Units and Dimensions

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. (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. (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
$\vec {P} + \vec {C} = \vec {S}$
$\vec {P} + b (\vec {Q} - \vec {P}) = \vec {S}$
$\Rightarrow \vec {S} = (1 - b) \vec {P} + b \vec {Q}$

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. (C)
(P) by $E = \frac{f}{2} kT$
$\Rightarrow [ K ] = \left[ M L ^ {2} T ^ {- 2} K ^ {- 1} \right]$
(Q) by $F = \eta A\frac{\Delta v}{\Delta x}$
$\Rightarrow [ \eta ] = [ M L ^ {- 1} T ^ {- 1} ]$
(R) by $E = h\nu$
$\Rightarrow [ h ] = \left[ M L ^ {2} T ^ {- 1} \right]$
(S) by $Q = \frac{kA(\Delta\theta)t}{\ell}\Rightarrow [k] = [MLT^{-3}K^{-1}]$

Ans. (B) $10\ m/s^{2}=5\frac{L}{T^{2}}$ $\frac{1}{2}\times(500kg)(400m/s)^{2}=2000\ ML^{2}T^{-2}$ $(500\ kg)\ (400\ m/s)=100\ MLT^{-1}$ Solve above equation to get L, T and M.

Ans. (C)$10\ m/s^{2}=5\frac{L}{T^{2}}$ $\frac{1}{2}\times(500kg)(400m/s)^{2}=2000\ ML^{2}T^{-2}$ $(500\ kg)\ (400\ m/s)=100\ MLT^{-1}$ Solve above equation to get L, T and M.

Ans. (A) $10\ m/s^{2}=5\frac{L}{T^{2}}$ $\frac{1}{2}\times(500kg)(400m/s)^{2}=2000\ ML^{2}T^{-2}$ $(500\ kg)\ (400\ m/s)=100\ MLT^{-1}$ Solve above equation to get L, T and M.

Ans. (4)
[Surface tension] = [force/length] = $M^{1}L^{0}T^{-2}$ Suppose [surface tension] = $E^{a} V^{b} T^{c}$ ∴ $M^{1}L^{0}T^{-2} = [M^{1}L^{2}T^{-2}]^{a} [L^{1}T^{-1}]^{b} [T]^{c}$ Matching dimensions of $M \Rightarrow a = 1$ Matching dimensions of $L \Rightarrow 2a + b = 0 \Rightarrow b = -2$ Matching dimensions of $T \Rightarrow -2a - b + c = -2 \Rightarrow c = -2$ ∴ [Surface tension] = $EV^{-2}T^{-2}$

Ans. (3)
We know that, a dimensionally incorrect equation is always incorrect.

Ans. (4) $\sin 1^{\circ} = 1\times \frac{\pi}{180}$ $\cos 2^{\circ} = 1$

Ans. (4)
$F = \frac {G M ^ {2}}{R ^ {2}}$
$\Rightarrow G = \left[ M ^ {- 1} L ^ {3} T ^ {- 2} \right]$
$E = h \nu$
$\Rightarrow h = \left[ M L ^ {2} T ^ {- 1} \right]$
$C = [ L T ^ {- 1} ]$
$t \propto G ^ {x} h ^ {y} C ^ {z}$
$[ T ] = [ M ^ {- 1} L ^ {3} T ^ {- 2} ] ^ {x} [ M L ^ {2} T ^ {- 1} ] ^ {y} [ L T ^ {- 1} ] ^ {z}$
$\left[ M ^ {0} L ^ {0} T ^ {1} \right] = \left[ M ^ {- x + y} L ^ {3 x + 2 y + z} T ^ {- 2 x - y - z} \right]$
On comparing the powers of $M, L, T$
$- x + y = 0 \quad \Rightarrow x = y$
$3 x + 2 y + z = 0 \Rightarrow 5 x + z = 0\tag{... (i}$
$- 2 x - y - z = 1 \quad \Rightarrow 3 x + z = - 1\tag{... (ii}$
On solving (i) & (ii)
$x = y = \frac {1}{2}, z = - \frac {5}{2}$
$t \propto \sqrt {\frac {G h}{C ^ {5}}}$

Ans. (4) $n_{1}u_{1}=n_{2}u_{2}$ $\frac{128kg}{m^{3}}=n_{2}\frac{50g}{(25cm)^{3}}$ $\frac{128\times1000g}{(100cm)^{3}}=n_{2}\frac{50g}{(25cm)^{3}}$ $n_{2}=40$

Ans. (2) $\frac{F}{A}=Y\frac{\Delta\ell}{\ell}$ $[Y]=\frac{F}{A}$ Now from dimension $[a]=LT^{-2}$ $[V]=LT^{-1}$ $[V]^{2}=[L][a]$ $L^{2}=\frac{V^{4}}{a^{2}}$ $[Y]=\frac{[F]}{[A]}=F^{1}V^{-4}a^{2}$

Ans. (2) $F=\alpha\beta e^{\left(\frac{-x^{2}}{\alpha KT}\right)}$ $\left[\frac{x^{2}}{\alpha KT}\right]=M^{0}L^{0}T^{0}$ $\frac{L^{2}}{[\alpha]ML^{2}T^{-2}}=M^{0}L^{0}T^{0}$ $\Rightarrow \quad [\alpha]=M^{-1}T^{2}$ $[F]=[α][β]$ $MLT^{-2}=M^{-1}T^{2}[β]$ $\Rightarrow \quad [\beta]=M^{2}LT^{-4}$

Ans. (4) $L=\mu F^{a}A^{b}T^{c}$ $\Rightarrow \quad [L]=[MLT^{-2}]^{a}[LT^{-2}]^{b}[T]^{c}$ $\Rightarrow \quad a=0,b=1,c=2$

Ans. (1) $n_{1}u_{1}=n_{2}u_{2}$ $n_{1}(10gm)(10cm)(0.1sec.)^{-2}=1(kg)(m)(sec.)^{-2}$ $\Rightarrow \quad n_{1}=0.1N$

Ans. (3) Let $x = \mu y^{\mathrm{a}}z^{\mathrm{b}}$ $\Rightarrow [ML^{2}T^{-5}] = [MT^{-1}]^{a}[LT^{-2}]^{b}$

Ans. (3)
If an equation is dimensionally correct then it may or may not be correct.
But if an equation is dimensionally incorrect then it will be incorrect for sure.
$\mathbf{E}\mathbf{x}:\quad S = u t + \frac{1}{2} a t^{2}$
$[ S ] = [ u t ] = [ a t ^ {2} ] = [ L ]$
This equation is dimensionally correct, but by this method we can't say that the exact equation will be $S = ut + \frac{1}{2} at^2$ .

Ans. (1)

Component = 18 cos 60° = 9 km/hr.

Ans. (4)
$\hat {A} = \frac {\overrightarrow {A}}{| \overrightarrow {A} |}$