11. Fluid Mechanics

7/29/2019 11. Fluid Mechanics

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Rise Academy, 607/608-A, Talwandi , Kota (Raj.) -324005. 212

Exercise # 11.3 g(H h)

because pressure varies with height.

D;ksafd nkc pkbZ ds lkFk ifjofrZr gksrk gSA

1.41

1

A

gm=

2

2

A

gm

Solving,gy djus ij m2

= 3.75 kg.

1.5 WA

> WB

as mass of water in A is more than in B

WA

> WBpwafd A dk nzO;eku B ls T;knk gS

PA

= PB

Area of A = Area of B

{ks=kQy A = {ks=kQy B

or;k PA

AreaA

= PB

AreaB

PA{ks=kQy

A= P

B{ks=kQy

B

or;k FA

= FB

1.8 (i) a = a0

( i j + k )

As there is no gravity; the pressure difference will be only due to the accelerat ion.

At point B the pseudo force is maximum hence pressure is maximum.

At point H the pseudo force is minimum hence pressure is minimum

;gk dksbZ xq:Ro ugh gS vr% nkc esa ifjorZu Roj.k ds dkj.k gksxkA

fcUnq B ij Nne~ cy vf/kdre gS blfy, nkc Hkh vf/kdre gSAfcUnq H ij Nne~ cy de gS blfy, nkc Hkh de gSA

1.9 )x(60sinxd2 sin 60 = ( + d) ( x) sin 60 + x sin 60

on solv ing gy djus ij

x =3

2.4 mg = 60 .................(i)

mg ivg = 40 .................(ii)

mg

gvmg

=

3

2or;k

0= 3

where 0

= density of the block and = density of the liquid.

;gk 0

= fi.M dk ?kuRo rFkk = nzo dk ?kuRo

2.5 U = mghU = (

b

)

vgh

FLUID MECHANICS


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2.6 103 5

4+ 13.5 103

5

1= 1

or;k = 3.5 103 kg/m3

2.7 [36 v l]g = [48 v2]g

g9

3636 i

= g

4848

0i

Solving gy djus ij, 0

= 3.

2.8 In stable equilibrium the object comes to its original state if disturbed.

lkE;koLFkk esa ;fn fdlh oLrq dks fopfyr djds NksM+ fn;k tk;s] rks ;g okil viuh izkjfEHkd fLFkfr dks izkIr dj ysrh

gSA2.9 As, weight = Buoyant force

Hkkj = mRiykod cymg = [100 6 0.6 g] + (100 1 4)g

m = 760 gm.

3.6 from equation of continuity,

lkaR;rk lehdj.k ls(A 3) = (A 1.5) + (1.5 A V)

V = 1 m/s2

3.7 From continuity equation, velocity at cross-section (1) is more than that at cross-section (2).

lkaR;rk lehdj.k ls vuqizLFk dkV {ks=k (1) dk osx vuqizLFk dkV (2) ds osx ls vf/kd gksxkA

Hence vr% ; P1

< P2

3.8* x = 2 )hH(H

x1

= 22070

x2

= 2 3060

x3

= 2 5040

x4

= 2 4050 or x3 = x4 = maximum vf/kdre


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EXERCISE # 2

PART-I

3. Since not touching,

D;ksfd lEidZ ugh gSA

So blfy, R = Fb = l(vg) = 40g.R R = 80g 40g = 40gHence R will be 40g more than R

vr% R R ls 40g T;knk gSA

4. By action reaction, Fb

is internal

f;k izfrf;k ls] Fbvkarfjd cy gSA

So, balance weight = (m1+ m

2)g.

vr%] larqfyr Hkkj = (m1+ m

2)g.

and extra mass = 10 g

rFkk vfrfjDr nzO;eku = 10g

5. Fth

=dt

dp2=

dt

dmV2 = V2 [ L ] = VL2

7.

