13) HC VERMA QN .NO 8 PAGE NO274 ...it starts as...."water is filled in a rectangular........................ sum 1 pl help...want assistance on fluids
1)a glass full of water has a bottom area 20cm2 and top area 20cm2 ht 20 cm ...volume half a litre considering the equilibrium of water find the resultant force exerted by the sides of the glass on the water... ...all standard data as usual ...
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12 Answers
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11(A) USE F=Vpg =6 X 10 X1000 =600000N
(B) WITH THE INCREASE IN DEPTH THE DENSITY OF THE FLUID CHANGES
SO
WITH IS δx TOTAL FORCE =δx X 2 X 1000=2000δxN
(C) Γ=R X F
SO TOTAL FORCE MULTIPLIED BY THE DISTANCE FROM THE EDGE GIVES THE TORQUE
i.e (1-x) X F = 2000δx X(1-x)Nm
(D) AT THE BOTTOM EDGE THE FORCE IS 1 x 10 x 1000=10000N
(E) TORQUE ABOUT THE BOTTOM EDGE
IS
Γ=∫px2dxg
=px3/3 X g
=10000/3Nm
621)a glass full of water has a bottom area 20cm2 and top area 20cm2 ht 20 cm ...volume half a litre considering the equilibrium of water find the resultant force exerted by the sides of the glass on the water... ...all standard data as usual ...
pressure at height H is given by P0+ÏgH
take a strip of height dh. in the tank. .
Its area is 2.pi.r. dh
force on this strip is given by 2.pi.r.dh(P0+ÏgH)
now integrate it from height 0 to H
24well here is another method to derive force exerted................
though calculus method is better
Assume height is h and width is w
Paverage=(Ptop+Pbottom)/2
=> Pavg=0+Ïgh/2
=> Pavg=Ïgh/2
Force exerted =Pavg*Area=Ïgh/2 * hw
=>Force=Ïgh2w/2[1][1]
62Eureka I have seen this method a couple of times
be very careful when using it
for instance this will not work when the cross section area changes with height!
24yuip sir..i know that....thats why i already wrote in the begining that calculus method is better