P1-Allens Made Physics Theory {PART-2}

JEE-Physics • Absolute  Pressure    : Sum  of  atmospheric  and    Gauge  pressure  is  called  absolute  pressure. P  =  P  +  Pgauge P  =  P  +  hg abs atm abs o The  pressure  which  we  measure  in  our  automobile  tyres  is  gauge  pressure. VARIATION  OF  PRESSURE  WITH  DEPTH y1 P1 y2 LA (i) Let  pressure  at  L  is  P   and  pressure  at  M  is  P 12 Then,    P A  =  P   A  +  gA  (y   –  y )   P   =  P   +  g(y   –  y ) 21 2 1     21 21 MA dP P2 Here  pressure  gradient  dy   =  g (ii) Pressure  is  same  at  two  points  in  the  same  horizontal  level. As  body  is  in  equilibrium,  P A  =  P A    P   =  P L M 12 12 P1 P2 Note  :  Pressure  P  is  independent  of  shape  of  container A A Example   Assuming  that  the  atmosphere  has  a  uniform  density  of  (1.3  kg/m3)  and  an  effective  height  of  10  km,  find  the force  exerted  on  an  area  of  dimensions  100  m  ×  80  m  at  the  bottom  of  the  atmosphere  . Solution F  =  PA  =  gh A  =  (1.3)  (9.8)  (104)  (100  ×  80)  =  12.74  ×  8×  107  N  =  1.0192  ×  109  N PRESSURE  EXERTED  BY  A  LIQUID  (EFFECT  OF  GR AVITY)  : Consider  a  vessel  containing  liquid.    As  the  liquid  is  in  equilibrium,  so  every  volume  element  of  the  fluid  is  also  in equilibrium.    Consider  one  volume  element  in  the  form  of  a  cylindrical  column  of  liquid  of  height  h  and  of  area  of  cross section  A.  The  various  forces  acting  on  the  cylindrical  column  of  liquid  are  : (i) Force,  F   =  P   A  acting  vertically  downward  on  the  top  face  of  the  column.    P   is  the  pressure  of  the  liquid  on 11 1 the  top  face  of  the  column  and  is  known  as  atmospheric  pressure. (ii)   Force,  F   =  P   A  acting  vertically  upward  at  the  bottom  face  of  the  cylindrical  column.    P   is  the  pressure  of  the 22 2 liquid  on  the  bottom  face  of  the  column. (iii) Weight,  W  =  mg  of  the  cylindrical  column  of  the  liquid  acting  vertically  downward.    Since  the  cylindrical  column node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 of  the  liquid  is  in  equilibrium,  so  the  net  force  acting  on  the  column  is  zero.  i.e.  F   +  W  –  F   =0 12 mg F1  P A  +mg  –  P A  =  0  P A  +  mg  =  P   A P    =   P +   ...(i) 1 21 A2 2 1   Now,  mass  of  the  cylindrical  column  of  the  liquid  is, h m  =  volume  ×  density  of  the  liquid  =    Area  of  cross  section  ×  height  ×  density  =  Ah mg     equation  (i)  becomes  P   =P   + A h g ,  P   =  P   +  hg  ...(ii) F2 21 A 21 P   is  the  absolute  pressure  at  depth  h  below  the  free  surface  of  the  liquid.    Equation  (ii),  shows  that  the  absolute 2 pressure  at  depth  h  is  greater  than  the  atmospheric  pressure  (P )  by  an  amount  equal  to  hg. 1 Equation  (ii)  can  also  be  written  as  (P –P )  =  hg    which  is  the  difference  of  pressure  between  two  points 21 separated  by  a  depth  h. 28 E

PRESSURE  IN  CASE  OF  ACCELER ATING  FLUID JEE-Physics (i) Liquid  placed  in  elevator    : a0 When  elevator  accelerates  upward  with  acceleration  a   thenpressure  in  the h 0 fluid,  at  depth  'h'  may  be  given  by, P  hg  a0  (ii) Free  surface  of  liquid  in  case  of  horizontal  acceleration  : h1 h2  a0 tan   ma0 = a0  1 2       mg g If  P  and  P  are  pressures  at  point  1  &  2  then    P –P   =  g  (h –  h )  =  gtan  =  a0 1 2 12 1 2 (iii) Rotating  Ve ssel Consider  a  cylindrical  vessel,  rotating  at  constant  angular  velocity  about  its  axis.  If  it  contains  fluid  then  after  an initial  irregular  shape,  it  will  rotate  with  the  tank  as  a  rigid  body.  The  acceleration  of  fluid  particles  located  at a  distance  r  from  the  axis  of  rotation  will  be  equal  to  2r,  and  the  direction  of  the  acceleration  is  toward  the  axis of  rotation  as  shown  in  the  figure.  The  fluid  particles  will  be  undergoing  circular  motion. Lets  consider  a  small  horizontal  cylinder  of  length  dr  and  cross-sectional  area  A  located  y  below  the  free  surface of  the  fluid  and  r  from  the  axis.  This  cylinder  is  accelerating  in  ground  frame  with  acceleration  2r  towards  the axis  hence  the  net  horizontal  force  acting  on  it  should  be  equal  to  the  product  of  mass  (dm)  and  acceleration. dm  =  Adr P A  –  P A  =  (Adr)2r 2 ! If  we  say  that  the  left  face  of  the  cylinder  is  y  below  the  free  surface  of  the  fluid  then  the  right  surface  is  y  +  dy below  the  surface  of  liquid.  Thus P  –  P  =  gdy 2 1 Thus  solving  we  get dy r2  dr g and,  therefore,  the  equation  for  surfaces  of  constant  pressure  is    y  2 r 2   +  constant 2g This  equation  means  that  these  surface  of  constant  pressure  are  parabolic  as  shown  in  figure. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 z axis of p1 p1 rotation p2 p2 p3 constant p4 p3  r2 2 pressure p4 lines r          r 2g ar=r 2  y x The  pressure  varies  with  the  distance  from  the  axis  of  rotation,  but  at  a  fixed  radius,  the  pressure  varies hydrostatically  in  the  vertical  direction  as  shown  in  figure. E 29

JEE-Physics Example An  open  water  tanker  moving  on  a  horizontal  straight  road  has  a  cubical  block  of  cork  floating  over  its  surface. If  the  tanker  has  an  acceleration  of  'a'  as  shown,  the  acceleration  of  the  cork  w.r.t.  container  is  (ignore  viscosity) a \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ Solution N arel ma mgsinmacos mgcos  masin mg marel  =  mgsin  –  macos    but  for  water  surface  tan    =  a/g      arel  =  0 Example An  open  rectangular  tank  1.5  m  wide,  2  m  deep  and  2m  long  is  half  filled  with  water.  It  is  accelerated horizontally  at  3.27  m/s2  in  the  direction  of  its  length.  Determine  the  depth  of  water  at  each  end  of  tank. [g  =  9.81  m/s2]  Solution 3 m a1  3.27 m/s2 Here  tan   g  3 ,  depth  at  corner  'A'  =  1–  1.5    tan     =  0.5  m depth  at  corner  'B'  =  1  +  1.5  tan     =  1.5  m 1m A B MEASUREMENT  OF  ATMOSPHERIC  PRESSURE node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 1. Mercury  Barometer  : Torricelli vaccum To  measure  the  atmospheric  pre ssure  experimental ly, hA torricelli  invented  a  mercury  barometer  in  1643. Mercury p  =h g Trough a E The  pressure  exerted  by  a  mercury  column  of  1mm  high  is called  1  Torr.                     1  Torr  =  1  mm  of  mercury  column 30

