Dimensional Analysis, Uses and Limitations Questions in English

(A) Given: $v_2 = \frac{n}{m^2} v_1$ and $a_2 = \frac{a_1}{mn}$.
We know that $a = \frac{v}{t}$,so $\frac{a_2}{a_1} = \frac{v_2}{v_1} \times \frac{t_1}{t_2}$.
Substituting the given ratios: $\frac{1}{mn} = \frac{n}{m^2} \times \frac{t_1}{t_2}$.
Rearranging for $t_2$: $t_2 = \frac{n}{m^2} \times mn \times t_1 = \frac{n^2}{m} t_1$.
Now,we know $v = \frac{S}{t}$,so $\frac{v_2}{v_1} = \frac{S_2}{S_1} \times \frac{t_1}{t_2}$.
Substituting the known values: $\frac{n}{m^2} = \frac{S_2}{S_1} \times \frac{t_1}{(n^2/m)t_1} = \frac{S_2}{S_1} \times \frac{m}{n^2}$.
Solving for $S_2$: $S_2 = S_1 \times \frac{n}{m^2} \times \frac{n^2}{m} = S_1 \times \frac{n^3}{m^3} = \left(\frac{n}{m}\right)^3 S_1$.

(D) Efficiency $\eta$ is a dimensionless quantity. The argument of the logarithmic function must also be dimensionless,so $\frac{\beta x}{kT} = 1$ (dimensionless).
Since $\eta$ is dimensionless,$[\eta] = [M^0 L^0 T^0]$.
From $\frac{\beta x}{kT} = 1$,we get $\beta = \frac{kT}{x}$.
Given $k = [M L^2 T^{-2} K^{-1}]$ and $T = [K]$,we have $\beta = \frac{[M L^2 T^{-2} K^{-1}][K]}{[L]} = [M L T^{-2}]$.
This is the dimension of force,so option $A$ is correct.
Since $\eta = \frac{\alpha \beta}{\sin \theta} \cdot \text{constant}$,and $\eta$ and $\sin \theta$ are dimensionless,$[\alpha \beta] = [1]$,which implies $[\alpha] = [\beta^{-1}] = [M^{-1} L^{-1} T^2]$.
Option $D$ states $[\alpha] = [\beta]$,which is incorrect.
For option $B$,$[\alpha^{-1} x] = [\beta x] = [M L T^{-2}][L] = [M L^2 T^{-2}]$,which is the dimension of energy. Thus,$B$ is correct.
For option $C$,$[\eta^{-1} \sin \theta] = [1] = [\alpha \beta]$,which is correct.

(D) Dimensional analysis requires that both sides of an equation have the same dimensions and that the argument of any exponential function must be dimensionless.
$(A) E = 2 E_0 \frac{L}{L_0}$
$LHS$: $[M L^2 T^{-2}]$
$RHS$: $[M L^2 T^{-2}] \times \frac{[L]}{[L]} = [M L^2 T^{-2}]$
Status: Dimensionally correct.
$(B) E = E_0 e^{-\frac{2 L}{L_0}}$
Exponent: $\frac{[L]}{[L]} = [1]$ (Dimensionless). Correct.
$LHS$: $[M L^2 T^{-2}]$
$RHS$: $[M L^2 T^{-2}] \times [1] = [M L^2 T^{-2}]$
Status: Dimensionally correct.
$(C) E = 2 L e^{-\frac{L}{E_0}}$
Exponent: $\frac{[L]}{[M L^2 T^{-2}]} = [M^{-1} L^{-1} T^2]$. Since the exponent is not dimensionless,the expression is invalid.
$LHS$: $[M L^2 T^{-2}]$
Pre-factor: $[L]$
Status: Dimensionally incorrect.
$(D) E = 2 \left( \frac{E_0}{L_0} \right) e^{-\frac{L}{L_0}}$
Exponent: $\frac{[L]}{[L]} = [1]$. Correct.
$LHS$: $[M L^2 T^{-2}]$
$RHS$: $\frac{[M L^2 T^{-2}]}{[L]} = [M L T^{-2}]$
Status: Dimensionally incorrect.
Therefore,equations $(C)$ and $(D)$ are incorrect.

