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Dimensions and Dimensional Formula Questions in English
Class 11 Physics · Units, Dimensions and Measurement · Dimensions and Dimensional Formula
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151
Easy
Write the definition and dimensional formula of the linear mass density of a string.
Solution
(N/A) Definition: The linear mass density (often denoted by $\mu$) of a string is defined as the mass per unit length of the string.
Mathematically,$\mu = \frac{M}{L}$,where $M$ is the mass of the string and $L$ is its length.
Dimensional Formula: Since mass has the dimension $[M]$ and length has the dimension $[L]$,the dimensional formula for linear mass density is:
$\mu = \frac{[M]}{[L]} = [M L^{-1} T^0]$.
Mathematically,$\mu = \frac{M}{L}$,where $M$ is the mass of the string and $L$ is its length.
Dimensional Formula: Since mass has the dimension $[M]$ and length has the dimension $[L]$,the dimensional formula for linear mass density is:
$\mu = \frac{[M]}{[L]} = [M L^{-1} T^0]$.
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Medium
Write the magnitude and dimensional formula of $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$.
Solution
(N/A) The expression $\frac{1}{\sqrt{\mu_0 \epsilon_0}}$ represents the speed of light in a vacuum,denoted by $c$.
Its magnitude is approximately $3 \times 10^8 \ m/s$.
Since it represents speed,its dimensional formula is $[M^0 L^1 T^{-1}]$.
Its magnitude is approximately $3 \times 10^8 \ m/s$.
Since it represents speed,its dimensional formula is $[M^0 L^1 T^{-1}]$.
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MediumMCQ
What is the dimensional formula of an electron volt $(eV)$, and what is its value in Joules $(J)$?
A
$[ML^2T^{-2}]$ and $1.6 \times 10^{-19} \ J$
B
$[MLT^{-2}]$ and $1.6 \times 10^{-19} \ J$
C
$[ML^2T^{-1}]$ and $1.6 \times 10^{-19} \ J$
D
$[ML^2T^{-2}]$ and $1.6 \times 10^{-20} \ J$
Solution
(A) An electron volt $(eV)$ is a unit of energy.
Energy is defined as the product of force and displacement, or work done.
The dimensional formula for energy is $[M^1L^2T^{-2}]$.
One electron volt is defined as the energy gained by an electron when it is accelerated through a potential difference of $1 \ V$.
$1 \ eV = q \times V = (1.6 \times 10^{-19} \ C) \times (1 \ V) = 1.6 \times 10^{-19} \ J$.
Energy is defined as the product of force and displacement, or work done.
The dimensional formula for energy is $[M^1L^2T^{-2}]$.
One electron volt is defined as the energy gained by an electron when it is accelerated through a potential difference of $1 \ V$.
$1 \ eV = q \times V = (1.6 \times 10^{-19} \ C) \times (1 \ V) = 1.6 \times 10^{-19} \ J$.
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Easy
Write the dimensional formula of the Rydberg constant $(R)$.
Solution
(N/A) The Rydberg formula for the wavelength $(\lambda)$ of light emitted in a hydrogen transition is given by:
$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Here, $\lambda$ is the wavelength, which has the dimension of length $([L])$.
$n_1$ and $n_2$ are integers, which are dimensionless.
Therefore, the dimension of the Rydberg constant $(R)$ is the inverse of the dimension of wavelength.
$[R] = \frac{1}{[L]} = [L^{-1}]$
In terms of fundamental dimensions, this is $[M^0 L^{-1} T^0]$.
$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Here, $\lambda$ is the wavelength, which has the dimension of length $([L])$.
$n_1$ and $n_2$ are integers, which are dimensionless.
Therefore, the dimension of the Rydberg constant $(R)$ is the inverse of the dimension of wavelength.
$[R] = \frac{1}{[L]} = [L^{-1}]$
In terms of fundamental dimensions, this is $[M^0 L^{-1} T^0]$.
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Medium
There is another useful system of units,besides the $SI/MKS$. $A$ system,called the $CGS$ (centimeter-gram-second) system. In this system,Coulomb's law is given by $\vec F = \frac{{Qq}}{{{r^2}}} \cdot \hat r$ where the distance $r$ is measured in $cm$ $(= 10^{-2} \ m)$,$F$ in dynes $(= 10^{-5} \ N)$ and the charges in electrostatic units $(esu)$,where $1 \ esu$ of charge $= \frac{1}{[3]} \times 10^{-9} \ C$. The number $[3]$ actually arises from the speed of light in vacuum which is now taken to be exactly given by $c = 2.99792458 \times 10^8 \ m/s$. An approximate value of $c$ then is $c = 3 \times 10^8 \ m/s$.
$(i)$ Show that the Coulomb law in $CGS$ units yields $1 \ esu$ of charge $= 1 \ (dyne)^{1/2} \ cm$. Obtain the dimensions of units of charge in terms of mass $M$,length $L$ and time $T$. Show that it is given in terms of fractional powers of $M$ and $L$.
$(ii)$ Write $1 \ esu$ of charge $= xC$,where $x$ is a dimensionless number. Show that this gives $\frac{1}{{4\pi \epsilon_0}} = \frac{{10^{-9}}}{{{x^2}}} \frac{N \ m^2}{C^2}$. With $x = \frac{1}{[3]} \times 10^{-9}$,we have $\frac{1}{{4\pi \epsilon_0}} = [3]^2 \times 10^9 \frac{N \ m^2}{C^2}$ or $\frac{1}{{4\pi \epsilon_0}} = (2.99792458)^2 \times 10^9 \frac{N \ m^2}{C^2}$ (exactly).
$(i)$ Show that the Coulomb law in $CGS$ units yields $1 \ esu$ of charge $= 1 \ (dyne)^{1/2} \ cm$. Obtain the dimensions of units of charge in terms of mass $M$,length $L$ and time $T$. Show that it is given in terms of fractional powers of $M$ and $L$.
$(ii)$ Write $1 \ esu$ of charge $= xC$,where $x$ is a dimensionless number. Show that this gives $\frac{1}{{4\pi \epsilon_0}} = \frac{{10^{-9}}}{{{x^2}}} \frac{N \ m^2}{C^2}$. With $x = \frac{1}{[3]} \times 10^{-9}$,we have $\frac{1}{{4\pi \epsilon_0}} = [3]^2 \times 10^9 \frac{N \ m^2}{C^2}$ or $\frac{1}{{4\pi \epsilon_0}} = (2.99792458)^2 \times 10^9 \frac{N \ m^2}{C^2}$ (exactly).
Solution
(A) $(i)$ In $CGS$ units,Coulomb's law is $F = \frac{Qq}{r^2}$.
For $1 \ esu$ of charge,$F = 1 \ dyne$ and $r = 1 \ cm$.
$1 \ dyne = \frac{(1 \ esu)^2}{(1 \ cm)^2} \implies 1 \ esu = (1 \ dyne)^{1/2} \ cm$.
Since $[F] = [M^1 L^1 T^{-2}]$ and $[L] = [L^1]$,the dimensions of $1 \ esu$ are $[M^1 L^1 T^{-2}]^{1/2} \times [L^1] = [M^{1/2} L^{3/2} T^{-1}]$.
Thus,the powers of $M$ and $L$ are $1/2$ and $3/2$ respectively,which are fractional.
$(ii)$ Let $1 \ esu = x \ C$. The force between two $1 \ esu$ charges at $1 \ cm$ distance is $1 \ dyne = 10^{-5} \ N$.
In $SI$ units,$F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$.
Substituting $q_1 = q_2 = x \ C$ and $r = 10^{-2} \ m$:
$10^{-5} \ N = \frac{1}{4\pi \epsilon_0} \frac{x^2}{(10^{-2} \ m)^2}$.
$\frac{1}{4\pi \epsilon_0} = \frac{10^{-5} \ N \times 10^{-4} \ m^2}{x^2} = \frac{10^{-9}}{x^2} \frac{N \ m^2}{C^2}$.
Substituting $x = \frac{1}{[3]} \times 10^{-9}$,we get $\frac{1}{4\pi \epsilon_0} = \frac{10^{-9}}{(1/[3] \times 10^{-9})^2} = [3]^2 \times 10^9 \frac{N \ m^2}{C^2}$.
For $1 \ esu$ of charge,$F = 1 \ dyne$ and $r = 1 \ cm$.
$1 \ dyne = \frac{(1 \ esu)^2}{(1 \ cm)^2} \implies 1 \ esu = (1 \ dyne)^{1/2} \ cm$.
Since $[F] = [M^1 L^1 T^{-2}]$ and $[L] = [L^1]$,the dimensions of $1 \ esu$ are $[M^1 L^1 T^{-2}]^{1/2} \times [L^1] = [M^{1/2} L^{3/2} T^{-1}]$.
Thus,the powers of $M$ and $L$ are $1/2$ and $3/2$ respectively,which are fractional.
$(ii)$ Let $1 \ esu = x \ C$. The force between two $1 \ esu$ charges at $1 \ cm$ distance is $1 \ dyne = 10^{-5} \ N$.
In $SI$ units,$F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$.
Substituting $q_1 = q_2 = x \ C$ and $r = 10^{-2} \ m$:
$10^{-5} \ N = \frac{1}{4\pi \epsilon_0} \frac{x^2}{(10^{-2} \ m)^2}$.
$\frac{1}{4\pi \epsilon_0} = \frac{10^{-5} \ N \times 10^{-4} \ m^2}{x^2} = \frac{10^{-9}}{x^2} \frac{N \ m^2}{C^2}$.
Substituting $x = \frac{1}{[3]} \times 10^{-9}$,we get $\frac{1}{4\pi \epsilon_0} = \frac{10^{-9}}{(1/[3] \times 10^{-9})^2} = [3]^2 \times 10^9 \frac{N \ m^2}{C^2}$.
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View Solution156
Difficult
List at least six physical quantities that have the dimensional formula $ML^2T^{-2}$.
Solution
(N/A) $(i)$ Work
$(ii)$ Torque
$(iii)$ Moment of force
$(iv)$ Couple
$(v)$ Potential energy
$(vi)$ Kinetic energy
$(ii)$ Torque
$(iii)$ Moment of force
$(iv)$ Couple
$(v)$ Potential energy
$(vi)$ Kinetic energy
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MediumMCQ
Are the dimensions of mass and weight the same?