Velocity of efflux of waterckgj fudyus okys ty dk osx (v) =

2

hg2 = gh

force on ejected water = Rate of change of momentum of ejected water.

ckgj fudyus okys ty ij cy = ckgj fudyus okys ty ds laosx esa ifjorZu dh nj= (av) (v)= av2

Torque of these forces about central line dsfUnz; js[kk ds lkis{k bu cyksa dk cyk?kw.kZ= (av2) 2R . 2

= 4av2

R = 4 agh R

10. Pressure exerted by fluid at closed end B is

nzo ds }kjk can fljs B ij yxus okyk nkcP = g

force exerted by fluid at closed end B is

nzo ds }kjk can fljs B ij yxus okyk cyF = PA = g A0


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11. For the given situation, liquid of density 2r should be behind that of r.

nh xbZ fLFkfr ds fy,] 2?kuRo okyk nzo ?kuRo okys nzo ls fiNs gksxkA

From right limb nk;h Hkqtk ds fy, :P

A= P

atm+

gh

PB

= PA

+ a

2 = P

atm+

gh +

a

2

PC

= PB

+ (2)a

2

= P

atm+

gh +

2

3

a .... (1)

But from left limb : ysfdu ck;h Hkqtk ds fy, :P

C= P

atm+ (2

)

gh .... (2)

From (1) and (2) : lehdj.k (1) o (2) ls

Patm

+ gh +

2

3

a = P

atm+ 2

gh h =

g2

a3 Ans.

12. No sliding pure rolling

dksbZ fQlyu ugh gS vr% 'kq) ykSVuh xfr gksxhATherefore, acceleration of the tube = 2a (since COM of cylinders are moving at 'a')

vr%] ufydk dk Roj.k = 2a D;ksfd csyu dk nzO;eku dsUnz a Roj.k ls xfreku gSA

PA = Patm + (2a) L (From horizontal limb {kSfrt Hkqtk ls)

Also blh izdkj ; PA

= Patm

+ g

H (From vertical limb m/okZ/kj Hkqtk ls) a =

L2

gH

Ans.

15. As long as W, pressure at the bottom of the pan would be same everywhere, according to the Pascals law.

tSls gh Wik=k dh lrg ij ik'dy ds fu;ekuqlkj] nkc lHkh txg leku gksxkA

on solving bls gy djus ij sin 43 .

16. The velocity of fluid at the hole isNsn ij nzo dk osx : V2

=)A/a(1

gh222

Using continuity equation at the two cross-sections (1) and (2) :

nksuksa dkV {ks=k (1) o (2) ij lkR;rk lehdj.k ls

V1A = V

2a V

1=

A

aV

2

acceleration (of top surface) ijh lrg dk Roj.k = dhdV

V 11


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Thus the increment in pressure at each point is P =A

F(by Pascals law)

bl izdkj izR;sd fcUnq ij nkc es a of) gS P =A

F(ikLdy ds fu;e ls)

27*. Let completely submerged in water, thenekuk ty esa iw.kZr% Mwck gqvk gS] rc

Fb = 1000 > mg(920) So, not possible vr%] ;g laHko ugh gSA

Let complete in oil ekuk iw.kZr% rsy esa gSAFb = (0.6) (4) (1000 + (1) (6) (100) = 840

Fb < mg So, not possible vr% ;g laHko ugh gSA

So, let 'x' part in oil and remaining in waterekuk x Hkkx rsy ds vanj gS vkSj 'ks"k ty esa gSA920 = [(1) (10 x) + (0.6) (x)] 100

9.2 = 10 x + 0.6 x

0.4 x = 0.8

x = 2 cm.

