Units and Measurement

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New answer posted•a year ago

Units and Measurement 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. α and β are constants. Then the dimensions of β will be:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

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

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

Units and Measurement Class 11th

In a vernier calipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be__________× 10^-²mm.

3 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

Let ‘x’ he the value of one division of main scale

$x = \frac{1}{20} c m = 0.05 c m$

Let y be value of one division on venire scale given

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New answer posted•a year ago

Units and Measurement Class 11th

The Vernier constant of Vernier calipers is 0.1 mm and it has zero error of (-0.05) cm. White measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The correct diameter will be__________ × 10^-2 cm.

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Least count of Vernier = 0.1mm

Reading of Vernier Scale = 5 × 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 × 10^-2 cm.

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New answer posted•a year ago

Units and Measurement 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

Units and Measurement Class 11th

The Vernier constant of Vernier calipers is 0.1 mm and it has zero error of (-0.05) cm. White measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The correct diameter will be__________ × 10^-² cm.

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Least count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 × 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 × 10^-2 cm.

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New answer posted•a year ago

Units and Measurement Class 11th

Given below are two statements : One is labeled as Assertion A and the other is labeled 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 = $\frac{F o r c e}{V e l o c i t y g r a d i e n t}$

Choose the correct answer from the options given below:

5 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

Pressure × time $= \frac{F t}{A} = [M L T^{- 2}] [T] [L^{- 2}] = [M L^{- 1} T^{- 1}]$

Coefficient of viscosity, $η = [M L^{- 1} T^{- 1}]$

$F_{v i s c o u s} = - η A \frac{Δ v}{Δ y}$

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New answer posted•a year ago

Units and Measurement Class 11th

Which of the following is a CORRECT statement?

37 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

(1) 2.3056 -> 4 decimals

10.138 -> 3 decimals

-7.4671 -> 4 decimals

_________________

4.9765

Answer should have 3 decimals.

(2) 2.38 × 1.0 = 2.38 -> answer should have 2 significant digits

(3) $\frac{8.05}{3.1} = 2.59 \approx 2.6$ answer should have 2SD

(4) 1.11 - 0.1 = 1.01 $\approx$ &nbs

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New answer posted•a year ago

Units and Measurement Class 11th

The Vernier constant of Vernier calipers is 0.1 mm and it has zero error of (-0.05) cm. White measuring diameter of a sphere, the main scale reading is 1.7 cm and coinciding vernier division is 5. The correct diameter will be__________ × 10^-² cm.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Least count of Vernier = 0.1mm

$\Rightarrow$ Reading of Vernier Scale = 5 × 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 × 10^-2 cm.

Upvote

New answer posted•a year ago

Units and Measurement 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

Units and Measurement Class 11th

A vernier calliper whose 10 vernier scale divisions coincide with 9 main scale divisions is used to take a reading. Each main scale division is 1 mm. Before taking the reading it is observed that the 4^th division of the vernier scale coincides with the main scale and zero of main scale is behind zero of vernier scale. When the reading is taken using this vernier calliper, the main scale reads 2.5 cm and the 7^th division of the vernier scale coincides with the main scale. Find the final value of the measurement.

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Zero error = 0.0 + 4 * 0.01 = 0.04 cm
Reading = 2.5a + 7 * 0.01 = 2.57 cm
Correct reading = 2.57 cm - 0.04 cm = 2.53 cm

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