irodov 106-235.pdf

Compact ISC Physics (XII)106

Q: 1.247: Method 1: Torque equation on whole system about centre

of pulley

2mRRmRmRgkmgm

22

12

212

)mm(R)kmm(g

2m

12

12

Acceleration of mass m1 :

2mmm

)kmm(gR12

121

displacement in time t : 21ta2

1S

Workdone of friction = 21t

2mmm

)kmm(gkmgskm 2

12

122

11

)mm(2mt)kmm(gkmW

21

212

21

Method 2: m2g - T1 = m2a1 -----(i) T1-km1g = m1a2 -----(ii) 2mRRTT

212 ----(i) a2 =

R -----(iv) from (i) and (ii) and (iv) :

2mmm

)kmm(g

12

121

21ta2

1S

)mm(2mt)kmm(gkm

s)gkm(W21

212

21

1

Ans.

Q: 1.248: F.B.D. of cylinder : Force equation : KN1 + N2 = mg ------(i) KN2 = N1 -----(ii)

Torque equation about centre of cylinder : 2

mRRKNRKN2

21 2mR]NN[KR

221

-------(iii) from (i) and (ii) : 2122 k1kmgNand

k1mgN

put in (iii) : )K1(R

)K1(kg22

. We know

: 2W02WW 202

02 )K1(kg4

)K1(Rw 220

)K1(gk8

)K1(Rw2

turnof.no22

0

T1

T1

m g2

km g2

a1m1 T2 T2

disc

a1

W0

R

CylinderN2

N1W

KN1

KN2

107

Q: 1.249 : Calculation of torque : Where df = friction force on differential region = This is scalar

sum because are going to calculate torque. df = k [dN] gxdx2Rmk 2

2RdxmgxK2

The differential torques : dxR

kmgx2xdfd2

2 kmgR3

2

R

dxkmgx2dR

02

2

Rkg

34kmgR

32

2mRI

2

We know W = W0 + t kg4WR3tt

Rkg

34W0 Ans

Q: 1.250: 2/1w 2/1cw 2/1kw . Time calculation : 1

0

t

0

0

w2/1 kdtw

dw

Cw2t

2/10

1

Angle calculation : dkwwdw 2/1 0

0

0

w

2/1 kddww0

C3

W2 23

0 . Man angular velocity

=

CW2C3

W2t 21

23

0

0

1

3w0 Ans

Q:1.251: Moment of inertial of chord :

22 xRMR)x(mI

ll

l

I = mR2. Toque about 0 : = I

2

2mR

2MRRmgx

l

)m2M(Rmgx2

l Ans.

Q:1.252: (a) Torque equation about ICOR : 2mR57R)Sinmg(

R7sing5

Torque about com: 2mR52Rfr

R7gSin5mR

52fr sinmg

3510fr ----- (i). At slipping : fr = K mg

W

Rx

df

dx

R0

x

lmgx

M

R

x

M

l-x

mg mgS

in

m,R

fr

ICOR

acm


Cos ------- (ii) from (i) and (ii): sinmg3510kmgCos tan35

10K tan72K

(b) 2ICOR WI21KE 22 tMR5

721KE

2

22

2

2222 t

14Sinmg5

R7x7Sing25tMR

107KE

Q:1.253: (a) Torque equation of ICOR which is attached with point of

string. 2mR23mgR R3

g2 3

gcm 2Ra . Force equation

on cylinder : mg-2T = m2g/3 T = mg/6

(b) ]ta[mgv.FP cm

t

3g2mg tmg3

2P 2 Ans.

Q:1.254: F.B.D. of cylinder in frame of lift : Torque about ICOR = 20 mR23R)wg(m

R3

)wg(2 0 )Wg(32RW 0

1 . In terms of vector : )wg(32w 0

1 . Torque about

com : 02

wgR32

2mRIFR

3)wg(mF 0 . In vector form :

3)wg(mF 0

Ans.

T Tcylinder

R

mg

Q:1.255: Torque about ICOR point P : (mgsin) r = (I+mr2)

2mrIsinmgr

acm = R = r

2

2cm

mrIsinmgra

Ans.

w1

m, R

cylinder

F

mg

w0

ICOR

F

mg + mw0

R

mgCos

acm

P

mgsin

T

109

Q:1.256: At time of slipping friction will act at maximum value KN then.Force equation in y dirn : KN = macm K(mg+F) = macm -------(i). Torque

equation about ICOR :

2mR

23FR

mR3F2

)i(inputm3F2Racm 3

F2ma)Fmg(k cm

k

32Fkmg k32

kmg3

k32kmgF

k32kmg3

32acm

k32kg2acm

Ans.Ans.

Q:1.257: (a) Force equation along x axis : F Cos fr = m acm -----(i) Torque equation about com:

RamRIaRfrFr cm2 cmamRRfrFr

------(ii) from (i) and (ii) : )1(mR)rRCos(Facm

(b) W = (F Cos) displacement 2cmta21)FCos( )1(m2

)R/rCos(tFW222

v

Q:1.258: Torque equation on sphere (A): 12

1 2mRRT2

2mRT2 11

-

----(i). Torque equation on sphere (B) : 22

1 2mRRT2

2mRT2 21

--

---(ii). From (i) and (ii): 1 = 2 . Force equation on (B): mg-2T1 = maB ------(ii)

Constraint : Acceleration of point P = R1 = aB-R2 . Then aB= R1 + Ra2 =2Ra1 . Put in (iii): mg-2T1 = m2Ra1 = 2mR1 -----(ii) from (i) : mR1 = 4T1 put in(iv) mg - 2T1 = 8T1 T1 = mg/10Q:1.259: Troque about point P for system :

)MRI()R3(m)RR2(mgMgR 22

2222 MRImR9)M3M(gR

MRImR9MgRmgR3

. Hence acceleration of particle :

aA = 3R 222

AMRImR9

)m3M(gR3a

2

A

RIm9M

)m3M(g3a

Ans.

acm

KN

cylinderm

mg F

y

x

r

m

F

fr

acm

I = mR 2

(A)

(B)

T1

T1

2

2

P

Q

m R1

m R1

1

2

mg

T2

mg

P

MR

m

mg


Q:1.260: (a) Cylinder Force equation on system : F = (m1 + m2) acm 21cm mm

Fa

. Torque

equation about frame attached with centre of sphere. 2RmFR

21 Rm

F2

1 . Acculeration of

point K : 121

cmk mF2

mmFRaa

)mm(m)m2m3(Fa

211

21k

(b) WF = Work done by F = Grain in K.E.

2k ta2

1F.E.K )mm(mt)m2m3(F

21.E.K

211

2212

Q:1.261: Acceleration of plank = acm - R. Then force equation on system : (Frm ground frame).F = m2 acm + m1 (acm - R) ----(i) Torque equation with frame attached with centre of sphere :

21

21cm1 Rm5

2RmFRRam ------(ii) from (i) and (ii)

21

cmm

72m

F72a Ans.

And acceleration of blank : w1 = acm - R -----(iii) put acm in (i) and calculate and put in ----(iii) :

211

m72m

Fw

Ans.

acm

m1

Fkm2B

A m1

Fm2

2macm

acm

F

m2

m1

R

x

ysphere m2

m1

R

Fm1acm

m2acm

Q:1.262: (a) Torque about bottom point is zero. Hence angular momenton will be consermed about

bottom point then. cmcm VRmLL

OwmR21L 0

2initial

-------(i)

)RW(MRmR21L 2final

wmR2

3 2 -------(ii)

from (i) and (ii) : wmR23wmR

21 2

02 3

w0w

m 1R

w

Vcm

mg

Nmcylinder

Rw0

kmgkSlipping

111

Torque equation about com :

2mRkmgR

2 kR

g2 . We know : w = w0 - t

kgRw

31tt

kRg2w

3w 0

00 (b)

20

22ffr )RW(m2

1wmR23

21kikW

220fr Rmw6

1W Ans.

Q:1.263: Suppose at angle ball have the sphere then energy equation : Loss of P.E. = Gain of K.E.

2cm

2 wI21cmmV

21)cos1()rR(mg 2cm

2 wI21cmmV

21)cos1()rR(mg

)cos1()rR(g7

10cmV2 ------(i) force equation in direction of N

: )rR(cmmVNCosmg

2

for leave contact N = 0 then.

rRcmmV0Cosmg

2

)rR(g

cmVCos2

. Put cos in ( i) :

)rR(g7

10Vcm

Q:1.264: Suppose at time t solid cylinder make angle with verticalenergy eqaation :

)RCosR(mgwmR23

21

RVmR

23

21 22

202

CosmgRmgRwmR43mV

43 222

0 -----(i) Force equation in dirn of normal reaction : mg Cos

- N = mRw2 mg Cos = N + mRw2 put mgCos value in equation (i) : mgRwmR47mV

43N 2220 -

-----(i). For stopping Bouncing N: will be minimum when W will be maximum and this condition willbe come when = . Hence we have to avoid bouncing at = then there will be no bouncing

before = . Now for minimum V0 put N = 0 and then gR74V

73WR 2o

22 . Put = and R2w2

value in (i) : CosgR37gR

74V 20 )4Cos7(3

gRV0 Ans.

m2

R

Vcmmg

m1rN

mg

wNR


Q:1.265: Since angular velocity(w) iscontinuously decreasing then there is chanceof minimum value of normal reaction at groundwhen particle is at maximum height or top ofhoop. Then force equation in y direction onsystem from reference attached with com ofhoop. 2 mg - N = mRW2 [Here w=final angualr velcoity]. Since N = 0 2mg = mRW2 w2 = 2g/R ----(i). Now energy equation : (from ICOR):

2222

02 w)R2(mmR221)R2(mg

RVmR2

21

Rg2)mR6(

21mgR2

2mV2 2

20

gR8V0

Q:1.266: Velocity of top point : VP = 2v. Hence kinetic energy of

crawler : 2)v2(

2m

21k

k = mv2 Ans.

Q:1.267: We know kinetic energy from fixed axis rotation is

given by: 2zzz2

yyy2

xx WI21WI

21WxI

21K . If

reference frame is not traslatory. And if reference frame istranslatory then we know to use kinetic energy of COM.

0WandRvWand

rvW zyx . Then 0WI2

1WI21K 2yyy

2xzz

222

22

RvMRmr

52

21

rvmr

52

21

2

22

Rr

721mv

107K Ans.

Q:1.268: (a) From this reference frame centrifugal force on particle of mass

(dm) is : 2icf wr)dm(dF 2

icf wr)dm(F i2 r)dm(w

mr)dm(mwF i2cf where

mr)dm(R ic

Position of CoM

c2

cf RmwF (b) Corolis force: ]wv[dm2dFcor

wv)dm(2wv)dm(2Fcor

= wmv)dm(m2

where

mV)dm(Vc

CoMofVelocityVc

cm

N

wmg

mg x

y

R

m

V0

w

m

v

m/2

m/2

x

y

0

z

v

R

m

y

0

z

xx

ir

Pdm

V

113

Q:1.269: Method : 1 dc = (dF) x Cos = w2x2sincos dm =

w2sincosx2 l

m dx

2

2

dxxcossinl

mw 22c

l

l

242sinmw2

c

l

2sinmw241 22

c l

Method : 2 sinw12

mL2l

dtdsinw

12m

dtLd 2

l

cosw)sinw(12

m 2l 2sinw

24m 22l

Ans.