JEE-Physics OPEN  TUBE  MANOMETER  : Open-tube  manometer  is  used  to  measure  the  pressure  gauge.  When  equilibrium  is  reached,  the  pressure  at the  bottom  of  left  limb  is  equal  to  the  pressure  at  the  bottom  of  right  limb. i.e.  p  +  y  g  =  p  +  y  g pa 1 a 2 p2+y2 g p  –  p  =  g  (y   –  y )  =  gy y=y2-y1 a 2 1 y2 p  –  p  =  g  (y   –  y )  =  gy p y1 a 2 1 p+y1 g p  =  absolute  pressure,  p  –  p   =  gauge  pressure. a Thus,  knowing  y  and    (density  of  liquid),  we  can  measure  the  gauge  pressure. Example The  manometer  shown  below  is  used  to  measure  the  difference  in  water  level  between  the  two  tanks.  Calculate this  difference  for  the  conditions  indicated Liquid (sp gravity = 0.9) 40cm Water Water Solution p   +  h   g  –  40 g  +  40g  =  p   +  h   g pa pa a1 1 a2 h2 h1 h  g  –  h  g  =  40  g  –  40  1g    as  1  =  0.9 Water 40cm Water 2 1 (h2 –  h1)  g  =  40g  –  36g h   –  h   =  4  cm 21 WATER  BAROMETER Let  us  suppose  water  is  used  in  the  barometer  instead  of  mercury.    hg  =  1.013  ×  105    or  h  =  1.013  105 g The  height  of  the  water  column  in  the  tube  will  be  10.3  m.  Such  a  long  tube  cannot  be  managed  easily,  thus water  barometer  is  not  feasible. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 Example y=25cm d2 p In  a  given  U-tube  (open  at  one-end)  find  out  relation  between  p  and  p . x=26cm Pa a y d2 d1 A Given  d2  =  2  ×  13.6  gm/cm3  d1  =  13.6  gm/cm3 P Solution x Pressure  in  a  liquid  at  same  level  is  same  i.e.  at  A  –  A–, A d1 pa  d2yg  xd1g  p In  C.G.S. p   +  13.6  ×  2  ×  25  ×  g  +  13.6  ×  26  ×  g  =  p a E p   +  13.6  ×  g  [50  +  26]  =  p a  2p   =  p  as  p  =  13.6  ×  g  ×  76 a a 31

JEE-Physics Patm Example Find  out  pressure  at  points  A  and  B.  Also  find  angle  ‘’. A h Patm B Solution PA  =  Patm  –  1  g  sin   Pressure  at  A  – Pressure  at  B P  =  P  +  2  gh   But  P   is  also  equal  to B atm B P  =  P  +  3  g  sin   B A Hence  - P  +  2  gh  =  P  +  3  g  sin   atm A P  +   gh = P  –   g sin  +   g sin atm 2 atm 1 3 sin   2 h ) . (3  1 Example a In  the  given  figure,  the  container  slides  down  with  acceleration  ‘a’  on  an  incline  of  angle  ‘’.  Liquid  is  stationary  with  respect  to  container.  Find  out (i)  Angle  made  by  surface  of  liquid  with  horizontal  plane. (ii)  Angle  if  a  =  g  sin  . Solution Consider  a  fluid  particle  on  surface.  The  forces  acting  on  it  are  shown  in  figure. ma sin ma a ma  Normal to liquid surface mg         ma cos  Normal to  liquid surface   =   ma cos liquid surface  (mg-ma sin liquid surface node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 mg ma cos  a cos  Resultant  force  acting  on  liquid  surface,  will  always  normal  to  it    tan    =  mg  ma sin    =  (g  a sin ) a cos  Thus  angle  of  liquid  surface  with  the  horizontal  is  equal  to      =  tan–1  (g  a sin )  a cos   g sin cos  (ii) If a  =  g sin ,  then     =  tan–1   g  g sin2  =  tan–1  g cos2                          =  tan–1  (tan  )     =  32 E

JEE-Physics Example An  L  shaped  glass  tube  is  kept  inside  a  bus  that  is  moving  with  constant  acceleration.  During  the  motion,  the level  of  the  liquid  in  the  left  arm  is  at  12  cm  whereas  in  the  right  arm,  it  is  at  8  cm  when  the  orientation  of  the tube  is  as  shown.  Assuming  that  the  diameter  of  the  tube  is  much  smaller  than  levels  of  the  liquid  and  neglecting effect  of  surface  tension,  acceleration  of  the  bus  find  the  (g  =  10  m/s2). 12cm 8cm 45° Solution tan   a  h2 h2  h1 tan 45  4 cm       a  =  2  m/s2 g tan 45  h1 20 cm FORCE  ON  SIDE  WALL  OF  VESSEL Force  on  the  side  wall  of  the  vessel  can  not  be  directly  determined  as  a  different  depths  pressures  are  different. To  find  this  we  consider  a  strip  of  width  dx  at  a  depth  x  from  the  surface  of  the  liquid  as  shown  in  figure,  and on  this  strip  the  force  due  to  the  liquid  is  given  as  :        dF  =  xg  ×  bdx h x dx a dF b This  force  is  acting  in  the  direction  normal  to  the  side  wall.  Net  force  can  be  evaluated  by  integrating  equation node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 h g b h 2 F F   dF   xgbdx 2 0 Average  pressure  on  side  wall The  absolute  pressure  on  the  side  wall  cannot  be  evaluated  because  at  different  depths  on  this  wall  pressure  is different.  The  average  pressure  on  the  wall  can  be  given  as  : F 1 gbh2 1 P av  bh  2  gh bh 2 Equation  shows  that  the  average  pressure  on  side  vertical  wall  is  half  of  the  net  pressure  at  the  bottom  of  the vessel. E 33

JEE-Physics TORQUE  ON  THE  SIDE  WALL  DUE  TO  FLUID  PRESSURE As  shown  in  figure,  due  to  the  force  dF,  the  side  wall  experiences  a  torque  about  the  bottom  edge  of  the  side which  is  given  as d  dF  h  x   xgb dx h  x  h This  net  force  is    d   gb hx  x2  dx 0     h3  h3   1 g b h 3 gb  2 3  6 Example Water  and  liquid  is  filled  up  behind  a  square  wall  of  side  .  Find  out A water  h1=5m square wall liquid 2 side =10m B h2=5m C (a)  Pressures    at  A,  B  and  C (b)  Forces  in  part  AB  and  BC (c)  Total  force  and  point  of  application  of  force.  (Neglect  atmosphere  pressure  in  every  calculation) Solution (a) As  there  is  no  liquid  above  ‘A’ A dx x So  pressure  at  A,  p   =  0 A B Pressure  at  B,  p  =  gh1 B Pressure  at  C,  p   =  gh   +  2gh C1 2 (b) Force  at  A  =  0 Take  a  strip  of  width  ‘dx’  at  a  depth  ‘x’  in  part  AB. C Pressure  is  equal  to  gx. Force  on  strip  =  pressure  ×  area dF  =  gx  dx h1  gxh12  1000 10 10  5  5  1.25  106 N  gxdx Total  force  upto  B :  F 02 2 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 In  part  BC  for  force  take  a  elementary  strip  of  width  dx  in  portion  BC.  Pressure  is  equal  to  = gh1  +  2g(x  –  h) 1 Force  on  elementary  strip  =  pressure  ×  area dF  =  [gh1  +  2g(x  –  h )]    dx 1 Total  force  on  part  BC F   [ gh1  2 g (x  h1 )] dx A h1  x2  h1 x  B dx =   g h 1 x  2 g  2  h1x    h1 =  gh1h2    +  2g   2  h12  h 1   h 2   2 1    C 34 E

