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Dimensional Analysis, Uses and Limitations Questions in English

Class 11 Physics · Units, Dimensions and Measurement · Dimensional Analysis, Uses and Limitations

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201
DifficultMCQ
Stokes' law states that the viscous drag force $F$ experienced by a sphere of radius $a$,moving with a speed $v$ through a fluid with coefficient of viscosity $\eta$,is given by $F=6 \pi \eta a v$. If this fluid is flowing through a cylindrical pipe of radius $r$,length $l$ and pressure difference of $p$ across its two ends,then the volume of water $V$ which flows through the pipe in time $t$ can be written as $\frac{V}{t}=k\left(\frac{p}{l}\right)^a \eta^b r^c$,where $k$ is a dimensionless constant. Correct values of $a, b$ and $c$ are
A
$a=1, b=-1, c=4$
B
$a=-1, b=1, c=4$
C
$a=2, b=-1, c=3$
D
$a=1, b=-2, c=-4$

Solution

(A) The given equation is $\frac{V}{t} = k \left( \frac{p}{l} \right)^a \eta^b r^c$.
The dimensions of the quantities are:
$[V/t] = [L^3 T^{-1}]$
$[p/l] = [M L^{-1} T^{-2} / L] = [M L^{-2} T^{-2}]$
$[\eta] = [M L^{-1} T^{-1}]$
$[r] = [L]$
Substituting these into the equation:
$[L^3 T^{-1}] = [M L^{-2} T^{-2}]^a [M L^{-1} T^{-1}]^b [L]^c$
$[L^3 T^{-1}] = M^{a+b} L^{-2a-b+c} T^{-2a-b}$
Equating the powers of $M, L,$ and $T$ on both sides:
For $M$: $a + b = 0 \Rightarrow b = -a$
For $T$: $-2a - b = -1$
Substituting $b = -a$ into the $T$ equation: $-2a - (-a) = -1 \Rightarrow -a = -1 \Rightarrow a = 1$.
Since $b = -a$,we get $b = -1$.
For $L$: $-2a - b + c = 3$
Substituting $a = 1$ and $b = -1$: $-2(1) - (-1) + c = 3 \Rightarrow -2 + 1 + c = 3 \Rightarrow -1 + c = 3 \Rightarrow c = 4$.
Thus,the values are $a=1, b=-1, c=4$.
202
MediumMCQ
If the buoyant force $F$ acting on an object depends on its volume $V$ immersed in a liquid,the density $\rho$ of the liquid,and the acceleration due to gravity $g$,then the correct expression for $F$ can be:
A
$V \rho g$
B
$\frac{\rho g}{V}$
C
$\rho g V^2$
D
$\sqrt{\rho g V}$

Solution

(A) Using the principle of dimensional homogeneity,we assume $F = k V^a \rho^b g^c$,where $k$ is a dimensionless constant.
The dimensions of the quantities are:
$[F] = [M L T^{-2}]$
$[V] = [L^3]$
$[\rho] = [M L^{-3}]$
$[g] = [L T^{-2}]$
Substituting these into the equation:
$[M L T^{-2}] = [L^3]^a [M L^{-3}]^b [L T^{-2}]^c$
$[M L T^{-2}] = [M^b L^{3a - 3b + c} T^{-2c}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M: b = 1$
For $T: -2c = -2 \Rightarrow c = 1$
For $L: 3a - 3b + c = 1$
Substituting $b=1$ and $c=1$ into the equation for $L$:
$3a - 3(1) + 1 = 1$
$3a - 2 = 1$
$3a = 3 \Rightarrow a = 1$
Thus,the expression is $F = k V^1 \rho^1 g^1$. Taking $k=1$,we get $F = V \rho g$.
203
MediumMCQ
The amount of heat energy $Q$ used to heat up a substance depends on its mass $m$,its specific heat capacity $s$,and the change in temperature $\Delta T$ of the substance. Using the dimensional method,find the expression for $s$. (Given that $[s] = [L^2 T^{-2} K^{-1}]$)
A
$Q m \Delta T$
B
$\frac{Q}{m \Delta T}$
C
$\frac{Q m}{\Delta T}$
D
$\frac{m}{Q \Delta T}$

Solution

(B) The heat energy $Q$ is given by the relation $Q = m^a s^b (\Delta T)^c$.
The dimensions of the quantities are:
$[Q] = [M L^2 T^{-2}]$
$[m] = [M]$
$[s] = [L^2 T^{-2} K^{-1}]$
$[\Delta T] = [K]$
Substituting these into the equation:
$[M L^2 T^{-2}] = [M]^a [L^2 T^{-2} K^{-1}]^b [K]^c$
$[M L^2 T^{-2}] = [M^a L^{2b} T^{-2b} K^{-b+c}]$
Comparing the powers of $M, L, T,$ and $K$ on both sides:
For $M$: $a = 1$
For $L$: $2b = 2 \Rightarrow b = 1$
For $T$: $-2b = -2 \Rightarrow b = 1$
For $K$: $-b + c = 0 \Rightarrow c = b = 1$
Thus,the expression is $Q = m^1 s^1 (\Delta T)^1$,which simplifies to $Q = m s \Delta T$.
Rearranging for $s$,we get $s = \frac{Q}{m \Delta T}$.
204
MediumMCQ
If $y$ represents pressure and $x$ represents velocity gradient,then the dimensions of $\frac{d^2 y}{d x^2}$ are
A
$[ML^{-1}T^{-2}]$
B
$[M^2L^{-2}T^{-2}]$
C
$[ML^{-1}T^0]$
D
$[M^2L^{-2}T^{-4}]$

Solution

(C) The dimensions of the derivative $\frac{d^2 y}{d x^2}$ are equivalent to the dimensions of $\frac{y}{x^2}$.
Given that $y$ represents pressure,its dimensional formula is $[ML^{-1}T^{-2}]$.
Given that $x$ represents the velocity gradient,its dimensional formula is $\frac{[v]}{[L]} = \frac{[LT^{-1}]}{[L]} = [T^{-1}]$.
Therefore,the dimensions of $\frac{d^2 y}{d x^2}$ are $\frac{[y]}{[x]^2} = \frac{[ML^{-1}T^{-2}]}{[T^{-1}]^2} = \frac{[ML^{-1}T^{-2}]}{[T^{-2}]} = [ML^{-1}T^0]$.
Thus,the correct option is $C$.
205
MediumMCQ
The dimensions of $\frac{\alpha}{\beta}$ in the equation $F = \frac{\alpha - t^2}{\beta v^2}$,where $F$ is force,$v$ is velocity,and $t$ is time,are ..........
A
$[MLT^{-1}]$
B
$[ML^{-1}T^{-2}]$
C
$[ML^3T^{-4}]$
D
$[ML^2T^{-4}]$

Solution

(C) Given the equation $F = \frac{\alpha - t^2}{\beta v^2}$.
According to the principle of homogeneity of dimensions,the dimensions of $\alpha$ must be the same as the dimensions of $t^2$.
Therefore,$[\alpha] = [T^2]$.
Now,the dimensions of the entire expression are $[F] = \frac{[\alpha]}{[\beta][v^2]}$.
Substituting the dimensions: $[MLT^{-2}] = \frac{[T^2]}{[\beta][LT^{-1}]^2}$.
$[MLT^{-2}] = \frac{[T^2]}{[\beta][L^2T^{-2}]}$.
Rearranging for $[\beta]$: $[\beta] = \frac{[T^2]}{[MLT^{-2}][L^2T^{-2}]} = \frac{[T^2]}{[ML^3T^{-4}]} = [M^{-1}L^{-3}T^6]$.
Now,calculate the dimensions of $\frac{\alpha}{\beta}$:
$\frac{[\alpha]}{[\beta]} = \frac{[T^2]}{[M^{-1}L^{-3}T^6]} = [ML^3T^{-4}]$.
206
EasyMCQ
Even if a physical quantity depends upon three quantities,out of which two are dimensionally same,then the formula cannot be derived by the method of dimensions. This statement
A
May be true
B
May be false
C
Must be true
D
Must be false

Solution

(C) The statement is correct.
Dimensional analysis relies on the principle of homogeneity of dimensions,which allows us to equate powers of fundamental quantities $(M, L, T)$ to solve for unknown exponents.
If a physical quantity $X$ depends on three quantities $a, b,$ and $c$,where $a$ and $b$ have the same dimensions,the relationship would be of the form $X = k \cdot a^x \cdot b^y \cdot c^z$.
Since $a$ and $b$ have the same dimensions,say $[M^p L^q T^r]$,the expression becomes $X = k \cdot ([M^p L^q T^r])^x \cdot ([M^p L^q T^r])^y \cdot c^z = k \cdot ([M^p L^q T^r])^{x+y} \cdot c^z$.
Here,we can only determine the value of $(x+y)$ and $z$ through dimensional analysis,but we cannot individually determine the values of $x$ and $y$.
Therefore,the unique formula cannot be derived using this method.
207
MediumMCQ
In a practical unit system,if the unit of mass becomes double and that of time becomes half,then $8 \text{ joule}$ will be equal to .............. unit of work.
A
$6$
B
$4$
C
$1$
D
$10$