A
Yes
B
No
C
Sometimes
D
Cannot be determined
Solution
(B) No,the dimensions of mass are $[M^1 L^0 T^0]$ (or simply $[M]$).
Weight is a force,defined as $W = mg$,where $m$ is mass and $g$ is acceleration due to gravity.
The dimensions of weight are $[M^1 L^1 T^{-2}]$.
Since the dimensions are different,mass and weight are not the same.
Weight is a force,defined as $W = mg$,where $m$ is mass and $g$ is acceleration due to gravity.
The dimensions of weight are $[M^1 L^1 T^{-2}]$.
Since the dimensions are different,mass and weight are not the same.
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MediumMCQ
Is it possible for a physical quantity to have dimensions but no units?
A
Yes
B
No
C
Sometimes
D
Depends on the system of units
Solution
(B) No,it is not possible. By definition,a dimension represents the physical nature of a quantity,and a unit is the standard measure used to quantify that dimension. If a quantity has dimensions,it must be measurable,and therefore it must have a corresponding unit of measurement.
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MediumMCQ
Find the dimensional formula of $\rho g v$,where $\rho = \text{density}$,$g = \text{acceleration}$,and $v = \text{velocity}$.
A
$[M L^{-1} T^{-3}]$
B
$[M L^{-2} T^{-2}]$
C
$[M L^{-3} T^{-1}]$
D
$[M L^{-1} T^{-2}]$
Solution
(A) The dimensional formula of density $\rho$ is $[M L^{-3}]$.
The dimensional formula of acceleration $g$ is $[L T^{-2}]$.
The dimensional formula of velocity $v$ is $[L T^{-1}]$.
Now,the dimensional formula of the product $\rho g v$ is given by:
$[\rho g v] = [\rho] [g] [v]$
$[\rho g v] = [M L^{-3}] [L T^{-2}] [L T^{-1}]$
$[\rho g v] = [M L^{-3+1+1} T^{-2-1}]$
$[\rho g v] = [M L^{-1} T^{-3}]$
The dimensional formula of acceleration $g$ is $[L T^{-2}]$.
The dimensional formula of velocity $v$ is $[L T^{-1}]$.
Now,the dimensional formula of the product $\rho g v$ is given by:
$[\rho g v] = [\rho] [g] [v]$
$[\rho g v] = [M L^{-3}] [L T^{-2}] [L T^{-1}]$
$[\rho g v] = [M L^{-3+1+1} T^{-2-1}]$
$[\rho g v] = [M L^{-1} T^{-3}]$
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Difficult
$A$ function $f(\theta)$ is defined as $f(\theta) = 1 - \theta + \frac{\theta^2}{2!} - \frac{\theta^3}{3!} + \frac{\theta^4}{4!} - \dots$. Why is it necessary for $f(\theta)$ to be a dimensionless quantity?
Solution
(N/A) The given function $f(\theta)$ is a power series expansion of the exponential function $e^{-\theta}$.
In any physical equation,terms added or subtracted must have the same dimensions.
Since the first term is a dimensionless constant $(1)$,all subsequent terms $(\theta, \frac{\theta^2}{2!}, \dots)$ must also be dimensionless.
Furthermore,the principle of homogeneity of dimensions states that the arguments of transcendental functions (like exponential,trigonometric,or logarithmic functions) must be dimensionless.
Therefore,for the expression to be physically meaningful,$f(\theta)$ must be a dimensionless quantity.
In any physical equation,terms added or subtracted must have the same dimensions.
Since the first term is a dimensionless constant $(1)$,all subsequent terms $(\theta, \frac{\theta^2}{2!}, \dots)$ must also be dimensionless.
Furthermore,the principle of homogeneity of dimensions states that the arguments of transcendental functions (like exponential,trigonometric,or logarithmic functions) must be dimensionless.
Therefore,for the expression to be physically meaningful,$f(\theta)$ must be a dimensionless quantity.
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Easy
The displacement of a progressive wave is represented by $y = A \sin(\omega t - kx)$,where $x$ is distance and $t$ is time. Write the dimensional formula of $(i)$ $\omega$ and $(ii)$ $k$.
Solution
(N/A) The argument of the trigonometric function,$(\omega t - kx)$,must be dimensionless because the sine function is a dimensionless ratio.
$(i)$ For $\omega t$ to be dimensionless:
$[\omega][t] = [M^0 L^0 T^0]$
$[\omega][T] = [M^0 L^0 T^0]$
$[\omega] = [T^{-1}] = [M^0 L^0 T^{-1}]$
$(ii)$ For $kx$ to be dimensionless:
$[k][x] = [M^0 L^0 T^0]$
$[k][L] = [M^0 L^0 T^0]$
$[k] = [L^{-1}] = [M^0 L^{-1} T^0]$
$(i)$ For $\omega t$ to be dimensionless:
$[\omega][t] = [M^0 L^0 T^0]$
$[\omega][T] = [M^0 L^0 T^0]$
$[\omega] = [T^{-1}] = [M^0 L^0 T^{-1}]$
$(ii)$ For $kx$ to be dimensionless:
$[k][x] = [M^0 L^0 T^0]$
$[k][L] = [M^0 L^0 T^0]$
$[k] = [L^{-1}] = [M^0 L^{-1} T^0]$
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MediumMCQ
What is the dimensional formula of the spring constant $k$?
A
$M^1 L^0 T^{-2}$
B
$M^1 L^1 T^{-2}$
C
$M^0 L^1 T^{-2}$
D
$M^1 L^0 T^{-1}$
Solution
(A) The spring force is given by Hooke's Law: $F = -kx$.
Here,$F$ is the force,$k$ is the spring constant,and $x$ is the displacement.
Rearranging for $k$,we get $k = F/x$.
The dimensional formula for force $F$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for displacement $x$ is $[L^1]$.
Therefore,the dimensional formula for $k$ is $[M^1 L^1 T^{-2}] / [L^1] = [M^1 L^0 T^{-2}]$.
Here,$F$ is the force,$k$ is the spring constant,and $x$ is the displacement.
Rearranging for $k$,we get $k = F/x$.
The dimensional formula for force $F$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for displacement $x$ is $[L^1]$.
Therefore,the dimensional formula for $k$ is $[M^1 L^1 T^{-2}] / [L^1] = [M^1 L^0 T^{-2}]$.
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MediumMCQ
Match Column-$1$ with Column-$2$.
| Column-$1$ | Column-$2$ |
| $(a)$ Wave vector | $(i)$ $M^1L^0T^{-3}$ |
| $(b)$ Linear mass density | $(ii)$ $M^1L^{-1}T^{-2}$ |
| $(c)$ Intensity of a wave | $(iii)$ $M^0L^{-1}T^0$ |
| $(d)$ Volume elasticity coefficient | $(iv)$ $M^1L^{-1}T^0$ |
A
$a-iii, b-iv, c-i, d-ii$
B
$a-i, b-ii, c-iii, d-iv$
C
$a-ii, b-iii, c-iv, d-i$
D
$a-iv, b-i, c-ii, d-iii$
Solution
(A) The dimensional formulas are calculated as follows:
$(a)$ Wave vector $(k = 2\pi / \lambda)$: The dimension is $[L^{-1}]$, which corresponds to $M^0L^{-1}T^0$ $(iii)$.
$(b)$ Linear mass density $(\mu = \text{mass} / \text{length})$: The dimension is $[ML^{-1}]$, which corresponds to $M^1L^{-1}T^0$ $(iv)$.
$(c)$ Intensity of a wave $(I = \text{Power} / \text{Area})$: Power is $ML^2T^{-3}$, so $I = (ML^2T^{-3}) / L^2 = MT^{-3}$, which corresponds to $M^1L^0T^{-3}$ $(i)$.
$(d)$ Volume elasticity coefficient $(B = \text{Stress} / \text{Strain})$: Stress is $ML^{-1}T^{-2}$ and strain is dimensionless, so $B$ is $M^1L^{-1}T^{-2}$ $(ii)$.
Therefore, the correct matching is $(a-iii, b-iv, c-i, d-ii)$.
$(a)$ Wave vector $(k = 2\pi / \lambda)$: The dimension is $[L^{-1}]$, which corresponds to $M^0L^{-1}T^0$ $(iii)$.
$(b)$ Linear mass density $(\mu = \text{mass} / \text{length})$: The dimension is $[ML^{-1}]$, which corresponds to $M^1L^{-1}T^0$ $(iv)$.
$(c)$ Intensity of a wave $(I = \text{Power} / \text{Area})$: Power is $ML^2T^{-3}$, so $I = (ML^2T^{-3}) / L^2 = MT^{-3}$, which corresponds to $M^1L^0T^{-3}$ $(i)$.
$(d)$ Volume elasticity coefficient $(B = \text{Stress} / \text{Strain})$: Stress is $ML^{-1}T^{-2}$ and strain is dimensionless, so $B$ is $M^1L^{-1}T^{-2}$ $(ii)$.
Therefore, the correct matching is $(a-iii, b-iv, c-i, d-ii)$.
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Easy
Give the dimensional formula of surface tension.
Solution
(N/A) Surface tension $(T)$ is defined as the force per unit length acting on the surface of a liquid.
Mathematically,$T = \frac{F}{L}$,where $F$ is the force and $L$ is the length.
The dimensional formula of force $(F)$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula of length $(L)$ is $[L^1]$.
Therefore,the dimensional formula of surface tension is $[T] = \frac{[M^1 L^1 T^{-2}]}{[L^1]} = [M^1 L^0 T^{-2}]$.
Mathematically,$T = \frac{F}{L}$,where $F$ is the force and $L$ is the length.
The dimensional formula of force $(F)$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula of length $(L)$ is $[L^1]$.
Therefore,the dimensional formula of surface tension is $[T] = \frac{[M^1 L^1 T^{-2}]}{[L^1]} = [M^1 L^0 T^{-2}]$.
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View Solution165
MediumMCQ
Give the dimensional formula of the coefficient of viscosity.
