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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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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.