JEE-Physics =  gh1h2  +  2g   [2  +  h12  –  2h1]  =  gh1h2  +g  (  –  h1)2 2 =  gh2  [h   +  h ]  =  gh22  =  1000  ×  10  ×  5  ×  10  ×  10  =  5  ×  106  N 1 2 (c) Total  force =  5  ×  106  +  1.25  ×  106    =  6.25  ×  106  N Taking  torque  about  A h1 Total  torque  of  force  in  AB      =   dF  x   gxdx.x 0                  gx3 h1  gh13  1000 10 10 125 1.25  107 N m   3 3   3 0 3 Total  torque  of  force  in  BC      =  dF  x On  solving  we  get         =  gh1h2[h1  +  h2 ]  +  gh22[h1  +  2h2 ] 2 3 10                               =  1000  ×  10  ×  5  ×  5  ×  10  [5  +  2.5]  +  1000  ×  10  ×  25  ×  10  [5  +  ] 3                             =  2.5  ×  7.5 ×  106 +  62.5 118.75   ×  106   ×  106 = 33 Total  torque                               =  11.875  107 1.25  107 13.125  107 += 33 3 Total  torque  =  total  force  ×  distance  of  point  of  application  of  force  from  top  =  F.x p 6.25  ×  106  x   =  13.125  107 p3 x   =  7m p PASCAL's  LAW If  the  pressure  in  a  liquid  is  changed  at  a  particular  point,  the  change  is  transmitted  to  the  entire  liquid  without being  diminished  in  magnitude.  Pascal's  law  is  stated  in  following  ways  – • The  pressure  in  a  fluid  at  rest  is  same  at  all  the  points  if  gravity  is  ignored. • A  liquid  exerts  equal  pressures  in  all  directions. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 • If  the  pressure  in  an  enclosed  fluid  is  changed  at  a  particular  point,  the  change  is  transmitted  to  every  point of  the  fluid  and  to  the  walls  of  the  container  without  being  diminished  in  magnitude. Applications  of  pascal's  law  hydraulic  jacks,  lifts,  presses,  brakes,  etc For  the  hydraulic  lift F1                                   F1 Q A2 Pressure  applied  =  A1 A1 P F2     Pressure  transmitted  =  A2     F1  =  F2 A1 A2 F1 A2     Upward  force  on  A is  F   =    ×    A   =    ×  F 2  2 A1 2 A1 1 E 35

JEE-Physics Ex.   A  vertical  U–tube  of  uniform  cross–section  contains  mercury  in both  arms.  A  glycerine  (relative  density  =  1.3)  column  of  length  10 cm  is  introduced  into  one  of  the  arms.  Oil  of  density  800  kg  m–3  is C h poured  into  the  other  arm  until  the    upper  surface  of  the  oil  and 0.1 h glycerine  are  at  the  same  horizontal  level.  Find  the  length  of  the B column.  Density  of  mercury  is  13.6  ×  103  kgm–3 10cm S o l .   Pressure  at  A  and  B  must  be  same          Pressure  at  A  =  P0  +  0.1  ×  (1.3  ×  1000)  ×  g A Pressure  at  B  =  P   +  h  ×  800  ×  g  +  (0.1  –  h)  ×  13.6  ×  1000  g 0  0.1  ×  1300  =  800  h  +  (0.1  –  h)  ×  13600  h  =  0.096  m  =  9.6  cm BUOYANCY  AND  ARCHIMEDE'S  PRINCIPLE • Buoyant  Force    :  If  a  body  is  partially  or  wholly  immersed  in  a  fluid,  it  experiences  an  upward  force  due  to  the fluid  surrounding  it.  This  phenomenon  of  force  exerted  by  fluid  on  the  body  is  called  buoyancy  and  force  is called    buoyant  force  or  upthrust. • Archimede's  Principle    :  It  states  that  the  buoyant  force  on  a  body  that  is  partially  or  totally  immersed  in  a liquid  is  equal  to  the  weight  of  the  fluid  displaced  by  it. Now  consider  a  body  immersed  in  a  liquid  of  density  . Top  surface  of  the  body  experiences  a  downward  force F   =  AP   =  A[h1g  +  P] ...(i) F1 h1 1 1 0 h2 Lower  face  of  the  body  will    experiences  a  upward  force L F   =  AP   =  A[h2g  +  P] ...(ii) 2 2 0 As  h   >  h . So  F   is  greater  than  F F2 21 21 So  net  upward  force  :  F  =  F  –  F  =  Ag[h2  –  h] 2 1 1     F  = AgL =  Vg [  V = AL] F L OATAT I O N When  a  body  of  density  ()  and  volume  (V)  is  completely    immersed  in  a  liquid  of  density  (),  the  forces  acting  on the  body  are  : (i) Weight  of  the  body  W  =  Mg  =  Vg  (directed  vertically  downwards  through  C.G.  of  the  body). (ii) Buoyant  force  or  Upthrust  Th  =  Vg  (directed  vertically  upwards  through  C.B.). The  apparent  weight  W   is  equal  to  W  –  Th. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 App The  following  three  cases  are  possible  : Case  I Density  of  the  body  ()  is  greater  than  that  of  liquid  ) In  this  case if>thenW  > Th So  the  body  will  sink  to  the  bottom  of  the  liquid. Case  II WApp = W –  Th =  Vg  –  Vg =  Vg  (1 – /)  =  W (1 – /). Density  of  the  body  is  equal  to  the  density  of  liquid  ( =  ) In  this  case if =then W = Th So  the  body  will  float  fully  submerged  in  the  liquid.  It  will  be  in  neutral  equilibrium. W   =  W  –  Th  =  0 App 36 E

JEE-Physics Case  III Density  of  the  body  is  lesser  than  that  of  liquid  ( <  ) In  this  case if<then  W  < Th So  the  body  will  float  partially  submerged  in  the  liquid.  In  this  case  the  body  will  move  up  and  the  volume  of liquid  displaced  by  the  body  (V )  will  be  less  than  the  volume  of  body  (V).  So  as  to  make  Th    equal  to  W in WApp  =  W  –  Th  =  0 The  above  three  cases  constitute  the  law  of  flotation  which  states  that  a  body  will  float  in  a  liquid  if  weight  of the  liquid  displaced  by  the  immersed  part  of  the  body  is  at  least  equal  to  the  weight  of  the  body. Rotatory  –  Equilibrium  in  Floatation  : When  a  floating  body  is  slightly  tilted  from  equilibrium  position,  the  centre  of  buoyancy  B  shifts.  The  vertical line  passing  through  the  new  centre  of  buoyancy  B'  and  initial  vertical  line  meet  at  a  point  M  called  meta  – centre.  If  the  metacentre  M  is  above  the  centre  of  gravity  the  couple  due  to  forces  at  G  (weight  of  body  W)  and at  B'  (upthrust)  tends  to  bring  the  body  back  to  its  original  position  (figure)  .  So  for  rotational  equilibrium  of floating  body  the  meta–centre  must  always  be  higher  than  the  centre    of  gravity  of  the  body. G Th M B M Th (A) B’ MG BW G B’ (B) W (C) However,  if  meta–centre  goes  below  centre  of  gravity,  the  couple  due  to  forces  at  G  and  B'  tends  to  topple  the floating  body.    This  is  why  a  wooden  log  cannot  be  made  to  float  vertical  in  water  or  a  boat  is  likely  to  capsize if  the  sitting  passengers  stand  on  it.  In  these  situations  centre  of  gravity  becomes  higher  than  meta  centre  and so  the  body  will  topple  if  slightly  tilted. Example A  cubical  block  of  wood  of  edge  3  cm  floats  in  water.  The  lower  surface  of  the  cube  just touches  the  free  end  of  a  vertical  spring  fixed  at  the  bottom  of  the  pot.  Find  the  maximum weight  that  can  be  put  on  the  block  without  wetting  it.  Density  of  wood  =  800  kg/m3  and spring  constant  of  the  spring  =  50  N/m.  Take  g  =  10  m/s2 Solution node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 The  specific  gravity  of  the  block  =  0.8.  Hence  the  height  inside  water  =  3  cm  ×  0.8  =  2.4  cm.  The  height outside  water  =  3  cm  –  2.4  =  0.6  cm.  Suppose  the  maximum  weight  that  can  be  put  without  wetting  it  is  W. The  block  in  this  case  is  completely  immersed  in  the  water. The  volume  of  the  displaced  water=  volume  of  the  block  =  27  ×  10–6  m3. Hence,  the  force  of  buoyancy  =  (27  ×  10–6  m3)  ×  1(1000  kg/m3)  ×  (10  m/s2)    =  0.27  N. The  spring  is  compressed  by  0.6  cm  and  hence  the  upward  force  exerted  by  the  spring =  50  N/m  ×  0.6  cm  =  0.3  N. The  force  of  buoyancy  and  the  spring  force  taken  together  balance  the  weight  of  the  block  plus  the  weight  W put  on  the  block.  The  weight  of  the  block  is W =  (27  ×  10–6  m)  ×  (800  kg/m3)  ×  (10  m/s2)  =  0.22  N. Thus, W  =  0.27  N  +  0.3  N  –  0.22  N  =  0.35  N. E 37