(B) Let the density $\rho$ be expressed as $\rho = k C^a G^b h^c$.
Dimensional formulas are:
$[\rho] = M L^{-3}$
$[C] = L T^{-1}$
$[G] = M^{-1} L^3 T^{-2}$
$[h] = M L^2 T^{-1}$
Substituting these into the equation:
$M^1 L^{-3} T^0 = (L T^{-1})^a (M^{-1} L^3 T^{-2})^b (M L^2 T^{-1})^c$
$M^1 L^{-3} T^0 = M^{-b+c} L^{a+3b+2c} T^{-a-2b-c}$
Comparing the exponents of $M, L,$ and $T$:
$1$) $-b + c = 1 \implies c = 1 + b$
$2$) $a + 3b + 2c = -3$
$3$) $-a - 2b - c = 0 \implies a = -2b - c$
Substitute $c = 1 + b$ into $(3)$: $a = -2b - (1 + b) = -3b - 1$
Substitute $a$ and $c$ into $(2)$: $(-3b - 1) + 3b + 2(1 + b) = -3$
$-3b - 1 + 3b + 2 + 2b = -3$
$2b + 1 = -3 \implies 2b = -4 \implies b = -2$
Now find $c$: $c = 1 + (-2) = -1$
Now find $a$: $a = -3(-2) - 1 = 6 - 1 = 5$
Thus,the dimension of density is $[\rho] = C^5 G^{-2} h^{-1}$.

(B) Given,the time period of oscillation $T$ is $T \propto p^\alpha S^\beta E^\gamma$ or $T = k p^\alpha S^\beta E^\gamma$.
Substituting the dimensions of $T, p, S$,and $E$:
$[M^0 L^0 T^1] = [ML^{-1} T^{-2}]^\alpha [ML^{-3}]^\beta [ML^2 T^{-2}]^\gamma$
$[M^0 L^0 T^1] = [M^{\alpha+\beta+\gamma} L^{-\alpha-3\beta+2\gamma} T^{-2\alpha-2\gamma}]$
Equating the powers of $M, L$,and $T$ on both sides,we get:
$1) \alpha + \beta + \gamma = 0$
$2) -\alpha - 3\beta + 2\gamma = 0$
$3) -2\alpha - 2\gamma = 1$
From equation $(3)$,$\alpha + \gamma = -\frac{1}{2}$.
Substituting this into equation $(1)$,$-\frac{1}{2} + \beta = 0 \implies \beta = \frac{1}{2}$.
Now,substitute $\beta = \frac{1}{2}$ into equation $(2)$:
$-\alpha - 3(\frac{1}{2}) + 2\gamma = 0 \implies -\alpha + 2\gamma = \frac{3}{2}$.
We have the system:
$i) \alpha + \gamma = -\frac{1}{2}$
$ii) -\alpha + 2\gamma = \frac{3}{2}$
Adding $(i)$ and $(ii)$,$3\gamma = 1 \implies \gamma = \frac{1}{3}$.
Substituting $\gamma = \frac{1}{3}$ into $(i)$,$\alpha + \frac{1}{3} = -\frac{1}{2} \implies \alpha = -\frac{1}{2} - \frac{1}{3} = -\frac{5}{6}$.
Thus,$\alpha = -\frac{5}{6}, \beta = \frac{1}{2}, \gamma = \frac{1}{3}$.

(B) The maximum mass $m_{\max}$ depends on $v$,$g$,and $\rho$. The dimensional formula for mass is $[M^1 L^0 T^0]$.
The dimensions of the variables are:
$[v] = [L T^{-1}]$
$[g] = [L T^{-2}]$
$[\rho] = [M L^{-3}]$
Given $m_{\max} = k v^x g^y \rho^z$,we write the dimensional equation:
$[M^1 L^0 T^0] = [L T^{-1}]^x [L T^{-2}]^y [M L^{-3}]^z$
$[M^1 L^0 T^0] = [M^z L^{x+y-3z} T^{-x-2y}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $z = 1$
For $T$: $-x - 2y = 0 \Rightarrow x = -2y$
For $L$: $x + y - 3z = 0$
Substituting $x = -2y$ and $z = 1$ into the $L$ equation:
$-2y + y - 3(1) = 0$
$-y = 3 \Rightarrow y = -3$
Now,$x = -2(-3) = 6$
Thus,the values are $x = 6, y = -3, z = 1$.

(A) In the expression $F = A \cos(Bx) + C \cos(Dt)$,the argument of the trigonometric function must be dimensionless.
Therefore,the dimensions of $Bx$ and $Dt$ must be equal to the dimension of a constant (dimensionless).
$[Bx] = [M^0 L^0 T^0] \implies [B] = [x^{-1}] = [L^{-1}]$.
$[Dt] = [M^0 L^0 T^0] \implies [D] = [t^{-1}] = [T^{-1}]$.
Now,we find the dimension of $\left(\frac{D}{B}\right)$:
$\left[\frac{D}{B}\right] = \frac{[T^{-1}]}{[L^{-1}]} = [L T^{-1}]$.
Since $[L T^{-1}]$ represents the dimension of velocity,the correct option is $A$.

