10. Surface Tension

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    EXERCISE # 11.3 cohesive force > adhesive forcellatd cy > vklatd cy

    is obtuse. vf/kd dks.k gSA

    1.5 After the portion A is punctured the thread has 2 options as shown in the figures.

    or

    Clearly, due to surface tension , the soap film wants to minimize the surface area which is happening in option

    (ii).

    Hence the thread will become concave towards A.tcA Hkkx dks fiu ls rksM+k tkrk gS] rks /kkxs ds fp=kkuqlkj nks fLFkfr;k lEHko gSa

    ;k

    Li"Vr;k] i"Bruko ds dkj.k] lkcqu dh fQYe i"B {ks=kQy dks de djus dk iz;kl djsx h tks fd fLFkfr (ii) esa lEHko gSAvr% /kkxk Adh rjQ vory gks tk;sxkA

    1.6 In the satellite, geff

    becomes zero but the surface tension still prevails. Hence the water will experience only

    surface Tension force which will push it fully outward.

    df=ke mixzg esa geff

    'kwU; gk s tkrk gS ys fdu i"B ruko vHkh Hkh jgrk gS] vr% ty ij dsoy i"B ruko cy yxsxk tks fd mlsdsoy ij dh vksj [khapsxkA

    1.7

    The small portion of film is approximately a straight part. Balancing forces on it:

    F denotes tension. T denotes surface tension.

    T 2 (d) is the surface tension force because 2 layers are formed.

    So 2 F sin (d) = T [2 R (2 d)]we get ; (sin (d) d. for small d)

    SURFACE TENSION

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    so F = T 2 R.

    fQYe dk NksVk Hkkx yxHkx lh/kh js[kk gS rFkk bl ij cyksa ds lUrqyu ls

    F ruko dks iznf'kZr djrk gS T, i"B ruko dks iznf'kZr djrk gS

    T 2 (d) cuh gqbZ nks lrgksa ds dkj.k i"B ruko ds dkj.kAvr% 2 F sin (d) = T [2 R (2 d)]NksVs dks.k ds fy;s d, sin (d) d.vr% F = T 2 R.

    1.8

    The FBD of disc is shown in the figure. The net upward surface tension force

    = FS

    cos = (T 2 r) cos .so F

    Scos + W = mg = W

    disc

    ;gk fp=k esa pdrh dk FBD cuk;k x;k gSA ij dh vksj yxus okyk dqy i"B ruko cy= F

    Scos = (T 2 r) cos .

    vr% FS

    cos + W = mg = Wdisc

    1.9 We know that surface energy

    US

    = T Area.

    Here. as 2 films are formed because of ring. so

    US

    = T 2 (A)

    = 5m

    N 2 0.02 m2. = 0.2 J

    ge tkurs gS fd i"B tk ZU

    S= T {ks=kQy

    ;gk, D;ksafd oy; ds dkj.k 2 fQYe curh gS vr%U

    S= T 2 (A)

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    = 5m

    N 2 0.02 m2. = 0.2 J

    1.10 In the shown diagram.

    PC

    = PB

    P0

    1r

    T2+ gh = P

    0

    2r

    T2

    Here, we may not know in advance which tube will rise above the other, but lets say the liquid level is higher in

    thinner tube.

    so 2T

    12 r

    1

    r

    1= gh. T = )rr(2

    rrgh

    12

    21

    as r2

    > r1

    ; so we assumed correctly

    fn;s x;s fp=k ls

    PC

    = PB

    P0

    1r

    T2+ gh = P

    0

    2r

    T2

    ;gk gesa igys ls ugh irk gS fd dkSulh uyh esa nzo Lrj, nwljs dh rqyuk esa ij tk;sxk ysfdu ekuk iryh V~;wc uyh esanzo dk ry T;knk gSA

    blfy, 2T

    12 r

    1

    r

    1= gh. T = )rr(2

    rrgh

    12

    21

    D;ksafd r2

    > r1

    ; vr% geus lgh ekuk gSA

    1.11 Since the contact angle in both cases remains the same.

    FS

    cos = Mg T 2 R cos = Mg .......(i)after doubling the radius

    T 2 (2R) cos = Mg .......(i i)= M = 2M.

    nksuksa fLFkfr;ks a esa lEidZ dks.k leku gS

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    FS

    cos = Mg T 2 R cos = Mg .......(i)f=kT;kvks dks nqxuk djus ijT 2 (2R) cos = Mg .......(i i)= M = 2M.