A
$M^1 L^{-1} T^{-1}$
B
$M^1 L^1 T^{-1}$
C
$M^1 L^{-1} T^{-2}$
D
$M^1 L^{-2} T^{-1}$
Solution
(A) The coefficient of viscosity $\eta$ is defined by Newton's law of viscosity: $F = -\eta A \frac{dv}{dx}$.
Rearranging for $\eta$,we get $\eta = \frac{F}{A (dv/dx)}$.
The dimensional formula for force $F$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for area $A$ is $[L^2]$.
The dimensional formula for velocity gradient $dv/dx$ is $[L T^{-1}] / [L] = [T^{-1}]$.
Substituting these into the formula: $\eta = \frac{[M^1 L^1 T^{-2}]}{[L^2] [T^{-1}]} = [M^1 L^{-1} T^{-1}]$.
Thus,the dimensional formula of the coefficient of viscosity is $M^1 L^{-1} T^{-1}$.
Rearranging for $\eta$,we get $\eta = \frac{F}{A (dv/dx)}$.
The dimensional formula for force $F$ is $[M^1 L^1 T^{-2}]$.
The dimensional formula for area $A$ is $[L^2]$.
The dimensional formula for velocity gradient $dv/dx$ is $[L T^{-1}] / [L] = [T^{-1}]$.
Substituting these into the formula: $\eta = \frac{[M^1 L^1 T^{-2}]}{[L^2] [T^{-1}]} = [M^1 L^{-1} T^{-1}]$.
Thus,the dimensional formula of the coefficient of viscosity is $M^1 L^{-1} T^{-1}$.
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EasyMCQ
Different physical quantities are given in Column-$I$ and their dimensional formulas are given in Column-$II$. Match them appropriately.
| Column-$I$ | Column-$II$ |
|---|---|
| $(a)$ Viscous force | $(i)$ $[M^1 L^1 T^{-2}]$ |
| $(b)$ Coefficient of viscosity | $(ii)$ $[M^1 L^{-1} T^{-1}]$ |
| $(iii)$ $[M^1 L^{-1} T^{-2}]$ |
A
$(a-i), (b-iii)$
B
$(a-i), (b-ii)$
C
$(a-iii), (b-ii)$
D
$(a-ii), (b-iii)$
Solution
(B) $1$. Viscous force $(F)$ is a type of force, and the dimensional formula for force is $[M^1 L^1 T^{-2}]$. Thus, $(a)$ matches with $(i)$.
$2$. According to Newton's law of viscosity, $F = \eta A \frac{dv}{dx}$, where $\eta$ is the coefficient of viscosity, $A$ is the area, and $\frac{dv}{dx}$ is the velocity gradient.
$3$. Rearranging for $\eta$: $\eta = \frac{F}{A (dv/dx)}$.
$4$. Substituting dimensions: $[\eta] = \frac{[M^1 L^1 T^{-2}]}{[L^2] [T^{-1}]} = [M^1 L^{-1} T^{-1}]$. Thus, $(b)$ matches with $(ii)$.
$2$. According to Newton's law of viscosity, $F = \eta A \frac{dv}{dx}$, where $\eta$ is the coefficient of viscosity, $A$ is the area, and $\frac{dv}{dx}$ is the velocity gradient.
$3$. Rearranging for $\eta$: $\eta = \frac{F}{A (dv/dx)}$.
$4$. Substituting dimensions: $[\eta] = \frac{[M^1 L^1 T^{-2}]}{[L^2] [T^{-1}]} = [M^1 L^{-1} T^{-1}]$. Thus, $(b)$ matches with $(ii)$.
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View Solution167
MediumMCQ
What is the dimensional formula for the ratio of angular momentum to linear momentum?
A
$M^{0}L^{1}T^{0}$
B
$M^{1}L^{1}T^{-1}$
C
$M^{0}L^{0}T^{1}$
D
$M^{1}L^{2}T^{-1}$
Solution
(A) Angular momentum $(L)$ is defined as $L = r \times p$,where $r$ is the position vector and $p$ is the linear momentum.
The dimensional formula for angular momentum $(L)$ is $[M^{1}L^{2}T^{-1}]$.
The dimensional formula for linear momentum $(p)$ is $[M^{1}L^{1}T^{-1}]$.
Taking the ratio of angular momentum to linear momentum: $\frac{L}{p} = \frac{[M^{1}L^{2}T^{-1}]}{[M^{1}L^{1}T^{-1}]}$.
Simplifying the dimensions: $\frac{M^{1-1}L^{2-1}T^{-1-(-1)}}{1} = M^{0}L^{1}T^{0}$.
Thus,the dimensional formula is $[L^{1}]$,which represents the dimension of length.
The dimensional formula for angular momentum $(L)$ is $[M^{1}L^{2}T^{-1}]$.
The dimensional formula for linear momentum $(p)$ is $[M^{1}L^{1}T^{-1}]$.
Taking the ratio of angular momentum to linear momentum: $\frac{L}{p} = \frac{[M^{1}L^{2}T^{-1}]}{[M^{1}L^{1}T^{-1}]}$.
Simplifying the dimensions: $\frac{M^{1-1}L^{2-1}T^{-1-(-1)}}{1} = M^{0}L^{1}T^{0}$.
Thus,the dimensional formula is $[L^{1}]$,which represents the dimension of length.
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MediumMCQ
The amount of solar energy received on the Earth's surface per unit area per unit time is defined as the solar constant. The dimension of the solar constant is:
A
$ML^{2}T^{-2}$
B
$MLT^{-2}$
C
$M^{2}L^{0}T^{-1}$
D
$MT^{-3}$
Solution
(D) The solar constant $S$ is defined as the energy $E$ incident per unit area $A$ per unit time $t$.
$S = \frac{E}{A \times t}$
The dimension of energy $E$ is $[ML^{2}T^{-2}]$.
The dimension of area $A$ is $[L^{2}]$.
The dimension of time $t$ is $[T]$.
Substituting these dimensions into the formula:
$S = \frac{[ML^{2}T^{-2}]}{[L^{2}] \times [T]}$
$S = \frac{[ML^{2}T^{-2}]}{[L^{2}T]}$
$S = [MT^{-3}]$
Thus,the dimension of the solar constant is $[MT^{-3}]$.
$S = \frac{E}{A \times t}$
The dimension of energy $E$ is $[ML^{2}T^{-2}]$.
The dimension of area $A$ is $[L^{2}]$.
The dimension of time $t$ is $[T]$.
Substituting these dimensions into the formula:
$S = \frac{[ML^{2}T^{-2}]}{[L^{2}] \times [T]}$
$S = \frac{[ML^{2}T^{-2}]}{[L^{2}T]}$
$S = [MT^{-3}]$
Thus,the dimension of the solar constant is $[MT^{-3}]$.
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EasyMCQ
Dimensions of stress are
A
$[M L^{-1} T^{-2}]$
B
$[M L T^{-2}]$
C
$[M L^{2} T^{-2}]$
D
$[M L^{0} T^{-2}]$
Solution
(A) Stress is defined as the restoring force per unit area.
$\text{Stress} = \frac{\text{Force}}{\text{Area}}$
Substituting the dimensional formulas for force $[M L T^{-2}]$ and area $[L^2]$:
$\text{Stress} = \frac{[M L T^{-2}]}{[L^2]}$
$\text{Stress} = [M L^{1-2} T^{-2}]$
$\text{Stress} = [M L^{-1} T^{-2}]$
$\text{Stress} = \frac{\text{Force}}{\text{Area}}$
Substituting the dimensional formulas for force $[M L T^{-2}]$ and area $[L^2]$:
$\text{Stress} = \frac{[M L T^{-2}]}{[L^2]}$
$\text{Stress} = [M L^{1-2} T^{-2}]$
$\text{Stress} = [M L^{-1} T^{-2}]$
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View Solution170
EasyMCQ
What is the dimension of Luminous flux?
A
$[cd^1]$
B
$[cd^1 T^{-1}]$
C
$[cd^1 L^{-2}]$
D
$[cd^1 L^1 T^{-1}]$
Solution
(A) Luminous flux is a measure of the perceived power of light.
In the International System of Units $(SI)$,the base unit for luminous intensity is the candela $(cd)$.
Since luminous flux is proportional to luminous intensity,its dimension is represented by the base dimension of luminous intensity.
Therefore,the dimension of Luminous flux is $[cd^1]$.
In the International System of Units $(SI)$,the base unit for luminous intensity is the candela $(cd)$.
Since luminous flux is proportional to luminous intensity,its dimension is represented by the base dimension of luminous intensity.
Therefore,the dimension of Luminous flux is $[cd^1]$.
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View Solution171
MediumMCQ
Match List-$I$ with List-$II$:
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $(a)$ $h$ (Planck's constant) | $(i)$ $[M L T^{-1}]$ |
| $(b)$ $E$ (kinetic energy) | $(ii)$ $[M L^2 T^{-1}]$ |
| $(c)$ $V$ (electric potential) | $(iii)$ $[M L^2 T^{-2}]$ |
| $(d)$ $P$ (linear momentum) | $(iv)$ $[M L^2 I^{-1} T^{-3}]$ |
Choose the correct answer from the options given below:
A
$(a) \rightarrow (iii), (b) \rightarrow (iv), (c) \rightarrow (ii), (d) \rightarrow (i)$
B
$(a) \rightarrow (ii), (b) \rightarrow (iii), (c) \rightarrow (iv), (d) \rightarrow (i)$
C
$(a) \rightarrow (i), (b) \rightarrow (ii), (c) \rightarrow (iv), (d) \rightarrow (iii)$
D
$(a) \rightarrow (iii), (b) \rightarrow (ii), (c) \rightarrow (iv), (d) \rightarrow (i)$
Solution
(B) We determine the dimensional formula for each quantity:
$1$. Planck's constant $(h)$: Since $E = h\nu$,$h = E / \nu$. The dimensions are $[M L^2 T^{-2}] / [T^{-1}] = [M L^2 T^{-1}]$. Thus,$(a) \rightarrow (ii)$.
$2$. Kinetic energy $(E)$: Energy has the dimensions of work,$[M L^2 T^{-2}]$. Thus,$(b) \rightarrow (iii)$.