Solution

(C) The dimensional formula for work is $[ML^2T^{-2}]$.
Using the principle of homogeneity of dimensions,$n_1[M_1L_1^2T_1^{-2}] = n_2[M_2L_2^2T_2^{-2}]$.
Given $n_1 = 8$,$M_2 = 2M_1$,$L_2 = L_1$,and $T_2 = \frac{1}{2}T_1$.
Substituting these values into the equation:
$n_2 = n_1 \left[\frac{M_1}{M_2}\right] \left[\frac{L_1}{L_2}\right]^2 \left[\frac{T_1}{T_2}\right]^{-2}$
$n_2 = 8 \times \left[\frac{M_1}{2M_1}\right] \times \left[\frac{L_1}{L_1}\right]^2 \times \left[\frac{T_1}{0.5T_1}\right]^{-2}$
$n_2 = 8 \times \frac{1}{2} \times 1 \times (2)^{-2}$
$n_2 = 8 \times \frac{1}{2} \times \frac{1}{4} = 1$.
Thus,$8 \text{ joule}$ is equal to $1$ unit of work in the new system.
208
MediumMCQ
In a new system of units,energy $(E)$,density $(d)$,and power $(P)$ are taken as fundamental units. Then the dimensional formula of the universal gravitational constant $G$ will be .......
A
$[E^{-1} d^{-2} P^2]$
B
$[E^{-2} d^{-1} P^2]$
C
$[E^2 d^{-1} P^{-1}]$
D
$[E^{-1} d^{-2} P^{-2}]$

Solution

(B) The dimensional formula of the universal gravitational constant $G$ is $[M^{-1} L^3 T^{-2}]$.
Let $G = [E^a d^b P^c]$.
The dimensions of the fundamental units are:
$E = [M L^2 T^{-2}]$
$d = [M L^{-3}]$
$P = [M L^2 T^{-3}]$
Substituting these into the equation:
$[M^{-1} L^3 T^{-2}] = [M L^2 T^{-2}]^a [M L^{-3}]^b [M L^2 T^{-3}]^c$
$[M^{-1} L^3 T^{-2}] = [M^{a+b+c} L^{2a-3b+2c} T^{-2a-3c}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
$1) a + b + c = -1$
$2) 2a - 3b + 2c = 3$
$3) -2a - 3c = -2 \Rightarrow 2a + 3c = 2$
From $(3)$,$c = (2 - 2a)/3$. Substituting $c$ and $b = -1 - a - c$ into $(2)$:
Solving the system of linear equations yields $a = -2, b = -1, c = 2$.
Therefore,$G = [E^{-2} d^{-1} P^2]$.
209
MediumMCQ
In the equation $y = x^2 \cos^2 \left( 2 \pi \frac{\beta \gamma}{\alpha} \right)$,the units of $x, \alpha, \beta$ are $m, s^{-1}$,and $(ms^{-1})^{-1}$ respectively. The units of $y$ and $\gamma$ are:
A
$m^2, ms^{-2}$
B
$m, ms^{-1}$
C
$m^2, m$
D
$m, ms^{-2}$

Solution

(A) The argument of a trigonometric function must be dimensionless.
Therefore,the term $\frac{\beta \gamma}{\alpha}$ must be dimensionless,i.e.,$[\beta][\gamma] = [\alpha]$.
Given units: $[x] = L$,$[\alpha] = T^{-1}$,and $[\beta] = (LT^{-1})^{-1} = L^{-1}T$.
Substituting these into the dimension equation: $(L^{-1}T) \cdot [\gamma] = T^{-1}$.
Solving for $[\gamma]$: $[\gamma] = T^{-1} \cdot L \cdot T^{-1} = LT^{-2}$.
The unit of $\gamma$ is $ms^{-2}$.
From the equation $y = x^2$,the unit of $y$ is the square of the unit of $x$.
Since $[x] = L$ (meters),$[y] = L^2$ (meters squared).
Thus,the units of $y$ and $\gamma$ are $m^2$ and $ms^{-2}$ respectively.
210
MediumMCQ
$A$ dimensionally consistent relation for the volume $V$ of a liquid of coefficient of viscosity $\eta$ flowing per second through a tube of radius $r$ and length $l$,having a pressure difference $P$ across its ends,is:
A
$V = \frac{\pi P r^4}{8 \eta l}$
B
$V = \frac{\pi \eta}{8 P r^4}$
C
$V = \frac{8 P \eta}{\pi r^4}$
D
$V = \frac{\pi P \eta}{8 r^4}$

Solution

(A) The volume flow rate $V$ (Poiseuille's Law) is given by the formula $V = \frac{\pi P r^4}{8 \eta l}$.
To verify this dimensionally:
Dimensions of $V = [L^3 T^{-1}]$.
Dimensions of $P = [M L^{-1} T^{-2}]$.
Dimensions of $r = [L]$.
Dimensions of $\eta = [M L^{-1} T^{-1}]$.
Dimensions of $l = [L]$.
Substituting these into the $RHS$: $\frac{[M L^{-1} T^{-2}] [L^4]}{[M L^{-1} T^{-1}] [L]} = \frac{[M L^3 T^{-2}]}{[M T^{-1}]} = [L^3 T^{-1}]$.
Since the dimensions of $LHS$ and $RHS$ match,the correct relation is $V = \frac{\pi P r^4}{8 \eta l}$.
211
DifficultMCQ
$A$ wave pulse can travel along a tense string like a violin string. $A$ series of experiments showed that the wave velocity $V$ of a pulse depends on the following quantities: the tension $T$ of the string,the cross-sectional area $A$ of the string,and the density $\rho$ (mass per unit volume) of the string. Obtain an expression for $V$ in terms of $T$,$A$,and $\rho$ using dimensional analysis.
A
$V = k \sqrt{\frac{T}{A \rho}}$
B
$V = k \sqrt{\frac{A \rho}{T}}$
C
$V = k \sqrt{\frac{T}{A \rho^2}}$
D
None of these

Solution

(A) Let the expression for velocity be $V = k T^a A^b \rho^c$,where $k$ is a dimensionless constant.
Writing the dimensions of each quantity:
$[V] = [LT^{-1}]$,$[T] = [MLT^{-2}]$,$[A] = [L^2]$,$[\rho] = [ML^{-3}]$.
Substituting these into the equation:
$[LT^{-1}] = [MLT^{-2}]^a [L^2]^b [ML^{-3}]^c = [M^{a+c} L^{a+2b-3c} T^{-2a}]$.
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $a + c = 0 \Rightarrow c = -a$.
For $T$: $-2a = -1 \Rightarrow a = 1/2$.
Thus,$c = -1/2$.
For $L$: $a + 2b - 3c = 1$.
Substituting $a = 1/2$ and $c = -1/2$: $1/2 + 2b - 3(-1/2) = 1 \Rightarrow 1/2 + 2b + 3/2 = 1 \Rightarrow 2 + 2b = 1 \Rightarrow 2b = -1 \Rightarrow b = -1/2$.
Substituting the values of $a, b, c$ back into the expression: $V = k T^{1/2} A^{-1/2} \rho^{-1/2} = k \sqrt{\frac{T}{A \rho}}$.
212
MediumMCQ
If $\varepsilon_0$ is the permittivity of free space,$e$ is the charge of a proton,$G$ is the universal gravitational constant,and $m_p$ is the mass of a proton,then the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p^2}$ is:
A
$[M^1 L^1 T^{-3} A^{-1}]$
B
$[M^0 L^0 T^0 A^0]$
C
$[M^1 L^3 T^{-3} A^{-1}]$
D
$[M^{-1} L^{-3} T^4 A^2]$

Solution

(B) The gravitational force between two protons separated by a distance $r$ is given by $F_g = \frac{G m_p^2}{r^2}$.
The electrostatic force between two protons separated by the same distance $r$ is given by $F_e = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}$.
Taking the ratio of the two forces,we get $\frac{F_e}{F_g} = \frac{\frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r^2}}{\frac{G m_p^2}{r^2}} = \frac{e^2}{4 \pi \varepsilon_0 G m_p^2}$.
Since the ratio of two forces is a dimensionless quantity,the dimensional formula for $\frac{e^2}{4 \pi \varepsilon_0 G m_p^2}$ is $[M^0 L^0 T^0 A^0]$.
213
MediumMCQ
The frequency $(v)$ of an oscillating liquid drop may depend upon the radius $(r)$ of the drop,the density $(\rho)$ of the liquid,and the surface tension $(s)$ of the liquid as: $v = r^{a} \rho^{b} s^{c}$. The values of $a, b,$ and $c$ respectively are:
A
$(-\frac{3}{2}, -\frac{1}{2}, \frac{1}{2})$
B
$(\frac{3}{2}, -\frac{1}{2}, \frac{1}{2})$
C
$(\frac{3}{2}, \frac{1}{2}, -\frac{1}{2})$
D
$(-\frac{3}{2}, \frac{1}{2}, \frac{1}{2})$

Solution

(A) The dimensional formula for frequency $(v)$ is $[T^{-1}]$.
The dimensional formula for radius $(r)$ is $[L]$.
The dimensional formula for density $(\rho)$ is $[ML^{-3}]$.
The dimensional formula for surface tension $(s)$ is $[MT^{-2}]$.
Given the relation: $v = r^{a} \rho^{b} s^{c}$.
Substituting the dimensions: $[M^{0}L^{0}T^{-1}] = [L]^{a} [ML^{-3}]^{b} [MT^{-2}]^{c}$.
$[M^{0}L^{0}T^{-1}] = M^{b+c} L^{a-3b} T^{-2c}$.
Comparing the powers of $M, L,$ and $T$ on both sides:
For $T$: $-2c = -1 \Rightarrow c = \frac{1}{2}$.
For $M$: $b + c = 0 \Rightarrow b = -c = -\frac{1}{2}$.
For $L$: $a - 3b = 0 \Rightarrow a = 3b = 3(-\frac{1}{2}) = -\frac{3}{2}$.
Thus,the values are $a = -\frac{3}{2}, b = -\frac{1}{2}, c = \frac{1}{2}$.
214
MediumMCQ
The equation of a circle is given by $x^2+y^2=a^2$,where $a$ is the radius. If the equation is modified to change the origin to a point other than $(0,0)$,find the correct dimensions of $A$ and $B$ in the new equation: $(x-At)^2+(y-\frac{t}{B})^2=a^2$. The dimensions of $t$ are given as $[T^{-1}]$.
A
$A=[L^{-1}T], B=[LT^{-1}]$
B
$A=[LT], B=[L^{-1}T^{-1}]$
C
$A=[L^{-1}T^{-1}], B=[LT^{-1}]$
D
$A=[L^{-1}T^{-1}], B=[LT]$