28. PV = constant vpj

(Assumed isothermal process) ekuk ;g izf;k lerkih; izf;k gSA

EXERCISE # 3

1.2 Pressure varies with height pkbZ ds lkFk nkc ifjorZu P = ghand is horizontal with acceleration rFkk {kSfrt esa Roj.k ds lkFk P = aso on (A) gh part is zero while average force ofax isvr% (A) esa gh okyk Hkkx 'kwU; gksxk tcfd vkSlr cy ax gSA

][2

a0 2

= )(2

a 2

=

2

)( 3a =

2

ma

In (B) a part is zero while average force ofgx is(B) esa a okyk Hkkx 'kwU; gS tcfd vkSlr cy gx gSA

22

g0

=

2

g(3)

= 2

)( 3

(g) = 2

ma

Similarly for other part. blh izdkj vU; ds fy, gksxk

1.3 (A) On ABCD avg pressure =

2

gh0 1

(A) ABCD ij vkSlr nkc =

2

gh0 1

So vr% F = ]h[2

gh1

=

2

gh21

(B) No contact of2 and not any pressure on ABCD due to 2


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(B) 2dk dgh ij Hkh lEidZ ugh gS rFkk

2ds dkj.k ABCD ij dksbZ nkc ugh gSaA

(C) On CDEF due to 1, at every point pressure is

1gh so average is also

1gh

so F = (1gh) (h) =

1gh2

(C) 1ds dkj.k CDEF ds izR;sd fcUnq ij nkc

1gh gS vr% vkSlr Hkh

1gh gksxkA

vr% F = (1gh) (h) =

1gh2

(D) On CDEF force due to liquid of density 2is

2

gh22

(D) 2ds dkj.k CDEF ij cy

2

gh22 gSA

2.4 Using equation of continuity A1v

1= A

2v

2

(12 cm2)vA

= (6 cm2) (8.0 m/s)

vA

= 4.0 m/s

2.5 Applying Bernoulli's principle between point A and C that are at same horizontal level

A2A pV.2

1

= atm2C pV.21

pA

= (1.01 105 N/m2) +2

1 13,600 (82 42)

= 4.27 105 N/m2

2.6 By applying Bernoulli 's equation between point B and C and using equation of continuity

vB

= 8.57 m/s

2.7 Apparent weight vkHkklh Hkkj (Wapp.

) = W V g

Since D;ksfd, Wapp. (Ram)

> Wapp. (Shyam) W

(Ram)> W

(Shyam)

Therefore, from given passage shyam has more fat than Ram.

vr% fn;s x;s vuqPNsn esa ';ke dk olk vo;o jke ls vf/kd gksxkA

2.8 V1

> V2

Wapp. (1)

< Wapp. (2)

(Since Wapp.

= W V g)

Hence (B)

2.9 Salt waver

> Fresh waver

Wapp. (s)

< Wapp. (F)

Hence (A)

2.10 Let 'V' be the total volume of the person ekuk O;f dk dqy vk;ru V gS rcThen ;

4

V(0.4 103) +

3103

4V

4

3= 165

V =1100

165

Reading on spring balance under water is :

dikuh ds vanj fLizax rqyk dk ikB~;kad

Wapp

= [165 10]

1100

165[103] [10]

= 150 N


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2.11 Just after the string is cut :

Lizhax dks dkVus ds rqjar ckn %

a =165

150= 0.91 m/s2

3.1 By pascal law assertion is correct and since reason is one of the assumption considered in fluid and due

to this reason is correct explanation of assertion.

ikLdy fu;e ds vk/kkj ij dFku lgh gS rFkk dkj.k nzO;;kfU=kd esa ,d dYiuk gS vr% blfy, dkj.k dFku dhlgh O;k[;k djrk gSA

3.2 From archimedes principle statement-2 is correct explanation of statement-1.

vkfdZehMht ds fl)kUr ls] oDrO;-2, oDrO;-1 dk lgh Li"Vhdj.k gSA

3.4 As the oil is poured till it covers the object completely, pressure in water at all points keeps on increasing. As

a result upward force on object exerted by water increases and the object moves up for the given duration.