    1.12 Water will rise to a height more than h when downward force (mgeff

    ) becomes lesser than mg.so in a lif t accelerating downwards, g

    effis (g a

    0). Hence capillary rise is more.

    On the poles geff

    is even more than g. Hence the capillary will even drop.

    tc uhps dh rjQ yxus okyk cy (mgeff

    ),mg ls de gksrk gS rks ty esa mWpkbZ ls T;knk WpkbZ rd p

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    2h height even if is very high. So h will be 2h if > h & will be h + only ifis lesser than h.

    tc ds'kuyh es ty h pkabZ rd ij mBrh gS vFkkZr i"B ruko nzo lrg tks 'h' pkbZ rd lgkjk nsrk gSAvc ;fn ds'kuyh ls ty ckgj fudyrk gS rks nzo xw:Ro ds dkj.k fuEu fcUnq ls ckgj fudyrk gSA

    Fs

    Fs

    ysfdu vc i"B ruko F ij dh vksj 2F cu tk;sxkA vr% 2F vf/kdre2h pkbZ rd lgkjk nsxk ;|fi cgqr cMk gSAvr% h 2h gksxk ;fn > h vkSj, ;fn , h ls de gS rks h' (h + ) ds cjkcj gksxkA

    1.16

    FsFs

    By balancing forces cy lUrqyu ls T (2 ) (cos) = d x h g

    we get ge izkIr djsxsa h = xdgcosT2

    .

    1.17* Force of cohesion keeps the molecules of a material bounded together and does not let them stick to the solid

    as force of adhesion is lesser.

    llatd cy fdlh iznkFkZ ds v.kqvks dks ,d nwljs ls ck/ks j[krs gS rFkk D;ksfd vkaltd cy de gS vr% bUgs Bksl ls tqMus ughnsrk gSA

    Hence

    vr% (A) (B) (C).

    1.18* Nature of liquid and material tube determine whether force of cohesion is more or force of adhesion is more.

    The inner radius also determines the rise of capillary as

    h =gr

    cosT2

    depends on radius r..

    If the length is not sufficient rise will be depends length also.

    nzo dk izdkj rFkk uyh ds iznkFkZ }kjk ;g fu/kkZfjr fd;k tkrk gS fd llatd cy T;knk gS ;k vkltd cy ds'kuyh dh vkUrfjdgS f=kT;k Hkh ds'kuyh esa nzo ds p

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    2.3

    Let (A) and (B) coalesce to form (C). ekuk (A) o (B) feydj (C) cukrs gSABy mole conservation : vr% eksy lj{k.k ls

    Pa

    . a3 + Pb

    . b3 = Pc

    . c3 ....... (i)

    Also vkSj Pa

    = P0

    +a

    4........(ii)

    Pb

    = P0

    +b

    4.......(iii)

    Pc

    = P0+

    c

    4.......(iv)

    Putting there values : izfrLFkkiu djus ij

    30

    30

    30 c

    c

    4Pb

    b

    4Pa

    a

    4P

    0cba4cbaP 2223330

    also rFkk c3 (b3 + a3) =4v3

    and vkSj c2 (a2 + b2) =4

    s.

    Putting there values : ekuks dk izfrLFkkiu djus ij

    P0

    4

    v3+ 4T

    4

    S= 0

    3P0V + 4ST = 0 Ans. (A).

    2.4 When charge is given to a soap bubble (whether positive or negative), these charges experience repulsive

    forces due to the other charges. Hence they tend to move out. Hence the size of bubble increases.

    tc lkcqu ds cqycqys dks vkos'k fn;k tkrk gS (pkgs /kukRed _.kkRed ) rks ;g vkos'k vU; vkos'kks ds dkj.k izfrd"kZ.kcy vuqHko djsxkA vr% ;g ckgj tkus dh izrfk j[ksxkA vr% cqycqys dk vkdkj c

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    get izkIr gksxk r = R/2.

    Now pressure difference in A =R

    4vc A esa nkckUrj =

    R

    4

    and that in B = 2/R

    4

    = 2 pressure difference in A.

    vkSj B esa nkckUrj =2/R

    4= 2 A esa nkckUrj

    2.7

    Pinside bubble PA =

    r

    T2

    and PA = Patm + gh.

    Pinside bubble = P + gh + rT2

    gy.

    PvkfUrjd cqycqys

    PA = rT2

    vkSj PA = Patm + gh.

    PvkUrfjd cqycqyk

    = P + gh +r

    T2

    2.8 n 3

    4r3 =

    3

    4R3 ......(i) { volumes are equal vk;ru leku gS}.

    and vkSj A = [4R2 n.4r2]where tgka W = (A) T.