Q:1.270: Torque of centrifuglal force

about 0: l

0

2 cosxwsinx(dm)d

cossin3

mw22l ------(i) Torque of

mg about 0 : sin2mgmg

l -----(ii)

from (i) and (ii) : for equilibniom: mg Sinmgcossin3mw

2

22ll

l2w2g3cos

Q:1.271: where k = coefficient of friction. x0 = Perpendicular distance b/w line of mg and N. HereN = mg. Since body is in rotational equilibrium then Thorque about COM will be zero. Kmg a/2 =

NX0 = mgx 2kax0 . Suppose at time t cube is displaced by x distance then Torque about origin

is: )2/ax(mg)2/axx(N 00 2a

2a0 mgmgx)xx(mg 2a2a mgmg2

kamg

2

mgka0 . Because of this torque angular

velocity of centre will be decreasingcontinuously.

x C

w

w (dm) xsin2

wwcoswsi

n

x

y

(dm) xsin w 2(centrifugal force)

x

0

w

dxmg

l/2

y

0

z

xx

x

a

a kmg

mg

a/2x0

N


Q:1.272: Method : 1 Suppose velocity of sleeve is Vr when its position fromaxis is x and angular velocity of rod is w then Centrifugal force: mxw2 = max

dxdVvmwmx rr2

l

0

2V

0rr xdxwdVV

r

l

0

22

r xdxw2

V -----(i).

Calculation of w : Using angular momentum conservation :

w3

MmxW3

M 220

2

ll 22

02

Mmx3WMW

ll

put in (i) :

l

ll

o222

2o

422r

)Mmx3(dxxWM

2V

Mm31WV 0r l Ans.

Method : 2 Energy conservation : 2r2

222

0

2mV

21w

3Mm

21W

3m

21

lll -----(i). Using angular

momentum conservation : w3

MmW3

M 220

2

lll ----(ii) put w in equation (i) :

2r2

2

02

222

02

mv21

3Mm

w3

M

3Mm

21w

3m

21

ll

llll

2r2

2

20

4220

2mV

3Mm9

wM3wm

ll

ll

Mm31wV 0r l

Q:1.273: Linear impluse equation :

mJcmcm VMVJ . Angular impluse

equation about com : l

llm

J6ww12

mJ2

2 .

F.B.D. of half part of rod : 2

x w42mforcelCentripitaF

l 22

2

mj36

42m

ll

22

xm2J9Fl

m VrM

lA

x

B

w0

Just before impact Just after impactJ

m, lCM

l/2m

w

l/2cmV

l/4

2

lx ,2mF

115

Q:1.274: (a) Top view : Angular momentum conservation about hinge : w3

m2

'mv2

mv2lll

---- (i) Equation of e : 'vw2v l ------ (ii) from (i) and (ii) : )m4m3(

mv12w

l

Vm4m3M4m3'v

AnsAns

(b) F.B.D. of rod : 2

22

m3M41

MV8W2

MFy

l

l

Q:1.275: (a) Angular momentum conservation :

w3

Mmv2

0ll lM

mv3w 0 -----(i). Using energy

conservation after collision :

)cos(MgW3

M21 22

lll )cos1(g

6w 2

l

)cos1(g6w 2 l

)cos1(g6w l

from (i) : )cos1(g6

Mmv3 0

ll

)cos1(g6m3

MV0 ll

22

0 sing32

mMV l Ans.

(b) 0cm mVV)Mm(P 0mVw2)Mm( l because = Hinge force put w and V0 then

22Sing

61MP l Ans.

2l

V V'

m m

M M

Just before impact Just after impact

Vm

l

lM

w

Fy

M l,

m,

m


(c) For momentom of system constant, hinge force will be zero. And forthis hinge velocity just after collision will be zero. Angular momentum conservation

about hinge. w3

MxmV2

0l

------(i). Using linear momentum conservation

: w2MmV0

l -----(ii) Divide both equation : 3

2x l Ans.

Q:1.276: (a) Using angular momentum conservation : w2MRwmR

2MR 2

02

2

0wMm21w

(b) Work done by force F = Change in K.E.

20

22

22

F WMR2MR

21w

2MR

21W

20

22

20

22wmR

2MR

21w

Mm21

2MR

21

Mm21

2wmR

W2

02

F Ans.

Q:1.277: (a) Since initially 0L

and no torque on system about centre of disc then

L will be constant always. 0ttanconsWI 11 0dt

dI1

0dI1 or

I11 + I22 = 0 (from ground frame) where 1 and 2 is angular displacement. Then

)'(Rm2Rm 2

12

2 121

12

1m2m'm2

m2

m'm

Ans.

(b) Using angular momentum conservation : RW)t('vRmw2Rm0 1

22

21

22

1 Rm2Rm

w)t('RVm R)m2m()t('Vm2W

12

1

dt)t('dV

R)m2m(m2

12

1

=

I dt)t('dV

)mm2(Rm2

2Rm

21

12

2

dt

)t('dVmm2Rmm

21

21

Ans.

V0

x

m

M

W

F

RR

mw0

M

R

0

m1

'

117

Q:1.278: (a) Angular momentum conservation about centre of disc:

wIIWIWI 212211

21

2211II

wIwIw

(b) Workdone by friction : 2212

222

11 w)II(21WI

21WI

21Work

w.w)II(21WI

21WI

21Work 21

222

211

221

21212

22

22

12

121

222

211

)II(

w.wII2WIWI)II(21WI

21WI

21

221

21

21 )w.w()II(

II21Work

Ans.

Q:1.279: Linear momentom conservation : mV =nm Vcm + mv' nVcm + v' = v ----(i). Equation

of restitution : 'vw2lVV cm -----(ii).

Angular momentum conservation about inertial

point (A): mcVrmLcm000

cm2

V2

mlw12

ml00 6lwVcm -----(iii) from (i), (ii)

and (iii) l)n4(v12w;

n4)n4(v'v

for v' = 0 n = 4 and for reverse direction : v' = () ive

n>4. 1800 rotation : Angularmomentum conservation about 00 axis. 0 + I0w0 = IW - I0w0 IwI2w 00 .

Work done by motor : 2

002

002 wI

21wI

21Iw

21Work

2002

IwI2I

21Iw

21Work

IwI2Work

20

20 Ans.

(b) wwIdtkdwI

dtLd

0000

0

0000 II

wIwI

0

20

20

IIwI

Ans.

w1I1

w2

I2

nm, l A

V

m

e = 1

Before collision

A

V

m

e = 1 A

v'

m

e = 1wVcm

y

x

zkwIL 00


Q:1.280: (a) 900 rotation: Moment of inertia ofmotor + platform = I. Angular velocity of shaft =w0. Moment of inertia of sphere = I0 . Angularmomentum about 00 axis will be conserved.Because torque about 00 is zero. Then

0

00000 II

wIwwIIwI0

. Work =

change in energy.

200

20

2200 wI2

1wI21Iw

21wI

21

20 w)II(21

)II(WI

21

)II(

WI)II(

21

0

20

20

20

20

20

0

Work done by motor =

0

20

20

IIWI

21

Ans.

Q:1.281: F.B.D. of rod : Here Ny = mg 20

x w2mN l 2

mg 0l . F.B.D. of hinge axis : Here

F2 = Ny = mg 20

x1 w2mNF l . Torque about point A : 2

mg2

F 01ll

2mg

2w

2m 020 lll ll

g2wg2w 2 Ans.

BM NA

0'

B'

P

0 Iinitial

Nw0

final

2l0

mg

A Bm,l

2l

w

0'

Nx

Nyw

mg

F2F1

NxNy

A

Q:1.282: (a) Resolved angular velocity : sinw12

mL2l

Ans.

(b) icossinw12mjsinw

12mL

212

2

i l

icossinw12

mjsinw12

mL2

122

1f

isinw12mlL 2

2

2sinw

12mL

21

(c) cossinw12mcosLW

dtLdL

dtLd 221

24

sinwm 2221 Ans.

C

wm,l

x

y

w

wsin wc

os

initial final

wsin

119

Q:1.283: (a) Here w = Angular precession velocity. Then gyroscopic torque Lx'w

= w IW Sin ---(i). Torque of gravity about hinge : = mg l sin ----(ii). Since

we know that gyroscopic torque is provided by gravity then mglsin = wIwsin Iwmgl'w

Ans.

(b) F.B.D. of rod : FH = centrifugal force. 2

H 'wsinmF l

F.B.D. of rod :

Q:1.284: Moment of gyroscopic forces : Lxw1

= 2n Iw

wnMRw2

mRn2 22

. Torque of psuedo force about origin: =

m(g+w) l -----(i) 2nRww

l from (i) and (ii) : m(g+w) l = nmR2w

2nR)wg(w

l Ans.

Q:1.285: gwtan 222222net wgmwmgmF . Now gyroscope motion : Torque

about fixed point L'wC

----(i) and 1sinIw'w

-----(ii) from (i) and (ii) : 122

1 sinwgmsinIw'w l

IW

wgm'w

22

l Ans.

Q:1.286: jww 11 kww . Moment of inertia about own axis :

2mR52I . Torque of gyro

scope : L'w

wmR52'w 2 . Since bearing distance is l then 'wwmR

52Fl 2

l5wwmR2F

12 Ans.

w'

l

mwImgFH

2w')sin(m l

wR

m(g+w)

w' = 2

0

mg Fnet

mw

vertical axis

equilibrium axis

equilibriuml

w'

w, I

22 wgm

wl

w1

m1R

y

x

z


Q:1.287: Here w is angular presession velocity and hence T2'w t'wsinm T

t2sinm

Tt2Cos

T2

dtd'w m

. Torque of gyroscope = L'w

Iw'w

w2

mrT

t2cosT

2 2m

. And this is equal to

bearing torque. t2coswmrT

2F 2m

l

1t2cosFmax then lTwmr

F2

mmax

Ans.

Q:1.288: Hence w = 2 and velocity of angular presession is w

= v/R . Then Torque of gyroscope is : L'w

IwRv

n2IRv

RnIv2

Ans.