$3$. Electric potential $(V)$: $V = W / q$. The dimensions are $[M L^2 T^{-2}] / [I T] = [M L^2 I^{-1} T^{-3}]$. Thus,$(c) \rightarrow (iv)$.
$4$. Linear momentum $(P)$: $P = m v$. The dimensions are $[M] [L T^{-1}] = [M L T^{-1}]$. Thus,$(d) \rightarrow (i)$.
Matching these,we get $(a) \rightarrow (ii), (b) \rightarrow (iii), (c) \rightarrow (iv), (d) \rightarrow (i)$.
$1$. Planck's constant $(h)$: Since $E = h\nu$,$h = E / \nu$. The dimensions are $[M L^2 T^{-2}] / [T^{-1}] = [M L^2 T^{-1}]$. Thus,$(a) \rightarrow (ii)$.
$2$. Kinetic energy $(E)$: Energy has the dimensions of work,$[M L^2 T^{-2}]$. Thus,$(b) \rightarrow (iii)$.
$3$. Electric potential $(V)$: $V = W / q$. The dimensions are $[M L^2 T^{-2}] / [I T] = [M L^2 I^{-1} T^{-3}]$. Thus,$(c) \rightarrow (iv)$.
$4$. Linear momentum $(P)$: $P = m v$. The dimensions are $[M] [L T^{-1}] = [M L T^{-1}]$. Thus,$(d) \rightarrow (i)$.
Matching these,we get $(a) \rightarrow (ii), (b) \rightarrow (iii), (c) \rightarrow (iv), (d) \rightarrow (i)$.
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View Solution172
DifficultMCQ
If $e$ is the electronic charge,$c$ is the speed of light in free space,and $h$ is Planck's constant,the quantity $\frac{1}{4 \pi \varepsilon_{0}} \frac{| e |^{2}}{h c}$ has dimensions of .......
A
$[M^{0} L^{0} T^{0}]$
B
$[L C^{-1}]$
C
$[M L T^{-1}]$
D
$[M L T^{0}]$
Solution
(A) The Coulomb force between two charges $e$ separated by distance $r$ is given by $F = \frac{1}{4 \pi \varepsilon_{0}} \frac{e^{2}}{r^{2}}$.
From this,the dimension of $\frac{e^{2}}{4 \pi \varepsilon_{0}}$ is $[F r^{2}] = [M L T^{-2} \cdot L^{2}] = [M L^{3} T^{-2}]$.
The energy of a photon is $E = \frac{hc}{\lambda}$,so the dimension of $hc$ is $[E \lambda] = [M L^{2} T^{-2} \cdot L] = [M L^{3} T^{-2}]$.
Dividing these two quantities,the dimensions are $\frac{[M L^{3} T^{-2}]}{[M L^{3} T^{-2}]} = [M^{0} L^{0} T^{0}]$.
Thus,the quantity is dimensionless.
From this,the dimension of $\frac{e^{2}}{4 \pi \varepsilon_{0}}$ is $[F r^{2}] = [M L T^{-2} \cdot L^{2}] = [M L^{3} T^{-2}]$.
The energy of a photon is $E = \frac{hc}{\lambda}$,so the dimension of $hc$ is $[E \lambda] = [M L^{2} T^{-2} \cdot L] = [M L^{3} T^{-2}]$.
Dividing these two quantities,the dimensions are $\frac{[M L^{3} T^{-2}]}{[M L^{3} T^{-2}]} = [M^{0} L^{0} T^{0}]$.
Thus,the quantity is dimensionless.
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View Solution173
MediumMCQ
Which of the following is not a dimensionless quantity?
A
Relative magnetic permeability $(\mu_{r})$
B
Power factor
C
Permeability of free space $(\mu_{0})$
D
Quality factor
Solution
(C) Relative magnetic permeability $(\mu_{r} = \mu / \mu_{0})$ is the ratio of two similar quantities,so it is dimensionless.
Power factor $(\cos \phi)$ is the ratio of resistance to impedance,so it is dimensionless.
Permeability of free space $(\mu_{0})$ has the $SI$ unit of $N A^{-2}$ or $T m A^{-1}$. Its dimensional formula is $[M L T^{-2} A^{-2}]$. Therefore,it is not a dimensionless quantity.
Quality factor $(Q)$ is defined as the ratio of energy stored to energy dissipated per cycle,making it dimensionless.
Thus,the correct option is $C$.
Power factor $(\cos \phi)$ is the ratio of resistance to impedance,so it is dimensionless.
Permeability of free space $(\mu_{0})$ has the $SI$ unit of $N A^{-2}$ or $T m A^{-1}$. Its dimensional formula is $[M L T^{-2} A^{-2}]$. Therefore,it is not a dimensionless quantity.
Quality factor $(Q)$ is defined as the ratio of energy stored to energy dissipated per cycle,making it dimensionless.
Thus,the correct option is $C$.
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View Solution174
MediumMCQ
If $E$ and $G$ respectively denote energy and gravitational constant,then $\frac{E}{G}$ has the dimensions of :
A
$[M][L^{-1}][T^{-1}]$
B
$[M^{2}][L^{-1}][T^{0}]$
C
$[M][L^{0}][T^{0}]$
D
$[M^{2}][L^{-2}][T^{-1}]$
Solution
(B) The dimensional formula for energy $E$ is $[M L^{2} T^{-2}]$.
The dimensional formula for the gravitational constant $G$ is $[M^{-1} L^{3} T^{-2}]$.
Therefore,the dimensions of $\frac{E}{G}$ are given by:
$\frac{[E]}{[G]} = \frac{[M L^{2} T^{-2}]}{[M^{-1} L^{3} T^{-2}]}$
$= [M^{1 - (-1)}] [L^{2 - 3}] [T^{-2 - (-2)}]$
$= [M^{2}] [L^{-1}] [T^{0}]$
The dimensional formula for the gravitational constant $G$ is $[M^{-1} L^{3} T^{-2}]$.
Therefore,the dimensions of $\frac{E}{G}$ are given by:
$\frac{[E]}{[G]} = \frac{[M L^{2} T^{-2}]}{[M^{-1} L^{3} T^{-2}]}$
$= [M^{1 - (-1)}] [L^{2 - 3}] [T^{-2 - (-2)}]$
$= [M^{2}] [L^{-1}] [T^{0}]$
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View Solution175
MediumMCQ
Identify the pair of physical quantities that have the same dimensions.
A
velocity gradient and decay constant
B
Wien's constant and Stefan constant
C
angular frequency and angular momentum
D
wave number and Avogadro number
Solution
(A) The dimension of velocity gradient is $[T^{-1}]$.
The decay constant $\lambda$ is defined by the relation $N = N_0 e^{-\lambda t}$,which implies that $\lambda t$ is dimensionless. Thus,the dimension of the decay constant is $[T^{-1}]$.
Since both velocity gradient and decay constant have the same dimensions $[T^{-1}]$,option $A$ is correct.
The decay constant $\lambda$ is defined by the relation $N = N_0 e^{-\lambda t}$,which implies that $\lambda t$ is dimensionless. Thus,the dimension of the decay constant is $[T^{-1}]$.
Since both velocity gradient and decay constant have the same dimensions $[T^{-1}]$,option $A$ is correct.
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View Solution176
MediumMCQ
The $SI$ unit of a physical quantity is pascal-second. The dimensional formula of this quantity will be ..............
A
$[ML^{-1}T^{-1}]$
B
$[ML^{-1}T^{-2}]$
C
$[ML^{2}T^{-1}]$
D
$[M^{-1}L^{3}T^{0}]$
Solution
(A) The unit given is pascal-second $(Pa \cdot s)$.
Pascal $(Pa)$ is the unit of pressure,which is defined as force per unit area: $Pa = \frac{N}{m^2} = \frac{kg \cdot m/s^2}{m^2} = kg \cdot m^{-1} \cdot s^{-2}$.
Therefore,the unit pascal-second is $(kg \cdot m^{-1} \cdot s^{-2}) \cdot s = kg \cdot m^{-1} \cdot s^{-1}$.
The dimensional formula for mass $(kg)$ is $[M]$,for length $(m)$ is $[L]$,and for time $(s)$ is $[T]$.
Substituting these into the unit expression,we get $[M][L]^{-1}[T]^{-1} = [ML^{-1}T^{-1}]$.
Pascal $(Pa)$ is the unit of pressure,which is defined as force per unit area: $Pa = \frac{N}{m^2} = \frac{kg \cdot m/s^2}{m^2} = kg \cdot m^{-1} \cdot s^{-2}$.
Therefore,the unit pascal-second is $(kg \cdot m^{-1} \cdot s^{-2}) \cdot s = kg \cdot m^{-1} \cdot s^{-1}$.
The dimensional formula for mass $(kg)$ is $[M]$,for length $(m)$ is $[L]$,and for time $(s)$ is $[T]$.
Substituting these into the unit expression,we get $[M][L]^{-1}[T]^{-1} = [ML^{-1}T^{-1}]$.
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View Solution177
MediumMCQ
Given below are two statements: One is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A$ : Product of Pressure $(P)$ and time $(t)$ has the same dimension as that of coefficient of viscosity.
Reason $R$ : Coefficient of viscosity $= \frac{\text{Force}}{\text{Area} \times \text{Velocity gradient}}$
Choose the correct answer from the options given below.
Assertion $A$ : Product of Pressure $(P)$ and time $(t)$ has the same dimension as that of coefficient of viscosity.
Reason $R$ : Coefficient of viscosity $= \frac{\text{Force}}{\text{Area} \times \text{Velocity gradient}}$
Choose the correct answer from the options given below.
A
Both $A$ and $R$ are true,and $R$ is the correct explanation of $A$.
B
Both $A$ and $R$ are true,but $R$ is $NOT$ the correct explanation of $A$.
C
$A$ is true but $R$ is false.
D
$A$ is false but $R$ is true.
Solution
(A) Step $1$: Analyze Assertion $A$. The dimensions of Pressure $(P)$ are $[M L^{-1} T^{-2}]$ and Time $(t)$ is $[T]$. Therefore,the dimension of $Pt$ is $[M L^{-1} T^{-2}] \times [T] = [M L^{-1} T^{-1}]$.