JEE-Physics Example A  wooden  plank  of  length  1  m  and  uniform  cross-section  is  hinged  at  one  end  to  the  bottom  of  a  tank  as  shown in  figure. FC  B  A mg 0 The  tank  is  filled  with  water  up  to  a  height  of  0.5  m.  The  specific  gravity  of  the  plank  is  0.5.  Find  the  angle   that  the  plank  makes  with  the  vertical  in  the  equilibrium  position.    (Exclude  the  case    =  0). Solution The  forces  acting  on  the  plank  are  shown  in  the  figure.  The  height  of  water  level  is    =  0.5  m.  The  length  of  the plank  is  1.0  m  =  2.  The  weight  of  the  plank  acts  through  the  centre  B  of  the  plank.  We  have  OB  =  .  The buoyant  force  F  acts  through  the  point  A  which  is  the  middle  point  of  the  dipped  part  OC  of  the  plank. We  have OC  OA  =  2 =  2 cos  . Let  the  mass  per  unit  length  of  the  plank  be  .  Its  weight  mg  =  2g.   The  mass  of  the  part  OC  of  the  plank  =   cos  . 1 2 The  mass  of  water  displaced  =  0.5 cos    =  cos  . The  buoyant  force  F  is,  therefore,  F  =  2g . cos  Now,  for  equilibrium,  the  torque  of  mg  about  O  should  balance  the  torque  of  F  about  O. So,  mg  (OB)  sin  =  F(OA)  sin  or,  (2)  =   2        or,  cos2 = 1   or,  cos  =  1  cos   2 cos  2 2 ,  or,  =  45°. GOLDEN  KEY  POINTS  • Buoyant  force  act  vertically  upward  through  the  centre  of  gravity  (C.G.)  of  the  displaced  fluid.  This  point  is called  centre  of  buoyancy  (C.B.).  Thus  centre  of  buoyancy  is  the  point  through  which  the  force  of  buoyancy  is supposed  to  act.  • Buoyant  force  or  upthrust  does  not  depend  upon  the  characteristics  of  the  body  such  as  its  mass,  size,  density, node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 etc.  But  it  depends  upon  the  volume  of  the  body  inside  the  liquid.  • It  depends  upon  the  nature  of  the  fluid  as  it  is  proportional  to  the  density  of  the  fluid. This  is  the  reason  that  upthrust  on  a  fully  submerged  body  is  more  in  sea  water  than  in  pure  water  • It  depends  upon  the  effective  acceleration. If  a  lift  is  accelerated  downwards  with  acceleration  a  (a<  g)  then Th  =  V   (g  –  a) in If  a  lift  is  accelerated  downwards  with  a  =  g  then Th  =  V   (g  –  a)  =  0 in If  a  lift  is  accelerated  upward  with  accelaration  a  then Th  =  V   (g  +  a) in  • If  a  body  is  weighed  in  air  (W ),  in  water  (W )  and  in  a  oil  (W ),  then AW O loss of weight in oil WA  WO E Specific  gravity  of  oil  =  loss of weight in water =  WA  WW 38

JEE-Physics Example A  body  weighs  160  g  in  air,  130  g  in  water  and  136  g  in  oil.  What  is  the  specific  gravity  of  oil? Solution loss of weight in oil 160 – 136 24 8 Specific  gravity  of  oil  =  loss of weight in water =  160 – 130 = 30 = 10   =  0.8 Example An  iceberg  is  floating  partially  immersed  in  sea–water.  The  density  of  sea–water  is  1.03  gm/cm3  and  that  of ice  is  0.92  gm/cm3.  What  is  the  fraction  of  the  total  volume  of  the  iceberg  above  the  level  of  sea–water  ? Solution In  case  of  flotation  weight  =  upthrust  i.e. mg = V g  Vg =  V g V  =  V  so  V  = V –  V  =   V 1   in in in  out in     f fout  Vout  1    1  0.92   0.11  0.106    f =  10.6  % V   1.03  1.03 out  Example   A  rubber  ball  of  mass  10  gm  and  volume  15  cm3  is  dipped  in  water  to  a  depth  of  10m.  Assuming  density  of water  uniform  throughout  the  depth  if  it  is  released  from  rest.  Find    (take  g  =  980  cm/s2) (a)  the  acceleration  of  the  ball,  and (b)  the  time  taken  by  it  to  reach  the  surface. Solution The  maximum  buoyant  force  on  the  ball  is  F   =  Vw g    15  ×  1×    980  dyne  =  14700  dyne. B The  weight  of  the  ball  is  mg  =  10  ×    g  =  10  ×    980  =  9800  dyne The  net  upward  force,  F  =  (15  ×  980  –  10  ×    980)  dyne  =  5  ×    980  dyne  =  4900  dyne (a)  Therefore,  acceleration  of  the  ball  upward  a  F  5  980   =  490  cm/s2  =  4.9  m/s2 m 10 (b)  Time  taken  by  it  reach  the  surface  is  t  2h 2 10  s  =  2.02  s a 4.9 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 E 39

JEE-Physics FLUID DYNAMICS When  a  fluid  moves  in  such  a  way  that  there  are  relative  motions  among  the  fluid  particles,  the  fluid  is  said  to  be flowing. TYPES  OF  FLUID  FLOW    :  Fluid  flow  can  be  classified  as  : l Steady  and  Unsteady  Flow Steady  flow  is  defined  as  that  type  of  flow  in  which  the  fluid  characteristics  like  velocity,  pressure  and  density  at a  point  do  not  change  with  time.  In  an  unsteady  flow,  the  velocity,  pressure  and  density  at  a  point  in  the  flow varies  with  time. l Streamline  Flow In  steady  flow  all  the  particles  passing  through  a  given  point  follow  the  same  path  and  hence  a  unique  line  of  flow. This  line  or  path  is  called  a  streamline.  Streamlines  do  not  intersect  each  other  because  if  they  intersect  each other  the  particle  can  move  in  either  direction  at  the  point  of  intersection  and  flow  cannot  be  steady. l Laminar  and  Turbulent  Flow Laminar  flow  is  the  flow  in  which  the  fluid  particles  move  along  well–defined  streamlines  which  are  straight  and parallel.  In  laminar  flow  the  velocities  at  different  points  in  the  fluid  may  have  different  magnitudes,  but  there directions  are  parallel.  Thus  the  particles  move  in  laminar  or  layers  gliding  smoothly  over  the  adjacent  layer. Turbulent  flow  is  an  irregular  flow  in  which  the  particles  can  move  in  zig–zag  way  due  to  which  eddies  formation take  place  which  are  responsible  for  high  energy  losses. l Compressible  and  Incompressible  Flow In  compressible  flow  the  density  of  fluid  varies  from  point  to  point  i.e.  the  density  is  not  constant  for  the  fluid whereas  in  incompressible  flow  the  density  of  the  fluid  remains  constant  throughout.  Liquids  are  generally incompressible  while  gases  are  compressible. l Rotational  and  Irrotational  Flow Rotational  flow  is  the  flow  in  which  the  fluid  particles  while  flowing  along  path–lines  also  rotate  about  their  own axis.  In  irrotational  flow  particles  do  not  rotate  about  their  axis.  So  they  have  no  net  angular  velocity. EQUATION  OF  CONTINUITY The  continuity  equation  is  the  mathematical  expression  of  the  law  of  conservation  of  mass  in  fluid  dynamics. A1 v1 A2 v2 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 v2 t v1 t In  the  steady  flow  the  mass  of  fluid  entering  into  a  tube  of  flow  in  a  particular  time  interval  is  equal  to  the  mass of    fluid  leaving  the  tube. m1  m2  1A1v1  = 2A2v2    (   1  =  2)    Av =  Av         Av =  constant t t 1 1  22   (Here   =  density  of  fluid,  v =  velocity  of  fluid,  A  =  Area  of  cross–section  of  tube) Therefore  the  velocity  of  liquid  is  smaller  in  the  wider  parts  of  a  tube  and  larger  in  the  narrower  parts. 40 E