Q:1.289 : Here w = 2n and angular precession

velocity is Rv'w . Here dirn of w is in

negative y dirn while direction of w is in zdirection. We know Torque of gyro scope =

Lx'w

n2IR

)Iw(R

vv

. Let gyroscopic force

is F then RnIv2Fl

RlnIv2F

w

w'

l

m, r, w2

mrI2

Radius of bearing = l

v

R

I, w

w'

v

R

w'

wI

ly

x

z

121

Q:1.290: If temperture is increased by T then increase in lengthof rod : l = l T ----(i) . Since no increament of length hence for F should decrease its length

in same amount. Using hooks law: strainstressE where E = youngs modules

ll

A/FE

ll

EAF

from (i) : ]T[EAF Pressure = TEA

F Ans

Q:1.291: (a) Top view = is very-2 small. Supposeinternal pressure is P then. F = P(area) = P (R) lAnd this force is balanced by tensile force T. Then2T Sin /2 = F T = RPl T = PRl ---- (i).If breaking strength is m then T = (lr) m

lPRRr:)i(from m

Here R = r

(b) Take a cone of apex angle . 4rP2/sinrPF

22

. And this force

is balance by T : 4rP

2T

4rPT

2Sin 22

4rPT

----

--(i). Now rrrrSin2T m2m ----(ii) From (i) and (ii) :

rr2P

4rPrr m

2

m

Ans

Q:1.292: rod ruptures where pressure is maximum and this point is P. To find force at point P :

F.B.D. of half of rod : Contrifugal force = 2w

42m

l

2w4

A2

ll. Then

8wAF

22l

m

22

8w

AFStress l

l

/22

w m l

/222 m

l

/2 m Ans. where m

= Braking strength of rod = Volume density of rod

1.6 Elastic deformation of solid body AF

A

Fl

l

/2T

F

T

R l

r

T

2

T

F

T

P

r

l

w

A AP

2w4lA

2l

F


Q:1.293: F.B.D. of arch making angle at centre. Here is very-2 small. m =mass of differential element = volume = (A) (r) where A = area of cross

section of wire 22 mrwTSin2 rw)Ar(T22

2 22 wArT

2rATw

AT

r1w 2

wireofstrengthBreakingwherer1w

2

r

n2

r2

1r2

12 Ans

Q:1.294: 2dR where R = radius of wire. Here 2T Cos

= mg -----------(i) Again we know Stress/strain = young

moldules = E )1ec(cos22

4dET

2

=

)1ec(cos4dE 2

----(ii) from (i) :

cos2mgT ---(iii) from

(ii) and (iii)

1

sin1

4dE

cos2mg 2

-----(iv). Here

cot2

h l ----(v) from (iv) and (v) which calculation is difficult: Ed2mg

23h l Ans

Q:1.295: Area of cross section = s young modules = E mF0a .

Since plank is accelerating and at each cross-section force willbe different. So stress will be different then. Now force

mFxmamF 0x l

lxFF 0 Now

dx)x(dEStrain

Stress sF

dxEsxFdx

ESF)x(d 0

l

l

l 00 xdxsE

F)x(d ES2

Fx 0l AnsAns

w

r22

m

T2

2T

lA A

P

mg

Th

l

T

mg

lF0m

S

dx

F

x

123

Q:1.296: Area of cross section = s. F.B.D. of dr part : - df (dm)

rw2

r2

F

0rwdrmdF

l l

r22 rw2mF

ll

2

2r2

12

mwFl

l Now E

dr)r(dE

SF

StrainStress

ll

l0 22

SE2mw drr1dr

SEF)r(d

2

Ew

31r

32l where ls

m AnsAns

Q:1.297: We know ES

F

ll Vold = Sl

SEF

SEF

llll SE

Fnew

lllll

SEFl ll

Also D

D

ll SE

RDD

SEDFDD

ll

Now

l

4DV

2

old

lllll

222

new DD1

4D)(

4)DD(V

lll 1

DD21

4D2

ll

DD21Vold

)21(SE

FSSEF)21(V

DD2VVVV oldoldnewold

lll

)21(EFV l Ans

Q:1.298: (a) area of cross section = s. F.B.D. of part x : Now

EStrainStress

E

dx)x(d

ST

where d (x) is elongation

in part dx.

l ll

ll0

22

E2g

SLm

SE2mgx)x(dE

smxgdx

rl

dr

w m, l

l

l

DD

F = 1000N

Cross Section = S

dxTl

x

weight of part x

gl

mxT

T

x

gl

mx


Eg

21xthen

Smwhere

2ll

(b) This part is same as Q: 1.297. ;

4DV

2

old

)(4

)DD(V2

new ll

lll 1

DD21D

42

old

oldnew

old VVV

VV

= ll

)21( . Then

take value DD

of using poissons ratio .

Q:1.299: (a) Here we have to find decreasement ofvolume due to x dirn, y dirn and z dirn pressure and addall three which will be answer. where S is surface areaS = l1l2

EP

1EP

EP

3new

33

3

33

llll

ll Also

EP

1

1

33

11

ll

ll

ll

. Similary :

E

P11new

1 ll

E

P12new

2 ll

2

321new EP1

EP1V

lll

EP21

EP1321 lll

EP

EP21

VV

old

new )12(EP1

VV

old

new )12(EP

VVV

old

oldnew

)21(EP

VV

VVV

old

newold

)21(

EP

VV

This is only due to force from x.

Similarly : )21(

EP

VV

This is only due to force from y )21(

EP

VV

This is only due

to force from z Total decreasement of forces are applied by all direction : allofsumVV

net

)21(EP3

Ans

(b) Compresibility : dP1

VdV

dP

dVV1

dP = Applied pressure

)21(EP3strainVolume

dPdV

PdP )21(PEP3

. Then )21(E3

AnsAns

l3

l2l1

P

P

z x

y

0

P

l3

l2l1

P

125

(c) Compressibility can not zero if we applied forces from all direction. Then > 0 0)21(E3

< 1/2

Q:1.300: This is a cantilever and bending moment of cantiliver is given by : RIEM (At point (x,y) where

radius is R). F.B.D. of cantiliver in dotted part b= width of cantiliver. Torque about point P :

2)x(bhg21

2)x1(mg l and I = Glometrical moment of inertia w.r.t. neurtal line See F.B.D. of cross

section of rod :

2

h

2h

32

12bhbdyyI . Then

R12EbhM

3 . This moment must be equal to torque of mg for

rotational equal 0xat)x(bhg21

R12Ebh 23 l

g6EhR 2

2

l

Q:1.301: (a) 22

dxhdEI)x(N . F.B.D. of length of (l-x) : Now N(x) = No EI

Ndx

ydNdx

ydEI 022

02

2

1CxEINo

dxdy

0dxdy0xat as shown in figure. Then C1 = 0 EI

xNdxdyNow o

22

o CEI2xNy . At x = 0 y = 0 C2 = 0 then EI2

NyxAtEI2xNy

20

20 ll from Q: 1.300:

12aIahbHereI

4

123bh

y

0

(x, y)

hl-x

bhP

M

bhg)xl(mg

y

x

h/2

h/2 dy

b

y

(x,y)

y

No

xl-x

No

N(x)


(b) (x,y)F

l, xP

F.B.D. of length (l-x): Here F1 = F l, x

F

N(x)

F1

N(x) = F

(l-x) )x(Fdx

ydEI 22

l )x(EIF

dxyd2

2 l

dx)x(EIF

dxdyd l 1

2C

EI2Fx

EIFlx

dxdy

0C0dxdy0xat 1

EI2Fx

EIxF

dxdy 2

l

132

CEI6

FxEI2xFy l at x = 0 y = 0

C2 = 0 EI6Fx

EI2xFy

32

l. Put x = l

EI2xF

EI6F

EI2xFy

333 lll

EI3xFy

3l AnsAns

Q:1.302: By symmetry F1 = F2 = F/2. F.B.D. of half of length : Then F3 must be equal to F1 F3 = F1 = F/2. Now

our question is as : Similar as Q: 1.301 : 203

00 HereEI3

Fy lll 2F0F 8EI23Fy

3

l

EI48Fy

3l

Q:1.303: (a) F.B.D. of (l-x): 2

)x(bhg)x(N2

l

22

2)x(

2bhg

dxydEI l 22

2)x(

EI2bhg

dxyd

l

21

3CxC

12)x(

EI2bhg

dxdy

l

EI6

bhglC0dxdy0x

3

1

. Again 12

)x(EI2

bhgy4

l

21 CxC at x = 0 y = 0 C2 = EI24bhg 4l

at x = l y = C1 l + C2 EI8bhgl

EI24bhgl

EI6bhgl 444

EI8bhg

23y

12bhIand

43 l

F1 F2F y

x

F3

F1y

x

F/2y

l l

l

(x,y)

xl-x

lN(x)

2)xl(

2)xl(

bhg)xl(

127

(b) Here F1 = mg/2 F.B.D. of half of portions Similar as part (a) 24

Ehg

25 l

Ans

Q:1.304: Torque about origin : )x(h31dxxh)x(N 33

x

2 llll

----(i). Now we know 2

2

dxhdEI)x(N

and 2233

dxhd

12hE)x(Nthen

12hI ll ------(ii) from (i) and (ii): )x(h3

1dx

yd12hE 332

23 lll

14

32 C4

xxEh4

dxdy

l 0C0dx

dy0xat 1 . Now 2523

2 C20x

2x

Eh4y

l at x = 0

y = 0 C2 = 0

20x

2xl

Eh4y

523

2

put x = l

2

5

Eh59y l Ans

Q:1.305: (a) Here 1 = shear strain. Take a differential length (ds) on circumference. If we rotate this element

by a angle then will be shear strain like above figure. Then 11

StressStress

1ds)r(F

ds)r(dF 1 . Then rdsrrdFd 1 Net torque = N then

rr2dsrrN 21r2

01

also we know

lll rr

r 111

lrrr2N 2

rr2N 3

l = Torsion angle

(b) Take differential ring at radius x. then dxxx2dN 2l

from part(a)

r

0

32 dxxx2dNl l4

r2N4

Nr2

4

l

F1 F1

mg

F1 l/2

F = mg/2

Because F + F1 = 0

x0

dx

)x()dxlh(a)dm(dF =

volumemaes

1

dFds

r

= = Torsion Angle = Shear coefficient

dxx


Q:1.306: from Q.No. 1.305:

2

2d

21d

dxx2dN 3l

16dd

42N

41

42

l

4142 dd322N

l Here = G then 4142 dd8x4

GN l

l32

ddGN4

14

2 AnsAns

Q:1.307: from Q.No. 1.305: l

4rN w.torquePower w.N

NW)Power( max l2wr4

GHere l2

wrG)Power(4

max

Q:1.308: dF on differential element of length dr : 2122t rrr)rdr2(mr)dm(a)dm(dF

r

2r

32

12

2drr

rr

m2rdFd 2122

42

42

14

21

22 rr2

rrmrrrr4

m2

. Note in integration

range take from r2 to r1 because elastic torque will be equal to torque of Pseudo force in region (r2-r)

Q:1.309: we know strainxstress21

volumeenergyelastic . Also we know stress = E strain

2volume

energyelastic )strain(E21

2E21

lAE21energyelastic 2

pmE

21 2

.

pmE

21energyelastic 2 Ans

Q:1.310:(a) g)x(rT 2 l g)x(rTStress 2

l we know

strainxstress21

volumeenergyelastic E

StressStrainEStrainStress

dxrxE

)stress(21energyelasticdU 2

2 dxr

Eg)xl(

21U 2

222

dx)x(Erg

21

0

2222

l l

l

0

3222

3)x(

E2gr

l 3222 grE6

1U l AnsAns

r2

r1

m

drr

dF

r

l

m A

l

r x

T

dx

129

(b) EStrainStress

dx)dx(dStrain

Eg)x(

l

0

0 Edx)x()dx(d

l

l l

02

2)x(

Eg

l

ll

E2

g 2ll E2

g 2lll

from (a) : 2

2g

E6lrU l

22E2

E6r

lll

22

E3r2U

lll Ans.Ans.

