(i) Liquid of density r flows at speed u along a flexible pipe bent into a semicircle of radius R. The cross sectional area of the pipe is A and its cross sectional radius is small compared to R. Three strings S1, S2 and S3 keep the pipe in place. S3 ties the two ends of the pipe and the other two string have their ends secured at A and B. Strings S1and S2 are perpendicular to the string S3. The entire system is in horizontal plane. Find the tension in the three strings.
(ii) A car of mass M is moving with a velocity V₀ on a smooth horizontal surface. Bullets, each of mass m, are fired horizontally perpendicular to the velocity of the car with a speed u relative to the car. After firing n bullets it was found that the car was travelling with velocity V₀ in a direction opposite to its original direction of motion. Assume that mu << MV0 and also that nm << M. Find n in terms of other given parameters.

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$T_{S!} = ρ A V^{2}$

$T_{S 2} = ρ A V^{2}$

$T_{S 3} = ρ A V^{2}$

Only 1 and 3 are correct

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

(i) Force on water inside the pipe in y direction = rate of change of y component of momentum of water
= (mass of water entering / leaving the pipe in unit time) × (2v)
$= (ρAv) (2 v) = 2 {ρAv}^{2} (in - ve y direction)$
Water applies same force on the pipe in y direction

$F_{y} = 2 {ρAv}^{2} ∴ T_{s 1} + T_{s 2} = 2 {ρAv}^{2}$
From symmetry $T_{s 1} = T_{s 2}$
$∴ T_{s 1} = T_{s 2} = {ρAv}^{2}$
Now consider a quarter of the pipe.
Change in x component of momentum of water in this segment in unit time is the x component of force
$F_{X} = (ρAv) (v) = {ρAv}^{2}$
Tension in s3 balances this force
$∴ T_{S 3} = {ρAv}^{2}$