    = 4[n2/3r2 n.r2] T = 4r2T. n2/3 [n1/31].Now vc R2 = n2/3 . r2 ; so blfy, W = 4R2T[n1/3 1].

    2.9

    PA = P0 + r

    4; PB = P0 + R

    4{P0 = atmospheric pressure}.

    Clearly PA > PB ; so air will f low from A to B.As r decreases; pressure wil l become more and hence more flow of air f rom A to B.

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    Ultimately bubble A collapses and B becomes bigger in size.

    PA = P0 + r

    4; PB = P0 + r

    4{P0 = ok;qe.Myh; nkc}.

    Li"Vrk PA > PB ; vr% ok;q A ls B dh vksj izokfgr gksxhAD;ksfd rde gksrh gS vr% nkc T;knk gksxh vksjA ls B dh vksj ok;q T;knk izokfgr gksxhAvr% cqycqyk A VwV tk;sxk vkSj B dk vkdkj c< tk;sxkA

    2.10

    R = 4 cm.

    r = 3 cm.

    Pr=

    r

    4; P

    R=

    R

    4{ outside is vacuum ckgj fuokZr gS}

    The two bubbles are coalescing; so conserving the no. the moles.

    tc nksuks cqycqyks feyk;k tkrk gS rc eksy lja{k.k ls

    T

    r3

    4.P 3r

    +

    T

    R3

    4.P 3R

    =

    T

    )'r(3

    4P 3final

    Putting eku j[kus ij Pfinal

    ='r

    4we get ge izkIr djsxs

    r = 22 Rr

    = 22 43 = 5 cm.

    2.11 Clearly the surface tension force on

    Hemisphere = FS

    = (2T). (2r) {2 layers are formed}.

    FS

    = 2 500m

    N

    2 3.14 5m.

    30,000 N 3000 kg.wt.

    Li"Vr;k v}Zxksys ij i"B ruko cy

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    v}Zxk sys ij i"B ruko cy = FS

    = (2T). (2r) {2 lrg cusxh}.

    FS

    = 2 500m

    N

    2 3.14 5m.

    30,000 N 3000 kg.wt.

    2.12P1

    Before r P2

    Afterr/2

    Lets say, initially, the pressure due to air inside the bubble is Pair

    .

    Pair

    P1

    =r

    T4..........(i)

    Finally, the radius becomes half ; so volume becomes8

    1th and hence pressure becomes 8P

    air.

    So, 8Pair

    P2

    =2/r

    T4.........(ii)

    Solving (i) and (ii)

    get P2

    = 8P1

    +r

    r24.

    P1rigys

    P2r/2

    ckn es a

    ekuk izkjEHk esa ok;q ds dkj.k cqycqys ds vUnj nkc Pair

    gSA

    Pair

    P1

    =r

    T4..........(i)

    vUr esa f=kT;k vk/kh gks tk;sxh vr% vk;ru8

    1gks tk;sxk vr% nkc 8P

    airgks tk;sxkA

    vr% 8Pok;q

    P2

    =2/r

    T4.........(ii)

    (i) vkSj (ii) dsk gy djus ij

    izkIr gksxkA P2 = 8P1 + rr24 .

    2.13 When the excess pressure at the hole becomes equal to the pressure of water height ;then only water will start

    coming out of the holes : [atm pressure on both sides is same].

    tc Nsn ij vkf/kD; nkc] ty dh WpkbZ ds nkc ds cjkcj gks tk;sxk rc dsoy ty Nsn ls ckgj vkuk izkjEHk djsxkA[nksuks rjQ ok;qe.Myh; nkc leku gksxk]

    hg =r

    2 h = rg

    2

    =1010

    2

    1.0

    m

    kg1000

    m

    N10702

    33

    3

    = 0.28 m.

