Dimensional analysis and its applications MCQ Quiz in मल्याळम - Objective Question with Answer for Dimensional analysis and its applications - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

CONCEPT:

• Dimensional formula: It is a compound expression showing which of the fundamental quantities are involved and how are they involved in making that physical quantity.

CALCULATION:

Given quantity \(A =\dfrac{mv}{qB}\)

Take dimensions of both sides

\([A] =\dfrac{[m][v]}{[q][B]}\)

\([X] =\dfrac{[M][LT^{-1}]}{[AT][MT^{-2}A^{-1}]}\)

\([X] =[L]\)

• So the correct answer is option 1.

The correct answer is option 1) i.e. ms-2

CONCEPT:​

• The dimensional formula is used to express any physical quantity in terms of fundamental quantities.
• Dimensional Analysis: It is a technique adopted to derive a relationship between various physical quantities using their dimensions and units of measurement. The two basic rules followed in the dimensional analysis are:
  • Only those quantities that have the same dimensions can be added and subtracted from each other.
  • Two physical quantities are equal if they have the same dimensions.​

EXPLANATION:

• The terms qx and rt² are added to each other. Hence, they will have the same dimensions.
• On the LHS is velocity whose dimensional formula is [L T-1]. Therefore, the dimensional formula of qx and rt2 will be [L T-1].

On equating,

[L T-1] = qx 

⇒ [L T-1] = q[L] ⇒ q = [T^-1]

[L T-1] = rt2

⇒ [L T-1]  = r[T²] ⇒ r = [L T-3]

Ratio = \(\frac{r}{q} = \frac{ [L T^{-3}]}{[T^{-1}]} = [LT^{-2}]\)

The units for length [L] and time [T] are meter (m) and second (s) respectively.

Substituting the units for dimensions [L] and [T],

\(\frac{r}{q} = ms^{-2}\)

The correct answer is option 1) i.e. ms-2, ms-3

CONCEPT:​

• The dimensional formula is used to express any physical quantity in terms of fundamental quantities.
• Dimensional Analysis: It is a technique adopted to derive a relationship between various physical quantities using their dimensions and units of measurement. The two basic rules followed in the dimensional analysis are:
  • Only those quantities that have the same dimensions can be added and subtracted from each other.
  • Two physical quantities are equal if they have the same dimensions.

EXPLANATION:

• LHS and RHS of the expression will have the same dimensions. 
• The terms αt2 and βt3 are added to each other. Therefore, they both will have the same dimensions which are equal to the dimension of x i.e. meter (m).

On equating,

αt² = m ⇒ α(s²) = m ⇒ α = ms^-2

βt3 = m ⇒ β(s3) = m ⇒ β = ms-3

The units of α and β will be ms-2 and ms-3 respectively.

CONCEPT:

• Universal Gravitational Constant (G): It is a relating force to mass and distance in Newton’s laws of gravitation.
  • Its value is 6.67408 × 10^-11 m³ kg^-1 s^-2 in SI Unit.
  • The law of Gravitation states, the force between two bodies is directly proportional to the product of their masses and inversely to the square of their distance between them.
  • \({F_{two\;bodies}} = G\frac{{{m_1}{m_2}}}{{{R^2}}}\)  [F = force between two bodies, m₁ = mass of one body, m₂ = mass of other body, R = distance between them, G = Gravitational Constant].
• Velocity: It is defined as displacement per unit time.
  • Its unit is represented by m/s or ms^-1.
  • Its dimensional is [LT^-1].
• Wavelength: It is the distance of two identical adjacent points in a wave.
  • Its unit is length and dimension is [L]

CALCULATION:

\({\rm{F}} = {\rm{G}}\frac{{G{M^2}}}{{{R^2}}} \Rightarrow G = \left[ {{M^{ - 1}}{L^3}{T^{ - 2}}} \right]\)

Mass = [M], Velocity = [LT^-1] and wavelength (λ) = [L]

G ∝ M^x V^y λ^z

[M^-1 L³ T^-2] ∝ [M¹]^x [LT^-1]^y [L]

[M^-1 L³ T^-2] ∝ [M]^x [L]^y+z [T]^-y

On comparing the power of M, L, T

x = -1, y + z = 3, -y = -2

On solving: x = -1, y = 2 and z = 1

G ∝ M^-1 V² λ¹

G ∝ (V² λ)/M

• Some Basic Dimensions are:

CONCEPT:

• Dimensional formula: It is a compound expression showing how and which of the fundamental quantities are involved in making that physical quantity.
• The dimension's principle of homogeneity says that an equation is dimensionally correct only if each term has the same dimension on both sides of the equation.

CALCULATION:

Given equation is x = at + bt2

The dimension of [x] will be equal to the dimension of [at] and [bt²]

[x] = [at]

[a] = [x/t]

$x$ has dimension of length and $t$ has dimension of time so unit of $a$ will be ms^-1.

[x] = [bt2]

[b] = [x/t2]

$x$ has dimension of length and $t$ has dimension of time so unit of b will be ms-2.

So the correct answer is option 4.

CONCEPT:

Principle of homogeneity of dimensions:

• According to this principle, a physical equation will be dimensionally correct if the dimensions of all

          the terms occurring on both sides of the equation are the same.

• This principle is based on the fact that only the physical quantities of the same kind can be

           added, subtracted, or compared.

• Thus, velocity can be added to velocity but not to force.

EXPLANATION:

Given that, v = u + at

• From the principle of dimensional homogeneity, the left-hand side of the equation dimensionally

          equal to the right-hand side of the equation.

• The dimension formula of velocity (v) = [LT^-1] 

Hence, [LT^-1] = u

• Therefore dimension of  'u' is [LT^-1].

For the second term

[LT-1] = aT

=> a = [LT^-1] / T = [LT^-2]

Therefore the dimension of a is [LT^-2]

CONCEPT:

Radioactivity:

• Radioactivity is a process by which the nucleus of an unstable atom loses energy by emitting radiation.
  • ​Two forces, namely the force of repulsion that is electrostatic and the powerful forces of attraction of the nucleus acts in the nucleus.
  • These two forces are considered extremely strong in nature.
  • The instability of the nucleus increases as the size of the nucleus increases because the mass of the nucleus becomes a lot to concentrate. That’s the reason why atoms of Plutonium, Uranium are extremely unstable and show radioactivity.
• The spontaneous breakdown of an atomic nucleus of a radioactive substance causing the emission of radiation from the nucleus is known as Radioactive decay.

​EXPLANATION:

• In radioactive decay, the number of remaining nuclei after decay time t is given as,

⇒ N = N_oe^-λt     -----(1)

Where N_o = initial amount of nuclei, and $\lambda =$ disintegration or decay constant

• We know that the exponential term is always unitless quantity. So the power of the exponential term is also a unitless quantity.

​The dimension of t is given as,

⇒ [t] = [M⁰ L⁰ T¹]     -----(2)

Let the dimension of disintegration constant is,

⇒ [λ] = [Mx Ly Tz]     -----(3)

The λt term is a unitless quantity. So,

⇒ [λt] = [M0 L0 T0]     -----(4)

by equation 2, equation 3, and equation 4,

⇒ [Mx Ly Tz][M0 L0 T1] = [M0 L0 T0]

⇒ [Mx Ly Tz+1] = [M0 L0 T0]     -----(5)

By equation 5,

⇒ x = 0,

⇒ y = 0,

⇒ z + 1 = 0

⇒ z = -1

Therefore the dimension of disintegration constant is,

⇒ [λ] = [M0 L0 T-1]     -----(6)

• By equation 6 it is clear that the possible units for the disintegration constant λ is 1/day.
• Hence, option 4 is correct.

CONCEPT:

Frequency

• It is defined as the number of cycles per unit of time.
• Its unit is Hz or sec^-1.

Force constant of spring

• It is defined as the force required for unit displacement of the spring.
• Its unit is N/m.

CALCULATION:

Given f = cm^xk^y     -----(1)

• The dimension of frequency,

\(\Rightarrow \left [ f \right ]=\left [ M^{0}L^{0}T^{-1} \right ]\)     -----(2)

• The dimension of mass,

\(\Rightarrow \left [ m \right ]=\left [ M^{1}L^{0}T^{0} \right ]\)     -----(3)

• The dimension of force constant of spring,

\(\Rightarrow \left [ k \right ]=\left [ M^{1}L^{0}T^{-2} \right ]\)     -----(4)

• The dimension of c is,

\(\Rightarrow \left [ c \right ]=\left [ M^{0}L^{0}T^{0} \right ]\)     -----(5)

By equation 1, equation 2, equation 3, equation 4, and equation 5,

\(\Rightarrow \left [ f \right ]=\left [ c \right ]\left [ m \right ]\left [ k \right ]\)

\(\Rightarrow \left [ M^{0}L^{0}T^{-1} \right ]=\left [ M^{0}L^{0}T^{0} \right ]\left [ M^{1}L^{0}T^{0} \right ]^x\left [ M^{1}L^{0}T^{-2} \right ]^y\)

\(\Rightarrow \left [ M^{0}L^{0}T^{-1} \right ]=\left [ M^{x+y}L^{0}T^{-2y} \right ]\)     -----(6)

By equating both the sides of equation 6,

\(\Rightarrow x+y=0\)     -----(7)

\(\Rightarrow -2y=-1\)

\(\Rightarrow y=\frac{1}{2}\)

\(\Rightarrow x=-\frac{1}{2}\)

• Hence, option 1 is correct.

Explanation:

We know that 

B = μ₀ni

\(\left[\frac{\mathrm{B}}{\mu_{0}}\right]=[\mathrm{ni}]=\mathrm{L}^{-1} \mathrm{~A}^{1}\)

Concept:

Thermal Conductivity:

• Thermal conductivity is a measure of a material's ability to conduct heat.
• It is defined as the quantity of heat transmitted through a unit thickness of a material in a direction normal to a surface of unit area due to a unit temperature gradient under steady state conditions.
• The SI unit of thermal conductivity is Watt per meter-Kelvin (W/m·K).
• The dimensional formula of thermal conductivity is derived from its unit.
• Thermal Conductivity, κ" id="MathJax-Element-56-Frame" role="presentation" style="position: relative;" tabindex="0">κκ" id="MathJax-Element-16-Frame" role="presentation" style="position: relative;" tabindex="0">κκ κ $\kappa$ :
  • SI Unit: W/m·K
  • Dimensional Formula: [MLT^-3K^-1]

Calculation:

Given,

Thermal conductivity unit: W/m·K

The dimensional formula of Watt (W) is [ML²T^-3].

The dimensional formula of meter (m) is [L].

The dimensional formula of Kelvin (K) is [K].

Therefore, the dimensional formula of W/m·K is:

⇒ [ML²T^-3] / [L] / [K]

⇒ [ML²T^-3] × [L^-1] × [K^-1]

⇒ [MLT^-3K^-1]

∴ The correct dimensional formula of thermal conductivity is [MLT^-3K^-1].