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1.

tSls gh rsy Mkyk tkrk gS o oLrq dks


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V

V= 3

R

R

or;kR

R=

3

1

V

V...........(1)

From equation dlehdj.k (1)

R

R=

3AK

g

5.2 The condition of floating is , rsjus ds fy, 'krZWeight = Upthrust Hkkj = iz.kksfnr cy (mRIykod cy)

V 1g = V

i

2g (

1= density of metal /kkrq dk ?kuRo ,

2= density of mercury ikjs dk ?kuRo )

V

Vi=

2

1

= fraction of volume of metal submerged in mercury ikjs es Mqcs gq, /kkrq ds vk;ru dk Hkkx ekuk

= x say gSNow, when the temperature is increased by T.vc tc rkieku T ls c


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6.6 In waterty esa a = g So, by symmetry in air and water in acceleration distance travelled = 19.6 m.

t = g/h22 = 8.9/6.1922 = 4 sec.

blfy,, ty vkSj ok;q esa Roj.k dh lekurk ds dkj.k r; dh xbZ nwjh = 19.6 m

7.2

l

GB

C 2L

2

l

Ycom

mFigure (1)

(M+m)g

Figure (2)

B

F

C

Let M = Mass of stick = R2L = Immersed length of the rod

G = COM of rod

B = Centre of buoyant force (F)

C = COM of rod + mass (m)

Ycom

= Distance of C from bottom of the rod

Mass m should be attached to the lower end because otherwise B will be below G and C wil l be above G

and the torque of the couple of two equal and opposite forces F and (M + m ) g will be counter clockwiseon displacing the rotational equilibrium. See the f igure 3 given alongside.

For vertical equilibrium

Mg + mg = F (upthrust )

(M+m)g

Figure (3)

B

F

C

m

or (R2 L g) + mg = ( R2 l) g

l =

2

2

R

mLR......(1)

Position of COM (of rod + m ) from bottom

Ycom = mM

2

LM

= m)LR(2

L)LR(

2

2

.... (2)

Centre of buoyancy (B) is at a height of2

lfrom the bottom.

We can see from f igure (2) that for rotational equilibrium of the rod, B should either lie above C or at the

same level of B.

Therefore2

l YCOM or

2

2

R2

mLR mLR

2

LLR

2

2

or m + LR2 LR2 or m LR2 )(

Minimum value of m is LR2 )( Ans.


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7.3 In elastic collision with the surface, direction of velocity is reversed but its magnitude remains the same. Therefore,

time of fall = time of rise.

lrg ls izR;kLFk VDdj esa osx dh fn'kk foifjr gks tkrh gS ijUrq bldk ifjek.k vifjofrZr jgrk gSA

vr% fxjus dk le; = mij tkus dk le;

or time of fall = t1/2;k fxjus dk le; = t

1/2

Hence, velocity of the ball just before it collides with liquid is

vr% nzo dh lrg ls Vdjkrs le; xsan dk osx

v = g2

t1....(1)

Retardation inside the liquid

nzo ds vUnj eanu

a =mass

weightupthrust =

nzO;ekuHkkjmRIykod cy

=Vd

VdggVdL =

d

dgdLg(V = volume of ballxsan dk vk;ru) ....(2)

Time taken to come to rest under this retardation will be

eanu ds dkj.k xsan ds fLFkjkoLFkk rd vkus esa yxk le;

)dd(2

dt

gd

dd2

gt

a2

gt

a

vt

L

1

L

11

Same will be the time to come back on the liquid surface.

Therefore,

nqckjk nzo lrg rd vkus esa yxk le; Hkh leku gksxk

vr%(a) t

2= time the ball takes to came back to the position from where it was released

xsan dks tgk ls NksMk x;k gS ml izkjfEHkd fLFkfr rd okil vkus esa fy;k x;k le;

= t1

+ 2t = t1

+dd

dt

L

1

=

dd

d1t

L1

or;k t2

=dd

dt

L

L1

(b) The motion of the ball is periodic but not simple harmonic because the acceleration of the ball is g in air

and gd

ddL

inside the liquid which is not proportional to the displacement, which is necessary and sufficient

condition for SHM.

xsan dh xfr vkofk gksxh ijUrq ljy vkoZk xfr ugh gksxh D;ksafd xsan dk gok esa Roj.k g gS rFkk nzo esa Roj.k gd

ddL

gS tks fd foLFkkiu ds lekuqikrh ugh gSA tks fd ljy vkoZk xfr ds fy, vko';d 'krZ gSA(c) When d

L= d, retardation or acceleration inside the liquid becomes zero (upthrust = weight). Therefore,

the ball will continue to move with constant velocity v = gt1/ 2 inside the liquid.

tc dL

= d, gS rc nzo ds vUnj eanu Roj.k'kwU; gS D;ksafd mRIykod cy] Hkkj ds cjkcj gS vr% xsan fu;r osxv = gt

1/ 2 ls nzo ds vUnj xfr djsxhA


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EXERCISE # 4

1. [Flower

Fupper

] byliquid

= Upthrust

F2

F1

= upthrust

F2

= F1

+ upthrust

F2

= gh (R2) + Vg

F2

F1

Upthrust

or F2

= g(V + R2h)

In this problem, we did not take the force due to air pressure on the cylinder. This is because force due to

air pressure is cancelled. At top and bottom of the cylinder the force due to air pressure is equal and

opposite.

[Fuhps

Fij

] nzo }kjk = mRIykou cy

F2

F1

= mRIykou cy

F2

= F1

+ mRIykou cy

F2

= gh (R2) + Vg

F2

F1

Upthrust

;k F2

= g(V + R2h)

bl iz'u esa csyu ij ok;q nkc ds dkj.k cy ugha fy;k x;k gSA D;ks fd ok;q nkc ds dkj.k cy ,d nwljs dks fujLr dj nsrs

gSA2. will decrease because the block moves up. h will decrease because the coin will displace the volume of

water (V1) equal to its own volume when it is in the water whereas when it is on the block it wil l displace

the volume of water (V2) whose weight is equal to weight of coin and since density of coin is greater than

the density of water V1

< V2.

?kVsxk D;ksfd CykWd ij tk;sxk h ?kVsxh D;ksfd flDdk Lo;a ds vk;ru ftruk ty (V1) foLFkkfir djsxk tc ;g ikuh esa

gksxkA fdUrq tc ;g CykWd ij gksxk rks vius Hkkj ds rqY; ty dk vk;ru (V2) foLFkkfir djsxkA pwafd flDds dk ?kuRo ty

ds ?kuRo ls vf/kd gS vr% V1< V

2.

3. (i )

Air

A

B

h

hA

hB

Liquid A is applying the hydrostatic force on cylinder from all the sides. So net force is zero.

(ii) In, equilibrium :

Weight of cylinder = Net upthrust on the cylinder

Let s be the area of crosssection of the cylinder, then

weight = (s) (h + hA

+ hB

) cylinder

g

and upthrust on the cylinder

= upthrust due to liquid A + upthrust due to liquid B

= shA

A

g+ sh

B

Bg


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Equating these two

s(h +hA

+ hB)

cylinderg = sh

A

Ag + sh

B

Bg

or (h +hA

+ hB)

cylinder= h

A

A+ h

B

B

Substituting hA

= 1.02 cm hB

= 0.8 cm, A

= 0.7 gm/cm3

B

= 1.02 gm/cm3 and cylinder

= 0.8 gm/cm3

in the above equation we get :

h = 0.25 cm Ans.

(iii) Net upward force = extra upthrust

= shB

g

Net acceleration a =cylinderofmass

force

or a =cylinderBA

B

)hhs(h

gsh

or a =

cylinderBA

Bg

)hhh(

h

Substituting the values of h, hA,h

B,

Band

cylinderwe get

a =6g (upwards) Ans.

Sol. (i)

Air

A

B

h

hA

hB

csyu ij lHkh vksj ls nzo A LFkSfrd nkc vkjksfir dj jgk gS vr% dqy cy 'kwU; gksxkA

(ii) lkE;koLFkk esa %

csyu dk Hkkj = csyu ij dqy ij dh fn'kk esa cyA

ekuk csyu dk dkV {ks=k s gS rks

Hkkj = (s) (h + hA

+ hB

) csyu

g

csyu ij ij dh fn'kk esa cy

= nzo A dss dkj.k cy + nzo B ds dkj.k cy= sh

A

Ag

+ sh

B

Bg

mi;qZDr nksuks dh rqY;rk ls

s(h +hA

+ hB)

csyug = sh

A

Ag + sh

B

Bg

;k (h +hA

+ hB)

csyu= h

A

A+ h

B

B

izfrLFkkfir djus ij hA

= 1.02 cm hB

= 0.8 cm, A

= 0.7 gm/cm3

B

= 1.02 gm/cm3 vksj csyu

= 0.8 gm/cm3

mi;qZDr lehdj.k esa izfrLFkkfir djus ij :

h = 0.25 cm Ans.

o p csyu dk eku izfrLFkkfir djus ij izkIr gksxk o =aij dh vksj


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(iii) ij dh fn'kk esa dqy cy = vfrfjDr mRIykou cy= sh

Bg

dqy Roj.k a =udk nzO;ekcsyu

cy

;k a =csyu

)hhs(h

gsh

BA

B;k a =

csyu

)hhh(gh

BA

B

h, hA, h

B,

Bo

csyudk eku izfrLFkkfir djus ij izkIr gksxkA

a =6

g(ij dh vksj) Ans.

4. From equation of continuity (Av = constant)

4

(8)2 (0.25) =

4

(2)2 (v)

Here, v is the velocity of water with which water comes out of the syringe (Horizontally).

Solving eq. (i), we getv = 4m/s

The path of water after leaving the syringe will be parabola. Substituting proper values in equation of trajectory.

y = x tan 22

2

cosu2

gx

we have, 1.25 = R tan0 0cos)4)(2(

)R)(10(22

2

(R = horizontal range)

Solving this equation, we get

R = 2m.

lkarR; dh lehdj.k ls (Av = fu;rkad)

4

(8)2 (0.25) =

4

(2)2 (v)

;gka v ty dk osx gS ftlls ty lhjhat ls ckgj fudyrk gS ({kSfrt) lehdj.k (i) dks gy djus ij ge ikrs gSv = 4m/s

lhjhat ls ckgj fudyus ij ty dk iFk ijoy; gksxkA iFk dh lehdj.k esa mi;qDr eku izfrLFkkfir djus ij

y = x tan 22

2

cosu2

gx

1.25 = R tan0

0cos)4)(2(

)R)(10(22

2

(R = {kSfrt ijkl)lehdj.k dks gy djus ij ge ikrs gS

R = 2m.

5. Apply continuity equation A1V

1= A

2V

2and Bernauli theorum

0p

+2

v2+ gh = constant at the top and at

the hole.

lkaR;rk lehdj.k ls A1V

1= A

2V

2vkSj cjuksyh izes; ls

0p

+2

v2+ gh = mijh lrg ij vkSj Nsn ij fu;r gS


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6. For circular motion of small element dx. we have

NksVs vYika'k dx dh okh; xfrdF = (dm)x2

(dp) A = (Adx). x2

or;k dP = 2 x.dx

2

1

P

P

L

0

2 dxxdP

P2 P1 =2

L22

gH =2

L22

H =g2

L22

7. A1

= r2 = area of base of cylinder in airA

2= 3r2 = area of base of cylinder in water

A3= 4r2 = cross-section area of cylinder

A1

= r2 = gok esa csyu ds vk/kkj dk {ks=kQy

A2

= 3r2 = ikuh esa csyu ds vk/kkj dk {ks=kQy

A3

= 4r2 = csyu ds vuqLFk dkV dk {ks=kQy

(P + h g)Aa 1 3r

r3

gHA3

(P )Aa 1{P + g (h + H)}Aa 1 2r

from the equilibrium of block (see diagram)

xqVds dh lkE;koLFkk ls fp=k nsf[k,( )