Q:1.311: (a) Make F.B.D. of bending: If radius is R then

2

Rr2 ll increase in length of AA = x

+ x length of neutral surface = R2 Rx

2Rx2strain

)dx()2R(Volume

)dx()2R()strain(E21dU 2

2

2R12

hEdx2RRxEU

32

. Now : for complete circle : 2 =

2 = = R = l/2 put then l/Eh61U 32

lm h

l Sh

R

A'

Neutral Surface

xdx

Here is very-very small

Q:1.312: l2r)strainxstress(21dE and

l

GrGStress and this

stress will be constant in next d rotation then. drGr21dE 2

l

drrGr

21 2

l

d2

rGE2

l l4rGE

24 AnsAns

Q:1.313: from Q: 1.312: l4

rGE24

l4rG4

drdE 24

drrGdE24

l

)rdr2(drrG

volumeEnergy

)rdr2(dE 24

ll

2

22 rG21

dVdE

l Ans

d

d

r

l d

lrrrd

d

R

drr


Q:1.314: We know dPdV

V1

. Here dP is pressure at depth x due to weight of water or hydrostatic

pressure then. then ]gx[VdV l . Also we know : strain)stress(2

1V

dV when B= Bulk modulus =

1

2)strain(B21

2)gx(1

21

2)gx(21

dVdE

. If x = h

then 2)gh(21

dVdE

Ans

1.7 Hydrodynamics

Q:1.315: Make F.B.D. of AA'BB'CC' : AA' 0

F1 CentreR

F2 . Since system is in circular

motion, (F1-F2) will provide required centripitalforce. And hence 2121 PPFF because F1and F2 will only arised due to pressure difference.Now, if we study motion of all particle in this differ-ential region then.With whatever mechanism, if particle at A go toward A its kinetic energy will be increased because of workdone 12A'A21 VVVVHenceFofthatthangreaterisF . Since velocity is increased, density ofstream line will be increased.

Streamline density variation analysics : Here P3 = P4 = P5 = P6 V3 = V4 =V5 = V6 . And hence density of stream line (Distance b/w two streamline) atline joining (3-4) and (5-6), at each point will be same. Now P1 > P2 V2 > V1.And density of streamline will be increased if we are going from (1) to (2).Using continutity equation at (1) and (3) V1 > V3 and continuty equation at (2)and (4): V2 > V4. Now if we take force concept alongle 2-4 then P2 > P4 and P1> P3

Q:1.316: Using bernaullis equation at S1 and S2 :

222

211 V2

1PV21P -------(i) Using continuity

equation : S1V1 = S2V2 122

1 SVSV . Put in (i)

2222

1

22

22

1 V21P

SVS

21P 122

1

222

2 PP1SSV

21

1SS

PP2V

21

22

1222

l

r

Cross Section = Sdx

x

0Centre

A

A'

C'C

B B' A A'

(5)

(4)

(6)

(3)

(2)

(1)

P2S2P1S1

x1x1

V2

h

V1

131

1SS

hg2

1SS

hg2V

21

22

21

22

2

. 101 gxPP )hx(gPP 102 hgPP 12 . Volume flow

rate 21

22

1122SS

hg2SSVSdtd

where dt

dQ s/volume 21

22

12SS

hg2SSdtdQ

Q:1.317: At point (1) point (2) pressure difference is only arised dueto dynamic pressure. Now bernaullis equation at (1) and (2) :

0PV21P 2

211

2112 V2

1PP ----(i). At line M-M pres-

sure will be same. Then)xx(gPhg)xx(gP 2120211 hgPP 012 .

Now put in (i) : 210 V21hg

hg2V 01 .

hg2SSVdtdrateflowVolume 01

Ans

Q:1.318: P0 = atmospheric pressure Bernaullis equation between (1) and

(2): 0V21PghV

21P 21011

2111 . Here V1


)xhx(4R 22 for 0dt)R(dThen:maxRmaxR

22 . Then 2

hx0x2h

22

max h4h

2hh4R

hR max h= 50 cm

Q:1.320: Bernaullis equation at point (1) and (2) along stream line

)xhh(gOPV21P 00

21 -----(i) Here gxPP 01

in (i) gx)hoh(gPv21gxP 0

20 0

2h

g2Vh

Q:1.321: Bernaullis equation between (1) and (2) : 232

1 V21Pgh)0(

21P

230 V2

1ghPP ---(i) Now continutity equation between (3) and (2) AA2V2 = A3V3

321 Vr2VR2 rgh2R

rVRV 1213 put in (i) : 2

21

0r

gh2R21ghPP

2

21

0r

R1ghPP Ans

h

h0

xV

free water surface

(2)

(1)

Q:1.322: Method 1: Here we know Wall forces = K.E. WF + Watmosphere = K.E. Here work done by atmo-

sphere = 0 Because Power due to atomosphere : e010 sVPSVPdtdwPower from continuity equation :

R1

(3)

r

R2

(1)

h

(3) (2)

= Thickness of orfficer R1

V1 Ve(2)

S

F(1)

s A

V1P0s

P0S

Ve

133

SV1 = sVe Power of atmosphere = 0 then WF = K.E. Bernaullis equation between (1) and (2)

along streamline. 202

11 Ve21P)V(

21P and

20101 Ve

PP2s/FPP

)s/F(2PP2Ve 01

Since F = constant Ve = Constant

)s/F(2

-----(i) . Now work done

in differential term : 2F Ve)dm(21dW 22F MVe2

1Ve)dm(W 2

F VeV21W -------(ii)

Also, using volume conservation : )s()tV(V e stVVe . Put in (i) :

22

3

Fts

V21W AnsAns

Method 2 : Work done by F : WF = F (displacement) = F (V1t) SFV)tSV(

SF

1 . Also from (i)

in method (2) 22

SF

stV

22Ve

22

2

22

2

Fts

V2

1Vts

V2

W Ans

Q:1.323: Suppose at time t, x length of water is inside tube then. gx2Ve .

Now using continuity equation: VesSV1 gx2SsV1

Again

0

dtg2Ss

dtdx

0

21

0dtg2

Ssx)dx(

g2Ssx2

0

h

21

g2Ssh2 g2S

s AnsAns

Q:1.324: Method : 1. Bernaullis equation w.r.t. rotatory axis :

.constgz)rw(21V

21P 22r equation b/w point (1) and (2):

0w21V

21P0w)h(

21V

21P 22220

22210 ll here

0V21 2

1 then 22222

2 w)h(21w

21V

21

ll

22222 )h(wV ll )h2h(w 2222 lll =

1h2hw 22 l 1

h2whV2

l AnsAns

Ve

h

x

Ss

V1

s


Method: 2 Bernaullis equation between (2) and (3): 0V21P0)0(

21P 230

22

2302 V2

1PP -------(i) Calculation P2-P0 : dF = (dm) xw2 2xw)dxS(S)dP(

h

2P

PxdxwdP

2

0 l 22202 )h(2

wPP ll )h2h(2w 22 l put in (1)

23

22V

211

n2

2hw

l 1

h2whV3

l Ans

Q:1.325: At differential element : dsdPgCos

dtdV

dPds)(cosgvdv

dPcos)ds(g

2

1

2

1

2

1

P

P

z

z

V

VdPdhgvdv

1212

21

22 PPgzgz

2V

2V

22

2212

11 gzV21PgzV

21P Ans

Q:1.326: We know force 2AVF

Then sgh2gh2SSVF 211

)hh(sg2))hh(g2(SF2 Fnet = F1-

F2 hsg2Fnet Ans

lh

B

w

(1) (2)

(3)A

w

x

dx

V2Z2

dh

g

Z1

V1

(1)

s

(2)

Pdtdv

dsdP

dtdv

dsdh

ds

)hh(g2V2 h

gh2V1

h

F1F2

135

Q:1.327: Here gx2V1 Now force is : 2AVdV gx2)dxb(

l

lhxdxgh2dF )h2(gbF ll Ans

Q:1.328: Using continuity equation : AQVAVQ 11 . F.B.D. of tube : A = r2 = Area of cross-

section 222

1 AVFalsoAVF . 222

21rQ

AQFFThen

Torque about O : 2

2

0r

Q

l

l

x

dxV1

b

h

Q:1.329: F.B.D. of water inside tube : where F = force due to change in momentum Fnet on tube

water : F)sS(PsPSghPF 000net FghsFnet -----(i) Calculation of F :

2121 VVQVVdtdm

dtdPF ----(ii) where Q = volume flow in one second. Here

sQV;

SQVAndgh2V 212 Put in (ii) :

sSsSQ

sQ

sQQ

dtdP 2

And

FsS

sSgh2sdtdPgh2ssVQ 22

Then S

)sS(ghsS

sSgh2sghSF2

2net

S)sS(ghF2

net

Fs)ghP( 0

)sS(P0 sP0

h S

V2

sV1 dm

V2

V1

dm

l

V10

V1

r

F1

F2

0


Q:1.330: Method : 1 (a) Dotted line express isobaric surface. Then F.B.D. of one of the dotted line.Bernaullis equation between point (1) and (2) from rotatory reference frame.

200 )xw(2

1gh0P00P . Then 22wx21gh . Here x = r then g2

wxy22

2

2r

g2wy

(b) Bernallis between (1) and (3): 200 )xw(210P00P 20 )xw(2

10PP Ans

Method : 2 Pressure equation from rotatory

frame : 12

0 P)xw(21P 2

20 P)xw(2

1P

(a) gy)xw(21PP 212 . Here P1 = P2 = P0

22

xg2

wy

Ans (b) 213 )xw(2

1PP 203 )xw(21PP

Q:1.331: Top view Viscous force at (dx) element at lower surface : hxw)dxx2(

dzdVAdF

Net viscous force at (dx) element from both lower and upper surface: dxxh

w4dF2dF 2net

. Torque

due to this force : dxhwx4xdFd

3

net

Power due to this torque dxhwx4wddP

3

R

0

32

net dxxhwx4)Power(

hRwx)Power(

42

net

x

y

w

0

y

x(1)

(2)

x

y

w

P0(1) (2)

x1

x2

(1)

(2)

y

x (3)

R

dx

h

h

w

x

w

dF

dxx

137

Q:1.332: Cross-sectional view viscous force on this differential

element: drdvrdx2dF where 2rdr = curve cross-sectional area.