Step $2$: Analyze the dimension of the coefficient of viscosity $(\eta)$. From Newton's law of viscosity,$F = \eta A \frac{dv}{dx}$,so $\eta = \frac{F}{A (dv/dx)}$. The dimensions are $[M L T^{-2}] / ([L^2] \times [T^{-1}]) = [M L^{-1} T^{-1}]$. Thus,Assertion $A$ is true.
Step $3$: Analyze Reason $R$. The formula for the coefficient of viscosity is $\eta = \frac{F}{A (dv/dx)}$. The statement provided in the original question was incomplete as it omitted the area term. With the correction,Reason $R$ is true and explains the dimension of viscosity. Therefore,both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
Step $2$: Analyze the dimension of the coefficient of viscosity $(\eta)$. From Newton's law of viscosity,$F = \eta A \frac{dv}{dx}$,so $\eta = \frac{F}{A (dv/dx)}$. The dimensions are $[M L T^{-2}] / ([L^2] \times [T^{-1}]) = [M L^{-1} T^{-1}]$. Thus,Assertion $A$ is true.
Step $3$: Analyze Reason $R$. The formula for the coefficient of viscosity is $\eta = \frac{F}{A (dv/dx)}$. The statement provided in the original question was incomplete as it omitted the area term. With the correction,Reason $R$ is true and explains the dimension of viscosity. Therefore,both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
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View Solution178
EasyMCQ
Which one of the following is a dimensionless physical quantity?
A
Angle
B
Stress
C
Force gradient
D
Velocity gradient
Solution
(A) physical quantity is dimensionless if it has no dimensions in terms of mass $(M)$,length $(L)$,and time $(T)$.
$1$. Angle is defined as the ratio of arc length to radius,i.e.,$\theta = \frac{s}{r}$. Since both are lengths,the dimensions are $[L]/[L] = [M^0 L^0 T^0]$,making it dimensionless.
$2$. Stress is defined as force per unit area,$[M L T^{-2}] / [L^2] = [M L^{-1} T^{-2}]$.
$3$. Force gradient is force per unit length,$[M L T^{-2}] / [L] = [M T^{-2}]$.
$4$. Velocity gradient is velocity per unit length,$[L T^{-1}] / [L] = [T^{-1}]$.
Therefore,Angle is the correct answer.
$1$. Angle is defined as the ratio of arc length to radius,i.e.,$\theta = \frac{s}{r}$. Since both are lengths,the dimensions are $[L]/[L] = [M^0 L^0 T^0]$,making it dimensionless.
$2$. Stress is defined as force per unit area,$[M L T^{-2}] / [L^2] = [M L^{-1} T^{-2}]$.
$3$. Force gradient is force per unit length,$[M L T^{-2}] / [L] = [M T^{-2}]$.
$4$. Velocity gradient is velocity per unit length,$[L T^{-1}] / [L] = [T^{-1}]$.
Therefore,Angle is the correct answer.
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View Solution179
EasyMCQ
What are the dimensions of the change in velocity?
A
$[M^0 L^0 T^0]$
B
$[M^0 L^1 T^{-1}]$
C
$[M^1 L^1 T^{-1}]$
D
$[M^0 L^1 T^{-2}]$
Solution
(B) The change in velocity is defined as $\Delta v = v_f - v_i$.
Since velocity has the dimensions of displacement divided by time,its dimensional formula is $[M^0 L^1 T^{-1}]$.
According to the principle of homogeneity of dimensions,only physical quantities of the same dimension can be subtracted from each other.
Therefore,the change in velocity also has the same dimensions as velocity,which is $[M^0 L^1 T^{-1}]$.
Since velocity has the dimensions of displacement divided by time,its dimensional formula is $[M^0 L^1 T^{-1}]$.
According to the principle of homogeneity of dimensions,only physical quantities of the same dimension can be subtracted from each other.
Therefore,the change in velocity also has the same dimensions as velocity,which is $[M^0 L^1 T^{-1}]$.
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View Solution180
MediumMCQ
The potential energy $U$ of a particle varies with distance $x$ from a fixed origin as $U = \frac{A \sqrt{x}}{x + B}$,where $A$ and $B$ are constants. The dimensions of $A$ and $B$ are respectively
A
$[ML^{5/2}T^{-2}], [L]$
B
$[MLT^{-2}], [L^2]$
C
$[ML^{3/2}T^{-2}], [L]$
D
$[L^2], [MLT^{-2}]$
Solution
(A) The potential energy $U$ has dimensions of work or energy,which is $[ML^2T^{-2}]$.
According to the principle of homogeneity of dimensions,only quantities with the same dimensions can be added or subtracted. In the denominator,$x + B$,$x$ is a distance,so $B$ must also have the dimensions of length.
Therefore,$[B] = [L]$.
Now,substitute the dimensions into the equation:
$[ML^2T^{-2}] = \frac{[A] \cdot [L^{1/2}]}{[L]}$
$[ML^2T^{-2}] = [A] \cdot [L^{-1/2}]$
$[A] = [ML^2T^{-2}] \cdot [L^{1/2}]$
$[A] = [ML^{5/2}T^{-2}]$
Thus,the dimensions of $A$ and $B$ are $[ML^{5/2}T^{-2}]$ and $[L]$ respectively. The correct option is $A$.
According to the principle of homogeneity of dimensions,only quantities with the same dimensions can be added or subtracted. In the denominator,$x + B$,$x$ is a distance,so $B$ must also have the dimensions of length.
Therefore,$[B] = [L]$.
Now,substitute the dimensions into the equation:
$[ML^2T^{-2}] = \frac{[A] \cdot [L^{1/2}]}{[L]}$
$[ML^2T^{-2}] = [A] \cdot [L^{-1/2}]$
$[A] = [ML^2T^{-2}] \cdot [L^{1/2}]$
$[A] = [ML^{5/2}T^{-2}]$
Thus,the dimensions of $A$ and $B$ are $[ML^{5/2}T^{-2}]$ and $[L]$ respectively. The correct option is $A$.
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View Solution181
MediumMCQ
The dimensions of the solar constant (energy falling on Earth per second per unit area) are:
A
$[M^0 L^0 T^0]$
B
$[MLT^{-2}]$
C
$[ML^2 T^{-2}]$
D
$[MT^{-3}]$
Solution
(D) The solar constant $S$ is defined as the energy incident per unit area per unit time.
$S = \frac{\text{Energy}}{\text{Area} \times \text{Time}}$
The dimensional formula for energy is $[ML^2 T^{-2}]$.
The dimensional formula for area is $[L^2]$.
The dimensional formula for time is $[T]$.
Substituting these into the formula:
$[S] = \frac{[ML^2 T^{-2}]}{[L^2] \times [T]} = \frac{[ML^2 T^{-2}]}{[L^2 T]} = [MT^{-3}]$.
Thus,the correct option is $D$.
$S = \frac{\text{Energy}}{\text{Area} \times \text{Time}}$
The dimensional formula for energy is $[ML^2 T^{-2}]$.
The dimensional formula for area is $[L^2]$.
The dimensional formula for time is $[T]$.
Substituting these into the formula:
$[S] = \frac{[ML^2 T^{-2}]}{[L^2] \times [T]} = \frac{[ML^2 T^{-2}]}{[L^2 T]} = [MT^{-3}]$.
Thus,the correct option is $D$.
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View Solution182
EasyMCQ
The focal power of a lens has the dimensions
A
$[L]$
B
$[ML^2T^{-3}]$
C
$[L^{-1}]$
D
$[ML^{-3}]$
Solution
(C) The power of a lens $(P)$ is defined as the reciprocal of its focal length $(f)$ in meters.
Mathematically,$P = \frac{1}{f}$.
The dimension of focal length $(f)$ is $[L]$.
Therefore,the dimension of power $(P)$ is $\frac{1}{[L]} = [L^{-1}]$.
Thus,the correct option is $C$.
Mathematically,$P = \frac{1}{f}$.
The dimension of focal length $(f)$ is $[L]$.
Therefore,the dimension of power $(P)$ is $\frac{1}{[L]} = [L^{-1}]$.
Thus,the correct option is $C$.
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View Solution183
MediumMCQ
Dimensional formula for pressure head is
A
$[M^0 L^0 T^0]$
B
$[ML^{-1} T^{-2}]$
C
$[M^0 L^1 T^{-2}]$
D
$[M^0 L^1 T^0]$
Solution
(D) Pressure head is defined as $h = \frac{P}{\rho g}$,where $P$ is pressure,$\rho$ is density,and $g$ is acceleration due to gravity.
Dimensional formula of pressure $P = \frac{\text{Force}}{\text{Area}} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
Dimensional formula of density $\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{[M]}{[L^3]} = [ML^{-3}]$.
Dimensional formula of acceleration due to gravity $g = [LT^{-2}]$.
Substituting these into the formula for pressure head:
$[h] = \frac{[ML^{-1}T^{-2}]}{[ML^{-3}] \cdot [LT^{-2}]}$
$[h] = \frac{[ML^{-1}T^{-2}]}{[ML^{-2}T^{-2}]}$
$[h] = [M^0 L^1 T^0]$.
Thus,the correct option is $D$.
Dimensional formula of pressure $P = \frac{\text{Force}}{\text{Area}} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
Dimensional formula of density $\rho = \frac{\text{Mass}}{\text{Volume}} = \frac{[M]}{[L^3]} = [ML^{-3}]$.
Dimensional formula of acceleration due to gravity $g = [LT^{-2}]$.
Substituting these into the formula for pressure head:
$[h] = \frac{[ML^{-1}T^{-2}]}{[ML^{-3}] \cdot [LT^{-2}]}$
$[h] = \frac{[ML^{-1}T^{-2}]}{[ML^{-2}T^{-2}]}$
$[h] = [M^0 L^1 T^0]$.
Thus,the correct option is $D$.
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View Solution184
MediumMCQ
The dimensional formula for pressure head is ............
A
$[M^0 L^0 T^0]$
B
$[ML^{-1} T^{-2}]$
C
$[M^0 L^1 T^{-2}]$
D
$[M^0 L^1 T^0]$
Solution
(D) Pressure head is defined as $h = \frac{P}{\rho g}$.