(A) The given equation is $F = \frac{X}{\text{Linear density}}$.
Rearranging for $X$,we get $X = F \times \text{Linear density}$.
The dimensional formula for Force $(F)$ is $[MLT^{-2}]$.
The dimensional formula for Linear density (mass per unit length) is $[ML^{-1}]$.
Substituting these into the equation for $X$:
$X = [MLT^{-2}] \times [ML^{-1}]$
$X = [M^{1+1} L^{1-1} T^{-2}]$
$X = [M^2 L^0 T^{-2}]$.

(B) Let mass $M$ be expressed as $M = C^a h^b G^c$.
The dimensions are:
$C = [LT^{-1}]$
$h = [ML^2T^{-1}]$
$G = [M^{-1}L^3T^{-2}]$
Substituting these into the equation:
$[M^1L^0T^0] = [LT^{-1}]^a [ML^2T^{-1}]^b [M^{-1}L^3T^{-2}]^c$
$[M^1L^0T^0] = [M^{b-c} L^{a+2b+3c} T^{-a-b-2c}]$
Comparing the powers of $M, L, T$ on both sides:
$b - c = 1$ $(i)$
$a + 2b + 3c = 0$ (ii)
$-a - b - 2c = 0$ (iii)
Adding (ii) and (iii): $b + c = 0$,so $b = -c$.
Substituting $b = -c$ into $(i)$: $-c - c = 1 \Rightarrow -2c = 1 \Rightarrow c = -1/2$.
Then $b = 1/2$.
Substituting $b = 1/2$ and $c = -1/2$ into (iii): $-a - 1/2 - 2(-1/2) = 0 \Rightarrow -a - 1/2 + 1 = 0 \Rightarrow a = 1/2$.
Thus,$M = C^{1/2} h^{1/2} G^{-1/2}$.

(B) The given quantity is $\frac{E J^2}{M^5 G^2}$.
We know the dimensional formulas for the given quantities are:
Dimensions of $E = [M L^2 T^{-2}]$
Dimensions of $J = [M L^2 T^{-1}]$
Dimensions of $M = [M]$
Dimensions of $G = [M^{-1} L^3 T^{-2}]$
Substituting these dimensions into the expression:
$\frac{[M L^2 T^{-2}] [M L^2 T^{-1}]^2}{[M]^5 [M^{-1} L^3 T^{-2}]^2} = \frac{[M L^2 T^{-2}] [M^2 L^4 T^{-2}]}{[M^5] [M^{-2} L^6 T^{-4}]} = \frac{[M^3 L^6 T^{-4}]}{[M^3 L^6 T^{-4}]} = [M^0 L^0 T^0]$
Since the resulting dimension is $[M^0 L^0 T^0]$,the quantity is dimensionless.
Among the given options,angle is a dimensionless quantity.

(B) According to the principle of homogeneity of dimensions,the dimensions of each term in a physical equation must be the same.
Given the equation $F = at + bt^2$,where $F$ is force and $t$ is time.
The dimension of force $F$ is $[MLT^{-2}]$.
For the first term: $[at] = [F]$
$[a] = [F] / [t] = [MLT^{-2}] / [T] = [MLT^{-3}]$.
For the second term: $[bt^2] = [F]$
$[b] = [F] / [t^2] = [MLT^{-2}] / [T^2] = [MLT^{-4}]$.
Therefore,the dimensions of $a$ and $b$ are $[MLT^{-3}]$ and $[MLT^{-4}]$ respectively.

(D) Given the wave equation: $y = a \sin (bt - cx)$.
In the trigonometric function $\sin(\theta)$,the argument $\theta$ must be dimensionless.
Therefore,both $bt$ and $cx$ are dimensionless.
$(a)$ $\frac{y}{a}$ is the ratio of two lengths,so it is dimensionless.
$(b)$ $bt$ is the argument of the sine function,so it is dimensionless.
$(c)$ $cx$ is the argument of the sine function,so it is dimensionless.
$(d)$ The dimensions of $b$ are $[T^{-1}]$ and the dimensions of $c$ are $[L^{-1}]$.
Thus,the dimensions of $\frac{b}{c} = \frac{[T^{-1}]}{[L^{-1}]} = [LT^{-1}]$,which represents the dimension of velocity.
Therefore,$\frac{b}{c}$ is a quantity with dimensions.