JEE-Physics BERNOULLI'S  THEOREM Bernoulli's  equation  is  mathematical  expression  of  the  law  of  mechanical  energy  conservation  in  fluid  dynamics. Bernoullis  theorem  is  applied  to  the  ideal  fluids. Characteristics  of  an  ideal  fluid  are  : (i)    The  fluid  is  incompressible. (ii)  The  fluid  is  non–viscous. (iii)  The  fluid  flow  is  steady. (iv)  The  fluid  flow  is  irrotational. Every  point  in  an  ideal  fluid  flow  is  associated  with  three  kinds  of  energies  : Kinetic  Energy 1 If  a  liquid  of  mass  (m)  and  volume  (V)  is  flowing  with  velocity  (v)  then    Kinetic  Energy=    mv2and  kinetic  energy 2 Kinetic Energy 1 m 1 per  unit  volume  =   v2  =  v2 volume 2 V 2 Potential  Energy If  a  liquid  of  mass  (m)  and  volume  (V)  is  at  height  (h)  from  the  surface  of  the  earth  then  its Potential Energy m Potential  Energy  =  mgh    and  potential  energy  per  unit  volume  =    = V gh  =  gh volume Pressure  Energy If  P  is  the  pressure  on  area  A  of  a  liquid  and  the  liquid  moves  through  a  distance  ()  due  to  this  pressure  then Pressure  energy  =  Work  done  =  force  x  displacement  =  pressure  x  area  x  displacement  =  PA  =  PV [  A  =  volume  V] Pressure  energy  per  unit  volume  =  Pressure energy   =P volume Theorem According  to  Bernoulli's  Theorem  ,  in  case  of    steady  flow  of  incompressible  and  non–viscous  fluid  through  a tube  of  non–uniform  cross–section,  the  sum  of  the  pressure,  the  potential  energy  per  unit  volume  and  the kinetic  energy  per  unit  volume  is  same  at  every  point  in  the  tube,  i.e.,  P  +  gh  1 v2    constant. 2 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 C P2 A2 v2 B v1 h2 P1 A1 h1 Consider  a  liquid  flowing  steadily  through  a  tube  of  non–uniform  area    of  cross–section  as  shown  in  figure.  If  P 1 and  P  are  the  pressures  at  the  two  ends  of  the   tub e  resp ec tively,  work  done  in  p ushi ng  the  volume  V   of 2 incompressible  liquid  from  point  B  to  C  through  the  tube W  =  P  (V)  =  (P1–P2)V ...(i) This  work  is  used  by  the  liquid  in  two  ways  : (i) In  changing  the  potential  energy  of  mass  m  (in  the  volume  V)    from E m g h 1   to   mgh2  i.e.,  U=mg  (h –h ) ...(ii) 2 1 41

JEE-Physics 1 1  1 m 2 2 2  (ii) 2 2 2 2 In  changing  the  kinetic  energy  from    m v 1   to    m v 2   ,  i.e.  K v 2  v 1 Now  as  the  liquid  is  non–viscous,  by  conservation  of  mechanical  energy,  W  =  U+K  i.e.,  P1  h2  1 v 2 2  P2 V=mg  h1 + 2 m 2  v 1 P –P   =  g(h2–h1)  +   1  v 2  v 2   [as    m / V ] 12 2 2 1 P   +  gh1  1 v 2  P2  gh2  1 v 2      P  +  gh+ 1 v2=  constant 1 2 1 2 2 2 This  equation  is  the  Bernoulli's  equation  and  represents  conservation  of  mechanical  energy  in  case  of  moving fluids. VENTURIMETER It  is  a  gauge  put  on  a  flow  pipe  to  measure  the  speed  of  flow  of  a  liquid  shown  in  figure. v1  A1 v2 A P1 A2 P2 B h m Let  the  liquid  of  density    be  flowing  through  a  pipe  of  area  of  cross  section  A .  Let  A  be  the  area  of  cross 1 2 section  at  the  throat  and  a  manometer  is  attached  as  shown  in  the  figure.  Let  v   and  P   be  the  velocity  of  the 11 flow  and  pressure  at  point  A,  v   and  P   be  the  corresponding  quantities  at  point  B. 22 Using  Bernoulli’s  theorem  :                 P1  gh1  1 v 2  P2  gh2  1 v 2  2 1  2 2 we  get              P1  gh  1 v 2  P2  gh  1 v 2     (Since  h   =  h   =  h)  or  (P –P )= 1 (v22  –  v 2) ....(i)  2 1  2 2 12 12 2 1 According  to  continuity  equation,  A v   =  A v     or  v2   A1   v 11 22  A 2  1 Substituting  the  value  of  v   in  equation  (i) 2  P1  P2  1    A1 2 v 2  v 2  1 v 2  A1 2  node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 2  A2 1 1  1  A2 we  have   2  1    Since  A   >  A ,  therefore, P  >  P     or  v 2  2(P1  P2 )  2 A 2 ( P1  P2 ) 12 12 1 2  A1 2   ( A 2  A 2 )   A2  1 2  1  where  (P   –  P )  =  mgh  and  h  is  the  difference  in  heights  of  the  liquid  levels  in  the  two  tubes. 1 2 v1  2m gh   A1 2  A2  1   The  flow  rate  (R)  i.e.,  the  volume  of  the  liquid  flowing  per  second  is  given  by  R  =  v A . E 11 42

JEE-Physics • Tor ricelli's  Law  of  Efflux  (Fluid  Outflow) As  shown  in  the  figure  since  the  area  of  cross–section  at  A  is  very  large                as  compared  to  that  at  orifice  B,  speed  at  A  i.e.    vA   0.  Also  the  two fluid  particles  at  A  and  B  are  at  same  pressure  P (atmospheric  pressure). 0  Applying  Bernoulli's  theorem  at  A  and  B. P0  gH  1 v 2  P0  g(H  h)  1 v 2   1 v 2  =  gh vB =  2gh 2 A 2 B 2 B Equation  is  same  as  that  of  freely  falling  body  after  falling  through  h  height  and  is  known  as  Torricelli's  law. Writing  equation  of  uniformly  accelerated  motion  in  vertical  direction H  –  h  =  0  +  1 gt2  (from  s   =  u t  +  1 a t2)    t  2(H  h) 2 y y 2y g Horizontal  range R  =  v t  =  2gh   ×   2(H  h) x g   =  2 h(H  h) Range  R  will  be  maximum  when  R2  is  maximum.  i.e.,  d R2 dH   =  0    4  dh   (Hh–h2) =  0     H–2h  =  0,  i.e.,  h  =  2 dh   R  =  2 H H  H  =  H maximum 2 2  Example   A  cylindrical  tank  1  m  in  radius  rests  on  a  platform  5  m  high.  Initially  the  tank  is  filled  with  water  upto  a  height of  5  m.  A  plug  whose  area  is  10–4  m2  is  removed  from  an  orifice  on  the  side  of  the  tank  at  the  bottom. Calculate A 5m A0 5m node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 (a)  initial  speed  with  which  the  water  flows  from  the  orifice (b)  initial  speed  with  which  the  water  strikes  the  ground Solution (a)  Speed  of  efflux  vH  2gh = 2  10  5   10  m/s (b)      As  initial  vertical  velocity  of  water  is  zero,  so  its  vertical  velocity  when  it  hits  the  ground    v = 2gh = 2  10  5   10  m/s v So  the  initial  speed  with  which  water  strikes  the  ground,  v  = v 2  v 2 =102  =  14.1  m/s H V E 43