$1$. Dimensional formula of Pressure $(P)$: $\frac{\text{Force}}{\text{Area}} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
$2$. Dimensional formula of Density $(\rho)$: $\frac{\text{Mass}}{\text{Volume}} = \frac{[M]}{[L^3]} = [ML^{-3}]$.
$3$. Dimensional formula of acceleration due to gravity $(g)$: $[LT^{-2}]$.
Substituting these into the formula:
$\text{Pressure head} = \frac{[ML^{-1}T^{-2}]}{[ML^{-3}] \times [LT^{-2}]}$
$= \frac{[ML^{-1}T^{-2}]}{[ML^{-2}T^{-2}]}$
$= [M^{1-1} L^{-1-(-2)} T^{-2-(-2)}]$
$= [M^0 L^1 T^0]$.
$1$. Dimensional formula of Pressure $(P)$: $\frac{\text{Force}}{\text{Area}} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]$.
$2$. Dimensional formula of Density $(\rho)$: $\frac{\text{Mass}}{\text{Volume}} = \frac{[M]}{[L^3]} = [ML^{-3}]$.
$3$. Dimensional formula of acceleration due to gravity $(g)$: $[LT^{-2}]$.
Substituting these into the formula:
$\text{Pressure head} = \frac{[ML^{-1}T^{-2}]}{[ML^{-3}] \times [LT^{-2}]}$
$= \frac{[ML^{-1}T^{-2}]}{[ML^{-2}T^{-2}]}$
$= [M^{1-1} L^{-1-(-2)} T^{-2-(-2)}]$
$= [M^0 L^1 T^0]$.
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View Solution185
EasyMCQ
If $G$ is the universal gravitational constant and $g$ is the acceleration due to gravity,then the dimensions of $\frac{G}{g}$ will be ...................
A
$[M^{-1} L^2]$
B
$[M^{-1} L]$
C
$[M^{-2} L]$
D
$[M^{-1} L^{-2}]$
Solution
(A) The dimensional formula for the universal gravitational constant $G$ is $[M^{-1} L^3 T^{-2}]$.
The dimensional formula for the acceleration due to gravity $g$ is $[L T^{-2}]$.
Now,calculate the dimensions of the ratio $\frac{G}{g}$:
$\frac{G}{g} = \frac{[M^{-1} L^3 T^{-2}]}{[L T^{-2}]}$
$= [M^{-1} L^{3-1} T^{-2 - (-2)}]$
$= [M^{-1} L^2 T^0]$
$= [M^{-1} L^2]$
Therefore,the correct option is $A$.
The dimensional formula for the acceleration due to gravity $g$ is $[L T^{-2}]$.
Now,calculate the dimensions of the ratio $\frac{G}{g}$:
$\frac{G}{g} = \frac{[M^{-1} L^3 T^{-2}]}{[L T^{-2}]}$
$= [M^{-1} L^{3-1} T^{-2 - (-2)}]$
$= [M^{-1} L^2 T^0]$
$= [M^{-1} L^2]$
Therefore,the correct option is $A$.
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View Solution186
EasyMCQ
In the relation: $\frac{dy}{dx} = 2\omega \sin(\omega t + \phi_0)$,the dimensional formula for $(\omega t + \phi_0)$ is:
A
$MLT$
B
$MLT^0$
C
$ML^0T^0$
D
$M^0L^0T^0$
Solution
(D) In the given trigonometric function $\sin(\omega t + \phi_0)$,the argument of the sine function,which is $(\omega t + \phi_0)$,must be a dimensionless quantity.
This is because trigonometric functions are defined only for dimensionless angles (radians).
Therefore,the dimensions of $(\omega t + \phi_0)$ are $[M^0L^0T^0]$,which represents a dimensionless quantity.
Thus,the correct option is $(d)$.
This is because trigonometric functions are defined only for dimensionless angles (radians).
Therefore,the dimensions of $(\omega t + \phi_0)$ are $[M^0L^0T^0]$,which represents a dimensionless quantity.
Thus,the correct option is $(d)$.
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View Solution187
MediumMCQ
Match List-$I$ with List-$II$.
Choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| $(A)$ Angular momentum | $(I)$ $[ML^2T^{-1}]$ |
| $(B)$ Torque | $(II)$ $[ML^2T^{-2}]$ |
| $(C)$ Stress | $(III)$ $[ML^{-1}T^{-2}]$ |
| $(D)$ Pressure gradient | $(IV)$ $[ML^{-2}T^{-2}]$ |
Choose the correct answer from the options given below:
A
$(A)-(I), (B)-(II), (C)-(III), (D)-(IV)$
B
$(A)-(III), (B)-(II), (C)-(III), (D)-(IV)$
C
$(A)-(III), (B)-(II), (C)-(IV), (D)-(I)$
D
$(A)-(IV), (B)-(II), (C)-(I), (D)-(III)$
Solution
(B) The dimensional formulas for the given physical quantities are as follows:
$1$. Angular momentum $(L = mvr)$: The dimensions are $[M][LT^{-1}][L] = [ML^2T^{-1}]$. Thus, $(A)-(III)$.
$2$. Torque $(\tau = r \times F)$: The dimensions are $[L][MLT^{-2}] = [ML^2T^{-2}]$. Thus, $(B)-(II)$.
$3$. Stress $(\sigma = \text{Force} / \text{Area})$: The dimensions are $[MLT^{-2}] / [L^2] = [ML^{-1}T^{-2}]$. Thus, $(C)-(III)$ is incorrect in the original table, let's re-evaluate: Stress is $[ML^{-1}T^{-2}]$.
$4$. Pressure gradient $(\Delta P / \Delta x)$: The dimensions are $[ML^{-1}T^{-2}] / [L] = [ML^{-2}T^{-2}]$. Thus, $(D)-(IV)$.
Correct matching: $(A)-(III), (B)-(II), (C)-(III), (D)-(IV)$. Note: There is a typo in the provided options. Based on standard physics, the correct mapping is $(A)-(III), (B)-(II), (C)-(III), (D)-(IV)$.
$1$. Angular momentum $(L = mvr)$: The dimensions are $[M][LT^{-1}][L] = [ML^2T^{-1}]$. Thus, $(A)-(III)$.
$2$. Torque $(\tau = r \times F)$: The dimensions are $[L][MLT^{-2}] = [ML^2T^{-2}]$. Thus, $(B)-(II)$.
$3$. Stress $(\sigma = \text{Force} / \text{Area})$: The dimensions are $[MLT^{-2}] / [L^2] = [ML^{-1}T^{-2}]$. Thus, $(C)-(III)$ is incorrect in the original table, let's re-evaluate: Stress is $[ML^{-1}T^{-2}]$.
$4$. Pressure gradient $(\Delta P / \Delta x)$: The dimensions are $[ML^{-1}T^{-2}] / [L] = [ML^{-2}T^{-2}]$. Thus, $(D)-(IV)$.
Correct matching: $(A)-(III), (B)-(II), (C)-(III), (D)-(IV)$. Note: There is a typo in the provided options. Based on standard physics, the correct mapping is $(A)-(III), (B)-(II), (C)-(III), (D)-(IV)$.
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View Solution188
EasyMCQ
The physical quantity that has the same dimensional formula as pressure is:
A
Force
B
Momentum
C
Young's modulus of elasticity
D
Coefficient of viscosity
Solution
(C) Pressure is defined as force per unit area,so its dimensional formula is $[M L^{-1} T^{-2}]$.
Young's modulus of elasticity $(Y)$ is defined as the ratio of stress to strain.
Since strain is a dimensionless quantity,the dimensional formula of $Y$ is the same as that of stress.
Stress is also defined as force per unit area,so its dimensional formula is $[M L^{-1} T^{-2}]$.
Therefore,the dimensional formula of Young's modulus of elasticity is the same as that of pressure.
Young's modulus of elasticity $(Y)$ is defined as the ratio of stress to strain.
Since strain is a dimensionless quantity,the dimensional formula of $Y$ is the same as that of stress.
Stress is also defined as force per unit area,so its dimensional formula is $[M L^{-1} T^{-2}]$.
Therefore,the dimensional formula of Young's modulus of elasticity is the same as that of pressure.
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View Solution189
MediumMCQ
Match List $I$ with List $II$:
Choose the correct answer from the options given below:
| List $I$ | List $II$ |
| $A$. Spring constant | $I$. $(T^{-1})$ |
| $B$. Angular speed | $II$. $(MT^{-2})$ |
| $C$. Angular momentum | $III$. $(ML^2)$ |
| $D$. Moment of inertia | $IV$. $(ML^2T^{-1})$ |
Choose the correct answer from the options given below:
A
$(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$
B
$(A)-(IV), (B)-(I), (C)-(III), (D)-(II)$
C
$(A)-(II), (B)-(III), (C)-(I), (D)-(IV)$
D
$(A)-(I), (B)-(III), (C)-(II), (D)-(IV)$
Solution
(A) $1$. Spring constant $(k)$: $F = kx \implies [k] = [F]/[x] = (MLT^{-2}) / L = MT^{-2}$. Thus,$A-II$.
$2$. Angular speed $(\omega)$: $\omega = \Delta\theta / \Delta t \implies [\omega] = [1] / T = T^{-1}$. Thus,$B-I$.
$3$. Angular momentum $(L)$: $L = mvr \implies [L] = M(LT^{-1})L = ML^2T^{-1}$. Thus,$C-IV$.
$4$. Moment of inertia $(I)$: $I = mr^2 \implies [I] = ML^2$. Thus,$D-III$.
Therefore,the correct matching is $(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$.
$2$. Angular speed $(\omega)$: $\omega = \Delta\theta / \Delta t \implies [\omega] = [1] / T = T^{-1}$. Thus,$B-I$.
$3$. Angular momentum $(L)$: $L = mvr \implies [L] = M(LT^{-1})L = ML^2T^{-1}$. Thus,$C-IV$.
$4$. Moment of inertia $(I)$: $I = mr^2 \implies [I] = ML^2$. Thus,$D-III$.
Therefore,the correct matching is $(A)-(II), (B)-(I), (C)-(IV), (D)-(III)$.