Solution

(B) The given equation is $(x-At)^2+(y-\frac{t}{B})^2=a^2$.
According to the principle of homogeneity of dimensions,the terms subtracted from $x$ and $y$ must have the same dimensions as length $[L]$.
For the term $At$:
$[At] = [L]$
$[A][T^{-1}] = [L]$
$[A] = [LT]$
For the term $\frac{t}{B}$:
$[\frac{t}{B}] = [L]$
$\frac{[T^{-1}]}{[B]} = [L]$
$[B] = [L^{-1}T^{-1}]$
Thus,the dimensions are $A=[LT]$ and $B=[L^{-1}T^{-1}]$.
215
DifficultMCQ
$(P+\frac{a}{V^2})(V-b)=RT$ represents the equation of state of some gases. Where $P$ is the pressure,$V$ is the volume,$T$ is the temperature and $a, b, R$ are the constants. The physical quantity,which has the same dimensional formula as that of $\frac{b^2}{a}$,will be
A
Bulk modulus
B
Modulus of rigidity
C
Compressibility
D
Energy density

Solution

(C) According to the principle of homogeneity of dimensions,the dimensions of terms added or subtracted must be the same.
Since $b$ is subtracted from $V$,the dimensions of $b$ are equal to the dimensions of volume $V$:
$[b] = [V] = [L^3]$
Similarly,$\frac{a}{V^2}$ is added to $P$,so the dimensions of $\frac{a}{V^2}$ are equal to the dimensions of pressure $P$:
$[\frac{a}{V^2}] = [P] \implies [a] = [P][V^2] = [ML^{-1}T^{-2}][L^6] = [ML^5T^{-2}]$
Now,we find the dimensions of $\frac{b^2}{a}$:
$[\frac{b^2}{a}] = \frac{[L^3]^2}{[ML^5T^{-2}]} = \frac{[L^6]}{[ML^5T^{-2}]} = [M^{-1}LT^2]$
We know that compressibility $K$ is the reciprocal of the bulk modulus $B$:
$[K] = \frac{1}{[B]} = \frac{1}{[ML^{-1}T^{-2}]} = [M^{-1}LT^2]$
Thus,the dimensional formula of $\frac{b^2}{a}$ is the same as that of compressibility.
216
DifficultMCQ
If the velocity of light $c$,universal gravitational constant $G$,and Planck's constant $h$ are chosen as fundamental quantities,the dimensions of mass in the new system are:
A
$[h^{1/2} c^{1/2} G^{-1/2}]$
B
$[h^1 c^1 G^{-1}]$
C
$[h^{-1/2} c^{1/2} G^{1/2}]$
D
$[h^{1/2} c^{-1/2} G^{1/2}]$

Solution

(A) Let the dimensional formula of mass be $M = h^x c^y G^z$.
The dimensional formulas are:
$h = [ML^2T^{-1}]$
$c = [LT^{-1}]$
$G = [M^{-1}L^3T^{-2}]$
Substituting these into the equation:
$[M^1 L^0 T^0] = [ML^2T^{-1}]^x [LT^{-1}]^y [M^{-1}L^3T^{-2}]^z$
$[M^1 L^0 T^0] = M^{x-z} L^{2x+y+3z} T^{-x-y-2z}$
Comparing the powers on both sides:
$1$) $x - z = 1$
$2$) $2x + y + 3z = 0$
$3$) $-x - y - 2z = 0$
Adding equations $(2)$ and $(3)$:
$(2x + y + 3z) + (-x - y - 2z) = 0 + 0$
$x + z = 0 \implies x = -z$
Substitute $x = -z$ into equation $(1)$:
$-z - z = 1 \implies -2z = 1 \implies z = -1/2$
Since $x = -z$,then $x = 1/2$.
Substitute $x = 1/2$ and $z = -1/2$ into equation $(3)$:
$-1/2 - y - 2(-1/2) = 0$
$-1/2 - y + 1 = 0 \implies y = 1/2$
Thus,the dimensions of mass are $[h^{1/2} c^{1/2} G^{-1/2}]$.
217
MediumMCQ
If force $(F)$,velocity $(V)$,and time $(T)$ are considered as fundamental physical quantities,then the dimensional formula of density will be:
A
$F V^{-2} T^2$
B
$F V^{-4} T^{-2}$
C
$F V^{-4} T^2$
D
$F^2 V^{-2} T^6$

Solution

(B) The dimensional formula for density $(\rho)$ is $[M L^{-3}]$.
Let the dimensional formula be expressed as $[\rho] = [F]^a [V]^b [T]^c$.
Substituting the dimensions: $[M L^{-3}] = [M L T^{-2}]^a [L T^{-1}]^b [T]^c$.
$[M L^{-3}] = [M^a L^{a+b} T^{-2a-b+c}]$.
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $a = 1$.
For $L$: $a + b = -3 \Rightarrow 1 + b = -3 \Rightarrow b = -4$.
For $T$: $-2a - b + c = 0 \Rightarrow -2(1) - (-4) + c = 0 \Rightarrow -2 + 4 + c = 0 \Rightarrow 2 + c = 0 \Rightarrow c = -2$.
Thus,the dimensional formula is $[F^1 V^{-4} T^{-2}]$.
218
MediumMCQ
In the equation $[X+\frac{a}{Y^2}][Y-b]= RT$,$X$ is pressure,$Y$ is volume,$R$ is universal gas constant and $T$ is temperature. The physical quantity equivalent to the ratio $\frac{a}{b}$ is
A
Energy
B
Impulse
C
Pressure gradient
D
Coefficient of viscosity

Solution

(A) According to the principle of homogeneity of dimensions,only physical quantities of the same dimension can be added or subtracted.
In the term $[X + \frac{a}{Y^2}]$,$X$ is pressure,so $\frac{a}{Y^2}$ must also have the dimensions of pressure.
$[X] = [ML^{-1}T^{-2}]$
$[Y] = [L^3]$
$[\frac{a}{Y^2}] = [ML^{-1}T^{-2}] \implies [a] = [ML^{-1}T^{-2}] \times [L^3]^2 = [ML^5T^{-2}]$
In the term $[Y - b]$,$Y$ is volume,so $b$ must have the dimensions of volume.
$[b] = [L^3]$
Now,the ratio $\frac{a}{b}$ has dimensions:
$\frac{[a]}{[b]} = \frac{[ML^5T^{-2}]}{[L^3]} = [ML^2T^{-2}]$
These are the dimensions of energy (or work/torque).
219
MediumMCQ
The speed of a wave produced in water is given by $v = \lambda^a g^b \rho^c$. Where $\lambda$,$g$,and $\rho$ are the wavelength of the wave,acceleration due to gravity,and density of water,respectively. The values of $a$,$b$,and $c$ are,respectively:
A
$\frac{1}{2}, \frac{1}{2}, 0$
B
$1, 1, 0$
C
$1, -1, 0$
D
$\frac{1}{2}, 0, \frac{1}{2}$

Solution

(A) The given equation is $v = \lambda^a g^b \rho^c$.
Using the dimensional analysis method,we write the dimensions of each quantity:
$[v] = [L T^{-1}]$
$[\lambda] = [L]$
$[g] = [L T^{-2}]$
$[\rho] = [M L^{-3}]$
Substituting these into the equation:
$[M^0 L^1 T^{-1}] = [L]^a [L T^{-2}]^b [M L^{-3}]^c$
$[M^0 L^1 T^{-1}] = [M^c L^{a+b-3c} T^{-2b}]$
Comparing the powers of $M$,$L$,and $T$ on both sides:
For $M$: $c = 0$
For $T$: $-2b = -1 \Rightarrow b = \frac{1}{2}$
For $L$: $a + b - 3c = 1$
Substituting $b = \frac{1}{2}$ and $c = 0$ into the equation for $L$:
$a + \frac{1}{2} - 3(0) = 1$
$a = 1 - \frac{1}{2} = \frac{1}{2}$
Thus,the values are $a = \frac{1}{2}$,$b = \frac{1}{2}$,and $c = 0$.
220
DifficultMCQ
The equation of state of a real gas is given by $(P+\frac{a}{V^2})(V-b)=RT$,where $P, V$ and $T$ are pressure,volume and temperature respectively and $R$ is the universal gas constant. The dimensions of $\frac{a}{b^2}$ are similar to that of:
A
$PV$
B
$P$
C
$RT$
D
$R$

Solution

(B) According to the principle of dimensional homogeneity,terms added or subtracted in an equation must have the same dimensions.
$1$. In the term $(P + \frac{a}{V^2})$,the dimensions of $P$ must be equal to the dimensions of $\frac{a}{V^2}$.
$[P] = [\frac{a}{V^2}] \Rightarrow [a] = [P][V^2] = [P][L^6]$.
$2$. In the term $(V - b)$,the dimensions of $V$ must be equal to the dimensions of $b$.
$[b] = [V] = [L^3] \Rightarrow [b^2] = [V^2] = [L^6]$.
$3$. Now,find the dimensions of $\frac{a}{b^2}$:
$[\frac{a}{b^2}] = \frac{[P][V^2]}{[V^2]} = [P]$.
Therefore,the dimensions of $\frac{a}{b^2}$ are the same as the dimensions of pressure $P$.
221
DifficultMCQ
If mass is written as $m=kc^{p} G^{-1 / 2} \,h^{1 / 2}$ then the value of $P$ will be : (Constants have their usual meaning with $k$ a dimensionless constant)
A
$1 / 2$
B
$1 / 3$
C
$2$
D
$-1 / 3$