JEE-Physics Example container  B In  a  given  arrangement of area A h2 (a)  Find  out  velocity  of  water  coming  out  of  ‘C’ h h1 A h3 (b)  Find  out  pressure  at  A,  B  and  C. area of cross section a Solution C (a) Applying  Bernoulli’s  equation  between  liquid  surface  and  point  ‘C’. liquid  p   +  1 v 2   =  p  –  gh3  +  1 v 2 a 2 1 a 2 2 through  continuity  equation Av  =  av   ,  v  =  av2     1 a2 v 2   =  – gh   +  1 v 2 1 21 A 2 A2 2 3 2 2 v 2   = 2gh3 ,  v   =  2gh3 B 2 a2 2 a2 h2 h1 1  A2 1  A2 h v1 A (b) Pressure  at  A  just  outside  the  tube p  =  p  +  gh h3 A atm 1 For  pressure  at  B  : p  +  0  +  0  =  p +  gh2  + 1 vB2 A B  2 C v2 1 2gh3  p  = P  –  gh2  –   a2  BA 2  1  A 2  Pressure  at  C  : p =  p C  atm Example   A  tank  is  filled  with  a  liquid  upto  a  height  H.  A  small  hole  is  made  at  the  bottom  of  this  tank.  Let  t   be  the  time 1 taken  to  empty  first  half  of  the  tank  and  t   time  taken  to  empty  rest  half  of  the  tank,  then  find  t1 . 2 t2 Solution Let  at  some  instant  of  time  the  level  of  liquid  in  the  tank is  y.  Velocity  of  efflux  at  this  instant of  time  v  =  2gy . Now,  at  this  instant  volume  of  liquid  coming  out  the  hole  per  second  is   dV1   dt  Volume  of  liquid  coming  down  in  the  tank  per  second  is   dV2   dt  dV1   =  dV2  dy        a 2gy =A    dy    ...(i) node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 dt dt  av=A   dt   dt  (Here  area  of  cross-section  of  hole  and  tank  are  respectively  a  and  A) Substituting  the  proper  limits  in  equation  (i), A 2A H 2A    H t1 dt 2g a 2g 2  A g   a    0     H / 2 y 1 / 2 dy     t1  =     y  =  a  H   =    2  1  ...(ii) a H / 2 2g  H Similarly,  t2dt   a A   0 y 1 / 2 dy     t  =  A   H 0 2g a g ...(iii) H/2 2 t1 From  equations  (ii)  and  (iii),  t2   =  2–1  =  0.414 44 E

JEE-Physics Example A  fixed  container  of  height  ' H '  with  large  cross-sectional  area  ' A '  is  completely  filled  with  water.  Two  small        orifice  of  cross-sectional  area  ' a '  are  made,  one  at  the  bottom  and  the  other  on  the  vertical  side  of  the  container    at  a  distance  H/2  from  the  top  of  the  container.  Find  the  time  taken  by  the  water  level  to  reach  a  height  of  H/ 2  from  the  bottom  of  the  container. Solution v  =  2g (h  H / 2) ; v  =  2 g h      By continuity equation 1 2  dh  dh a v1  A    h H/2 d t    =  a  (v   +  v )    A  d t    =  a  2 g (h  H / 2)  2 g h H/2 1 2   A H/2 dh  t dt 2A H h 0 3a g or  a 2 g H h H /2       t  = 2 1 v2 a Example A  cylindrical  vessel  filled  with  water  upto  a  height  of  2  m  stands  on  a  horizontal  plane.  The  side  wall  of  the vessel  has  a  plugged  circular  hole  touching  the  bottom.  Find  the  minimum  diameter  of  the  hole  so  that  the vessel  begins  to  move  on  the  floor  if  the  plug  is  removed.  The  coefficient  of  friction  between  the  bottom  of  the vessel  and  the  plane  is  0.4  and  total  mass  of  water  plus  vessel  is  100  kg. Solution From  Torricelli's  theorem,  velocity  of  efflux  =  2gh Momentum  per  second  carried  by  water  stream =  density  ×    volume  coming  out  per  second  ×    velocity    =    ×  av  ×  v  =  av2 Hence  force  on  cylindrical  vessel  =  a2gh Cylinder  starts  to  move  when  reaction  force  is  just    equal  to  maximum  force  of  friction. M 0.4  100 i.e.,  Mg=a2gh    a  =  2h   =  2  103  2   =  0.01  m2 d2 0.01  4 Area  of  circular  hole  =    =  0.01  m2    d  =    =  0.113m 4 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 EXAMPLES  OF  BERNOULLI'S  THEOREM l Magnus  Effect  (Spinning  Ball) Tennis  and  cricket  players  usually  experience  that  when  a      ball  is  thrown  spinning  it  moves  along  a  curved path.  This  is  called  swing  of  the  ball.  This  is  due  to  the  air  which  is  being  dragged  round  by  the  spinning  ball. When  the  ball  spins,  the  layer  of  the  air  around  it  also  moves  with  the  ball.  So,  as  shown  in  figure  the  resultant velocity  of  air  increases  on  the  upper  side  and  reduces  on  the  lower  side. force on ball speed of air  spin flow increases  \\pressure reduced speed of air  flow decreases\\pressure increased motion of a spin ball Hence  according  to  Bernoulli's  theorem  the  pressure  on  the  upper  side  becomes  lower  than  that  on  the  lower side.  This  pressure  difference  exerts  a  force  on  the  ball  due  to  which  it  moves  along  a  curved  path.  This  effect is  known  as  Magnus–effect. E 45

JEE-Physics l Motion  of  the  Ping–Pong  Ball When  a  ping–pong  ball  is  placed  on  a  vertical  stream  of  water–    fountain,  it  rises  upto  a air certain  height  above  the  nozzle  of  the  fountain  and  spins  about  its  axis. The  reason  for  this  is  that  the  streams  of  water  rise  up  from  the  fountain  with  very  large velocity  so  that  the  air–pressure  in  them  decreases.  Therefore,  whenever  the  ball  goes  out from    the    streams,  the  outer  air  which  is  at  atmospheric  pressure  pushes  it  back  into  the streams  (  in  the  region  of  low  pressure).  Thus  the  ball  remains  in  stable  equilibrium  on  the fountain. If  we  blow  air  at  one  end  of  a  narrow  tube,  the  air  emerges  from  the  other  end  at  high  speed  and  so  the node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 pressure  falls  there.  If  a  ping–pong  ball  is  left  free  slightly  below    this  end,  it  does  not  fall  down  due  to  the  large pressure  (atmospheric)  below  the  ball.  Similarly,  if  we  blow  air  in  between  two  ping–pong  balls  suspended  by light  threads  near  each  other,  the  balls  come  close  to  each  other  due  to  the  decrease  of  air  pressure  between them.  Same  is  the  reason  that  when  air  is  blown    below  a  pan  of  a  physical  balance  the  pan  is  depressed  down. l Aerofoil This  is  a  device  which  is  shaped  in  such  a  way  so  that  the    relative  motion  between  it  and  a  fluid  produces  a force  perpendicular  to  the  flow.  As  shown  in  the  figure  the  shape  of  the  aerofoil  section  causes  the  fluid  to  flow faster  over  the  top  surface  then  over  the  bottom  i.e.  the  streamlines  are  closer  above  than  below  the  aerofoil. By  Bernoullis  theorem  the  pressure  above  is  reduced  and  that  underneath  is  increased. high speed, reduced pressure aerofoil lift low speed, increased  pressure principle of an aerofoil Thus  a  resultant  upward  force  is  created  normal  to  the  flow  and  it  is  this  force  which  provides  most  of  lift upward  force  for  an  aeroplane.    Examples  of  aerofoils  are  aircraft  wings,  turbine  blades  and  propellers. l Pull–in  or  Attraction  Force  by  Fast  Moving  Trains If  we  are  standing  on  a  platform  and  a  train  passes  through  the  platform  with  very  high  speed  we  are  pulled towards  the  train.  This  is  because  as  the  train  comes  at  high  speed,  the  pressure  between  us  and  the  train decreases.  Thus  the  air  behind  us  which  is  still  at  atmospheric  pressure  pushes  us  towards  the  train.  The  reason behind  flying–off  of  small  papers,  straws  and  other  light  objects  towards  the  train  is  also  the  same. 46 E