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    2.14 Look at a very small element at the junction of 3 bubbles.

    rhuks cqycqyks ds feyku fcUnq ij ,d cgqr NksVs vo;o dks ns[krs gSA

    All 3 forces of same magnitude ( surface tension is same) are acting along the tangential directions on the

    small element.

    leku ifjek.k ds rhuks cy ( i"Bruko leku gS) NksVs vko;o ij f+=kHkqth; fn'kk esa dk;Z djsxsA

    Now by LAMIEs theorem vc ykeh izes; ls 1=

    2=

    3=

    3

    360= 120

    2.15 Energy released = (A) { = surface tension}mRlftZr tkZ = (A) { = i"B ruko }Let us say n no. of small drops coalesced.

    ekuk NksVh cwns tks feyrh gS mudh la[;k n gSA

    n. 3a3

    4 =

    3b3

    4

    b = a.n1/3 n =3

    a

    b

    A = 4b2 n.4a2 {this is ve, hence energy is released} {;g _.kkRed gS vr% tkZ mRlftZr gksxh}= 4a2 (n2/3 n)

    U = 4a2T (n n2/3). = 4a2T

    23

    a

    b

    a

    b

    This U converts to K.E. ;g fLFkfrt tkZ xfrt tkZ esa ifjorhZr gks tk;sxhA

    Hence vr% 3b3

    4.

    2

    1 V2 = 4a2T 2

    2

    a

    b

    a

    ab.

    V =

    b

    1

    a

    1T6

    2.16* When ever two drops coalesce to make a bigger drop. surface area is reduced, hence energy is released.

    tc dHkh Hkh nks cwns feydj ,d cMh cwan cukrh gS i"B {ks=kQy ?kV tkrk gS rFkk tkZ mRlftZr gksrh gSA

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    EXERCISE # 2

    7. p 1r

    T2= Pa ...(1)

    p 2r

    T2+ hg = Pa ....(2)

    PaPa

    P2T

    r1

    P

    .

    P

    p 1r

    T2= p

    2r

    T2+ hg

    hg =2r

    T2

    1r

    T2

    h =8.910

    107523

    3

    33 101

    2

    105.0

    2=

    8.910

    3003

    = 3.1 102 m

    14. PV = P1V1 + P2V2

    r

    T4

    3

    4r3 =

    31

    1

    r3

    4

    r

    T4 +

    32

    2

    r3

    4

    r

    T4

    r2 = 2221 rr

    r = 5 cm

    EXERCISE # 3

    1. r1

    = 1.44 103 m. r2

    = 0.72 103 m.

    Equating pressures at points (B) & (C)

    fcUnq (B) & (C) ij nkc cjkcj j[kus ij

    PA

    2r

    2+ (0.2) g = P

    C. and rFkk P

    B

    1r

    2= P

    C.

    so blfy, PB

    PA

    = 2

    21 r

    1

    r

    1+ 0.2 g

    = 2 72 103m

    N

    72.0

    10

    44.1

    10 33

    + (0.2) 103 938

    =72.044.1

    )72.0(144

    + 1960

    = 100 + 1960 = 1860 N/m2.

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    2.

    rh = r1

    21 rrL rh = 0.5 10

    3 1.0

    )25.05.0( 103 8 102

    = 0.3 103m ;hr

    T2 0= gh T

    0

    =2

    103.01088.91014

    1 324

    T0 = 0.084 N/m For Tube B uyh B ds fy,

    For Temp 0Cr

    T2 0= gh1 r =

    1

    0

    gh

    T2

    = 24 1068.91014

    1

    084.012

    0C rkieku ds fy,r

    T2 0= gh1 r =

    1

    0

    gh

    T2

    = 24 1068.91014

    1

    084.012

    r = 0.40 103 m For temp 50Cr

    T2 50= gh

    r = 0.40 103 m 50C rkieku ds fy,r

    T2 50= gh

    T50 = 2

    ghr=

    2

    104.0105.58.91014

    1 324

    T50 = 0.077 N/m2 =

    050

    TT 050

    =

    50

    084.0077.0 = 1.4 104 N/mC

    Ans. 1.4 104 N/(m C)

    3.

    We know that force ge tkurs gS fd cy =dt

    V)dm(

    where tgkadt

    dm= rate of mass transferred. LFkkukUrfjr nzO;eku dh nj

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    Now vcdt

    dmis also eqval to cjkcj gS

    dt

    dm=

    dt

    )Vdt()b( 2 = b2 V.

    so force blfy, cy = V2b2

    = pressure exerted by air on walls ok;q }kjk fnokjksa ij vkjksfir nkc = 222

    b

    bV

    = V2.