Solution

(A) Given the formula for mass: $m = k c^{P} G^{-1/2} h^{1/2}$.
We use the dimensional formulas for each constant:
$c$ (speed of light) $= [L T^{-1}]$
$G$ (gravitational constant) $= [M^{-1} L^3 T^{-2}]$
$h$ (Planck's constant) $= [M L^2 T^{-1}]$
Substituting these dimensions into the equation:
$[M^1 L^0 T^0] = [L T^{-1}]^{P} [M^{-1} L^3 T^{-2}]^{-1/2} [M L^2 T^{-1}]^{1/2}$
$[M^1 L^0 T^0] = [L^P T^{-P}] [M^{1/2} L^{-3/2} T^1] [M^{1/2} L^1 T^{-1/2}]$
Combining the powers of $M$,$L$,and $T$ on the right side:
$M: 1/2 + 1/2 = 1$
$L: P - 3/2 + 1 = P - 1/2$
$T: -P + 1 - 1/2 = -P + 1/2$
Comparing the powers of $L$ on both sides:
$P - 1/2 = 0 \implies P = 1/2$.
Thus,the value of $P$ is $1/2$.
222
DifficultMCQ
$A$ force is represented by $F = ax^2 + bt^{1/2}$,where $x$ is distance and $t$ is time. The dimensions of $b^2/a$ are:
A
$[ML^3 T^{-3}]$
B
$[MLT^{-2}]$
C
$[ML^{-1} T^{-1}]$
D
$[ML^2 T^{-3}]$

Solution

(A) According to the principle of homogeneity of dimensions,each term in the equation must have the same dimensions as force $F$.
$1$. For the term $ax^2$:
$[ax^2] = [F] = [MLT^{-2}]$
$[a] = [F] / [x^2] = [MLT^{-2}] / [L^2] = [ML^{-1}T^{-2}]$
$2$. For the term $bt^{1/2}$:
$[bt^{1/2}] = [F] = [MLT^{-2}]$
$[b] = [F] / [t^{1/2}] = [MLT^{-2}] / [T^{1/2}] = [MLT^{-5/2}]$
$3$. Calculating the dimensions of $b^2/a$:
$[b^2/a] = [b]^2 / [a] = ([MLT^{-5/2}])^2 / [ML^{-1}T^{-2}]$
$[b^2/a] = [M^2 L^2 T^{-5}] / [ML^{-1}T^{-2}] = [M^{2-1} L^{2-(-1)} T^{-5-(-2)}] = [ML^3 T^{-3}]$
223
DifficultMCQ
Consider two physical quantities $A$ and $B$ related to each other as $E = \frac{B - x^2}{At}$,where $E, x,$ and $t$ have dimensions of energy,length,and time respectively. The dimension of $AB$ is
A
$L^{-2} M^1 T^0$
B
$L^2 M^{-1} T^1$
C
$L^{-2} M^{-1} T^1$
D
$L^0 M^{-1} T^1$

Solution

(B) According to the principle of homogeneity of dimensions,quantities added or subtracted must have the same dimensions.
Since $B$ is subtracted by $x^2$,the dimension of $B$ must be equal to the dimension of $x^2$.
$[B] = [x^2] = L^2$.
Now,the equation is $E = \frac{B - x^2}{At}$. Rearranging for $A$,we get $A = \frac{B - x^2}{Et}$.
Substituting the dimensions: $[A] = \frac{[L^2]}{[E][t]}$.
Given $[E] = M^1 L^2 T^{-2}$ and $[t] = T^1$,we have $[A] = \frac{L^2}{(M^1 L^2 T^{-2})(T^1)} = \frac{L^2}{M^1 L^2 T^{-1}} = M^{-1} T^1$.
Finally,the dimension of $AB$ is $[A][B] = (M^{-1} T^1)(L^2) = L^2 M^{-1} T^1$.
224
DifficultMCQ
The equation of a stationary wave is given by $y = 2a \sin \left( \frac{2 \pi nt}{\lambda} \right) \cos \left( \frac{2 \pi x}{\lambda} \right)$. Which of the following statements is $NOT$ correct?
A
The dimensions of $nt$ are $[L]$.
B
The dimensions of $n$ are $[LT^{-1}]$.
C
The dimensions of $n/\lambda$ are $[T^{-1}]$.
D
The dimensions of $x$ are $[L]$.

Solution

(C) In the argument of the sine function,$\frac{2 \pi nt}{\lambda}$ must be dimensionless. Since $2\pi$ is dimensionless,the dimensions of $\frac{nt}{\lambda}$ must be $[M^0 L^0 T^0]$.
This implies $[nt] = [\lambda] = [L]$. Thus,option $A$ is correct.
Given $[nt] = [L]$ and $[t] = [T]$,we have $[n] = [L/T] = [LT^{-1}]$. Thus,option $B$ is correct.
For the cosine function,$\frac{2 \pi x}{\lambda}$ must be dimensionless,so $[x] = [\lambda] = [L]$. Thus,option $D$ is correct.
Now,consider the dimensions of $n/\lambda$. Since $[n] = [LT^{-1}]$ and $[\lambda] = [L]$,we have $[n/\lambda] = [LT^{-1}] / [L] = [T^{-1}]$.
Option $C$ states the dimensions are $[T]$,which is incorrect. Therefore,option $C$ is the correct answer.
225
DifficultMCQ
Applying the principle of homogeneity of dimensions,determine which one is correct. Where $T$ is time period,$G$ is gravitational constant,$M$ is mass,and $r$ is the radius of the orbit.
A
$T^2 = \frac{4 \pi^2 r}{GM^2}$
B
$T^2 = 4 \pi^2 r^3$
C
$T^2 = \frac{4 \pi^2 r^3}{GM}$
D
$T^2 = \frac{4 \pi^2 r^2}{GM}$

Solution

(C) According to the principle of homogeneity,the dimensions of the $LHS$ must be equal to the dimensions of the $RHS$.
For option $C$: $T^2 = \frac{4 \pi^2 r^3}{GM}$
Dimension of $LHS = [T^2]$
Dimension of $RHS = \frac{[L]^3}{[M^{-1} L^3 T^{-2}] [M]} = \frac{[L^3]}{[L^3 T^{-2}]} = [T^2]$
Since the dimensions of $LHS$ and $RHS$ are equal,option $C$ is correct.
226
DifficultMCQ
What is the dimensional formula of $ab^{-1}$ in the equation $(P+\frac{a}{V^2})(V-b)=RT$,where letters have their usual meaning?
A
$[M L^5 T^{-2}]$
B
$[M L^2 T^{-2}]$
C
$[M^{-1} L^5 T^3]$
D
$[M^6 L^7 T^4]$

Solution

(B) According to the principle of homogeneity of dimensions,only physical quantities of the same dimension can be added or subtracted.
In the term $(V-b)$,since $V$ is volume,the dimension of $b$ must be equal to the dimension of $V$.
Therefore,$[b] = [L^3]$.
In the term $(P + \frac{a}{V^2})$,the dimension of $\frac{a}{V^2}$ must be equal to the dimension of pressure $P$.
$[P] = [M L^{-1} T^{-2}]$.
So,$[a] = [P] \times [V^2] = [M L^{-1} T^{-2}] \times [L^3]^2 = [M L^{-1} T^{-2}] \times [L^6] = [M L^5 T^{-2}]$.
Now,we need the dimensional formula of $ab^{-1} = \frac{a}{b}$.
$\frac{[a]}{[b]} = \frac{[M L^5 T^{-2}]}{[L^3]} = [M L^2 T^{-2}]$.
227
MediumMCQ
$A$ force defined by $F = \alpha t^2 + \beta t$ acts on a particle at a given time $t$. The factor which is dimensionless,if $\alpha$ and $\beta$ are constants,is:
A
$\alpha t / \beta$
B
$\alpha \beta t$
C
$\alpha \beta / t$
D
$\beta t / \alpha$

Solution

(A) According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.
Therefore,$[F] = [\alpha t^2] = [\beta t]$.
From $[\alpha t^2] = [\beta t]$,we can write:
$\frac{[\alpha]}{[\beta]} = \frac{[t]}{[t^2]} = \frac{1}{[t]}$.
Multiplying both sides by $[t]$,we get:
$\frac{[\alpha][t]}{[\beta]} = 1$.
This implies that the quantity $\frac{\alpha t}{\beta}$ is dimensionless.
228
AdvancedMCQ
$A$ length-scale $(l)$ depends on the permittivity $(\varepsilon)$ of a dielectric material,the Boltzmann constant $(k_B)$,the absolute temperature $(T)$,the number density $(n)$ of certain charged particles,and the charge $(q)$ carried by each of the particles. Which of the following expression$(s)$ for $l$ is(are) dimensionally correct?
$(A)$ $l=\sqrt{\left(\frac{n q^2}{\varepsilon k_B T}\right)}$
$(B)$ $l=\sqrt{\left(\frac{\varepsilon k_B T}{n q^2}\right)}$
$(C)$ $l=\sqrt{\left(\frac{q^2}{\varepsilon n^{2 / 3} k_B T}\right)}$
$(D)$ $l=\sqrt{\left(\frac{q^2}{\varepsilon n^{1 / 3} k_B T}\right)}$
A
$A, B$
B
$B, C$
C
$C, A$
D
$B, D$