(A) Let the Planck's constant $h$ be expressed as $h = k G^x L^y E^z$,where $k$ is a dimensionless constant.
Dimensional formulas are:
$[h] = [M^1 L^2 T^{-1}]$
$[G] = [M^{-1} L^3 T^{-2}]$
$[L] = [M^1 L^2 T^{-1}]$
$[E] = [M^1 L^2 T^{-2}]$
Substituting these into the equation:
$[M^1 L^2 T^{-1}] = [M^{-1} L^3 T^{-2}]^x [M^1 L^2 T^{-1}]^y [M^1 L^2 T^{-2}]^z$
$[M^1 L^2 T^{-1}] = [M^{-x+y+z} L^{3x+2y+2z} T^{-2x-y-2z}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
$(i)$ $-x + y + z = 1$
(ii) $3x + 2y + 2z = 2$
(iii) $-2x - y - 2z = -1$
Adding $(i)$ and (iii):
$(-x + y + z) + (-2x - y - 2z) = 1 - 1$
$-3x - z = 0 \implies z = -3x$
Substitute $z = -3x$ into $(i)$:
$-x + y - 3x = 1 \implies y - 4x = 1 \implies y = 1 + 4x$
Substitute $y$ and $z$ into (ii):
$3x + 2(1 + 4x) + 2(-3x) = 2$
$3x + 2 + 8x - 6x = 2$
$5x + 2 = 2 \implies 5x = 0 \implies x = 0$
Thus,the dimension of $G$ in the formula for $h$ is $0$.

(C) The formula for conversion between two systems is $n_2 = n_1 \left[ \left( \frac{M_1}{M_2} \right)^a \left( \frac{L_1}{L_2} \right)^b \left( \frac{T_1}{T_2} \right)^c \right]$.
The dimensions of force are $[M L T^{-2}]$,so $a=1, b=1, c=-2$.
Given $n_1 = 100$,$M_1 = 1 \ g$,$L_1 = 1 \ cm$,$T_1 = 1 \ s$.
In the new system,$M_2 = 1 \ kg = 1000 \ g$,$L_2 = 1 \ m = 100 \ cm$,$T_2 = 1 \ min = 60 \ s$.
Substituting these values:
$n_2 = 100 \left[ \left( \frac{1 \ g}{1000 \ g} \right)^1 \left( \frac{1 \ cm}{100 \ cm} \right)^1 \left( \frac{1 \ s}{60 \ s} \right)^{-2} \right]$
$n_2 = 100 \left[ \frac{1}{1000} \times \frac{1}{100} \times (60)^2 \right]$
$n_2 = 100 \times \frac{1}{1000} \times \frac{1}{100} \times 3600$
$n_2 = 3.6$.

(A) The given equation is $V = -\frac{GM}{r} + \frac{A}{r^2}$.
According to the principle of homogeneity of dimensions,the dimensions of each term on both sides must be the same.
Therefore,the dimension of $\frac{A}{r^2}$ must be equal to the dimension of $\frac{GM}{r}$.
$[V] = [\frac{GM}{r}] = [\frac{A}{r^2}]$
From this,we get $[A] = [\frac{GM}{r}] \times [r^2] = [GM] \times [r]$.
We know that the gravitational potential $V$ has the dimensions of energy per unit mass,which is $[L^2 T^{-2}]$.
Also,$\frac{GM}{r}$ represents the gravitational potential,so $[\frac{GM}{r}] = [L^2 T^{-2}]$.
Since the velocity of light $c$ has dimensions $[L T^{-1}]$,then $c^2$ has dimensions $[L^2 T^{-2}]$.
Thus,$[\frac{GM}{r}] = [c^2]$.
Substituting $[r] = \frac{[GM]}{[c^2]}$ into the expression for $[A]$:
$[A] = [GM] \times \frac{[GM]}{[c^2]} = \frac{G^2 M^2}{c^2}$.

(A) The dimensions of the variables are: $[t] = T^1$,$[\rho] = ML^{-3}$,$[r] = L^1$,$[\eta] = ML^{-1}T^{-1}$,and $[\sigma] = ML^{-3}$.
Given the relation $t = A \rho^{a} r^{b} \eta^{c} \sigma^{d}$,we equate the dimensions on both sides:
$T^1 = (ML^{-3})^a (L)^b (ML^{-1}T^{-1})^c (ML^{-3})^d$
$T^1 = M^{a+c+d} L^{-3a+b-c-3d} T^{-c}$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $a+c+d = 0$
For $L$: $-3a+b-c-3d = 0$
For $T$: $-c = 1 \implies c = -1$
Substituting $c = -1$ into the $M$ equation: $a+d-1 = 0 \implies a+d = 1$.
Substituting $c = -1$ into the $L$ equation: $-3a+b+1-3d = 0 \implies b - 3(a+d) + 1 = 0$.
Since $a+d = 1$,we get $b - 3(1) + 1 = 0 \implies b - 2 = 0 \implies b = 2$.
Finally,calculating the value: $\frac{b+c}{a+d} = \frac{2 + (-1)}{1} = \frac{1}{1} = 1$.