    when the thrust of this pressure becomes equal to the excess pressure

    tc ;g vkf/kD; nkc ds cjkcj gks tkrk gSA

    V2 =r

    T4 r

    final= 2V

    T4

    cqycqyk ufydk ls vyx gks tk;sxk] ;fn gok ds dkj.k B ij iz.kksn cy] nkc vkf/kD; ds dkj.k cy ds cjkcj gksxkA

    pAv2 =

    r

    T4AA

    (A = B ij cqycqys dk {ks=kQy] tgk gok Vdjkrh gS)

    r =

    2pv

    T4

    Alternate

    When force due to surface tension on bubbles is equal to the Force due to blowing air bubble leave contact

    with ring (separate from ring)

    tc i"B ruko ds dkj.k cqycqys ij yxus okyk cy dk eku, cgrh gqbZ gok ds dkj.k cqycqys ij yxus okyk cy (rccqycqyk oy; ls vyx gks tk;sxkA)

    F = 2 (2bT) sin (sin =R

    b)

    F = 4Tb

    R

    b= b2 v2

    R = 2v

    T4

    4.

    balancing forces on the wire in vertical direction :

    rkj ij /oZ fn'kk esa cyksa dks larqfyr djus ij

    2 (T) sin = ()g. = T =

    sin2

    g=

    y2

    )y(g 22

    so bl fy, T = y2g

    22 y y2g

    (for small y) ( y NksVk gSA)

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    5. Pushing force ncko cy =

    (Area) ({kS=kQy)

    = )Rh2(2

    )hgp()p( 00

    = 2p0

    Rh + g h2 RPulling force [khapko cy = (T) (2R)

    Net force ifj.kkeh cy = TR2RghRhp2 20

    6. Pressure inside tube = P = P0

    +r

    T4

    P2

    < P1

    (since r2

    > r1)

    Hence pressure on side 1 will be greater

    than side 2. So air from end 1

    flows towards end 2.

    Ans. (B)

    V~;wc ds vUnj nkc = P = P0

    +r

    T4

    P2

    < P1

    (since r2

    > r1)

    vr% 1 rjQ dh Hkqtk esa nkc Hkqtk 2ls vf/kd gksxk vr% mkj (B) gSAfljs 1 ls fljs 2 dh rjQ gok dk cgko gksxkAAns. (B)

    8. PA

    = P0

    +Ar

    T4 P

    A= 8 +

    02.0

    04.04

    PA

    = 16 N/m2

    PB

    = P0

    +Br

    T4= 8 +

    04.0

    04.04

    PB

    = 12 N/m2

    for bubble A, PV = nRT

    (16)3

    4 (0.02)3 = n

    ART ....(1)

    for bubble B

    (12)

    3)04.0(

    34 = n

    BRT ....(2)

    dividing eqn (i) and (2)

    6

    1

    n

    n

    B

    A ; 6nn

    A

    B Ans. 6

    PA

    = P0

    +Ar

    T4 P

    A= 8 +

    02.0

    04.04

    PA

    = 16 N/m2

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    PB

    = P0

    +Br

    T4= 8 +

    04.0

    04.04

    PB

    = 12 N/m2

    cqycqys A ds fy,, PV = nRT

    (16)3

    4 (0.02)3 = n

    ART ....(1)

    cqycqys B ds fy,

    (12)

    3)04.0(

    3

    4= n

    BRT ....(2)

    lehdj.k (1) esa (2) dk Hkkx nsus ij

    6

    1

    n

    n

    B

    A

    6nn

    A

    B Ans. 6

    9.

    R

    F

    r

    R

    Due to surface tension, vertical force on drop = Fv = T2r sin = T2rRr

    = R

    r2T 2

    10. Equating forces on the drop :

    R

    r2T 2= gR

    3

    4 3 (Assume drop as a complete sphere)

    R =

    4/12

    g2

    Tr3

    =

    4/1

    3

    8

    10102

    102511.03

    = 14.25 104 m = 1.425 103 m

    11. Surface energy of the drop

    U = TA

    = 0.11 4 (1.4 103)2

    = 2.7 106 J

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    Rise Academy, 607/608-A, Talwandi , Kota (Raj.) -324005. 211

    12.

    2ra =

    2

    2

    2Kq

    a+

    2

    2

    2Kq

    2a

    1

    2

    a3 =2Kq

    r

    11

    2 2

    a =

    1/3

    21

    Kq 12 2

    r

    N = 3

    PART - II1. The excess pressure inside the soap bubble in inversely proportional to radius of soap bubble i.e. P 1/r, r being

    the radius of bubble. It follows that pressure inside a smaller bubble is greater than that inside a bigger bubble.

    Thus, if these two bubbles are connected by a tube, air will f low from smaller bubble to bigger bubble and the

    bigger bubble grows at the expense of the smaller one.

    2. Water fi lls the tube entirely in gravity less condition.

    4. W = TA =

    = T2 )rr(4 2122

    = 0.4 mJ

    5.