Equating the forces, we get cyksa dks cjkcj djus ij

(Pa+ gh

1)A

3+

3

gH AA

3= (P

a)A

1+ [P

a+ g(h

1+ H)]A

2

On solving gy djus ij

h1=

3

5H

8. (A)

PaA

3+

3

A

3Hg = P

aA

3+ gh

2

P Aa 3

3 gHA3

(P )Aa 3g h A2 2

from the equilibrium of block (see diagram)

h2

= 4H/9.

9. For h2

< 4h/9 cylinder does not move up because further bouyant force decreases while the weight of block

remains same.

h2 < 4h/9 ds fy, csyu ij dh vksj xfr ugha djsxk D;ksafd mRIykou cy ?kV tk;sxk vkSj xqVds dk Hkkj ogha jgsxkA


17/22


10. Since it is open from the top, the pressure will be P0

pwafd ;s ij ls [kqyk gS] nkc P0 gksxk A

11. Resultant force on the pistion is zero (Let pressure in air be P.)

From the equilibrium of the piston

Mg

(P P) R0 2

(P0 P) R2 = Mg

P = P0

2R

Mg

From the conservation of moles of air :

P1V

1= P

2V

2, it follows that

P0

. 2L = Px

x =P

L2.P0=

20

0

R

MgP

L2P

fiLVu ij ifj.kkeh cy 'kwU; gS ekuk ok;q esa nkc P gS)fiLVu dh lkE;koLFkk ls

Mg

(P P) R0 2

(P0

P) R2 = Mg

P = P0

2R

Mg

ok;q ds eksyksa ds laj{k.k ls :P

1V

1= P

2V

2

P0

. 2L = Px

x =P

L2.P0=

20

0

R

Mg

P

L2P

12. Pressure in air inside cylinder = Pressure at point A = P0+(L

0 H)g

PV = constant in the air inside the cylinder

_ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ _

P +(L H) g0 0

L H0

A

P0L

0= [P

0+ (L

0 H)g] (L

0 H)

P0

(L0 H) + g (L

0 H)2 P

0L

0= 0

csyu ds vUnj ok;q esa nkc = fcUnq A ij nkc A = P0+(L

0 H)g

PV = csyu ds vanj ok;q ds fy, fu;r _ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ __ _ _ _

P +(L H) g0 0

L H0

A

P0L

0= [P

0+ (L

0 H)g] (L

0 H)

P0

(L0 H) + g (L

0 H)2 P

0L

0= 0


18/22


13. As the stream falls down, its speed wil l increase and cross-section area wil l decrease.

Thus it will become narrow.

Similarly as the stream will go up, speed will decrease and cross-section area will increase.

Thus it will become broader.

Hence statement-1 is correct and statement-2 is correct explanation also.

tSls gh /kkjk uhps fxjrh gS bldh pky c


19/22


(r)

(A) force on Y due to X = 220 )mg(]g)mm[(

(B) As the system moves down, gravitational P.E. of X decreases

(C) As the system moves down, total mechanical energy of (X + Y) also decreases

(D) P 0

(s)

(A) force on Y due to X = Buoyancy force which is less than mg

(B) As the sphere moves down, that volume of water comes up, so gravitational P.E. of X increases.

(C) As there is no nonconservative force, so total mechanical energy of X + Y remains conserved.