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View Solution190
DifficultMCQ
Given below are two statements :
$Statement$ $(I)$ : Planck's constant and angular momentum have same dimensions.
$Statement$ $(II)$ : Linear momentum and moment of force have same dimensions.
In the light of the above statements,choose the correct answer from the options given below :
$Statement$ $(I)$ : Planck's constant and angular momentum have same dimensions.
$Statement$ $(II)$ : Linear momentum and moment of force have same dimensions.
In the light of the above statements,choose the correct answer from the options given below :
A
Statement $I$ is true but Statement $II$ is false
B
Both Statement $I$ and Statement $II$ are false
C
Both Statement $I$ and Statement $II$ are true
D
Statement $I$ is false but Statement $II$ is true
Solution
(A) The dimensional formula for Planck's constant $(h)$ is $[h] = ML^2 T^{-1}$.
The dimensional formula for angular momentum $(L)$ is $[L] = ML^2 T^{-1}$.
Since both have the same dimensions,Statement $I$ is true.
The dimensional formula for linear momentum $(P)$ is $[P] = MLT^{-1}$.
The dimensional formula for moment of force (torque,$\tau$) is $[\tau] = ML^2 T^{-2}$.
Since these dimensions are different,Statement $II$ is false.
Therefore,Statement $I$ is true but Statement $II$ is false.
The dimensional formula for angular momentum $(L)$ is $[L] = ML^2 T^{-1}$.
Since both have the same dimensions,Statement $I$ is true.
The dimensional formula for linear momentum $(P)$ is $[P] = MLT^{-1}$.
The dimensional formula for moment of force (torque,$\tau$) is $[\tau] = ML^2 T^{-2}$.
Since these dimensions are different,Statement $II$ is false.
Therefore,Statement $I$ is true but Statement $II$ is false.
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View Solution191
DifficultMCQ
The dimensional formula of angular impulse is:
A
$[M L^2 T^{-1}]$
B
$[M L^2 T^{-2}]$
C
$[M L T^{-1}]$
D
$[M L^2 T^{-1}]$
Solution
(D) Angular impulse is defined as the change in angular momentum.
Dimensional formula of angular impulse $=$ Dimensional formula of angular momentum.
Angular momentum $L = mvr$.
Dimensions of $m = [M]$,$v = [L T^{-1}]$,and $r = [L]$.
Therefore,$[L] = [M] \times [L T^{-1}] \times [L] = [M L^2 T^{-1}]$.
Thus,the correct option is $D$.
Dimensional formula of angular impulse $=$ Dimensional formula of angular momentum.
Angular momentum $L = mvr$.
Dimensions of $m = [M]$,$v = [L T^{-1}]$,and $r = [L]$.
Therefore,$[L] = [M] \times [L T^{-1}] \times [L] = [M L^2 T^{-1}]$.
Thus,the correct option is $D$.
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View Solution192
DifficultMCQ
If $G$ is the gravitational constant and $u$ is the energy density,then which of the following quantities has the same dimensions as $\sqrt{uG}$?
A
Pressure gradient per unit mass
B
Force per unit mass
C
Gravitational potential
D
Energy per unit mass
Solution
(B) The dimension of energy density $u$ is $[M^1 L^{-1} T^{-2}]$.
The dimension of gravitational constant $G$ is $[M^{-1} L^3 T^{-2}]$.
Therefore,the dimension of $uG$ is $[M^1 L^{-1} T^{-2}] \times [M^{-1} L^3 T^{-2}] = [M^0 L^2 T^{-4}]$.
Taking the square root,the dimension of $\sqrt{uG}$ is $[L^1 T^{-2}]$.
This dimension $[L T^{-2}]$ corresponds to acceleration.
Force per unit mass is given by $F/m = ma/m = a$,which has the dimension $[L T^{-2}]$.
Thus,option $B$ is correct.
The dimension of gravitational constant $G$ is $[M^{-1} L^3 T^{-2}]$.
Therefore,the dimension of $uG$ is $[M^1 L^{-1} T^{-2}] \times [M^{-1} L^3 T^{-2}] = [M^0 L^2 T^{-4}]$.
Taking the square root,the dimension of $\sqrt{uG}$ is $[L^1 T^{-2}]$.
This dimension $[L T^{-2}]$ corresponds to acceleration.
Force per unit mass is given by $F/m = ma/m = a$,which has the dimension $[L T^{-2}]$.
Thus,option $B$ is correct.
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View Solution193
DifficultMCQ
Given below are two statements :
Statement $(I)$ : Dimensions of specific heat are $\left[L^2 \,T^{-2} \,K^{-1}\right]$.
Statement $(II)$ : Dimensions of gas constant are $\left[ML^2 \,T^{-1} \,K^{-1}\right]$.
Statement $(I)$ : Dimensions of specific heat are $\left[L^2 \,T^{-2} \,K^{-1}\right]$.
Statement $(II)$ : Dimensions of gas constant are $\left[ML^2 \,T^{-1} \,K^{-1}\right]$.
A
Statement $(I)$ is incorrect but statement $(II)$ is correct.
B
Both statement $(I)$ and statement $(II)$ are incorrect.
C
Statement $(I)$ is correct but statement $(II)$ is incorrect.
D
Both statement $(I)$ and statement $(II)$ are correct.
Solution
(C) The formula for heat energy is $\Delta Q = mS \Delta T$,where $S$ is the specific heat capacity.
Therefore,$S = \frac{\Delta Q}{m \Delta T}$.
The dimensions are $[S] = \frac{[ML^2 T^{-2}]}{[M][K]} = [L^2 T^{-2} K^{-1}]$.
Thus,Statement $(I)$ is correct.
The ideal gas equation is $PV = nRT$,where $R$ is the universal gas constant.
Therefore,$R = \frac{PV}{nT}$.
The dimensions are $[R] = \frac{[ML^{-1} T^{-2}][L^3]}{[mol][K]} = [ML^2 T^{-2} mol^{-1} K^{-1}]$.
Comparing this with the given statement,Statement $(II)$ is incorrect.
Therefore,$S = \frac{\Delta Q}{m \Delta T}$.
The dimensions are $[S] = \frac{[ML^2 T^{-2}]}{[M][K]} = [L^2 T^{-2} K^{-1}]$.
Thus,Statement $(I)$ is correct.
The ideal gas equation is $PV = nRT$,where $R$ is the universal gas constant.
Therefore,$R = \frac{PV}{nT}$.
The dimensions are $[R] = \frac{[ML^{-1} T^{-2}][L^3]}{[mol][K]} = [ML^2 T^{-2} mol^{-1} K^{-1}]$.
Comparing this with the given statement,Statement $(II)$ is incorrect.
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View Solution194
MediumMCQ
If $\varepsilon_0$ is the permittivity of free space and $E$ is the electric field,then $\varepsilon_0 E^2$ has the dimensions
A
$[M^0 L^{-2} T A]$
B
$[M L^{-1} T^{-2}]$
C
$[M^{-1} L^{-3} T^4 A^2]$
D
$[M L^2 T^{-2}]$
Solution
(B) The energy density $u$ of an electric field is given by the formula $u = \frac{1}{2} \varepsilon_0 E^2$.
Energy density is defined as energy per unit volume.
The dimensions of energy are $[M L^2 T^{-2}]$ and the dimensions of volume are $[L^3]$.
Therefore,the dimensions of energy density are $\frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}]$.
Since $\varepsilon_0 E^2$ is proportional to energy density,it shares the same dimensions.
Thus,the dimensions of $\varepsilon_0 E^2$ are $[M L^{-1} T^{-2}]$.
Energy density is defined as energy per unit volume.
The dimensions of energy are $[M L^2 T^{-2}]$ and the dimensions of volume are $[L^3]$.
Therefore,the dimensions of energy density are $\frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}]$.
Since $\varepsilon_0 E^2$ is proportional to energy density,it shares the same dimensions.
Thus,the dimensions of $\varepsilon_0 E^2$ are $[M L^{-1} T^{-2}]$.
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View Solution195
DifficultMCQ
The dimensional formula of latent heat is:
A
$[M^0 L^2 T^{-2}]$
B
$[MLT^{-2}]$
C
$[M^0 L^2 T^{-1}]$
D
$[ML^2 T^{-2}]$
Solution
(A) Latent heat $(L)$ is defined as the heat energy $(Q)$ required per unit mass $(m)$ for a phase change.
$L = \frac{Q}{m}$
Since heat energy $(Q)$ has the dimensions of work or energy,its dimensional formula is $[ML^2 T^{-2}]$.
Mass $(m)$ has the dimensional formula $[M]$.
Therefore,the dimensional formula of latent heat is:
$L = \frac{[ML^2 T^{-2}]}{[M]} = [M^0 L^2 T^{-2}]$
$L = \frac{Q}{m}$
Since heat energy $(Q)$ has the dimensions of work or energy,its dimensional formula is $[ML^2 T^{-2}]$.
Mass $(m)$ has the dimensional formula $[M]$.
Therefore,the dimensional formula of latent heat is:
$L = \frac{[ML^2 T^{-2}]}{[M]} = [M^0 L^2 T^{-2}]$
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View Solution196
DifficultMCQ
The de-Broglie wavelength associated with a particle of mass $m$ and energy $E$ is $h / \sqrt{2 m E}$. The dimensional formula for Planck's constant $h$ is:
A
$[ML^{-1} T^{-2}]$
B
$[ML^2 T^{-1}]$
C
$[MLT^{-2}]$
D
$[M^2 L^2 T^{-2}]$
Solution
(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{\sqrt{2mE}}$.
Rearranging for $h$,we get $h = \lambda \sqrt{2mE}$.
The dimensional formula for wavelength $\lambda$ is $[L]$.
The dimensional formula for mass $m$ is $[M]$.
The dimensional formula for energy $E$ is $[ML^2 T^{-2}]$.
Substituting these into the expression for $h$:
$[h] = [L] \cdot \sqrt{[M] \cdot [ML^2 T^{-2}]}$
$[h] = [L] \cdot \sqrt{[M^2 L^2 T^{-2}]}$
$[h] = [L] \cdot [MLT^{-1}]$
$[h] = [ML^2 T^{-1}]$.