JEE-Physics l Sprayer  or  Atomizer This  is  an  instrument  used  to  spray  a  liquid  in  the  form  of  small        droplets  (fine  spray).  It  consists  of  a  vertical tube  whose  lower  end  is  dipped  in  the  liquid  to  be  sprayed,  filled  in  a  vessel. spray Rubber  air flowing out with  bulb high velocity The  upper  end  opens  in  a  horizontal  tube.  At  one  end  of  the  horizontal  tube  there  is  a  rubber  bulb  and  at  the other  end  there  is  a  fine  bore  (hole).  When  the  rubber  bulb  is  squeezed,  air  rushes  out  through  the  horizontal tube  with  very  high  velocity  and  thus  the  pressure  reduces  (according  to  Bernoulli's  theorem).  So  the  liquid  rises and  comes  out  through  narrow  end  in  form  of  droplets.  It  is  used  in  spray  gun,  perfumes,  deodorant  and  etc. l Bunsen's  Burner  (Jet) It  is based  on  the  working of  jet.  It  consists  of  a  long  tube  having  a    fine  nozzle flame gas +air O  at  the  bottom.  The  burning  gas  enters  the  tube  through  O  and  burns  in  a gas flame  at  the  top  of  the  tube. air nosel To  produce  a  non–luminous  flame,  the  air  of  the  atmosphere  is  mixed  with the  gas.  Since  the  nozzle  O  is  fine,  the  gas  enters  with  a  large  velocity  and  so the  pressure  inside  the  tube  is  lowered  than  the  outer  atmospheric  pressure. Therefore,  air  from  outside  rushes  into  the  tube  through  a  hole  and  is  mixed with  the  burning  gas. l Filter  Pump  (Nozzle) water MA It  is  based  on  the  working  of  nozzle.  It  consists  of  a  wide  tube  M  N  in    the upper  part  of  which there  is another  tube  A.   The upper  end  of  A  is  connected vessel to  a  water  tank,  while  its  lower  end  has  a  fine  bore  through  which  water air comes  out  in  the  form  of  a  jet.  The  vessel  which  is  to  be  evacuated  is  connected to  the  tube  M  N  as  shown. water jet N The  velocity  of  the  emerging  water  jet  is  very  large.  Therefore,  the  pressure of  the  air  near  the  jet  becomes  less  than  the  pressure  in  the  vessel.  Hence  air water and air from  the  vessel  rushes  into  the  tube  M  N  and  is  carried  out  along  with  the water  jet.  Thus  partial  vacuum  is  created  in  the  vessel. node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 l Blow i ng–off  of  Ti n  Roof  Tops  i n  Wi nd  Storm When  wind  blows  with  a  high  velocity  above  a  tin  roof,  it  causes  lowering  of  pressure  above  the  roof,  while  the pressure  below  the  roof  is    still  atmospheric.  Due  to  this  pressure–difference  the  roof  is  lifted  up. wind P v large  P0 So P<P0 E 47

JEE-Physics VISCOSITY Viscosity  is  the  property  of  the  fluid  (liquid  or  gas)  by  virtue  of  which  it  opposes  the  relative  motion  between  its  adjacent layers.  It  is  the  fluid  friction  or  internal  friction.  The  internal  tangential  force  which  try  to  retard  the  relative  motion between  the  layers  is  called  viscous  force. NEWTON'S  LAW  OF  VISCOSITY Suppose  a  liquid  is  flowing  in  streamlined  motion  on    a  horizontal  surface  OX.  The  liquid  layer  in  contact  with  the surface  is  at  rest  while  the  velocity  of  other  layers  increases  with  increasing  distance  from  the  surface  OX.  The highest  layer  flows  with  maximum  velocity. R A v+vx y P y S O Av Q X horizontal surface  Let  us  consider  two  parallel  layers  PQ  and  RS  at  distances  y  and  y  +  y  from  OX.  Thus  the  change  in  velocity in  a  perpendicular  distance  y  is  v .  The  rate  of  change  of  velocity  with  distance  perpendicular  to  the  direction x v x of  flow  i.e.  y ,  is  called  velocity–gradient.  According  to  Newton,  the  viscous  force  F  acting  between  two  layers of  a  liquid  flowing  in  streamlined  motion  depends  upon  following  two  factors  : (i) F    Contact–area  of  the  layers    i.e.  F    A (ii) F    Velocity–gradient   v x  v x .  Therefore  F   A v x F  =  A v x where    is     between  layers  i.e.    F   y y y  y  a  constant  called  coefficient  of  viscosity  of  the  liquid. v x node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 In  above  formula    if  A  =  1  and  y   =  1,  then  F  =  .  i.e.  the  coefficient  of  viscosity  of  a  liquid  is  defined  as  the viscous  force  per  unit  area  of  contact  between  two  layers  having  a  unit  velocity  gradient.   •  SI  UNITS N s :  m 2 or  deca  poise •  CGS  UNITS :  dyne–s/cm2  or  poise    (1  decapoise  =  10  poise) •  Dimension :  M1L–1T–1 Example   A  plate  of  area  2  m2  is  made  to  move  horizontally  with    a  speed  of  2  m/s  by  applying  a  horizontal  tangential force  over  the  free  surface  of  a  liquid.  If  the  depth  of  the  liquid  is  1m  and  the  liquid  in  contact  with  the  plate  is stationary.  Coefficient  of  viscosity  of  liquid  is  0.01  poise.  Find  the  tangential  force  needed  to  move  the  plate. 48 E

JEE-Physics Solution v 2  0 m / s v=2m/s   Velocity  gradient  =  y   =  1  0   =2  m   =  2s–1 F From,  Newton's  law  of  viscous  force, 1m F v  A =  (0.01  ×  10–1)  (2)  (2)  =  4  ×  10–3  N y Example A  man  is  rowing  a  boat  with  a  constant  velocity  v   in  a  river  the  contact  area  of  boat  is  ‘A’  and  coefficient  of 0 viscosity  is  .  The  depth  of  river  is  ‘D’.  Find  the  force  required  to  row  the  boat. Solution F  –  F   =  m  a FT ares T D F v0 As  boat  moves  with  constant  velocity  a  =  0  so  F  =  F T But  F  =    A  dv     but  dv   =  v0  0   =  v0 T dz dz D D then  F  =  F   =  Av 0 30° TD Example A  cubical  block  (of  side  2m)  of  mass  20  kg  slides  on  inclined  plane  lubricated  with  the oil  of  viscosity    =  10–1  poise  with  constant  velocity  of  10  m/s.  Find  out  the  thickness of  layer  of  liquid.  (g  =  10  m/s2) Solution dv dv v mgsin F F  =  F  =    A  =  mg  sin  = V h F' dz dz h 10  20  ×  10  ×  sin  30°  =    ×  4  ×  h node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65  h  =  4  ×  10–3  m  =  4  mm Example As  per  the  shown  figure  the  central  solid  cylinder  starts  with  initial  angular  velocity  0.  Find  out  the  time  after which  the  angular  velocity  becomes  half. 0 mass,m R2   0 R1 liquid,  E 49