(B) Using dimensional analysis,we define the Planck units based on fundamental constants $h$,$G$,and $c$:
$1$. Planck length $(L)$ is given by $l_p = \sqrt{\frac{Gh}{c^3}}$,which corresponds to $A-IV$.
$2$. Planck time $(S)$ is given by $t_p = \sqrt{\frac{Gh}{c^5}}$,which corresponds to $B-II$.
$3$. Planck mass $(M)$ is given by $m_p = \sqrt{\frac{hc}{G}}$,which corresponds to $C-I$.
$4$. By elimination,Kelvin $(K)$ corresponds to $D-III$.
Thus,the correct matching is $A-IV, B-II, C-I, D-III$.

(B) The dimensions of the given quantities are:
$[L] = ML^2T^{-2}A^{-2}$
$[C] = M^{-1}L^{-2}T^4A^2$
$[R] = ML^2T^{-3}A^{-2}$
Now,let us evaluate the dimensions of the expression $\frac{R^2}{L}$:
$[R^2] = (ML^2T^{-3}A^{-2})^2 = M^2L^4T^{-6}A^{-4}$
$[L] = ML^2T^{-2}A^{-2}$
Therefore,the dimensions of $\frac{R^2}{L}$ are:
$\frac{[R^2]}{[L]} = \frac{M^2L^4T^{-6}A^{-4}}{ML^2T^{-2}A^{-2}} = ML^2T^{-4}A^{-2}$
This matches the given dimensional formula. Thus,the correct expression is $\frac{R^2}{L}$.

(D) The dimension of distance $x$ is $[L]$.
The dimension of potential energy $V$ is $[ML^2T^{-2}]$.
From the principle of homogeneity,in the denominator $(x + B)$,the dimension of $B$ must be equal to the dimension of $x$. Therefore,$[B] = [L]$.
The given equation is $V = \frac{A\sqrt{x}}{x + B}$.
Substituting the dimensions: $[ML^2T^{-2}] = \frac{[A][L^{1/2}]}{[L]}$.
$[ML^2T^{-2}] = [A][L^{-1/2}]$.
Therefore,$[A] = [ML^2T^{-2}] \times [L^{1/2}] = [ML^{5/2}T^{-2}]$.
Now,the dimension of $AB$ is $[AB] = [ML^{5/2}T^{-2}] \times [L] = [ML^{7/2}T^{-2}]$.

(B) The dimensional formulas are: $[H] = [M^1 L^0 T^{-2} A^{-1}]$,$[x] = [L^1]$,$[\epsilon] = [M^{-1} L^{-3} T^4 A^2]$,$[E] = [M^1 L^1 T^{-3} A^{-1}]$,and $[t] = [T^1]$.
Substituting these into the equation $H = x^p \epsilon^q E^r t^{-s}$:
$[M^1 L^0 T^{-2} A^{-1}] = [L]^p [M^{-1} L^{-3} T^4 A^2]^q [M^1 L^1 T^{-3} A^{-1}]^r [T]^{-s}$
$[M^1 L^0 T^{-2} A^{-1}] = M^{-q+r} L^{p-3q+r} T^{4q-3r-s} A^{2q-r}$
Comparing the powers on both sides:
For $A$: $2q - r = -1$ $(i)$
For $M$: $-q + r = 1$ (ii)
Adding $(i)$ and (ii): $q = 0$. Substituting $q=0$ in (ii),$r = 1$.
For $L$: $p - 3q + r = 0 \Rightarrow p - 0 + 1 = 0 \Rightarrow p = -1$.
For $T$: $4q - 3r - s = -2 \Rightarrow 4(0) - 3(1) - s = -2 \Rightarrow -3 - s = -2 \Rightarrow s = -1$.
Since the equation is $H = \frac{x^p \epsilon^q E^r}{t^s}$,the exponent of $t$ is $-s$. If the given form is $t^{-s}$,then $s = -1$. However,if the form is $t^s$,then $s = 1$. Given the options,$p=-1, q=1, r=1, s=1$ matches option $B$.