Units and Measurement

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:

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Contributor-Level 10

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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a month ago

In a typical combustion engine the work done by a gas molecule is given by W = $α^{2} β e^{- \frac{β x^{2}}{k T}},$ where x is the displacement, k is the Boltzmann constant and T is the temperature. If and are constants, dimensions of will be

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Contributor-Level 10

$W = α^{2} β e^{- \frac{- β x^{2}}{k T}}$

$[β x^{2}] = [k T] = [M L^{2} T^{- 2}]$

$β = M T^{- 2}$

$[W] = [α^{2} β] \Rightarrow [M L^{2} T^{- 2}] = [α^{2}] [M T^{- 2}] \Rightarrow [α^{2}] = [L^{2}] \Rightarrow [α] = [L]$

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a month ago

The period of oscillation of a simple pendulum is $T = 2 π \sqrt[]{\frac{L}{g}} .$ Measured value of 'L' is 1.0m from meter scale having a minimum division of 1mm and time of one-complete oscillation is 1.95s measured from stopwatch of 0.01s resolution. The percentage error in the determination of 'g' will be:

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Vishal Baghel

Contributor-Level 10

$T = 2 π \sqrt[]{\frac{L}{g}} o r T^{2} = 4 π^{2} (\frac{L}{g})$

$\Rightarrow g = 4 π^{2} (\frac{L}{T^{2}})$

$\Rightarrow \frac{Δ g}{g} % = (\frac{Δ L}{L} + \frac{2 Δ T}{T}) % = [\frac{1 m m}{1 m} + \frac{2 * 0.01}{1.95}] * 100 % = (0.001 + 0.0102) * 100 = 1.13 %$

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New answer posted

a month ago

The pitch of the screw gauge is 1mm and there are 100 divisions on the circular scale. When nothing is put in between the jaws, the zero of the circular scale lies 8 divisions below the reference line. When a wire is placed between the jaws, the first linear scale division is clearly visible while 72^nd division on circular scale coincides with the reference line, The radius of the wire is:

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Vishal Baghel

Contributor-Level 10

$L C = \frac{1}{100} m m = 0.01 m m$

Zero error = +0.08 mm.

Diameter = 1 + 72 * 0.01 – 0.08 = 1.64 mm

Radius = 0.82 mm

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a month ago

If e is the electronic charge, c is the speed of light in free space and h is Planck's constant, the quantity $\frac{1}{4 π ε_{0}} \frac{{| e |}^{2}}{h c}$ has dimensions of

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Contributor-Level 10

$\frac{1}{4 π ε_{0}} \frac{{| e |}^{2}}{h c} = \frac{1}{4 π ε_{0}} \frac{{| e |}^{2}}{r^{2}} * \frac{r^{2}}{\frac{h c}{λ} * λ} = \frac{F r * r}{E * λ}$

Dimension of 'Fr' and 'E' are same dimensionless

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a month ago

A force defined by $F = α t^{2} + β t$ acts on a particle at a given time $t$ . The factor which is dimensionless, if $α$ and $β$ are constants, is:

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Contributor-Level 10

From principle of homogeneity

$[F] = [α t^{2}] = [β t]$

$[α] = \frac{[F]}{[t^{2}]} and [β] = \frac{[F]}{[t]}$

$∴ [α] [t] = [β]$

$∴ \frac{α t}{β} = dimensionless$

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a month ago

In a vernier callipers, $(N + 1)$ divisions of vernier scale coincide with $N$ divisions of main scale. If 1 MSD represents 0.1 mm, the vernier constant (in cm) is:

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Contributor-Level 10

$V . C = M S D - V S D$ … (i)

given : $(N + 1) VSD = N MSD$

$VSD = (\frac{N}{N + 1}) MSD$ … (ii)

From (1) and (2)

$V . C = (MSD) - \frac{N}{N + 1} (MSD)$

$= MSD (1 - \frac{N}{N + 1}) = \frac{MSD}{N + 1}$

$= \frac{0.01}{N + 1} = \frac{1}{100 (N + 1)}$

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a month ago

A wire of length 'l' and resistance $100 Ω$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

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Contributor-Level 10

Divided into 10 parts

$R = \frac{ρ l}{A}$

$R^{'} = \frac{ρ l}{10 A} = \frac{R}{10}$

$R_{S} = 5 * \frac{R}{10} [series]$

$R_{S} = 50$

$R_{P} = \frac{R}{50} [parallel]$

$R_{eq} = R_{S} + R_{P}$

$= 52 Ω$

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a month ago

In an experiment to determine the Young's modulus of wire of a length exactly 1m, the extension in the length of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm when a load of 1kg is applied. The diameter of the wire is measured as 0.4mm with an uncertainty of $\pm$ 0.02mm. The error in the measurement of Young's modulus $(Δ Y)$ is found to be x * 10¹⁰Nm^-2. The value of x is _________.

(take g = 10 ms^-2)

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Contributor-Level 10

L = 1 m

$Δ L = 0.4 * 1 0^{- 3} m$

$m = 1 k g$

d = 0.4 * 103 m

$\frac{F}{A} = y \frac{Δ L}{L}$

$y = \frac{F L}{A Δ L} = \frac{(m g) * 1}{(\frac{π d^{2}}{4}) * 0.4 * 1 0^{- 3}}$

$Δ y = 0.1 * 0.199 * 1 0^{12} = 1.99 * 1 0^{10}$

= 1.99

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a month ago

A screw gauge of pitch 0.5mm is used to measure the diameter of uniform wire of length 6.8cm, the main scale reading is 1.5mm and circular scale reading is 7. The calculated curved surface area wire to appropriate significant figures is:

[Screw gauge has 50 divisions on its circular scale]

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