Solution

(D) To determine the dimensionally correct expression,we analyze the dimensions of the given physical quantities:
$[\varepsilon] = [M^{-1} L^{-3} T^4 A^2]$
$[k_B T] = [Energy] = [M L^2 T^{-2}]$
$[q^2] = [A^2 T^2]$
$[n] = [L^{-3}]$
First,consider the expression $\frac{\varepsilon k_B T}{q^2}$:
$\left[\frac{\varepsilon k_B T}{q^2}\right] = \frac{[M^{-1} L^{-3} T^4 A^2] [M L^2 T^{-2}]}{[A^2 T^2]} = [L^{-1}]$
Now,check the options:
For $(B)$: $l = \sqrt{\frac{\varepsilon k_B T}{n q^2}} \Rightarrow [l] = \sqrt{\frac{L^{-1}}{L^{-3}}} = \sqrt{L^2} = [L]$. This is dimensionally correct.
For $(D)$: $l = \sqrt{\frac{q^2}{\varepsilon n^{1/3} k_B T}} \Rightarrow [l] = \sqrt{\frac{1}{L^{-1} (L^{-3})^{1/3}}} = \sqrt{\frac{1}{L^{-1} L^{-1}}} = \sqrt{L^2} = [L]$. This is dimensionally correct.
Options $(A)$ and $(C)$ result in dimensions of $[L^{-2}]$ and $[L^{3/2}]$ respectively,which are incorrect for a length scale. Thus,$(B)$ and $(D)$ are correct.
229
EasyMCQ
Young's modulus of elasticity $Y$ is expressed in terms of three derived quantities,namely,the gravitational constant $G$,Planck's constant $h$,and the speed of light $c$,as $Y = c^\alpha h^\beta G^\gamma$. Which of the following is the correct option?
A
$\alpha = 7, \beta = -1, \gamma = -2$
B
$\alpha = -7, \beta = -1, \gamma = -2$
C
$\alpha = 7, \beta = -1, \gamma = 2$
D
$\alpha = -7, \beta = 1, \gamma = -2$

Solution

(A) The dimensional formula for Young's modulus $Y$ is $[M^1 L^{-1} T^{-2}]$.
The dimensional formulas for the given constants are:
Speed of light $c = [L T^{-1}]$
Planck's constant $h = [M L^2 T^{-1}]$
Gravitational constant $G = [M^{-1} L^3 T^{-2}]$
Given the relation $Y = c^\alpha h^\beta G^\gamma$,we equate the dimensions:
$[M^1 L^{-1} T^{-2}] = [L T^{-1}]^\alpha [M L^2 T^{-1}]^\beta [M^{-1} L^3 T^{-2}]^\gamma$
$[M^1 L^{-1} T^{-2}] = [M^{\beta - \gamma} L^{\alpha + 2\beta + 3\gamma} T^{-\alpha - \beta - 2\gamma}]$
Comparing the powers of $M, L,$ and $T$:
$1$) $\beta - \gamma = 1$
$2$) $\alpha + 2\beta + 3\gamma = -1$
$3$) $-\alpha - \beta - 2\gamma = -2$
Adding equations $(2)$ and $(3)$:
$(\alpha + 2\beta + 3\gamma) + (-\alpha - \beta - 2\gamma) = -1 + (-2)$
$\beta + \gamma = -3$
Now,solving $\beta - \gamma = 1$ and $\beta + \gamma = -3$:
Adding them gives $2\beta = -2 \Rightarrow \beta = -1$.
Substituting $\beta = -1$ into $\beta - \gamma = 1$ gives $-1 - \gamma = 1 \Rightarrow \gamma = -2$.
Substituting $\beta = -1$ and $\gamma = -2$ into equation $(2)$:
$\alpha + 2(-1) + 3(-2) = -1$
$\alpha - 2 - 6 = -1$
$\alpha - 8 = -1 \Rightarrow \alpha = 7$.
Thus,the correct values are $\alpha = 7, \beta = -1, \gamma = -2$.
230
AdvancedMCQ
$A$ dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing fixed positive ions surrounded by free electrons can be treated as neutral plasma. Let $N$ be the number density of free electrons, each of mass $m$. When the electrons are subjected to an electric field, they are displaced relatively away from the heavy positive ions. If the electric field becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency $\omega_p$, which is called the plasma frequency. To sustain the oscillations, a time-varying electric field needs to be applied that has an angular frequency $\omega$, where a part of the energy is absorbed and a part of it is reflected. As $\omega$ approaches $\omega_p$, all the free electrons are set to resonance together and all the energy is reflected. This is the explanation of high reflectivity of metals.
$1.$ Taking the electronic charge as $e$ and the permittivity as $\varepsilon_0$, use dimensional analysis to determine the correct expression for $\omega_p$.
$(A) \sqrt{\frac{N e}{m \varepsilon_0}}$ $(B) \sqrt{\frac{m \varepsilon_0}{N e}}$ $(C) \sqrt{\frac{N e^2}{m \varepsilon_0}}$ $(D) \sqrt{\frac{m \varepsilon_0}{N e^2}}$
$2.$ Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons $N \approx 4 \times 10^{27} \ m^{-3}$. Take $\varepsilon_0 \approx 10^{-11}$ and $m \approx 10^{-30}$, where these quantities are in proper $SI$ units.
$(A) 800 \ nm$ $(B) 600 \ nm$ $(C) 300 \ nm$ $(D) 200 \ nm$
Give the answer for question $1$ and $2$.
A
$(B, D)$
B
$(C, B)$
C
$(A, C)$
D
$(B, A)$

Solution

(B) $1.$ The dimension of angular frequency is $[\omega] = T^{-1}$.
The dimension of permittivity is $[\varepsilon_0] = M^{-1} L^{-3} T^4 A^2$. Since $I = Q/T$, we use $[e] = Q = AT$.
$[m] = M$, $[N] = L^{-3}$, $[e^2] = Q^2$.
Checking option $(C)$: $\left[ \frac{N e^2}{m \varepsilon_0} \right] = \frac{L^{-3} Q^2}{M (M^{-1} L^{-3} T^2 Q^{-2})} = \frac{L^{-3} Q^2}{L^{-3} T^2} = Q^2 T^{-2} = (AT)^2 T^{-2} = A^2 = T^{-2}$ (in terms of charge units).
More simply, using $[\varepsilon_0] = \frac{Q^2 T^2}{M L^3}$, we get $\left[ \frac{N e^2}{m \varepsilon_0} \right] = \frac{L^{-3} Q^2}{M (Q^2 T^2 / M L^3)} = T^{-2}$.
Thus, $\sqrt{\frac{N e^2}{m \varepsilon_0}}$ has dimensions of $T^{-1}$, which is $\omega$.
$2.$ At resonance, $\omega = \omega_p = \sqrt{\frac{N e^2}{m \varepsilon_0}}$.
Given $N = 4 \times 10^{27}$, $e = 1.6 \times 10^{-19}$, $m = 10^{-30}$, $\varepsilon_0 = 10^{-11}$.
$\omega_p = \sqrt{\frac{4 \times 10^{27} \times (1.6 \times 10^{-19})^2}{10^{-30} \times 10^{-11}}} = \sqrt{\frac{4 \times 10^{27} \times 2.56 \times 10^{-38}}{10^{-41}}} = \sqrt{10.24 \times 10^{30}} = 3.2 \times 10^{15} \ rad/s$.
Frequency $f = \frac{\omega}{2\pi} = \frac{3.2 \times 10^{15}}{2 \times 3.14} \approx 0.5 \times 10^{15} \ Hz$.
Wavelength $\lambda = \frac{c}{f} = \frac{3 \times 10^8}{0.5 \times 10^{15}} = 6 \times 10^{-7} \ m = 600 \ nm$.
231
AdvancedMCQ
In electromagnetic theory, electric and magnetic phenomena are related to each other. Therefore, the dimensions of electric and magnetic quantities must also be related. In the questions below, $[E]$ and $[B]$ stand for dimensions of electric and magnetic fields respectively, while $[\varepsilon_0]$ and $[\mu_0]$ stand for dimensions of the permittivity and permeability of free space respectively. $[L]$ and $[T]$ are dimensions of length and time respectively. All quantities are in $SI$ units.
$(1)$ The relation between $[E]$ and $[B]$ is:
$(A)$ $[E] = [B][L][T]$
$(B)$ $[E] = [B][L]^{-1}[T]$
$(C)$ $[E] = [B][L][T]^{-1}$
$(D)$ $[E] = [B][L]^{-1}[T]^{-1}$
$(2)$ The relation between $[\varepsilon_0]$ and $[\mu_0]$ is:
$(A)$ $[\mu_0] = [\varepsilon_0][L]^2[T]^{-2}$
$(B)$ $[\mu_0] = [\varepsilon_0][L]^{-2}[T]^2$
$(C)$ $[\mu_0] = [\varepsilon_0]^{-1}[L]^2[T]^{-2}$
$(D)$ $[\mu_0] = [\varepsilon_0]^{-1}[L]^{-2}[T]^2$
Give the answers for questions $(1)$ and $(2)$.
A
$C, D$
B
$C, A$
C
$C, B$
D
$B, C, D$

Solution

$(C, D)$ $(1)$ The Lorentz force on a charge $q$ moving with velocity $v$ is given by $F = qE + q(v \times B)$. For the dimensions to be consistent, the magnitude of the electric force $qE$ must be equal to the magnitude of the magnetic force $qvB$.
Thus, $qE = qvB$, which implies $E = vB$.
Since the dimension of velocity $v$ is $[L][T]^{-1}$, we have $[E] = [L][T]^{-1}[B]$. This matches option $(C)$.
$(2)$ The speed of light $c$ in free space is related to permittivity $\varepsilon_0$ and permeability $\mu_0$ by the relation $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
Squaring both sides, we get $c^2 = \frac{1}{\mu_0 \varepsilon_0}$, or $\mu_0 = \frac{1}{\varepsilon_0 c^2}$.
Since the dimension of $c$ is $[L][T]^{-1}$, the dimension of $c^2$ is $[L]^2[T]^{-2}$.
Therefore, $[\mu_0] = [\varepsilon_0]^{-1} ([L]^2[T]^{-2})^{-1} = [\varepsilon_0]^{-1}[L]^{-2}[T]^2$. This matches option $(D)$.
232
EasyMCQ
Let us consider a system of units in which mass and angular momentum are dimensionless. If length has dimension of $L$,which of the following statement$(s)$ is/are correct?
$(1)$ The dimension of force is $L^{-3}$
$(2)$ The dimension of energy is $L^{-2}$
$(3)$ The dimension of power is $L^{-5}$
$(4)$ The dimension of linear momentum is $L^{-1}$
A
$1, 2, 4$
B
$1, 2, 3$
C
$1, 2$
D
$1, 3$