Now drdvr2

dxdF

in laminor flow CttanconsdxdF

then

CdrdVr2 r

drcdV2 1CrlnCV2 ----(i). Using

boundary condition : 01 VVRr then 110 CRlnCV2

------(ii) and 2Rr 21 RlnCCthen0V from (ii) and (iii):

21R

R0 RlnCCand

lnV2C

21

put in (i) 22R

1R0

Rrln

ln

VV

Q:1.333: (a) rdx2drdwrdF

. Torque due to this force :

rdx2drdwrrdFd 2

drdwr2

dxd 3

. For laminor

flow this torque is constant then Cdxd

now dw2rdrc

drdwr2C 3

3

12 CrCw2 ----(i). To find C, C1 use boundary condition: (i) 21 wwRr

(ii) 0wRr 2 then 121

2 CRCw2 -----(ii) 12

2C

RC02 ---(iii) from (ii) and (iii) find

C1 and C2 and put in (i):

22

12

12

1

22

21

2 r1

R

1

RR

RRww Ans

Q:1.334: (a)

2

2

0 Rr1vv

2

2

0 Rr1V)rdr2(v)rdr2(dQ

drRrrV2dQ

R

02

3

0

2

42

0R4

R2

RV2Q 2

RVQ

20 AnsAns

R2

R1

V0

dxdr

r

x dr

dx

R2

R1dx

dr

rw2

r

dr


(b)

drdwr2

dxd 3

22

12

12

2

22

21

2r1

R1

RRRR

ww

32

12

2

22

212

r2

RRRRw

drdw

3

12

12

2

22

212

Rrdrdw

R2

RRRRw

1 2122212

22

RRR

Rw2

. Then

2

12

21

2223

1 RRR

Rw2R2

dxd

21

22

22

212

RR

RRw4dxd

(c) from option (a) : 20RV21Q ---- (i) And from poiseuilles law :

l

8

PPRQ 214

. Comparing

(i) and (ii) :

l

8

PPRRV21 21

42

0 20

21R

V4PP

l Ans

Q:1.335: If we see carefully, we get that heights aredifferent in pipe (1) and (2). While velocity of flow atbottom of both pipes are same because cross-sectionarea of horizontal pipe is same at each point, then usingcontinuity equation, we can say that velocity of flowis same at each cross-section of horizontal pipe. Thisis because of friction loss and bent loss in pipes stillwe can say that dotted line show isobaric surface. Height

above isobaric surface provide, velocity at efflux. Then )hh(g2)AC(g2V 3e . Calculation

of h : Using similar tringle properties. cm3020x23

2h3h

23

hh 22

ll

h3-h = 35 cm - 30 cm

= 5 cm s/m1105g2V 2e 2v)dm(

21dk

2

2

20

Rr1V)rdr2(

21

l

drRr1rVK

2R

02

220

l 20

2 VR61K l Ans

R

l

vr

,r dr

l l l

h1

h3h2

pipe (2) pipe (1)c

h

B Vc

139

Q:1.336: We know Reynolds no. for circular cross-section is

defined as

lvRe where l = length of charactersitic and for

circular tube with full of water. D = 2R l = D

vDRe

22

11

21

11

2

1RvRv

DVDV

ReRe

------(i) Also we know :

222

211 RVRVQ put in (i) : 2

1

22

2

1V

R

RV

)0(0

x0

1

2

2

12

1

22

2

1er

er

R

R

R

R

R

R

Re

Re

x

2

1 eReRe Ans

Q:1.337: Maximum on value of Reynolds no. for glycerin for laminor flow :

DVR maxg

1

111g r2vR

. Reynold no. for water :

2

122w r2vR

We know : Reynolds no. for turbulent flow

> that of laminor 1

111

2

122 r2vr2v

122

22112 r

rvv

122

22112 r

rvv AnsAns

Q:1.338: F.B.D. of sphere : = density of glycerin at maximum velocity, terminal velocity will be

attained then. gR34gR

34rV60F 33net

2lmax gr92V

Rynold no. :

DVR max 2l grP9

2r2x215.0

2l2 grr89 ----(i). Put density of

33363 m/kg1026.1m101026.1 l 33lead m/kg103.11 put in (i) : get mm25r

D = 5 mm

R

R1

v1

R2

v2

x0

D

r

gR34 3

maxrV6

mg


Q:1.339: Where 0 = density of olive oil. Since 0gR34

03 Because mm10

23R 3 which is

very-2 small Then rv6mgFnet we know rv6mgdtdvmFnet

dt)rv6mg(mdV

1max t

0

1001V

0dt

rV6mgmdV

1

max

t0

100V

)rV6mg(r6

lnm

1max t

mg100rV6mg

lnr6

m

.

Calculation of Vmax : At time of 0rV6mg0F maxnet r6mgVmax put in (i)

nln18

dt2

Ans

1.8 Relativistic Mechanics

Q:1.340: We know 22

cv0 1 ll

21

2

2

00cv11l lll

21

2

2

0 cv11

l

l

0llll

Here1

cv1

2

02

2

02

2 21cv1

ll

, 0

22 c2V

ll

----(i )

1005.05.0100

0

ll

ll

, put in (i) : 22 c

1005.02V

V = 0.1C

Q:1.341: (a) Here 2

3a60sina1 l length l appeared from reference frame (x-y): 22

cv11app ll

22app cv123a

l Now app22

0 2a ll

2222 cv1

4a3

4a

2

2

cv34

2a

rV6gR34 3 0

mg

l0V

a

acB

A

al1

x

v

y

600

a/2c B

A

lappl0l0

141

Perimeter (P) = 22

cv00 34aaa ll

2c

2v341aP Ans

a3)21(aPcvfor

(b) Length AC : 22AC c/v1a l . There will no change of length of BD while there is change

in length of AD: 22

cvAD 12

al . Now appeared length AB : 2

2

cv2AD

2AD 42

a43a ll

Now perimeler (P) : 22

22

cv

cvAB 4a4aAC2 l

2c

2v2c

2v 14aP Ans

v


Q:1.343: (a) BD = radius of cone = r0 length of AD = height of cone = l0 00rtan

l . Then

lateral surface area secrS 000 l from this reference frame length along x axis doest not appear

to change. Hence from this reference frame : BD = r0 And length l0 will be appear as l : 2c

cv0 1 ll

220

00

cv1

rr'tan

ll

put value of and v in above function : 1 = 590

A

BC

D

r0

l0

C54v

m4S45 200

(b) A1B1 = l sec 1 Lateral surface area : 10 secr l 21122200 tan1cv1r l2

1

22

222

00cv1

tan1cv1r

l

21

22

222220

cv1cvSeccv1

SecS

21

2122

222220

cv1

cvSeccv1SecS

21

2

22

0 ccosv1SS

. Put value of v and and

S0 S = 3.3 m2

Q:1.344: Time measured by moving dock : t; Actual time of moving clock : t-t

we know 22 cv1

ttt

22 cv1ttt

222

cv1t

tt

22

2222

tt)tt2(

t)tt(tcv t

ttt2cv

v = 0.6 108 m/s

Q:1.345: l0

v

v1 = 0 v

t l ------(i) tv

l Now . Proper length = l0 =

length of rod appeared by that frame from which rod will be appear to stationary. Then

y

x

v

0

A'

B' C'

'

r0

l

D'

221

cv1

tantan

143

22022

0cv1

cv1

llll now 22

01

cv1vvt

ll Then

2

22

2

22

21

'tt

tc1

tc1

tt

l

l

2

222'tt1tcl

2

'tt1tc

l

l0

v

v1 = 0

y

x

v

y

x

Q:1.346: Assume velocity of particle w.r.t laboratory frame is v then 220

cv1

tt

2022t

tcv1

20

2

2

t

t1

cv

20t

t1tcv

. Distance travel in laboratory

frame is : 2

0t

t1tCcetanDis

Ans

Q:1.347: (a) We know 220

cv1

tt

2022t

tcv1

22220 cv1tt 220 cv1tt 220 cv1Vt l

(b) Actual distance travel is : distance = Actual velocity x proper time = 220 cv1tv l Ans

Q:1.348: tC43

l -----(i) We know 220 cv1 ll 220

cv1

ll

2222 cv1

tv

cv1

tC4/3

220 cv1

tv

l

Ans

Q:1.349: We know proper length of rod does not change w.r.t. reference frame then 212201 cv1x l

y

xk

l

v = 0.99c

vl

c99.0lt

3/4 C 3/4 C

l

TARGET


-------(i) 212202 cv1/x l -------(ii) dividing both : 22

2

1 cv1xx

2

122xx

1cv

2

1xx

1cv put in (i) : 210 xx l Ans

Q:1.350: Time to move from A to 0: r

2200

Vcvr1

t

ll

2

22

02

0rc

Vr1tV ll

22

20

0r

tc1

t2V

l

l

Ans

Q:1.351: )vc(2tVtCt A2AA

ll

)vc(2tVtCt B2BB

ll Here tA < tB means later count occur first. AB ttt

)vc(2)vc(2

ll 2222 vc

Vvc

vcvc2

ll )cv1(C

t 22c

v

l. Assume v/c

)1(Ct 2

l Ans

Q:1.352: (a) Since rod is moving with speed V with respectto this frame it length will be appear less than proper length

Then xA - l = VtA )tt(VxxVtx

BABA

BB

l

)tt(V)xx( BABA l now proper length : 22 cv10 ll

22

BABA0

cv1

ttVxx

l

AnsAns

v

0 1 2 3 4

x1

y

x

vx2 y'

x'v

l0 l0

l0

2

2r

0 c

v1l

Ax

y

0

Vr

l/2 l/2

0restv v

BA

x(0,0,0)

y

BV

( , 0, 0)lA

145

(b) Here 0BA xx l 0BA xx l 0BA xx l Then

23BA0

0cv1

ttV

ll

220BA cv11V

tt l or

23BA0

0cv1

ttV

ll

220BA cv11V

tt l AnsAns

Q:1.353: (a) Reading of clock at B: 220 cv1VV)B('t

ll

(b) Appeanent length of AA from A: 220'AA cv1 ll 220 cv1

V)A(t l . When B will be

at B then distance traul by point A in frame of rod AB will do Then V)'A(t 0l Ans

Q:1.354: Lorentz transformation of time 22

2

cv1

cxVt't

.

If take t =0 for k frame then 22

2

cv1

cxV't

. 0't0xIf 0't0xIf

Q:1.355: Since both show zero reading at origin. If clock (k) reads time t and clock (k) reads time

t then. according to lorentz transformation. 22

2

cv1

cxVt't

according to queation : t = t. Then

tcv1

cxVt't22

2

differentiate w.r.f. time 1

cv1

cVVxt

22

2

22

2

x cv11VcV we know cv

2x 11

cV Ans

Q:1.356: Suppose a shot is made and it hit the target after time

t then from k frame (mouing with V velocity) 22

2

cv1

cvxt't

22

2

cv1

cVdt't

. If means target is hitted before shot is fired then. 0cdV't0't 2

A' l0 B'V

A

l0Bl0

x

VK'

K

target

x = 0t = 0

x = dt = t

K frame (stationary)


2cdV't tcdV 2 x2 V

tdifcV

td

. Then

2x cVV . It is possible only of

one of the Vx > C or V > c which is not possible. It prove that target will be after shot made.

