A. $${\text{p}} = {\text{T}} \times {\text{r}}$$
B. $${\text{p}} = \frac{{\text{T}}}{{\text{r}}}$$
C. $${\text{p}} = \frac{{\text{T}}}{{2{\text{r}}}}$$
D. $${\text{p}} = \frac{{2{\text{T}}}}{{\text{r}}}$$
Answer: Option D
This Question Belongs to Civil Engineering >> Hydraulics And Fluid Mechanics
A. 22.5 m/sec.
B. 33 m/sec.
C. 40 m/sec.
D. 90 m/sec.
A. the weight of the body
B. more than the weight of the body
C. less than the weight of the body
D. weight of the fluid displaced by the body
The difference of pressure between the inside and outside of a liquid drop is
A. $${\text{p}} = {\text{T}} \times {\text{r}}$$
B. $${\text{p}} = \frac{{\text{T}}}{{\text{r}}}$$
C. $${\text{p}} = \frac{{\text{T}}}{{2{\text{r}}}}$$
D. $${\text{p}} = \frac{{2{\text{T}}}}{{\text{r}}}$$
A. cannot be subjected to shear forces
B. always expands until it fills any container
C. has the same shear stress.at a point regardless of its motion
D. cannot remain at rest under action of any shear force
👍
Force inside drop=Force outside
Excess pressure x Area= Tension x length of meniscus
P x π x r²= T x 2 x π x r
P= 2T/r
Surface Tention in case of drop
T=PR/2
P=2T/R
Here in case Of bubble
brusting force = resistance force
P xπR^2 = T x 2πR
P=2T/R
T=four times of surface tension
what you mean T
How it comes