Units and Measurement

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New answer posted•a year ago

Units and Measurement Class 11th

Distance travelled by linear scale of screw gauge during two full rotation of circular scale is 1mm and circular scale has 50 divisions. In an experiment to measure thickness of a plate, six divisions of main scale are clearly visible and 28th division of circular scale coincides with reference line. Moreover when studs touch each other, zero of circular scale lies 4 division below the reference line, thickness of the plate will be:

16 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Pitch p = 0.5 mm
Device has +ve zero error
R = 6p + 28p/50 - 4p/50 = 6p + 24p/50; 3mm + (24/50) * 0.5 = 3.24 mm

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New answer posted•a year ago

Units and Measurement 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

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. 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

Units and Measurement Class 11th

If $Z = \frac{A^{2} B^{3}}{C^{4}},$ $\frac{^{}^{}}{^{}}$ then the relative error in Z will be:

2 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$z = \frac{A^{2} B^{3}}{C^{4}}$

The relative error in z is given by

$\frac{Δ z}{z} = \frac{2 Δ A}{A} + \frac{3 Δ B}{B} + \frac{4 Δ C}{C}$

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New answer posted•a year ago

Units and Measurement Class 11th

If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

E ∝ F? A? T?
[M¹L²T? ²] ∝ [M¹L¹T? ²]? [LT? ²]? [T]?
a = 1
a + b = 2 ⇒ b = 1
-2a - 2b + c = -2
⇒ c = 2
a = 1 b = 1 c = 2
E ∝ [F] [A] [T²]

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New answer posted•a year ago

Units and Measurement Class 11th

Consider the following statements (A) and (B) and identify the correct answer.
(A) A zener diode is connected in reverse bias, when used as a voltage regulator.
(B) The potential barrier of p-n junction lies between 0.1 V to 0.3 V.

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Least count = 1 mm / 100 = 0.01 mm
Diameter = main scale reading + circular scale reading
Diameter = 0 + 52 * 0.01 mm
= 0.52 mm = 0.052 cm

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New answer posted•a year ago

Units and Measurement Class 11th

A screw gauge gives the following readings when used to measure the diameter of a wire
Main scale reading: 0 mm
Circular scale reading: 52 divisions
Given that 1 mm on main scale corresponds to 100 divisions on the circular scale. The diameter of the wire from the above data is:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Least count = (1 mm) / 100 = 0.01 mm.

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New answer posted•a year ago

Units and Measurement Class 11th

A metal wire has mass $(0.4 \pm 0.002) g$ , radius $(0.3 \pm 0.001) m m$ $_{}$ and length $(5 \pm 0.02) c m$ . The maximum possible percentage error in the measurement of density will nearly be :

20 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$m = ρ π r^{2} 1$

$\begin{array}{r} ρ = \frac{m}{π r^{2} 1} \\ \frac{Δ ρ}{ρ} = \frac{Δ m}{m} + \frac{2 Δ r}{r} + \frac{Δ l}{1} \\ \frac{Δ ρ}{ρ} \times 100 = \frac{0.002}{0.4} \times 100 + \frac{2 \times 0.001}{0.3} \times 100 + \frac{0.02}{5} \times 100 \\ = \frac{0.2}{0.4} + \frac{0.2}{0.3} + \frac{2}{5} \\ = 0.5 + 0.67 + 0.4 \\ = 1.57 \\ = 1.6 % \end{array}$

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New answer posted•a year ago

Units and Measurement Class 11th

The errors in the measurement which arise due to unpredictable fluctuations in temperature and voltage supply are

3 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

As the factors controlling temperature and voltage supply are beyond prediction and control so the error occurred due to unpredictable fluctuations of temperature and voltage would be random errors.

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New answer posted•a year ago

Units and Measurement 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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