Q:1.357: (a) We know in variant formula : 212212

212'12

'12

2 ttcxtc in frame k both events occure

at same point then 0x '12 then 21221222122122122 xxttcxxtc'tc =

212212 xxctct = 162516 22 282'

12103

16t

ns13s103.1103

4t 88112

(b) in frame k if both occure simultaneously then 0t112 16xtct212

212

22'12 m4t

112

Q:1.358: We know 2

x

x'x

c

VV1

VVV

2x

22y

y

cVV1

cv1V'V

Then net velocity appear from k

frame : 22net y'Vx'VV

2

2222x

net cVVx1

cV1VyVVV

Ans

x

y

K

Vy

Vx

x'

y'

K'

V

Q:1.359: v1 = 0.5C v2

v2 = 0.75C Velocity of approach is taken from laboratory frame hence

Velocity of approach = V1 + V2 = 0.5C + 0.75C = 1.25C Ans

Relative velocity : (Vr) : V1 V2

K'

221

21x

c

VV1

VV'V

221

21x cVV1

VV'V

Q:1.360: 2222c

)V(rel 1

V2

cV1

V2)V(1

)V(VV

now l0 l0V

V

147

apparent length is : 22

0c

relV1 ll

2222

0c1

V21

ll 222

01

41

ll

2

2

011

ll Ans

Q:1.361: 2By

Bx

Ay

1AxVV0V

0VVV

From reference frame A : velocity

components are as shown : 121

1'Bx VcVo1

V0V

22122

1

2212'

By cV1VcV01

cV1VV

Then

2By

2Bxnet 'V'VV 2212221net cV1VVV

2212

22

1net cVV

VVV

AnsAns

Q:1.362: To find velocity of particle in k frame assumecomponents of velocity in k frame is Vx and Vythen using lorenz transformation :

2x

22

y2x

xx cVV1

cV1Vy'V;

cVV1

VV'V

Then

from k : VVcVV1

VV0 x2

x

x

22y2

22cV1'VV

cVV1

cV1Vy'V

. Now

2

22222222

net c'VV'VV)cV1('VVV . Hence we say that Vnet is velocity from k frame

then. For proper time t0, we have to choose another frame k which is attached with particle. ThenReference frame k will be appear to move with Vnet speed then time internal will be appear to increase

frame this frame (k) then. 220

c/netV1

tt

. Distance travel in k frame :

V1A B

V2y

x0

V' y

x

K

y

xV

K'

V'

'V'V

0'V

y

x

y

x

K

2

1212

netC

VVV-VV2

2

netV

K

x


2net20

netnetc/V1

tVtVDist

2

21212

0c

V212

cc/VVvV1

tVVv222

212

Distance = )c/v1()c/v1(

)c/v1(VVt2222

22120

2

Q:1.363:

22x

x

cV)cosv(1

VcosV

c

VV1

VVx'V

----(i) 22

2x

22y

y c/v1VcosVsinV

c

VV1

c/v1V'V

22 c/v1VcosV

sinV'Vx'Vy'tan

Ans

Q:1.364: Here in k frame length of rod will be contracted because

it is moving with V velocity then 220 c/v1 ll Velocity of

rod : Vx = V 2

2

21

21

c/ov1

)c/v(1'v

c/VVx1

)c/V(1VyVy

22y c/v1'vV . Time difference of measuring for A and B:

2

21

c/v1

c/vx'tt

20y2

20 c/'vVtVyVy

c/v1

c/vt ll

220

20

c/v1

c/'vVytan

l

ll 222 c/v1c

'vVtan

Ans

y'

x'

V

y

x

v

Vy'

Vx'

V'

'

K'

x'

y'

AK'

B

l

V

v'

(0,0,0,0) ( ,0,0,0)l

O'

x

y

V=VxAK

B Vyl

B

A Dl

y

149

Q:1.365: (a) We know 2x

xc/VV1

VV'Vx

now differential

equation :

2

22

c/VVx1

c/V)dVx(VVxdVxc/VVx1'dVx

22222 c/VVx1c/Vc/VVxc/VVx1dVx

2222 )c/VVx1()c/V1(

'dt)dVx(

'dt'dVx

------(i) Also we know 22

2

c/V1

c/xVt't

22

2

c/V1

c/V)dx(dt'dt

2

22

2/322

222

2222

cVxV1

cVxV1

c/V1dt

dVx

)c/VVx1()c/Vdt(

)c/V1(c/V1)dVx('dt'dVx

vVx

c/VVx1

c/V1w'w 322

2/322

Ans

Q:1.366: Co-moving frame(k) is that frame in which particle is appear toinstanteously rest. But its accn of particle in this frame may or may not

zero. now we know : 2/322

2/322

cV1

cV1w'w

from Q:1.365 Here Vx = V then

2/3222/322

2/322cV1w

cV1

cV1w'w

2/322 cV1'ww

dtcV1'wdv 2/322 t

0

x

02/322

dt'w)cV1(

dv

2

ct'w1

t'wV

Ans

t

02/12

x

0 )c/t'w(1

dtt'wdx

1

ct'w1

'wcx

22

2

t'wc

21

cvc

x

y

K

v wx'

Vk'

y'

x

VK

y

Vx' = 0 earth

y'

w'Vx' = 0


Q: 1.365 (b) we know 2cVVx1VVx'Vx

0aV01

V0'Vx 'x

2

22

cVVx1

cV1Vy'Vy

22

2

22cV1Vy

cVVx1

cV1Vy'Vy

22 cV1'dt

dVy'dt'dVy

from option (a) :

22

2

cV1

cV)dx(dt'dt

222 cV1cV)dx(dt

dVy'dt'dVy

222222

cV1cVVx1

w

cVdtdx1

)cV1()dt/dVy(

22 cV1w'w . Here Vx = 0

Q:1.367:

0

20 dt)c/v(1 -----(i) from Q: 1.366

2

ct'w1

t'wV

put in (i)

0 2

2

0 dtc

ct'w1

)w(1

022

2dt

)t'w(c

)w(1

022

2dt

t'wc

c

2

020 c

'w1C

c'wln'w

c

ct'w1

dt

2

0 c'w1

c'wln

'wc

Ans

Q:1.368: We know 2c

2v

0

1

mm

Here cv 20

1

mm

)1()1(

m0

C

C100

01.0C

cv

10001.1 )1(2

mm 0

1 + = 2 70)1(2

1mm

0

AnsAns

Q:1.369: Density is defined as volumem

volumemaesrest 0

321

0mlll

-----(i) '''m

321

0lll

'Here 22 ll '33 ll 22

cv11 1' ll

22

cv321

0

1

m

lll ---- (ii) from (i)/(ii):

x

y

K

v w

x'

Vk'

y'

151

22

cv0 1

Here

1

1)1( 00 Then 22

cv1

11

)2()1(

CV

Ans

Q:1.370: Assume mass of proton is m0 then P = mv where v = velocity of proton

2c

2v

0

1

vmP

Squaring both side : 2

220

2220

2

220

P

Cm1

C

PCm

PC

CPm

Pv

Now

2

220P

Cm1

11CV1

CVC

21

2

220

P

Cm11

CVC

Ans

Q:1.371: We know Newtonium momentum : P = m0V -----(i) Reletivistic momentum:

V1

mP2

2c

2v

0

------(ii) 2c2v

)ii()i( 1

21: 2c

2v141

C23

cv

2

2 C

23v

Q:1.372: Classical mechanics : if KKw = 22

022

0 c)6.0(m21c)8.0(m

21

= 20 c)36.64(.2

m

20 c28.0x2

m 20cm14.0w . Relativistic mechanics :

2ii

2ff cmcmw

222

0220 c6.06.01

mc8.0

082.1

m

= m0 0.42 c2 w = 0.42c2

Q:1.373: m0c2 = Rest mass energy mc2 = Total energy Kinetic energy = mc2 - m0c

2 according

to question : m0c2 = mc2 - m0c

2 m = 2m0 0

2c

2v

0 m21

m

411 2c

2v 43

2c

2v

c23v Ans

x'V

y' y'

V = 0

l2

l2l3


Q:1.374: Using classical mechanics : 2c0vm21T where Vc = velocity calculated by chassical mechanics

0c m

T2V . Relativistic mechanics : T = mc2 - m0c2

20

2c

2v

20 cm

1

cmT

1cv

2)12/1()2/1(

c2v111

cmT

4

4

2

221

2c

2v2

0

1cv

83

c2v1

cmT

4

2

2

2

20

=

22

cv

2

2

431

c2v

approximately 20

2

2

cmT

c2v

then

2

02

2

20 cm

T231

c2v

cmT

1

20

20

2

2

cmT

431

cmT2

cv

2

02

0 cmT

231

cmT2

2

00m

Tcm

T2312V Now

0

02

04

30m

T

c

c

mT2

mT2

cmT1

VVV 3

4cm

T2

0 at max Ans

Q:1.375: We know 224202 cPcmE -----(i). Let us kinetic energy is T then. TcmE 20 put

in (i) 222022

0 cPcmTcm 2242

02

02

4c

2c0m cPcmTcm2T 2022 cm2TTcP

20cm2TTc1P Ans

Q:1.376: From question no. 1.375 : 20cm2TTc1P of no. of particle collide per second is

n then ne = I eI

momentum transfer per second is : 20cm2TTCeIp

dtdPF . Power

= Energy radiate or absorb persecond Power = n (kinetic energy of one particle) = T TeIPower

153

Q:1.377: Sphere Reference frame Velocity of gas particle in frame of sphere 2c

vx

xx V1

VV'V

Here

Vx = 0 Vx = -v V = v . Momentum transfer in one collision :

222222 cv1

mv2

cv1

)mv(

cv1

)v(mdP

. Since volume will be decreased by a factor of 22 cv1

But actually no. of particles of gas does not change but its occupied volume is decreased and hence

apparent no. of particles per unit volume is increased 22app cv1

. no. of particles collides

per second : 2

22app)dA(

cv1N

V)dA(

cv1cv1

mv2dtdPdF

2222

.

22

2

22

2

cv1mv2

cv1mv2

dAdFessurePr

Gas Refrence frame : We know 2x

xc'VVx1'VV'V

v'Vx (velocity of sphere w.r.t. gas frame after

collision) 2c'Vv1

'Vvv

vVx (velocity of sphere after collision which does not change)

'vv2'vVc

'vv2c

'vv1V 2224

222

v2c

v21cv'V 2

3

4

4

2

2

3

cv1

v2'V

V = v

Before collision

mmV'=

After collision

V = vm V'

v

V = v

Before collision

After collision

v

vv

dA dA


momentum transfer in one collision = 2c

2v2

1

Vm

222

22 cv1mv2PV

cv1mV2

dAdF

Ans

Since in gas frame mass of gas particle and its density does not change

Q:1.378: At time t, letus velocity of sphere is V then F = ma dtdV

cv1

m

220

dvcv1dtmF t

0

22t

00 2222

0 tFcm

Fctv

Also

22220 tFcm

Fctdtdsv

t

0 2t2F2c20m

dtts

0FCds

F

cmtcFcmS

202222

0 AnsAns

Q:1.379: Given 222 tcax 22222

2222222

2

tca)tc2(tcc)tca(

dtdV

tcatc

dtdxv

222

222

tcatca

dtdV

Momentum : P = mv v

1

m

2c

2v

0

2c

2v

2c

2v002c

2v

1

1dtdvm

dtdVm1

FdtdP

put value of dv/dt : a

cmF

20 Ans

Q:1.380: (a) and (b) : we know

2/1

2c

2v0

2c

2v

0 1Vdtdm

1

Vm

dtd

dtPdF

dtdv

cV21V

21

dtVd1mF 2

2/3

2c

2v2/1

2c

2v0

We know

v.aadtdvand

dtdva t

Then

v.aV1

c

V

1

amF 2c

2v2

2c

2v0

V

F

m

155

v.a

cv1c

V

cv1

amF 23222220

If av.v.av||a

then

232222

0

22

0

23

22

0

22

0

cv1

acvm

cv1

am

cv1

vvam

cv1

amF

22

22

22

0

cv1

cv1

cv1

amF

23220

cv1

amF

Ans

Q:1.381: Assume velocity of particle in K frame is Vx then we know

2

2

20

2

20

x

cVx1

cmEand

cVx1

mP

22

20

220

xVxc

cmEandVxc

VxcmP

.

Also we know from invarient theorem : (ds)2 = c2(dt)2 - (dx)2 = constant = c2(dt)2 - (Vxdt)2

dtdsVxc)dt(Vxc)ds( 222222 constant in any inertial frame of references

Then )ii(dsdtcmE)i(dt

)ds(VxcmP 30

0x

Vxdt = dx Also

22000

xcv1

vdtdxds

cm'dxds

cmds

'dt'Vxcm'P

220

22x

20

20

xVxc

cm

1

V

1

P

ds1

cVdtmdsdx

1

cm'P

2

2cV

222

30

2

2c

v

2Px

x1

E

1

Px

Vxc

cm

11'P

2

2cVPx

x1

E'P

proved

ds'dtcm'E 30

2

230

1

cV)dx(dtds

cm 2

230

2

30

1

cVdxds

cmdsdt

1

cm'E


dsdx

1

CVm

1

E2

02

dsVxdt

1

CVm

1

E2

02

20

2 1

VdsCVxdtm

1

E

22 1

PxV

1

E'E

21

PxVE'E

Ans

Q:1.382: We know 420222 cmCPE For photon. rest mass m0 = 0 then 222 CPE E =

PC From K frame : CPx c/Px form k frame = C'Px'

C

2

2

1

cEVPx

'

2

2

2 1

)1(

1

cV

'

11' Ans For

11

2' 5

3cV

AnsAns

Q:1.383: From Q 1.381 We know dsdxcmP 0x

dsdtcmE 30 ds

dycmP 0y dsdzcmP 0z

Then dsdzcmP 20x -----(i) ds

dycmCP 20y -----(ii) dsdzcmCP 20z -----(iii). Squaring and add

((i)+(ii)+(iii)) : 222242

022z

2y

2x dzdydx

)ds(

cmcPPP 2

2

42022 )dl(

)ds(

cmcP

2

242

0

262

0222

)ds()dl(cm

dsdtcmCPE

courcm

dsdldtccmCPE 4202

22242

0222

420

222 cmCPE Ans

Q:1.384: We know P = mV Vc

mc2

2 Also we know 2c

EVP )cm2T(TPC 20 Now

EPC

CV

For system : system

systemcmE

CPc

V 2

02

0

20

20

cmcm)cmT(

C)cm20(0)cm2T(T

cV

20

20

20cm

cwmTT

cmT

)cm2T(TCC

V

2

0

cm

cm2T

TCC

V

Ans

157

(a) Since maes of both particles are same hence magnitude of momentum of both particles from comframe will be same. Velocity of each particle from com frame = 0- Vcm = Vcm. Energy in com frame:

)cm2T(cm2

cmVc1

cm2E 202

0

2

2

20

. Using equation : 20cm2T~E~

202020 cm2cm2Tcm2T~

1

cm2

Tcm2cm2T~ 20

20

20

cm22cm

0cm22

20 V

cV1

mVcv1

cm2P~

Q:1.385: m0 m0 m0 V We know 220222 CMiantvarInCPE . WeWe

assume energy does not losses. 2202022020 Cmcm2TTcmTcm

Tcm2m2Ccm 20020 Tcm2m2C1m 2000 Ans. Also we know from Q: 1.384

2cEVP

E)PC(

EPCv

22

2

0

20

cm2T

cm2TTCv

2

0cm2T

TCv

AnsAns

Q:1.386: Use invariant equation : 222 CPE Invariant

0Tcm22cm2TTTcm2 22020112120 . Becausemomentum of system from com = 0.

420

20

2201

21

420

201

21 cm4cTm8T4cmT2Tcm4cmT4T

201

20

2 cmT2cTm8T4

20

20

1cm2

)cm2T(T4T

2

0

201

cm2

cm2TT2T

Ans

Q:1.387: 20321 cmEEE 0PPP 321 . Take system of (m 2 + m3)

2212120221232 cPEcmcPPEE Invariant 42322212120 cmmcPEcm .

m0 m0 m0 m0=

T1 TT

Inertial K frame COM frame


Because this invariant is same in all frame of reference hence from com frame of (m2 + m3). In

variant = 4232 cmm Now 423222112021420 c)mm(cPEcm2Ecm 42

3212

042

142

0 c)mm(Ecm2cmcm

423242120120 cmmcmmEcm2

20

3221

20

1 cm2mmmmmaxE

Ans

Q:1.388: Suppose velocity of ejected mass is Vx from a frame having

zero velocity then we know 2cvx

xx V1

VV'V

2

x2c

vx

x

cuv1

uVVV1

VVu

. Change in momentum of (dm) mass : dP = -(dm) V + dm Vx

2

2cv

2 cuv1

uudm

cuv1

uvvdm Here uc

uv2 then

2cuv1

udt

)dm(dtdP

Force on

gas particle then force on rocket : dtdvm

cuv1

udtdmF

2

v

02c

uvm

0mdv1

mdmu

C

u2

0

c/u20

mm1

mm1cv

m0

v = 0 = 0

m1

m3

E3

E1

E2

m2

u dm

vm

159

2.1 Equation of gas state processQ: 2.1: Suppose gas is at pressure P, volume V and tamp. T. Then we know PV=

nRT. Also RTMmPV where m = mass of total gas M = molar mass. Now VV

= const. and T = const M

RT)dm(VdP . Then VRT

)dP(Mdm ----(i). Calculation of M/RT : We Know

RTMmPV RTPM

P

RTM

. Given at NTP (normal tempreature and pressure) : T0 =

00C = 273 K P = 1 atm = P0 = 1105 N/m2. Then

0P

RTM

put in (i) PVP

mV)dP(P

dm 00

Q: 2.2: Total no. of mole of gas is n and due to heating n mole

of gas goes in other chamber. 1

111 RT

VPnnRTVP

2

222 RT

VPnnTR)nn(VP ---(i) 22 nRTV)PP(

2

2RT

V)PP(n put value of n and n in (i) : 2

2

2

0

1

1RT

VPRT

V)PP(RT

VP

21

1

22 T

PTP

T11P

21

212 T

PT2TPP . Increase of pressure of Vessel B :

P

TTP

21

2P

T2TPPP

1

21

1

212

Q: 2.3:V;m,t;P

He;H2 Let mass of H2 gas is m1 and that of He is m2. no. of moles of H2 : n2 = m/

2. no. of moles of He : n2 = m2/4. Also m1 + m2 = m ----(i). RtnnPV 21 Rt4m

2mPV 21

RtPV

4m

2m 21 Rt

PV22

mm 21 ----(ii) Now (i) - (ii) : RtPV2m

2m2

RtPV2m2m2

put in (i) : mRtPV4mmm 21

RtPV

mm

mm

RtPV

RtPV

2m

4m

RtPV

RtPV2m2

mRtPV4

mm

1

2

2

1

Ans

Part Two

Thermodynamics and Molecular Physics

P.V.T

P1 P = 0t1

n

V

(A) (B)Initial

FinalP2

t2n- n

V

P2- Pt2 n

ValueV


Q: 2.4:)m(CO

P;T)m(N

22

012 We know PM = RT----------(i) Where M = Molecular weight of mixture.

= Density of mixture. Calculation of M : no. of mole of N2 : (n1) = m1/M1 no. of mole of CO2

: (n2) = m2/M2 where M1 and M2 molecular weight of N2 and CO2. Then

2

2

1

121

21

2211

Mm

Mm

mmnn

MnMnM

Put in (i) : RT

Mm

Mm

mmP

2

2

1

1

21

2

2

1

1

210

Mm

MmRT

mmP

Ans

Q: 2.5: (a) V, T1, v1, v2, v3. Suppose molar maes of O2, N2, CO2 are M1, M2 and M3. Then PV =

nRT PV = (v1 + v2 + v3) RT VRTP 321 vvv Ans

(b) Total maes of mixture: M = v1M1 + v2M2 + v3M3 molar mass of mixuture = moleof.nototalmassTotal

321

332211 MMMMvvv

vvv

Ans

Q: 2.6: Here v)1('v)1'( ----(i) F.B.D. of piston : P2-P1 = mg/A Similar P21-P1

1 = mg/A

now 01 RTvP vRT

P 01 V

RTP 02

11'v

RT'P'P:Similar11V

RTPP 120

12

Since A/mg'P'PPP 1212 then

'n11

'VRT

n11

VRT0

'n)1'n()1'n(

V)1(RT

'n1'n

'VRT

n1n

VRT0

02

2

T1'n

1n'nT

Ans

n = 1T0

Tn = 1

V

T0n = 1

Tn = 1

V V'

n' V'

FinalInitial

P1A

P2A mgP2 = P + mg/A1

161

Q: 2.7: This question is based on operation of an engine. In this engine, first, piston pull right sideand during pulling piston, value is opened and gas is filled in vacent space. then value is closed andgas between value and piston is removed. And piston moves left way. After that piston is again pullright way and value is opened and gas comes with piston. This process continues.

First stroke right PV = P1 (V + V) VVPVP1

------(i)

First stroke left : P1, V

Second stroke right )VV(PVP 21 22

222

)VV(

PVP)VV(PVV

PV

-------(ii)

In 3rd stroke : 33

3)VV(

PVP

In n th stroke :

P

)VV(

PVP n

nn

1

)VV(

Vn

n

ln

)VV(Vlnn

VV1ln

ln

VVVln

lnn Ans

Q: 2.8: Suppose at time t, pressure is P and in next dt timedP pressure is released. Then PV = (P+dP) (V-Cdt) PV =PV + VdP + CPdt + C(dP) (dt) -VdP = (CP + CdP) dt CP

+ CdP = CP -VdP = CPdt t

0

P

0PP

dP dtCV P + dP

< P because P is decreasing function hence dP = (-) ive.

tvcP/Pln 0

v

ct0ePP

Ans

V Piston

Valve

P, V

V, P0

Isothermal

Value piston


Q: 2.9: from Q: 2.8: vct

0 ePP

v

ct0

0 ePP

vctln

ln

cvt AnsAns

Q: 2.10: Let assume pressure inside gas chamber is P F.B.D. of piston. Force

equation : P0S+ P(S-S) + m0g + (m-m0) g = PS + P0(S-S) Smg

0PP

----(i) . After increasing T of tempreature, final temp. is T + T. since from

(i) pressure is constant, process will be isobaric. Then constTV

TTVV

TV

VV1

TTTT

VVTT Vinitial = Sl1 + (S - S) l2

Change in volume : Vinitial = Sl1 + (S -S) l2. If l2 increase, l2 decrease in

same amount. Then l2 l l2 -l Then Vinitial = Sl = V. Then SlVTT

----(i) Also, PV = nRT . Here n =1 RTVSmgP0

Now

SmgP

R1

VT

0 put in (i)

SmgP

R)S(T 0l

R0 )SPmg(T l AnsAns

Q: 2.11: (a) P = P0-V2 -------(i) No. of mole of gas = 1. We know PV = nRT V

RTP put in

(i) 20 VPVRT

RV

RVPT

30 ------(iii). For T maximum, 0dV

dT 0

RV3

RP0

20

RPV 0 put in (ii):

3P

3P

RRP

RP

T 0000max

3P

3P2T 00max AnsAns

(b) P = P0 e-V V0ePV

RT V0 VeRP

T -----(i). Now 0dVdT

for maximum Temp. Calculate

V and put in (i) ReP

T 0max Ans

Q: 2.12: 20 VTT ------(i) We know PV = nRT = RTT. If v is increasing, T will increase and hence

P will increase. Hence calculation of Pmin: RPVT put in ----(i) 20 VTR

PV ------------(ii)

P0S

l1

m0

l2

P

P0 S- S

m-m0

m0g

(m-m0) g P )0 (S- S

P ) (S- S

PS

P0 S

163

VVTRP 10 To 0

dVdP:Pmin

020

TVVTR0 put in (ii)

00

00 T2

TT

TRP

0

T2R

P

0min TR2P

Ans

Q: 2.13: We know PM = RT (dP) M = R(dT) -(gdx) M

= R dT RgM

dxdT

RMg

dhdT

dxdT

Ans

Q: 2.14: Here CconstPn

n

1

CPP n

Also PM =

RT RTCPPM

n1

MRTCP n

1n

11

Differentiate :

dTMRc)dP(P1 n

1n

1

n1

-----(i) Also we know dP = - gdh put in

(i) dTMR)gdh(cP

n1n 21n/1

MgcP

n)1n(

dhdT 2121

MgCPn

)1n( 2121

n1

n2

1 PMgPn

1n

Mgn

1n

Mgn

)1n(dhdT

Ans

Q: 2.15: We know gdzdPgdzdP Also PM = RT

gdzdPRTPM In question dz = dh dPgdh

RTPM

h

0

P

0PP

dP dhRTMg

RTMgh

0ePP

. Above earth surface :

kg1078MK300C27T

3.8Rs/m8.9gm105h

3

0

2

3

. Put in above equation

Atm5.0m/N105.0P 25

N2 = const

dx

x

g

dx

x

s

T = constM = constdh

x


Below earthsurface : kg1078n

K300C27T

3.8Rs/m8.9gm105h

3

0

2

3

put Atm2m/N102P 25

Q: 2.16: RT)d(M)dP(RTPM ----------(i) Also we know

: gdzdP ---------(ii) then from (i) and (ii): gdzM

RT)d(

0

h

0

d dzRTMg

hz

dxdz

RTMgh

e0

(a) e0 Then RT

Mgh

0 e0e

RTMghee 1RT

Mgh1

Mg

RTh1 AnsAns

(b)

0

0 )1(0 RT

2Mgh

e)1( 00

MgRTh

MgRT)1(ln

MgRTh 22

Q: 2.17: Suppose pressure at height x is P then in (dx) volumePV = nRT in differential volume Psdx = (dN) RT We Know

MmN RTM

dmdxPs

-----(i)

MdmdN . Also we known

gdxdP Sdxdm

dVdm

dxgSxdmdP

gSdPdm

sdmgdP

-----(ii) put i (i)

P

0PP

dPx

0dx

RTMg

gSdP

MRTdxPS

RTxMg

0ePP

from (ii) : P

0Pg

SdPm

0dm

RTxMg

00 e1g

SPPP

gSm

when x = h

RTxMg

0 e1gSP

m Ans

T = constgM = const = const

dh

h

dxh

s

x

M, T, g

P0

165

Q: 2.18: We know

dmxdm

xcm . Also we know from Q: 1.215

RTMgx0ePP

And RTMdm)sdx(P

dxePRTMSdxPS

RTMdm RTMgx0

dxe

dxex

e

ex

0

RTMgx0

RTMgx

RTMgxRT

0MSP

RTMgxRT

x0MSP

cm

MgRTxcm

Q: 2.19: (a) T = T0 (1-ah) RTMdmPsdx -----(i) And also

)dm(sgdxg

sdxdmdxgdP

)dp(g

sdm -------(i)

dP

gs

MRTdxPs ax1(TTAlso 0

P

0PP

dPx

0)ax1(0TR

Mgdx 0

x

00P/Pln)ax1(ln

)a(RTMg

0

0P/Pln)ax1(ln

RTaMg

P = P0 (1-ax) Assume : hxandnaRT/Mg 0 P = P0 (1-ah)n Ans

(b) Similiar as above : n0

)ah1(PP

Ans

Q: 2.20: Force equation of dr element 2rw)dm(dF 2rw

Sdm

SdPdP

2rwS

dmdP

.

dPrw

sdm2

also we know RTM

dm)Sdr(P

dPrw

SMRT)dr(PS 2

P

0PP

dPr

0

2 RTrdrMw

0PP

22lnRT

2rMw

RT22r2MW

0 ePP Ans

dx

s

M, T, g

P0

x

dx

s

M, g

P = P0 at h = 0

x

P0 r

M, T

s

This end isopen in air

dx


Q: 2.21: T,?P

M = (12+32) 103 kg = 44 103 kg. Idial gas equation : 3333

mkg

m10kg500 500

We know PM = RT atm2801044

3002.8500MRTP 3

Vanderwall equation : (for one

mole gas) : RT)bV(V

aP 2

V

M

M2

2 VRTbPM

MaPP

2

2

Ma

bM

RTP

2

2

Ma

bMRTP

Ans

Q: 2.22: (a) Using idial gas equation : VRTPRTPV 1 . Using vander wall

equation : RT)bv(v

aP 2

22 V

abV

RTP

. Here

2

21P

PP

)1(PP 21 )1(V

abV

RTV

RT2

2V

a)1(bVR)1(

VRT

)vbVnV(RV

V)bV(a)1(

VR

bVR1

v/a)1(T 2

2

)bnv(VR

)1)(bV(aT

Ans

(b) Put value of this temp in vanderwall equation. 22 V

abV

RTP

Ans

Q: 2.23: 2221 P,TP,T

V1n

121 RT)bV(V

aP

----(i) 222 RT)bV(V

aP

-

-----(ii) from (i) and (ii) : 1T2T1P2T2P1T2Va

Ans 1P2P1T2TRVb Ans

V

n=1

167

Q: 2.24: We know bulk modulus of a gas is given by : V

dvdPB . While compressibility is given

by :

dPdV

V1

dPV

dv in vanderwall equation : RT)bv(

v

aP 2

if process is

isothermal: T = const. differentiate (i): 0)dV(V

aP)bv(dVV

adP 23

0V

aP)bV(V

adVdP

23

)bV(V

)bV(aaVPV

Va

)bV(

P

dVdP

3

3

3

2Va

)Vb(a)aVPV()bV(V

)bV(a)avPV()bV(V

V1x 3

2

3

3

----(ii) put value of P from (i) in (ii)

])bv(a2RTV[)bv(V

23

22

Ans

Q: 2.25: Using ideal gas : PV = RT; V= RT P-1 2PRTPV

RTV

PVRT1x 21 from Q :

2.24: 23

22

)vV(a2RTV

)bV(Vx

According to equation : x > x1 RT

V

)vV(a2RTV

)bV(V23

22

Rb

aT

2.2 The first law of thermodynamics, Heat capacityQ: 2.26: We know that room is an open thermodynamic system, in whichno. of molecules may change suppose no. of mole of gas is n. Then PV

= nRT. Also we know Internal energy is given by: 1

PV1

nRTTnCU V

vv .

Here v

pcC

constant and room pressure and volume is also constant. The Internal energy also const.

U = Const if P and V are const. And not depend on T because when T increase, no. of moles willbe decreased in room because PV = Const. Hence in room internal energy (U) = Const.

VP

U

Win

dow


Q: 2.27: Suppose no. of moles of gas = n. Directional kinetic energy of

gas = 2V)nM(21

. When Vessel sudden stop, then after long time this directional

kinetic energy of gas is converted into randm kinetic energy when thermodynamic

equililnium will be achieved and then TnCV)nM(21

V2

R)1(

2MVT

2

v 2MVR2

)1(T v

Ans

Q: 2.28: Method 1: When value is opened and thermodynamics equilibrium is attained then. no.

of moles will be constant. Then

RTVVP

RTVP

RTVP 21

2

22

1

11

TVVP

TVP

TVP 21

2

22

1

11 ------(i). Also

we know, in whole system : WU0Q 0Q Because vessell is insulated. And also. 0W Because gas does not work on atmosphere. Because vessel closed. Then U System

= 0 then 22112v21v1 TnTnTCnTcn 0TTTVPTT

TVP

22

221

1

11 ------(ii)

122211

112221TVPTVPVPVPTTT

put in (i) Ans

21

2211VV

VPVPP

Ans

Method : 2 From method : 1, we see that therewill be no change of internal inergy of system. HenceInitial Internal Energy = final internal energy.

Also 1PVU

v then

1VVP

1VP

1VP 212211

vvv 21

2211VV

VPVPP

. Also we know PV = nRTT

RTRT

VPRT

VPVVVV

VPVP

2

22

1

1121

21

2211

122211

212211TVPTVPTTVPVPT

Ans

Q: 2.29: Method: 1: We know Suppose initial temp. is T1 and final is (T1-T) then.

1rVPU;

1VPU 2f1i

v

12 PP1rVU

----( i). 11 nRTVP VnRTP 11

1

1T

VPnRalso VnRTP 22

V

nRTV

nRT1rTVU 12 T1

nR

v

M

V

PCC

V

Vessel (1) (2)

Value111 T,P,V 222 T,P,V

TP

V,V 21

111 T,P,V 222 T,P,V21 VV

,T,P

Sealed vessel

V T

169

0PHere)1(T

TVPU 11

1

v TT1 = T0 )1(T

TVPU0

0

v Ans U = Increase in P.E. Q = U

+ w Since vessel is realed then w = 0 )1(TTVPUQ

0

0

v Ans

Method : 2 T)1(RnTncU v

v ----(i) Here P = P0 P0V = nR T0 V = v 0

0T

VPnR put

in (i) )1(TTVPU

0

0

v Ans Also Q = U + w w = 0 then )1(T

TVPUQ

0

0

Ans

Q: 2.30: We know : Q = U + W = A1VP

v . Also W = PU + AA

A1AQ

v

1AQ

vv

Q: 2.31: Isobaric process: We know Q =

irodov 106-235.pdf

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