Alternatively,using $E = h\nu$,where $\nu$ is frequency $[T^{-1}]$:
$[h] = [E] / [\nu] = [ML^2 T^{-2}] / [T^{-1}] = [ML^2 T^{-1}]$.
Rearranging for $h$,we get $h = \lambda \sqrt{2mE}$.
The dimensional formula for wavelength $\lambda$ is $[L]$.
The dimensional formula for mass $m$ is $[M]$.
The dimensional formula for energy $E$ is $[ML^2 T^{-2}]$.
Substituting these into the expression for $h$:
$[h] = [L] \cdot \sqrt{[M] \cdot [ML^2 T^{-2}]}$
$[h] = [L] \cdot \sqrt{[M^2 L^2 T^{-2}]}$
$[h] = [L] \cdot [MLT^{-1}]$
$[h] = [ML^2 T^{-1}]$.
Alternatively,using $E = h\nu$,where $\nu$ is frequency $[T^{-1}]$:
$[h] = [E] / [\nu] = [ML^2 T^{-2}] / [T^{-1}] = [ML^2 T^{-1}]$.
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View Solution197
MediumMCQ
Which two of the following five physical parameters have the same dimensions?
$(a)$ Energy density
$(b)$ Refractive index
$(c)$ Dielectric constant
$(d)$ Young's modulus
$(e)$ Magnetic field
$(a)$ Energy density
$(b)$ Refractive index
$(c)$ Dielectric constant
$(d)$ Young's modulus
$(e)$ Magnetic field
A
$(a), (d)$
B
$(a), (e)$
C
$(b), (d)$
D
$(c), (e)$
Solution
(A) The dimensions of the given physical parameters are as follows:
$(a)$ Energy density: $\text{Energy} / \text{Volume} = [ML^2T^{-2}] / [L^3] = [ML^{-1}T^{-2}]$
$(b)$ Refractive index: $\text{Ratio of speeds} = \text{Dimensionless} = [M^0L^0T^0]$
$(c)$ Dielectric constant: $\text{Ratio of permittivities} = \text{Dimensionless} = [M^0L^0T^0]$
$(d)$ Young's modulus: $\text{Stress} / \text{Strain} = [ML^{-1}T^{-2}] / [M^0L^0T^0] = [ML^{-1}T^{-2}]$
$(e)$ Magnetic field: $\text{Force} / (\text{Charge} \times \text{Velocity}) = [MLT^{-2}] / ([IT] \times [LT^{-1}]) = [MT^{-2}I^{-1}]$
Comparing the dimensions,we see that $(a)$ Energy density and $(d)$ Young's modulus have the same dimensions,$[ML^{-1}T^{-2}]$.
Therefore,the correct option is $(a), (d)$.
$(a)$ Energy density: $\text{Energy} / \text{Volume} = [ML^2T^{-2}] / [L^3] = [ML^{-1}T^{-2}]$
$(b)$ Refractive index: $\text{Ratio of speeds} = \text{Dimensionless} = [M^0L^0T^0]$
$(c)$ Dielectric constant: $\text{Ratio of permittivities} = \text{Dimensionless} = [M^0L^0T^0]$
$(d)$ Young's modulus: $\text{Stress} / \text{Strain} = [ML^{-1}T^{-2}] / [M^0L^0T^0] = [ML^{-1}T^{-2}]$
$(e)$ Magnetic field: $\text{Force} / (\text{Charge} \times \text{Velocity}) = [MLT^{-2}] / ([IT] \times [LT^{-1}]) = [MT^{-2}I^{-1}]$
Comparing the dimensions,we see that $(a)$ Energy density and $(d)$ Young's modulus have the same dimensions,$[ML^{-1}T^{-2}]$.
Therefore,the correct option is $(a), (d)$.
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View Solution198
EasyMCQ
The quantity having dimension $-1$ in the length is:
A
force
B
pressure
C
gravitational constant
D
all of these
Solution
(B) The dimensional formula for force is $[F] = [M^1 L^1 T^{-2}]$. The dimension of length is $1$.
The dimensional formula for pressure is $[P] = [M^1 L^{-1} T^{-2}]$. The dimension of length is $-1$.
The dimensional formula for the gravitational constant is $[G] = [M^{-1} L^3 T^{-2}]$. The dimension of length is $3$.
Therefore,the quantity having dimension $-1$ in length is pressure.
The dimensional formula for pressure is $[P] = [M^1 L^{-1} T^{-2}]$. The dimension of length is $-1$.
The dimensional formula for the gravitational constant is $[G] = [M^{-1} L^3 T^{-2}]$. The dimension of length is $3$.
Therefore,the quantity having dimension $-1$ in length is pressure.
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View Solution199
DifficultMCQ
$A$ temperature difference can generate $e.m.f.$ in some materials. Let $S$ be the $e.m.f.$ produced per unit temperature difference between the ends of a wire,$\sigma$ the electrical conductivity and $\kappa$ the thermal conductivity of the material of the wire. Taking $\text{M, L, T, I}$ and $K$ as dimensions of mass,length,time,current and temperature,respectively,the dimensional formula of the quantity $Z=\frac{S^2 \sigma}{\kappa}$ is :-
A
$\left[M^0 L^0 T^0 I^0 K^{-1}\right]$
B
$\left[M^0 L^0 T^0 I^0 K^0\right]$
C
$\left[M^1 L^2 T^{-2} I^{-1} K^{-1}\right]$
D
$\left[M^1 L^2 T^{-4} I^{-1} K^{-1}\right]$
Solution
(A) The dimension of $e.m.f.$ is $[M L^2 T^{-3} I^{-1}]$. Since $S$ is $e.m.f.$ per unit temperature difference,$[S] = [M L^2 T^{-3} I^{-1} K^{-1}]$.
Electrical conductivity $\sigma$ is the reciprocal of resistivity $\rho$. Since $R = \rho \frac{l}{A}$,$\rho = \frac{R A}{l}$. The dimension of resistance $R$ is $[M L^2 T^{-3} I^{-2}]$. Thus,$[\rho] = [M L^3 T^{-3} I^{-2}]$ and $[\sigma] = [M^{-1} L^{-3} T^3 I^2]$.
Thermal conductivity $\kappa$ is defined by $Q = \frac{\kappa A (T_2 - T_1) t}{d}$,so $[\kappa] = \frac{[Energy] [Length]}{[Area] [Temperature] [Time]} = [M L T^{-3} K^{-1}]$.
Now,calculate the dimension of $Z = \frac{S^2 \sigma}{\kappa}$:
$[Z] = \frac{[M L^2 T^{-3} I^{-1} K^{-1}]^2 [M^{-1} L^{-3} T^3 I^2]}{[M L T^{-3} K^{-1}]}$
$[Z] = \frac{[M^2 L^4 T^{-6} I^{-2} K^{-2}] [M^{-1} L^{-3} T^3 I^2]}{[M L T^{-3} K^{-1}]}$
$[Z] = \frac{[M L T^{-3} K^{-2}]}{[M L T^{-3} K^{-1}]} = [K^{-1}] = [M^0 L^0 T^0 I^0 K^{-1}]$.
Electrical conductivity $\sigma$ is the reciprocal of resistivity $\rho$. Since $R = \rho \frac{l}{A}$,$\rho = \frac{R A}{l}$. The dimension of resistance $R$ is $[M L^2 T^{-3} I^{-2}]$. Thus,$[\rho] = [M L^3 T^{-3} I^{-2}]$ and $[\sigma] = [M^{-1} L^{-3} T^3 I^2]$.
Thermal conductivity $\kappa$ is defined by $Q = \frac{\kappa A (T_2 - T_1) t}{d}$,so $[\kappa] = \frac{[Energy] [Length]}{[Area] [Temperature] [Time]} = [M L T^{-3} K^{-1}]$.
Now,calculate the dimension of $Z = \frac{S^2 \sigma}{\kappa}$:
$[Z] = \frac{[M L^2 T^{-3} I^{-1} K^{-1}]^2 [M^{-1} L^{-3} T^3 I^2]}{[M L T^{-3} K^{-1}]}$
$[Z] = \frac{[M^2 L^4 T^{-6} I^{-2} K^{-2}] [M^{-1} L^{-3} T^3 I^2]}{[M L T^{-3} K^{-1}]}$
$[Z] = \frac{[M L T^{-3} K^{-2}]}{[M L T^{-3} K^{-1}]} = [K^{-1}] = [M^0 L^0 T^0 I^0 K^{-1}]$.
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View Solution200
EasyMCQ
The damping force of an oscillator is directly proportional to the velocity. The unit of the constant of proportionality is:
A
$kg \cdot m \cdot s^{-2}$
B
$kg \cdot s^{-1}$
C
$kg \cdot m \cdot s^{-1}$
D
$kg \cdot s^{-1}$
Solution
(B) The damping force $F$ is given by the relation $F = -bv$,where $b$ is the constant of proportionality (damping constant) and $v$ is the velocity.
To find the unit of $b$,we rearrange the formula: $b = \frac{F}{v}$.
The $SI$ unit of force $F$ is Newton $(N)$,which is equivalent to $kg \cdot m \cdot s^{-2}$.
The $SI$ unit of velocity $v$ is $m \cdot s^{-1}$.
Substituting these units into the formula for $b$:
$b = \frac{kg \cdot m \cdot s^{-2}}{m \cdot s^{-1}} = kg \cdot s^{-1}$.
Therefore,the unit of the constant of proportionality is $kg \cdot s^{-1}$.
To find the unit of $b$,we rearrange the formula: $b = \frac{F}{v}$.
The $SI$ unit of force $F$ is Newton $(N)$,which is equivalent to $kg \cdot m \cdot s^{-2}$.
The $SI$ unit of velocity $v$ is $m \cdot s^{-1}$.
Substituting these units into the formula for $b$:
$b = \frac{kg \cdot m \cdot s^{-2}}{m \cdot s^{-1}} = kg \cdot s^{-1}$.
Therefore,the unit of the constant of proportionality is $kg \cdot s^{-1}$.
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View SolutionUnits, Dimensions and Measurement — Dimensions and Dimensional Formula · Frequently Asked Questions
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