Tags Units and Measurement

Units and Measurement

Get insights from 122 questions on Units and Measurement, answered by students, alumni, and experts. You may also ask and answer any question you like about Units and Measurement

Follow Ask Question

122

Questions

Discussions

Active Users

Followers

New answer posted

2 weeks 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:

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

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

$\frac{k t}{β x}$ = 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

0 0

New answer posted

2 weeks 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.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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.

0 0

New answer posted

2 weeks 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:

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

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

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

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

0 0

New answer posted

3 weeks ago

Units and Measurement Class 11th

Which of the following is a CORRECT statement?

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

(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 $\approx$ 1.0 but is an intermediate step so we keep 1 digit extra. 1.01 * 9.0 = 9.09 $\approx$ 9.1 -> both 1.01 and 9.0 have 2SD

0 0

New answer posted

3 weeks ago

Units and Measurement Class 11th

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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.

0 0

New answer posted

3 weeks 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:

0 Follower 1 View

Vishal Baghel

Contributor-Level 10

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

m = FTV¹

0 0

New answer posted

3 weeks 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.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

0 0

New answer posted

3 weeks 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:

0 Follower 1 View

Vishal Baghel

Contributor-Level 10

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

0 0

New answer posted

3 weeks 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:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

0 0

New answer posted

3 weeks 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:

0 Follower 2 Views

alok kumar singh

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, $[α β^{2}] = M L^{2} T^{- 2}$

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

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Units and Measurement

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience