DIMENSIONAL ANALYSIS AND ITS APPLICATIONS

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New answer posted•a year ago

DIMENSIONAL ANALYSIS AND ITS APPLICATIONS Class 11th

An expression for a dimensionless quantity P is given by $P = \frac{α}{β} l o g_{e} (\frac{k t}{β t});$ where a and b are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of a will be:

5 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$P = \frac{α}{β} l o g_{e} (\frac{k t}{β x})$

$\frac{k t}{β x}$ = Dimensionless

$β = \frac{k t}{x} = \frac{[M L^{2} T^{- 2} k^{- 1}] [k]}{[L]}$

$\frac{α}{β} =$ dimensionless

a = dimensionless of b

a = MLT^-2

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New answer posted•a year ago

DIMENSIONAL ANALYSIS AND ITS APPLICATIONS Class 11th

If velocity [V], time [T] and force [F] are chosen as the base quantities, the dimensions of the mass will be:

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Using F = MA = $m \frac{V}{T}$ $\frac{}{}$

m = FTV¹

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New answer posted•a year ago

DIMENSIONAL ANALYSIS AND ITS APPLICATIONS Class 11th

If mass M, area A' and velocity V are chosen as fundamental units, then the dimension of coefficient of viscosity will be:

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

F = ηAdv / dx
η = F (dx/dv) (1/A) = m × a × time × (1/area) = MV/A

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New answer posted•a year ago

DIMENSIONAL ANALYSIS AND ITS APPLICATIONS Class 11th

The work done by a gas molecule in an isolated system is given by $W = α β^{2} e^{- \frac{x^{2}}{α k T}}$ , where x is the displacement, k is the Boltzmann constant and T is the temperature. a and b are constants. Then the dimensions of b will be:

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Since, $\frac{x^{2}}{α k T}$ should be dimensionless.

So, dimension of $α, [α] = \frac{L^{2}}{M L^{2} T^{- 2}} = M^{- 1} T^{2}$

Dimension of $α β^{2}$ should be that of W.

So, $[α β^{2}] = M L^{2} T^{- 2}$

$\Rightarrow [β^{2}] = \frac{M L^{2} T^{- 2}}{M^{- 1} T^{2}} = M^{2} L^{2} T^{- 4} \Rightarrow [β] = M L T^{- 2}$

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New answer posted•a year ago

DIMENSIONAL ANALYSIS AND ITS APPLICATIONS Class 11th

If E and G respectively denotes energy and gravitational constant, then E/G has the dimensions of:

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

The dimensional formula for energy E is [ML²T? ²].
The dimensional formula for the gravitational constant G is [M? ¹L³T? ²].
The ratio E/G has dimensions: [ML²T? ²] / [M? ¹L³T? ²] = [M²L? ¹T? ].

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New answer posted•a year ago

DIMENSIONAL ANALYSIS AND ITS APPLICATIONS Class 11th

A quantity X is given by (IFv²MLL⁴) in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of:

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

x = (IFV²)/ (WL? )
I = [ML²]
F = [MLT? ²]
V² = [L² T? ²]
W = [ML² T? ²]
Q? = [L? ]
X = [ML? ¹ T? ²]

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New answer posted•a year ago

DIMENSIONAL ANALYSIS AND ITS APPLICATIONS Class 11th

Dimensional formula for thermal conductivity is (here K denotes the temperature):

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

K = (Q?)Δx / AΔT
⇒ (ML²T?²)(L) / (L²)(θ)(T)
⇒ M¹L¹T?³θ?¹

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New answer posted•a year ago

DIMENSIONAL ANALYSIS AND ITS APPLICATIONS Class 11th

The dimension of stopping potential $V_{0}$ $_{}$ in photoelectric effect in units of Planck's constant ' $h$ ', speed of light ' $c$ ' and Gravitational constant ' $G$ ' and ampere $A$ is:

5 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

Bonus $V = K (h)^{a} (I)^{b} (G)^{c} (C)^{d} (V$

is voltage $)$

We know, $\begin{array}{r} [h] = {M L}^{2} T^{- 1} \\ [l] = A \\ [G] = M^{- 1} L^{3} T^{- 2} \\ [C] = L T^{- 1} \\ [V] = M L^{2} T^{- 3} A^{- 1} \end{array}$

$M L^{2} T^{- 3} A^{- 1} = {(M L^{2} T^{- 1})}^{a} (A)^{b} {(M^{- 1} L^{3} T^{- 2})}^{c} {(L T^{- 1})}^{d}$

$M L^{2} T^{- 3} A^{- 1} = M^{a - c} L^{2 a + 3 c + d} T^{- a - 2 c - d} A^{b}$

$a - c = 1$

$2 a + 3 c + d = 2$

$- a - 2 c - d = - 3$

$b = - 1$

On solving,

$c = - 1$ $a = 0$

$d = 5, b = - 1$

$V = K (h)^{?} (I)^{- 1} (G)^{- 1} (C)^{5}$

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New answer posted•a year ago

DIMENSIONAL ANALYSIS AND ITS APPLICATIONS Class 11th

The work done by a gas molecule in an isolated system is given by $W = α β^{2} e^{- \frac{x^{2}}{α k T}}$ , where x is the displacement, k is the Boltzmann constant and T is the temperature. a and b are constants. Then the dimensions of b will be:

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Since, $\frac{x^{2}}{α k T}$ should be dimensionless.

So, dimension of $α, [α] = \frac{L^{2}}{M L^{2} T^{- 2}} = M^{- 1} T^{2}$

Dimension of $α β^{2}$ should be that of W.

So, $[α β^{2}] = M L^{2} T^{- 2}$

$\Rightarrow [β^{2}] = \frac{M L^{2} T^{- 2}}{M^{- 1} T^{2}} = M^{2} L^{2} T^{- 4} \Rightarrow [β] = M L T^{- 2}$

Upvote

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