Question
A steel ball of diameter $3.2 \mathrm{~mm}$ falls under gravity through an oil of density $920 \mathrm{~kg} \mathrm{~m}^{-3}$ and viscosity $1.64 \mathrm{Ns} \mathrm{m}^{-2}$. The density of steel. may be taken as $7820 \mathrm{~kg} \mathrm{~m}^{-3}$. What is the terminal velocity of the ball ?(A) $3.45 \mathrm{cms}^{-1}$(B) $4.25 \mathrm{cms}^{-1}$(C) $1.85 \mathrm{~cm}^{-1}$(D) $2.39 \mathrm{cms}^{-1}$
Step 1
The diameter is given as $3.2 \mathrm{~mm}$, so the radius $r$ is half of the diameter, which is $1.6 \mathrm{~mm}$ or $1.6 \times 10^{-3} \mathrm{~m}$. Show more…
Show all steps
Your feedback will help us improve your experience
Prem Bijarniya and 75 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Using the equation of the previous problem, find the viscosity of motor oil in which a steel ball of radius 0.8 mm falls with a terminal speed of 4.32 cm/s. The densities of the ball and the oil are 7.86 and 0.88 g/mL, respectively.
A wide jar is filled with glycerine having specific gravity $1.26$, in this jar, a steel ball of radius $0.25 \mathrm{~cm}$ has been dropped. After some time it has been observed that ball is taking equal interval of time $(1.8 \mathrm{~s})$ to cover equal successive distances, of $20 \mathrm{~cm} .$ [Take, $\left.\rho_{\text {steel }}=7.8 \times 10^{3} \mathrm{~kg}-\mathrm{m}^{3}, g=9.81 \mathrm{~ms}^{-2}\right]$. The viscosity of glycerine is [in $\mathrm{N}-\mathrm{sm}^{-2}$ ] (a) $0.802$ (b) $1.67$ (c) $0.76$ (d) $0.963$
Experimental Physics
Round 1
A steel ball of diameter $3 \mathrm{~mm}$ falls through glycerine and covers a distance of $25 \mathrm{~cm}$ in $10 \mathrm{~S} .$ The specific gravity of steel and glycerine are $7.8$ and $1.26$ respectively. The viscosity of glycerine is about........ pascal-sec (A) $1.3$ (B) $1.5$ (C) $0.8$ (D) $1.0$
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD