Units and Dimensions

Ans. (AB)
Given $L=x^{\alpha}$ ...(1) $LT^{-1}=x^{\beta}$ ...(2) $LT^{-2}=x^{p}$ ...(3) $MLT^{-1}=x^{q}$ ...(4) $MLT^{-2}=x^{r}$ ...(5)
⇒ $\frac{(1)}{(2)}T=x^{\alpha-\beta}$ From (3) $\frac{x^{\alpha}}{x^{2(\alpha-\beta)}}=x^{p}$ ⇒ $\alpha+p=2\beta\Rightarrow(A)$ From (4) $M=x^{q-\beta}$ From (5)
⇒ $x^{q}=x^{r}x^{\alpha-\beta}$ ⇒ $\alpha+r-q=\beta$ ...(6)
Replacing value of 'α' in equation (6) from (A) $2\beta-p+r-q=\beta$ ⇒ $p+q-r=\beta\Rightarrow(B)$
Replacing value of ' $\beta'$ in equation (6) from (A)
$2 \alpha + 2 r - 2 q = \alpha + p$
$\alpha = p + 2 q - 2 r$
Answer is (AB).

Ans. (ABD)
Mass = $M^{0}L^{0}T^{0}$ MVr = $M^{0}L^{0}T^{0}$ $M^{0}\frac{L^{1}}{T^{1}}.L^{1}=M^{0}L^{0}T^{0}$ $L^{2}=T^{1}$ ...(1)
Force = $M^{1}L^{1}T^{-2}$ (in SI)
= $M^{0}L^{1}L^{-4}$ (In new system from equation (1))
= $L^{-3}$ Energy = $M^{1}L^{2}T^{-2}$ (In SI)
= $M^{0}L^{2}L^{-4}$ (In new system from equation (1))
= $L^{-2}$ Power = $\frac{Energy}{Time}$ = $M^{1}L^{2}T^{-3}$ (in SI)
= $M^{0}L^{2}L^{-6}$ (In new system from equation (1))
= $L^{-4}$ Linear momentum = $M^{1}L^{1}T^{-1}$ (in SI)
= $M^{0}L^{1}L^{-2}$ (In new system from equation (1))
= $L^{-1}$

Ans. (AC)
Using $C = \frac{1}{\sqrt{\mu_0 \in_0}}$ and $R = \sqrt{\frac{\mu_0}{\in_0}}$ we can check the correctness.
(A) $\mu_0I^2 = \varepsilon_0V^2$
$\frac {\mu_ {0}}{\epsilon_ {0}} = \frac {V ^ {2}}{I ^ {2}} = R ^ {2}$
$\therefore \quad R ^ {2} = R ^ {2} \text { correct }$
(B) $\varepsilon_0I = \mu_0V$
$\frac {\epsilon_ {0}}{\mu_ {0}} = \frac {V}{I}$
$\frac {1}{R ^ {2}} = R \mathrm{notcorrect}$
(C) $I = \varepsilon_0cV$
$\frac {I}{V} = \in_ {0} c = \frac {\in_ {0}}{\sqrt {\mu_ {0} \in_ {0}}}$
$\frac {1}{R} = \sqrt {\frac {\epsilon_ {0}}{\mu_ {0}}} = \frac {1}{R} \text {correct}$
(D) $\mu_0cI = \varepsilon_0V$
$\frac {\mu_ {0} C}{\epsilon_ {0}} = \frac {V}{I}$
$\frac {\mu_ {0}}{\epsilon_ {0}} \frac {1}{\sqrt {\mu_ {0} \epsilon_ {0}}} = R$
$\therefore \quad \frac {1}{\in_ {0}} \sqrt {\frac {\mu_ {0}}{\in_ {0}}} = R \Rightarrow \frac {R}{\in_ {0}} = R \mathrm{incorrect}$

Ans. (BC)
$2 \big | \hat {\imath} + x \hat {j} + 3 \hat {k} \big | = \big | 4 \hat {\imath} (4 x - 2) \hat {j} + 2 \hat {k} \big |$
$4 (1 + x ^ {2} + 9) = (1 6 + (4 x - 2) ^ {2} + 4)$
$1 2 x ^ {2} - 1 6 x - 1 6 = 0$
$x = \frac {4 \pm \sqrt {1 6 + 4 8}}{6}$
$x = 2, - \frac {2}{3}$

Ans. (BD)
$(\vec{A} + \vec{B} - \vec{C} + \vec{D})$ can not be zero if sum of magnitude of three smallest vectors is less than the magnitude of bigger vector.

Ans. (CD)
$\vec{A}+\vec{B}=5\hat{i}$
$\vec{A} + \vec{B} = 5\hat{i}$
$\vec{A} + \vec{B} = \hat{i} + 4\hat{j}$

Ans. (BD)
$\vec {S} = [ \vec {E} \times \vec {B} ] \frac {1}{\mu_ {0}}$
$\vec{S}$ is poynting vector which denotes flow of energy per unit area per unit time.
$S = \frac {\mathrm{watt}}{m ^ {2}}$
Hence B, D are correct.