A water drop of radius 1 μ m falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 * 10-5 Nsm-2 and its density is negligible as compared to that of water 106 gm-3. Terminal velocity of the water drop is: (Take acceleration due to gravity = 10 ms-2)

NLM 1 ρω43πr3g=6πηrVT ρω43πr2g= 6πηvT VT=43ρωπr2g6πη =29*103* (10−6)2*101.8*10−5 =0.1234*10−3 vT = 123.4 * 10-6m/sec

A water drop of radius $1 μ m$ falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 * 10^-⁵ Nsm^-² and its density is negligible as compared to that of water 10⁶ gm^-³. Terminal velocity of the water drop is:

(Take acceleration due to gravity = 10 ms^-²)

Option 1 - 145.4 × 10-6 ms-1

Option 2 - 118.0 × 10-6 ms-1

Option 3 - 132.6 × 10-6 ms-1

Option 4 - 123.4 × 10-6 ms-1

3 Views|Posted 6 months ago

Asked by Shiksha User

1 Answer

Sort by:

Answered by

Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 4

Detailed Solution:

NLM 1

$ρ_{ω} \frac{4}{3} π r^{3} g = 6 π η r V_{T}$

$\frac{ρ_{ω} 4}{3} π r^{2} g =$ $6 π η v_{T}$

$V_{T} = \frac{4}{3} \frac{ρ_{ω} π r^{2} g}{6 π η}$

$= \frac{2}{9} * 1 0^{3} * \frac{{(1 0^{- 6})}^{2} * 10}{1.8 * 1 0^{- 5}}$

$= 0.1234 * 1 0^{- 3}$

vT = 123.4 * 10^-6m/sec

Upvote

Similar Questions for you

If p is the density and $η$ is coefficient of viscosity of fluid which flows with a speed v in the pipe of diameter d, the correct formula for Reynolds number Re is:

Vishal Baghel

Kindly consider the following answer

$R e = \frac{ρ v d}{η}$

Two capacitors having C₁ and C₂ respectively are connected as shown in figure. Initially, capacitor C₁ is charged to a potential difference V volt by a battery. The battery is then removed and the charged capacitor C₁ is not connected to uncharged capacitor C₂ by closing the switch S. The amount of charge on the capacitor C₂, after equilibrium, is

Payal Gupta

After switch ‘S’ is closed

$Q_{1} + Q_{2} = C_{1} V$ - (1)

Using KVL

$\frac{Q_{1}}{C_{1}} - \frac{Q_{2}}{C_{2}} = 0$

$Q_{1} = Q_{2} \frac{C_{1}}{C_{2}}$ - (2)

from (1) & (2)

$\Rightarrow Q_{2} [\frac{C_{1} + C_{2}}{C_{2}}] = C_{1} V \Rightarrow Q_{2} = (\frac{C_{1} C_{2}}{C_{1} + C_{2}}) V$

Vishal Baghel

$\frac{4}{3} π R^{3} = 729 \frac{4}{3} π r^{3}$

R³ = 729r³

$R = (729) \frac{1}{3} (r) (\frac{1}{3})$

R = 9r

$Δ u = T (4 π r^{2}) \times 729 - T \times 4 π R^{2}$

$= T \times 4 π \times 8 R^{2}$

$\begin{array}{l} = 75.39 \times 1 0^{- 5} J \\ Δ u = 7.5 \times 1 0^{- 4} J \end{array}$

Consider a cylindrical tank of radius 1m is filled with water. The top surface of water is at 15m front the bottom of the cylinder. There is a hole on the wall of a cylinder at a height of 5m from the bottom. A force of 5 × 10⁵N is applied an the top surface of water using a piston. The speed of ifflux from the hole will be:

(given atmospheric pressure P_A = 1.01 × 10⁵Pa, density of water $ρ_{w} = 1000 k g / m^{3}$  and gravitational acceleration g = 10m/s²)

Vishal Baghel

$ρ_{ω} = 1000 k g / m^{3}$

G = 10 m/s²

Using bernaulli’s eqn

$\frac{5 \times 1 0^{5}}{π} + ρ g \times 10 = ρ_{0} + \frac{1}{2} ρ V_{e}^{2}$

$= \sqrt[]{318} ? 17.8 m / s$

A pressure-pump has a horizontal tube of cross sectional area 10cm2 for the outflow of water at a speed of 20 m/s. The force exerted on the vertical wall just in front of the tube which stops water horizontally flowing out of the tube, is:

[given : density of water = 1000 kg/m³]

Vishal Baghel

A = 10 cm², V = 20 m/s

$F = \frac{d p}{d t} = ρ A V^{2} = 1 0^{3} \times 10 \times 1 0^{- 4} \times 400$

= 400 N.

Taking an Exam? Selecting a College?

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

On Shiksha, get access to

66K

Colleges

1.2K

Exams

6.8L

Reviews

1.8M

Answers

Learn more about...

Physics NCERT Exemplar Solutions Class 12th Chapter Ten 2025

View Exam Details

Share Your College Life Experience

Didn't find the answer you were looking for?

Search from Shiksha's 1 lakh+ Topics

Ask Current Students, Alumni & our Experts

Quick Actions

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Have a question related to your career & education?

See what others like you are asking & answering