Solution

(A) Given that mass $(M)$ and angular momentum $(L_{ang} = Mvr)$ are dimensionless,we have:
$M = M^0 L^0 T^0$
$L_{ang} = M^1 L^2 T^{-1} = M^0 L^0 T^0$
Since $M$ is dimensionless,$L^2 T^{-1} = 1$,which implies $T = L^2$.
$(1)$ Force $(F = M L T^{-2})$: Since $M$ is dimensionless and $T = L^2$,$F = L^1 (L^2)^{-2} = L^1 L^{-4} = L^{-3}$. (Correct)
$(2)$ Energy $(E = M L^2 T^{-2})$: $E = L^2 (L^2)^{-2} = L^2 L^{-4} = L^{-2}$. (Correct)
$(3)$ Power $(P = E/T = M L^2 T^{-3})$: $P = L^2 (L^2)^{-3} = L^2 L^{-6} = L^{-4}$. (Incorrect)
$(4)$ Linear momentum $(p = M L T^{-1})$: $p = L^1 (L^2)^{-1} = L^1 L^{-2} = L^{-1}$. (Correct)
Thus,statements $(1), (2),$ and $(4)$ are correct.
233
AdvancedMCQ
Sometimes it is convenient to construct a system of units so that all quantities can be expressed in terms of only one physical quantity. In one such system, dimensions of different quantities are given in terms of a quantity $X$ as follows: $[position] = [X^\alpha]$; $[speed] = [X^\beta]$; $[acceleration] = [X^p]$; $[linear momentum] = [X^q]$; $[force] = [X^r]$. Then -
$(A)$ $\alpha + p = 2\beta$
$(B)$ $p + q - r = \beta$
$(C)$ $p - q + r = \alpha$
$(D)$ $p + q + r = \beta$
A
$A, B$
B
$A, C$
C
$A, D$
D
$B, C$

Solution

(A) Let the dimensions of the fundamental quantity $X$ be $[X] = [M^a L^b T^c]$.
Given:
$[L] = [X^\alpha] = [M^{a\alpha} L^{b\alpha} T^{c\alpha}] \implies a\alpha = 0, b\alpha = 1, c\alpha = 0 \implies a=0, c=0, b=1/\alpha$.
$[LT^{-1}] = [X^\beta] = [M^{a\beta} L^{b\beta} T^{c\beta}] \implies b\beta = 1, c\beta = -1 \implies \beta = \beta(b) = 1/\alpha \implies \beta = 1/\alpha \implies b\beta = 1$.
From $[LT^{-2}] = [X^p]$, we have $b p = 1$ and $c p = -2$. Since $b = 1/\alpha$, $p = \alpha$. Also $c = -2/p = -2/\alpha$.
From $[MLT^{-1}] = [X^q]$, we have $a q = 1, b q = 1, c q = -1$.
From $[MLT^{-2}] = [X^r]$, we have $a r = 1, b r = 1, c r = -2$.
Using these relations, we find:
$1$) $\alpha + p = 2\beta$ is correct because $[L] [LT^{-2}] = [L^2 T^{-2}] = [LT^{-1}]^2$.
$2$) $p + q - r = \beta$ is correct as $[LT^{-2}] [MLT^{-1}] / [MLT^{-2}] = [LT^{-1}]$.
Thus, options $(A)$ and $(B)$ are correct.
234
AdvancedMCQ
To find the distance $d$ over which a signal can be seen clearly in foggy conditions,a railway engineer uses dimensional analysis and assumes that the distance depends on the mass density $\rho$ of the fog,intensity (power/area) $S$ of the light from the signal,and its frequency $f$. The engineer finds that $d$ is proportional to $S^{1/n}$. The value of $n$ is:
A
$3$
B
$4$
C
$5$
D
$6$

Solution

(A) Let the distance $d$ be given by $d = k \rho^a S^b f^c$,where $k$ is a dimensionless constant.
The dimensions are:
$[d] = [L]$
$[\rho] = [M L^{-3}]$
$[S] = [Power/Area] = [M L^2 T^{-3} / L^2] = [M T^{-3}]$
$[f] = [T^{-1}]$
Substituting these into the equation:
$[L]^1 = [M L^{-3}]^a [M T^{-3}]^b [T^{-1}]^c$
$[L]^1 = M^{a+b} L^{-3a} T^{-3b-c}$
Comparing the powers of $M, L, T$ on both sides:
For $M: a + b = 0 \Rightarrow a = -b$
For $L: -3a = 1 \Rightarrow a = -1/3$
Thus,$b = 1/3$
For $T: -3b - c = 0 \Rightarrow c = -3b = -3(1/3) = -1$
Since $d \propto S^b$ and $b = 1/3$,we have $d \propto S^{1/3}$.
Comparing this with $d \propto S^{1/n}$,we get $n = 3$.
235
DifficultMCQ
Planck's constant $h$,speed of light $c$,and gravitational constant $G$ are used to form a unit of length $L$ and a unit of mass $M$. Then the correct option$(s)$ is(are):
$(A)$ $M \propto \sqrt{c}$
$(B)$ $M \propto \sqrt{G}$
$(C)$ $L \propto \sqrt{h}$
$(D)$ $L \propto \sqrt{G}$
A
$(A, B, C)$
B
$(A, B, D)$
C
$(A, C, D)$
D
$(B, C, D)$

Solution

(C) The dimensions of the constants are:
$h = [M L^2 T^{-1}]$
$c = [L T^{-1}]$
$G = [M^{-1} L^3 T^{-2}]$
To find the mass $M$,we assume $M = k h^a c^b G^d$. Substituting dimensions:
$[M] = [M L^2 T^{-1}]^a [L T^{-1}]^b [M^{-1} L^3 T^{-2}]^d$
$[M] = M^{a-d} L^{2a+b+3d} T^{-a-b-2d}$
Comparing powers:
$a - d = 1 \implies a = 1 + d$
$2a + b + 3d = 0$
$-a - b - 2d = 0 \implies b = -a - 2d = -(1+d) - 2d = -1 - 3d$
Substituting into the second equation: $2(1+d) + (-1-3d) + 3d = 0 \implies 2 + 2d - 1 = 0 \implies d = -1/2$.
Then $a = 1 - 1/2 = 1/2$ and $b = -1 - 3(-1/2) = 1/2$.
Thus,$M \propto h^{1/2} c^{1/2} G^{-1/2} = \sqrt{\frac{hc}{G}}$.
Similarly for length $L = k h^x c^y G^z$:
$[L] = [M L^2 T^{-1}]^x [L T^{-1}]^y [M^{-1} L^3 T^{-2}]^z$
$x - z = 0 \implies x = z$
$2x + y + 3z = 1$
$-x - y - 2z = 0 \implies y = -x - 2z = -3z$
$2z - 3z + 3z = 1 \implies 2z = 1 \implies z = 1/2$.
Thus $x = 1/2, y = -3/2, z = 1/2$.
$L \propto h^{1/2} c^{-3/2} G^{1/2} = \sqrt{\frac{hG}{c^3}}$.
From these relations:
$M \propto \sqrt{h}, M \propto \sqrt{c}, M \propto 1/\sqrt{G}$
$L \propto \sqrt{h}, L \propto \sqrt{G}, L \propto 1/\sqrt{c^3}$
Therefore,options $(A), (C), (D)$ are correct.
236
MediumMCQ
In terms of potential difference $V$,electric current $I$,permittivity $\varepsilon_0$,permeability $\mu_0$,and speed of light $c$,the dimensionally correct equation$(s)$ is(are):
$(A)$ $\mu_0 I^2 = \varepsilon_0 V^2$
$(B)$ $\varepsilon_0 I = \mu_0 V$
$(C)$ $I = \varepsilon_0 cV$
$(D)$ $\mu_0 cI = V$
A
$(B, D)$
B
$(B, C)$
C
$(A, C)$
D
$(A, D)$

Solution

(C) We know that the energy density in an electric field is $u_E = \frac{1}{2} \varepsilon_0 E^2$ and in a magnetic field is $u_B = \frac{1}{2} \frac{B^2}{\mu_0}$.
Since $u_E = u_B$,we have $\varepsilon_0 E^2 = \frac{B^2}{\mu_0}$,which implies $\frac{E^2}{B^2} = \frac{1}{\mu_0 \varepsilon_0} = c^2$.
Using $V = Ed$ and $B = \frac{\mu_0 I}{2\pi r}$,we can relate these quantities.
Alternatively,using dimensional analysis:
$[\varepsilon_0] = M^{-1} L^{-3} T^4 A^2$,$[\mu_0] = M L T^{-2} A^{-2}$,$[V] = M L^2 T^{-3} A^{-1}$,$[I] = A$,$[c] = L T^{-1}$.
For $(A)$: $[\mu_0 I^2] = (M L T^{-2} A^{-2})(A^2) = M L T^{-2}$. $[\varepsilon_0 V^2] = (M^{-1} L^{-3} T^4 A^2)(M^2 L^4 T^{-6} A^{-2}) = M L T^{-2}$. Thus,$(A)$ is dimensionally correct.
For $(C)$: $[I] = A$. $[\varepsilon_0 c V] = (M^{-1} L^{-3} T^4 A^2)(L T^{-1})(M L^2 T^{-3} A^{-1}) = A$. Thus,$(C)$ is dimensionally correct.
Therefore,the correct options are $(A)$ and $(C)$.
237
DifficultMCQ
In a particular system of units,a physical quantity can be expressed in terms of the electric charge $e$,electron mass $m_e$,Planck's constant $h$,and Coulomb's constant $k = \frac{1}{4 \pi \epsilon_0}$,where $\epsilon_0$ is the permittivity of vacuum. In terms of these physical constants,the dimension of the magnetic field is $[B] = [e]^\alpha [m_e]^\beta [h]^\gamma [k]^\delta$. The value of $\alpha + \beta + \gamma + \delta$ is . . . . .
A
$3$
B
$4$
C
$5$
D
$6$

