^ {2} \cos^ {2} \theta$
$(1 + n ^ {2}) \cos \theta = n ^ {2} - 1$
So, $\cos\theta=\frac{n^{2}-1}{n^{2}+1}$
$\theta = \cos^ {- 1} \left[ \frac {n ^ {2} - 1}{n ^ {2} + 1} \right]$

Ans. (2)
$\vec {r} _ {G} = \frac {a}{2} \hat {\imath} + \frac {a}{2} \hat {k}$
$\vec {r} _ {H} = \frac {a}{2} \hat {j} + \frac {a}{2} \hat {k}$
$\vec {r} _ {H} - \vec {r} _ {G} = \frac {a}{2} (\hat {j} - \hat {\imath})$

Ans. (3)
$\left| \vec {A} _ {1} \right| = 3; \left| \vec {A} _ {2} \right| = 5; \left| \vec {A} _ {1} + \vec {A} _ {2} \right| = 5$
$\left| \vec {A} _ {1} + \vec {A} _ {2} \right| = \sqrt {\left| \vec {A} _ {1} \right| ^ {2} + \left| \vec {A} _ {2} \right| ^ {2} + 2 \left| \vec {A} _ {1} \right| \left| \vec {A} _ {2} \right| \cos \theta}$
$5 = \sqrt {9 + 2 5 + 2 \times 3 \times 5 \cos \theta}$
$\begin{array}{r l} & {\cos \theta = - \frac {9}{2 \times 3 \times 5} = - \frac {3}{1 0}} \\ & {\Rightarrow \quad (2 \vec {A} _ {1} + 3 \vec {A} _ {2}). (3 \vec {A} _ {1} - 2 \vec {A} _ {2}) = 6 | \vec {A} _ {1} | ^ {2} + 9 \vec {A} _ {1}. \vec {A} _ {2} - 4 \vec {A} _ {1} \vec {A} _ {2} - 6 | \vec {A} _ {2} | ^ {2}} \\ & {\qquad = 5 4 + 5 \times 3 \times 5 (- \frac {3}{1 0}) - 6 \times 2 5} \\ & {\qquad = 5 4 - 1 5 0 - \frac {4 5}{2} = - 1 1 8. 5} \end{array}$

Ans. (C)
We have $\frac{E}{C} = B$ $\therefore [B] = \frac{[E]}{[C]} = [E]L^{-1}T^{1}$ $\Rightarrow [E] = [B][L][T^{-1}]$

Ans. (D)
We have,
$\begin{array}{r l} & C = \frac {1}{\sqrt {\mu_ {0} \epsilon_ {0}}} \\ \therefore & [ C ^ {2} ] = \left[ \frac {1}{\mu_ {0} \epsilon_ {0}} \right] \\ \Rightarrow & L ^ {2} T ^ {- 2} = \frac {1}{[ \mu_ {0} ] [ \epsilon_ {0} ]} \\ \Rightarrow & [ \mu_ {0} ] = [ \epsilon_ {0} ] ^ {- 1} [ L ] ^ {- 2} [ T ] ^ {2} \end{array}$

Ans. (C)
Let vector from point P to point S be $\vec{C}$
$\Rightarrow \vec {C} = b | \vec {R} | \hat {R} = b | \vec {R} | \left(\frac {\vec {R}}{| \vec {R} |}\right) = b \vec {R} = b (\vec {Q} - \vec {P})$
From triangle rule of vector addition
$\vec {P} + \vec {C} = \vec {S}$
$\vec {P} + b (\vec {Q} - \vec {P}) = \vec {S}$
$\Rightarrow \vec {S} = (1 - b) \vec {P} + b \vec {Q}$