Which physical quantity has unit of pascal - second? (a) Force (b) Energy (c) Coefficient of viscosity (d) velocity

Short Answer

Expert verified
The physical quantity with the unit of Pascal-second (Pa·s) is the Coefficient of Viscosity. The correct answer is option (c).

Step by step solution

01

Analyze Option (a) - Force

Force is measured in Newtons (N), which is equal to kg·m/s². Here, we can see that the unit of Force does not include Pascal or seconds. So, option (a) is not the correct answer.
02

Analyze Option (b) - Energy

Energy is measured in Joules (J), which is equal to N·m or kg·m²/s². Here, we can see that the unit of Energy also does not include Pascal or seconds. So, option (b) is not the correct answer.
03

Analyze Option (c) - Coefficient of viscosity

Coefficient of viscosity (η) is a measure of a fluid's resistance to flow. Its unit is given by: η = \( \frac{Force \times Time}{Area} \) The unit of force is Newton (N) and the unit of area is m². Thus the unit of coefficient of viscosity becomes \( \frac{N \cdot s}{m^2} \) Now, we know that 1 Pascal (Pa) is equal to \( \frac{N}{m^2} \), so we can rewrite the unit of coefficient of viscosity as: \( \frac{Pa \cdot N \cdot s}{N} \) Cancelling out the N, we get: \( Pa \cdot s \) Thus, the unit of coefficient of viscosity is Pascal-second (Pa·s). So, option (c) is the correct answer.
04

Analyze Option (d) - Velocity

Velocity is measured in meters per second (m/s). Here, we can see that the unit of Velocity does not include Pascal. So, option (d) is not the correct answer.
05

Conclusion

As shown in Step 3, the physical quantity with the unit of Pascal-second (Pa·s) is the Coefficient of Viscosity. The correct answer is option (c).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The dimensional formula of magnetic flux is ............ (a) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{-1}\) (b) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{1} \mathrm{~A}^{2}\) (c) \(\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{-2} \mathrm{~A}^{2}\) (d) \(\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^{1} \mathrm{~A}^{2}\)

What is the number of significant figures in \(5.50 \times 10^{3}\) ? (a) 2 (b) 7 (c) 3 (d) 4

Test if the following equation are dimensionally correct \((\mathrm{S}=\) surface tension \(\rho=\) density \(\mathrm{P}=\) pressure \(\mathrm{v}=\) volume \(\mathrm{n}=\) coefficient of viscosity \(\mathrm{r}=\) radius \()\) (a) \(\mathrm{h}=[(2 \mathrm{~S} \cos \theta) /(\rho \mathrm{rg})]\) (b) \(\mathrm{v}=\sqrt{(p / \rho)}\) (c) $\mathrm{v}=\left[\left(\pi \mathrm{pr}^{4} \mathrm{t}\right) /(8 \mathrm{n} \ell)\right]$ (d) all correct

If $\mathrm{P}=\left[\left(\mathrm{A}^{2} \mathrm{~B}\right) /\left(\mathrm{C}^{3}\right)\right]\( where percentage error in \)\mathrm{A}, \mathrm{B}\( and \)\mathrm{C}$ are respectively \(\pm 2 \% \pm 3 \%\) and \(\pm 5 \%\) then total percentage error in measurement of \(\mathrm{p}\) (a) \(18 \%\) (b) \(14 \%\) (c) \(21 \%\) (d) \(12 \%\)

The periodic time of simple pendulum is $\mathrm{T}=2 \pi \sqrt{(\ell / \mathrm{g})}\( relative error in the measurement of \)\mathrm{T}\( and \)\ell$ are \(\pm \mathrm{a}\) and \(\pm \mathrm{b}\) respectively find relative error in the measurement of \(g\) (a) \(a+b\) (b) \(2 \mathrm{~b}+\mathrm{a}\) (c) \(2 \mathrm{a}+\mathrm{b}\) (d) \(a-b\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free