Solution

(B) The dimensional formula for magnetic field $B$ is $[M^1 T^{-2} A^{-1}]$.
The dimensions of the constants are:
$[e] = [A^1 T^1]$
$[m_e] = [M^1]$
$[h] = [M^1 L^2 T^{-1}]$
$[k] = [M^1 L^3 T^{-4} A^{-2}]$
Equating the dimensions:
$[M^1 T^{-2} A^{-1}] = [A^1 T^1]^\alpha [M^1]^\beta [M^1 L^2 T^{-1}]^\gamma [M^1 L^3 T^{-4} A^{-2}]^\delta$
$[M^1 T^{-2} A^{-1}] = M^{\beta + \gamma + \delta} L^{2\gamma + 3\delta} T^{\alpha - \gamma - 4\delta} A^{\alpha - 2\delta}$
Comparing the powers of $M, L, T, A$ on both sides:
$1) \beta + \gamma + \delta = 1$
$2) 2\gamma + 3\delta = 0$
$3) \alpha - \gamma - 4\delta = -2$
$4) \alpha - 2\delta = -1$
From $(4)$,$\alpha = 2\delta - 1$. Substituting into $(3)$:
$(2\delta - 1) - \gamma - 4\delta = -2 \implies -\gamma - 2\delta = -1 \implies \gamma + 2\delta = 1$.
Solving with $(2)$ $(2\gamma + 3\delta = 0)$:
$2(1 - 2\delta) + 3\delta = 0 \implies 2 - 4\delta + 3\delta = 0 \implies \delta = 2$.
Then $\gamma = 1 - 2(2) = -3$.
Then $\alpha = 2(2) - 1 = 3$.
Then $\beta = 1 - (-3) - 2 = 2$.
Thus,$\alpha + \beta + \gamma + \delta = 3 + 2 - 3 + 2 = 4$.
238
AdvancedMCQ
$A$ dimensionless quantity is constructed in terms of electronic charge $e$,permittivity of free space $\varepsilon_0$,Planck's constant $h$,and speed of light $c$. If the dimensionless quantity is written as $e^\alpha \varepsilon_0^\beta h^\gamma c^\delta$ and $n$ is a non-zero integer,then $(\alpha, \beta, \gamma, \delta)$ is given by
A
$(2n, -n, -n, -n)$
B
$(n, -n, -2n, -n)$
C
$(n, -n, -n, -2n)$
D
$(2n, -n, -2n, -2n)$

Solution

(A) For the quantity to be dimensionless,$e^\alpha \varepsilon_0^\beta h^\gamma c^\delta = M^0 L^0 T^0 A^0$.
Substituting the dimensions:
$[e] = AT$
$[\varepsilon_0] = M^{-1} L^{-3} T^4 A^2$
$[h] = M L^2 T^{-1}$
$[c] = L T^{-1}$
$(AT)^\alpha (M^{-1} L^{-3} T^4 A^2)^\beta (M L^2 T^{-1})^\gamma (L T^{-1})^\delta = M^0 L^0 T^0 A^0$
Comparing powers of $M, L, T, A$:
$M: -\beta + \gamma = 0 \Rightarrow \gamma = \beta$
$A: \alpha + 2\beta = 0 \Rightarrow \alpha = -2\beta$
$L: -3\beta + 2\gamma + \delta = 0 \Rightarrow -3\beta + 2\beta + \delta = 0 \Rightarrow \delta = \beta$
$T: \alpha + 4\beta - \gamma - \delta = 0 \Rightarrow -2\beta + 4\beta - \beta - \beta = 0$ (Consistent)
Let $\beta = -n$,then $\alpha = 2n, \gamma = -n, \delta = -n$.
Thus,$(\alpha, \beta, \gamma, \delta) = (2n, -n, -n, -n)$.
239
MediumMCQ
The position of a particle moving on the $x-$axis is given by $x(t) = A \sin t + B \cos^2 t + Ct^2 + D$,where $t$ is time. The dimension of $\frac{ABC}{D}$ is $-$
A
$L$
B
$L^3 T^{-2}$
C
$L^2 T^{-2}$
D
$L^2$

Solution

(C) The principle of homogeneity of dimensions states that only quantities with the same dimensions can be added or subtracted. Since $x(t)$ represents position,its dimension is $[L]$.
$1$. For the term $A \sin t$: The argument of a trigonometric function must be dimensionless. Thus,$[A] = [x] = [L]$.
$2$. For the term $B \cos^2 t$: Similarly,$[B] = [x] = [L]$.
$3$. For the term $Ct^2$: Since $[Ct^2] = [x] = [L]$,we have $[C] [T^2] = [L]$,which implies $[C] = [L T^{-2}]$.
$4$. For the term $D$: Since $D$ is added to $x(t)$,$[D] = [x] = [L]$.
Now,calculate the dimension of $\frac{ABC}{D}$:
$\left[ \frac{ABC}{D} \right] = \frac{[L] \times [L] \times [L T^{-2}]}{[L]} = [L^2 T^{-2}]$.
240
MediumMCQ
In a measurement,it is asked to find the modulus of elasticity per unit torque applied on the system. The measured quantity has the dimension of $[M^a L^b T^c]$. If $b = 3$,the value of $c$ is . . . . . . .
A
$1$
B
$2$
C
$0$
D
$7$

Solution

(C) The modulus of elasticity $(Y)$ has dimensions of stress,which is $[M^1 L^{-1} T^{-2}]$.
Torque $( au)$ has dimensions of force $\times$ distance,which is $[M^1 L^2 T^{-2}]$.
The quantity to be measured is $\frac{Y}{\tau}$.
Dimensions of $\frac{Y}{\tau} = \frac{[M^1 L^{-1} T^{-2}]}{[M^1 L^2 T^{-2}]} = [M^0 L^{-3} T^0]$.
Comparing this with $[M^a L^b T^c]$,we get $a = 0$,$b = -3$,and $c = 0$.
Note: The question states $b = 3$,which implies a magnitude comparison or a specific convention. Given the standard dimensional analysis,$c$ remains $0$.
241
MediumMCQ
The expression given below shows the variation of velocity $(v)$ with time $(t)$,$v=At^2+\frac{Bt}{C+t}$. The dimension of $ABC$ is
A
$[M^0 L^2 T^{-3}]$
B
$[M^0 L^1 T^{-3}]$
C
$[M^0 L^1 T^{-2}]$
D
$[M^0 L^2 T^{-2}]$

Solution

(A) Given the equation $v = At^2 + \frac{Bt}{C+t}$.
According to the principle of homogeneity of dimensions,the dimensions of each term added must be the same as the dimension of velocity $[LT^{-1}]$.
For the term $\frac{Bt}{C+t}$,the dimension of $C$ must be equal to the dimension of $t$,so $[C] = [T]$.
Now,for the term $\frac{Bt}{C+t}$,the dimension is $\frac{[B][T]}{[T]} = [B]$. Since this must equal $[LT^{-1}]$,we get $[B] = [LT^{-1}]$.
For the term $At^2$,the dimension is $[A][T^2]$. Since this must equal $[LT^{-1}]$,we get $[A] = [LT^{-3}]$.
Finally,the dimension of $ABC$ is $[A][B][C] = [LT^{-3}] \cdot [LT^{-1}] \cdot [T] = [L^2 T^{-3}]$.
242
MediumMCQ
The equation for a real gas is given by $(P + \frac{a}{V^2})(V - b) = RT$,where $P, V, T$ and $R$ are the pressure,volume,temperature,and gas constant,respectively. The dimension of $ab^{-2}$ is equivalent to that of
A
Planck's constant
B
Compressibility
C
Strain
D
Energy density

Solution

(D) The given equation is $(P + \frac{a}{V^2})(V - b) = RT$.
According to the principle of homogeneity of dimensions,dimensions of terms added or subtracted must be the same.
Therefore,$[P] = [\frac{a}{V^2}] \implies [a] = [P][V^2]$.
Since $[P] = ML^{-1}T^{-2}$ and $[V] = L^3$,we have $[a] = (ML^{-1}T^{-2})(L^6) = ML^5T^{-2}$.
Also,$[b] = [V] = L^3$.
Now,the dimension of $ab^{-2}$ is $[a][b]^{-2} = (ML^5T^{-2})(L^3)^{-2} = (ML^5T^{-2})(L^{-6}) = ML^{-1}T^{-2}$.
This dimension $ML^{-1}T^{-2}$ is the same as the dimension of pressure or energy density (Energy/Volume = $ML^2T^{-2} / L^3 = ML^{-1}T^{-2}$).
243
MediumMCQ
Given a charge $q$,current $I$ and permeability of vacuum $\mu_0$. Which of the following quantity has the dimension of momentum?
A
$qI / \mu_0$
B
$q \mu_0 I$
C
$q^2 \mu_0 I$
D
$q \mu_0 / I$

Solution

(B) The dimension of momentum $P$ is $[MLT^{-1}]$.
Given dimensions are: $[q] = [AT]$,$[I] = [A]$,and $[\mu_0] = [MLT^{-2}A^{-2}]$.
Let the dimension of the quantity be $[P] = [q]^x [\mu_0]^y [I]^z$.
Substituting the dimensions: $[MLT^{-1}] = [AT]^x [MLT^{-2}A^{-2}]^y [A]^z$.
$[MLT^{-1}] = [M^y L^y T^{x-2y} A^{x-2y+z}]$.
Comparing the powers of $M, L, T,$ and $A$ on both sides:
For $M$: $y = 1$.
For $L$: $y = 1$.
For $T$: $x - 2y = -1 \Rightarrow x - 2(1) = -1 \Rightarrow x = 1$.
For $A$: $x - 2y + z = 0 \Rightarrow 1 - 2(1) + z = 0 \Rightarrow -1 + z = 0 \Rightarrow z = 1$.
Thus,the required quantity is $[q^1 \mu_0^1 I^1] = [q \mu_0 I]$.
244
MediumMCQ
If $\mu_0$ and $\varepsilon_0$ are the permeability and permittivity of free space,respectively,then the dimension of $\left(\frac{1}{\mu_0 \varepsilon_0}\right)$ is
A
$L^2 / T^2$
B
$L / T^2$
C
$T^2 / L^2$
D
$L^2 / T$

Solution

(A) The speed of light $c$ in free space is given by the relation $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.
Squaring both sides,we get $c^2 = \frac{1}{\mu_0 \varepsilon_0}$.
The dimension of speed $c$ is $[L T^{-1}]$.
Therefore,the dimension of $c^2$ is $[L T^{-1}]^2 = [L^2 T^{-2}]$.
Thus,the dimension of $\left(\frac{1}{\mu_0 \varepsilon_0}\right)$ is $L^2 / T^2$.
245
MediumMCQ
In an electromagnetic system,the quantity representing the ratio of electric flux and magnetic flux has dimensions of $M^{P} L^{Q} T^{R} A^{S}$. The values of $Q$ and $R$ are:
A
$(3, -5)$
B
$(-2, 2)$
C
$(-2, 1)$
D
$(1, -1)$

Solution

(D) The ratio of electric flux $\phi_{E}$ to magnetic flux $\phi_{M}$ is given by $\frac{\phi_{E}}{\phi_{M}} = \frac{E \cdot A}{B \cdot A} = \frac{E}{B}$.
Since $E = c \cdot B$ (where $c$ is the speed of light),the ratio $\frac{E}{B} = c$.
The dimensions of speed $c$ are $[L T^{-1}]$.
Comparing this with $M^{P} L^{Q} T^{R} A^{S}$,we have $M^{0} L^{1} T^{-1} A^{0}$.
Thus,$Q = 1$ and $R = -1$.
246
DifficultMCQ
The dimensional formulas for acceleration,velocity,and length are $\alpha \beta^{-2}$,$\alpha \beta^{-1}$,and $\alpha \gamma$ respectively. What is the dimensional formula for the coefficient of friction?
A
$\alpha \beta \gamma$
B
$\alpha^0 \beta^{-1} \gamma^0$
C
$\alpha^{-1} \beta^0 \gamma^0$
D
$\alpha^0 \beta^0 \gamma^0$

Solution

(D) Given dimensional formulas:
$[a] = \alpha \beta^{-2} = LT^{-2}$
$[v] = \alpha \beta^{-1} = LT^{-1}$
$[\ell] = \alpha \gamma = L$
From $[v] = \alpha \beta^{-1} = LT^{-1}$ and $[a] = \alpha \beta^{-2} = LT^{-2}$,we can divide the two:
$\frac{[v]}{[a]} = \frac{\alpha \beta^{-1}}{\alpha \beta^{-2}} = \frac{LT^{-1}}{LT^{-2}} = T$
Thus,$[\beta] = T$.
Substituting $[\beta] = T$ into $[v] = \alpha \beta^{-1} = LT^{-1}$:
$\alpha T^{-1} = LT^{-1} \Rightarrow [\alpha] = L$.
Substituting $[\alpha] = L$ into $[\ell] = \alpha \gamma = L$:
$L \gamma = L \Rightarrow [\gamma] = 1$ (dimensionless).
The coefficient of friction is a ratio of two forces (frictional force and normal force),so it is a dimensionless quantity,represented by $[M^0 L^0 T^0]$.
Checking the options:
Option $(D)$ is $\alpha^0 \beta^0 \gamma^0 = 1$,which is dimensionless. Therefore,the correct option is $(D)$.
247
MediumMCQ
Given that the displacement of an oscillating particle is given by $y = A \sin(Bx + Ct + D)$. The dimensional formula for $ABCD$ is: (Here $x$ is position,$t$ is time)
A
$[M^0 L^{-1} T^0]$
B
$[M^0 L^0 T^0]$
C
$[M^0 L^{-1} T^{-1}]$
D
$[M^0 L^0 T^{-1}]$

Solution

(D) The argument of the sine function,$(Bx + Ct + D)$,must be dimensionless.
Therefore,each term must be dimensionless.
For $Bx$: $[B][x] = [M^0 L^0 T^0] \implies [B] = [L^{-1}]$.
For $Ct$: $[C][t] = [M^0 L^0 T^0] \implies [C] = [T^{-1}]$.
For $D$: $[D] = [M^0 L^0 T^0]$ (dimensionless).
For $A$: Since $y$ is displacement,$[A] = [L]$.
Now,calculate the dimensions of $ABCD$:
$[ABCD] = [A][B][C][D] = [L] \cdot [L^{-1}] \cdot [T^{-1}] \cdot [1] = [T^{-1}] = [M^0 L^0 T^{-1}]$.
248
DifficultMCQ
$A$ gas bubble formed from an explosion under water oscillates with a period $T$,proportional to $P^{a} d^{b} E^{c}$,where $P$ is the static pressure,$d$ is the density of water,and $E$ is the energy of explosion. Then $a, b, c$ are,respectively:
A
$1, 1, 1$
B
$\frac{1}{3}, \frac{1}{2}, \frac{-5}{6}$
C
$\frac{-5}{6}, \frac{1}{2}, \frac{1}{3}$
D
$\frac{1}{2}, \frac{-5}{6}, \frac{1}{3}$

Solution

(C) The period $T$ is given by $T \propto P^{a} d^{b} E^{c}$.
Dimensional formulas are: $[T] = [T^1]$,$[P] = [M^1 L^{-1} T^{-2}]$,$[d] = [M^1 L^{-3}]$,and $[E] = [M^1 L^2 T^{-2}]$.
Substituting these into the equation: $[T^1] = [M^1 L^{-1} T^{-2}]^{a} [M^1 L^{-3}]^{b} [M^1 L^2 T^{-2}]^{c}$.
Equating the powers of $M, L,$ and $T$ on both sides:
For $M: a + b + c = 0$ (Eq. $1$)
For $L: -a - 3b + 2c = 0$ (Eq. $2$)
For $T: -2a - 2c = 1$ (Eq. $3$)
From Eq. $3$,$a + c = -1/2$,so $c = -1/2 - a$.
Substitute $c$ into Eq. $1$: $a + b + (-1/2 - a) = 0 \implies b = 1/2$.
Substitute $b = 1/2$ into Eq. $2$: $-a - 3(1/2) + 2c = 0 \implies -a + 2c = 3/2$.
Substitute $c = -1/2 - a$ into this: $-a + 2(-1/2 - a) = 3/2 \implies -a - 1 - 2a = 3/2 \implies -3a = 5/2 \implies a = -5/6$.
Finally,$c = -1/2 - (-5/6) = -3/6 + 5/6 = 2/6 = 1/3$.
Thus,$a = -5/6, b = 1/2, c = 1/3$.
249
MediumMCQ
If $z = xP + G$,where $P$ is pressure and $G$ is the universal gravitational constant; then the dimensional formulas for $x$ and $z$ respectively are (here,$G = \frac{Fr^2}{m_1 m_2}$,$P = \frac{\text{Thrust}}{\text{Area}}$).
A
$MLT^{-2}, M^2 L^3 T$
B
$MLT, M^{-1} L^{-1} T$
C
$M^{-2} L^4 T^0, M^{-1} L^3 T^{-2}$
D
$M^2 L^4 T^0, M^1 L^3 T^2$

Solution

(C) According to the principle of homogeneity of dimensions,the dimensions of each term in an equation must be the same.
Given the equation $z = xP + G$,it implies that $[z] = [xP] = [G]$.
First,find the dimensions of $G$ (universal gravitational constant):
$[G] = \frac{[F][r^2]}{[m_1][m_2]} = \frac{[MLT^{-2}][L^2]}{[M][M]} = [M^{-1} L^3 T^{-2}]$.
Since $[z] = [G]$,the dimension of $z$ is $[M^{-1} L^3 T^{-2}]$.
Next,find the dimensions of $P$ (pressure):
$[P] = \frac{[Thrust]}{[Area]} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1} T^{-2}]$.
Since $[z] = [xP]$,we have $[x] = \frac{[z]}{[P]} = \frac{[M^{-1} L^3 T^{-2}]}{[ML^{-1} T^{-2}]} = [M^{-2} L^4 T^0]$.
Therefore,the dimensions of $x$ and $z$ are $[M^{-2} L^4 T^0]$ and $[M^{-1} L^3 T^{-2}]$ respectively.
250
MediumMCQ
In a nuclear power plant,the energy released depends on the mass of the uranium sample $(m)$,the length of the oscillator $(\ell)$,and the frequency $(f)$ of oscillation as $E = m^{x} \ell^{y} f^{z}$. Then,what is the value of $x + y + z$?
A
$1$
B
$5$
C
$2$
D
None of these

Solution

(B) The dimensional formula for energy $E$ is $[M^1 L^2 T^{-2}]$.
Given the relation $E = m^x \ell^y f^z$,we write the dimensional equation:
$[M^1 L^2 T^{-2}] = [M]^x [L]^y [T^{-1}]^z$
$[M^1 L^2 T^{-2}] = [M^x L^y T^{-z}]$
Comparing the powers of $M, L,$ and $T$ on both sides:
For $M$: $x = 1$
For $L$: $y = 2$
For $T$: $-z = -2 \Rightarrow z = 2$
Therefore,$x + y + z = 1 + 2 + 2 = 5$.

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