(D) p 0

(t)

(A) As the sphere is moving with constant velocity

B + fv= Mg

so force on Y due to X is B + fv= mg

(B) As the sphere moves down, that volume of water comes up, so gravitational P.E. of X will increase

(C) Increase in mechanical energy

= wfr= ve

(D) p

= 0

(p)

Y ij X ds dkj.k yxus okyk dqy cy X = 22 )sinMg()cosMg( = Mg

(B) D;ksafd urry fLFkj gSA vr% Xdh xq:Roh; fLFkfrt tkZ fu;r gksxhA(C) D;ksafd xfrt tkZ fu;r gS vkSj Y dh fLFkfrt tkZ ?kVsxhA vr% (X + Y) dh ;kaf=kd tkZ ?kVsxhA(q)

(A) X ds dkj.k Y ij yxus okyk cy Mg ls T;knk gksxk] tksfd (Mg + frd"kZ.k cy) ds rqY; gksxkA

(B) D;ksafd fudk; ij dh vksj xfreku gS vr% X dh fLFkfrt tkZ c


20/22


(r)

(A) X ds dkj.k, Y ij yxus okyk cy X = 220 )Mg(]g)mM[(

(B) D;ksafd fudk;] uhps dh vksj xfreku gS vr% Xdh xq:Roh; fLFkfrt tkZ ?kVsxhA(C) D;ksafd fudk; uhps dh vksj xfreku gS vr% (X + Y) dh dqy ;kaf=kd tkZ Hkh ?kVsxhA(D)

P 0

(s)

(A) X ds dkj.k Y ij yxus okyk cy, mRIykou cy gksxk tksfd Mg ls de gksxkA(B) D;ksafd xksyk uhps dh vksj xfreku gS vr% ikuh dk vk;ru ij dh vksj tk;sxk] vr% Xdh xq:Roh; fLFkfrt tkZ c


21/22


300

)H500(105 + 2000 = 105

5 107 H 105 + 6 = 300

H = 206 mm

fall in height = 6 mm

P0

= ok;qe.Myh; nkcP + 200 103 1000 10 = P

0....(1)

P0(500 H) = P . (300 mm)

P = mm300mm)H500(P0

....(2)

lehdj.k (1) o (2)

300

)H500(P0 + 2000 = P

0

300

)H500(105 + 2000 = 105

5 107 H 105 + 6 = 300 H = 206 mm

apkbZ esa deh = 6 mm Ans. 6

19. dF

> dA

dB

> dF

& dAVg

+ d

BVg

= d

F(2V)g

dA

+ dB

= 2dF

20. Weight of water inside cylinder + weight of cylinder = buoyancy force

mWg + mg =

w.

2

vg

wv

f+ v

s=

w.2

v

vf=

2

v

c. v

s

now for any value ofc

vfis less than v/2 so.


22/22


PART - II

1. As pwafd P.E. = K.E.

mgh =2

1mv

2

h=20m

v

v = gh2

= 20102 [Here : g = 10 m/s2]

= 20 m/s

2. Since solid ball f loats in between the two liquids hence 1<

3<

2

pwafd xsan nksuks nzoksa ds chp rSjrh gS vr% 1<

3<

2

3. For equilibrium, weight should be balanced by buoyant force.

density of oil < density of water

and ball should be in between oil and water.?kuRo ds inkFkZ ls ,d xsan cuh gS tgk rsy < < ikuh vkSj rsyvkSj ikuh e'k% rsy ,oa ikuh ds ?kuRo n'kkZrs gSaA rsy ,oa

ikuh vfeJ.kh; gSA bl rsy vkSj ikuh ds feJ.k esa mi;qZ xsan ;fn lkE;koLFkk esa gS] rc fuEUkfyf[kr esa ls dkSulk fp=k bldh

lkE;koLFkk fLFkfr dks n'kkZrk gS \

(1) (2) (3) (4)

lkE;koLFkk ds fy, Hkkj] mRIykou cy ls larqfyr gksrk gSA

rsy ?kuRo < ty ?kuRo

o xsan dh fLFkfr rsy o ty ds e/; gksuh pkfg,A

Documents

11. Fluid Mechanics