JEE-Physics Solution v=0 dv dv R1  0 F =   A  dz   , where  dz =  R 2  R1 2 R1 R1 2 R 3   R  F  =   R 2  R1 1 2 and     =  FR   =         -R R2 1 R2  R1 1  R1 R1  2   R 3   M R 2   d 2  R 3  solid 1 1  dt  1 cylinder    =     =  R2  R1 2 R2  R1 liquid,   0 2 d 4R1  tm (R 2  R1 )n2 or          –   =  m (R 2  R1 ) dt  t =  4 R1 0 0 DEPENDENCY  OF  VISCOSITY  OF  FLUIDS • On  Temperature  of  Fluid (a)    Since  cohesive  forces  decrease  with  increase  in  temperature  as  increase  in  K.E.  Therefore  with  the  rise  in temperature,  the  viscosity  of  liquids  decreases. (b)  The  viscosity  of  gases  is  the  result  of  diffusion  of  gas  molecules  from  one  moving  layer  to  other  moving  layer. Now  with  increase  in  temperature,  the  rate  of  diffusion  increases.  So,  the  viscosity  also  increases.  Thus,  the viscosity  of  gases  increases  with  the  rise  of  temperature. • On  Pressure  of  Fluid (a)  The  viscosity  of  liquids  increases  with  the  increase  of  pressure. (b)  The  viscosity  of  gases  is  practically  independent  of  pressure. • On  Nature  of  Fluid STOKE'S  LAW  AND  TERMINAL  VELOCITY Stoke  showed  that  if  a  small  sphere  of  radius  r  is  moving  with  a  velocity  v  through  a  homogeneous  medium (liquid  or  gas),  coefficient  of  viscosity    then  the  viscous  force  acting  on  the  sphere  is  Fv  =  6rv.  It  is  called Stoke's  Law. • Termi nal  Velocit y node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 When  a  solid  sphere  falls  in  a  liquid  then  its  accelerating  velocity  is  controlled  by  the  viscous  force  of  liquid  and hence  it  attains  a  constant  velocity  which  is  known  as  terminal  velocity  (v ). T As  shown  in  figure  when  the  body  moves  with  constant  velocity  i.e.  terminal  velocity  (zero  acceleration)  the  net upward  force  (upthrust  Th  +  viscous  force  F )  balances  the  downward  force  (weight  of  body  W) v 2 r2 (  ) Th= 4  r3 g g 3 4 4 Therefore   Th  +  F  =  W  3 r3g  +  6rvT  =  3 r3g   v  =  9 v T where  r  =  radius  of  body,    =  density  of  body,    =  density  of  liquid,  Fv = 6rvT density=    =  coefficient  of  viscosity W= 4   rg 3 50 E

JEE-Physics • Some  applications  of  Stoke’s  law  : (a)  The  velocity  of  rain  drops  is  very  small  in  comparison  to  the  velocity  of  a  body  falling  freely  from  the           height  of  clouds. (b)  Descending  by  a  parachute  with  lesser  velocity. (c)  Determination  of  electronic  charge  with  the  help  of  Milikan's  experiment. Example A  spherical  ball  is  moving  with  terminal  velocity  inside  a  liquid.  Determine  the  relationship  of  rate  of  heat  loss with  the  radius  of  ball. Solution 2 g r 2 (0   ) 2   Rate  of  heat  loss  =  power  =  F  ×  v    =  6    rv  ×  v    =  6  rv2    =6pr   9   Therefore  rate  of  heat  loss    r5 Example A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is 1.8 ×  10–5   kg  m–1 s–1.  What will  be  the  terminal  velocity  of  the  drop.  Density  of  air  can  be  neglected. Solution 2  15  104 2  103  9.8  vT  2 r2 (  )g   1000   2.72  104  m/s 9    9  1.8  105 REYNOLDS  NUMBER  (R ) e The  type  of  flow  pattern  (laminar  or  turbulent)  is  determined  by  a  non–dimensional  number  called  Reynolds number  (R ).  Which  is  defined  as    R   = vd   where    is  the  density  of  the  fluid  having  viscosity    and    flowing  with ee  node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 mean  speed  v.  d  denotes  the  diameter  of  obstacle  or  boundary  of  fluid  flow. Although  there  is  not  a  perfect  demarkation  for  value  of  R   for  laminar  and  turbulent  flow  but  some  authentic e references  take  the  value  as Re < 1000 >2000 between 1000 to 2000 Type of flow laminar often turbulent may be laminar or turbulent on  gradually  increasing  the  speed  of  flow  at  certain  speed  transition  from  laminar  flow  to  turbulent  flow  takes place.  This  speed  is  called  critical  speed.  For  lower  density  and  higher  viscosity  fluids  laminar  flow  is  more probable. E 51

JEE-Physics SOME WORKED OUT EXAMPLES Example#1 The  pressure  of  water  in  a  water  pipe  when  tap  is  opened  and  closed  is  respectively  3  ×  105  Nm–2  and 3.5  ×  105  Nm–2.  With  open  tap,  the  velocity  of  water  flowing  is (A)  10  m/s (B)  5  m/s (C)  20  m/s (D)  15  m/s Solution Ans.  (A)  Popen  1 v2  Pclosed  v  2 Pclosed  Popen  2  3.5  3  105 2  103 =  10  m/s Example#2 A  large  cylindrical  tank  of  cross-sectional  area  1m2  is  filled  with  water.  It  has  a  small  hole  at  a  height  of  1m  from the  bottom.  A  movable  piston  of  mass  5  kg  is  fitted  on  the  top  of  the  tank  such  that  it  can  slide  in  the  tank freely.  A  load  of  45  kg  is  applied  on  the  top  of  water  by  piston,  as  shown  in  figure.  The  value  of  v  when  piston is  7m  above  the  bottom  is  (g  =  10  m/s2) 45kg v (A)  120  m/s (B) 10 m/s (C) 1 m/s (D) 11 m/s Solution Ans. (D) 1 v2  gh  Mg  v  2gh  2Mg  2 10  6  2  50 10  120  1  121  11m / s 2A A 103 1 Example#3 An  open  vessel  full  of  water  is  falling  freely  under  gravity.  There  is  a  small  hole  in  one  face  of  the  vessel,  as shown  in  the  figure.  The  water  which  comes  out  from  the  hole  at  the  instant  when  hole  is  at  height  H  above  the ground, strikes the ground at a distance of x from P.  Which of the following is correct for the situation described? (A)  The  value  of  x  is  2 2hH 2h/3 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 3 4hH Hg (B)  The  value  of  x  is  \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 3 Ground P (C)  The  value  of  x  can't  be  computed  from  information  provided. (D)  The  question  is  irrevalent  as  no  water  comes  out  from  the  hole. Solution Ans.  (D) As  vessel  is  falling  freely  under  gravity,  the  pressure  at  all  points  within  the  liquid  remains  the  same  as  the atmospheric  pressure.  If  we  apply  Bernoulli's  theorem  just  inside  and  outside  the  hole,  then P   +  v 2  geff y  p outside  v 2  geff y inside inside outside 2 2 v =  0 p =  p   =  p [atmospheric  pressure] inside    ,  inside outside 0 Therefore,  v   =  0.  i.e.,  no  water  comes  out. E outside 52

JEE-Physics Example#4 A  cuboid  (a  ×  a  ×  2a)  is  filled  with  two  immiscible  liquids  of  density  2  &    as  shown  in  the  figure.  Neglecting atmospheric  pressure,  ratio  of  force  on  base  &  side  wall  of  the  cuboid  is B Aa a C 2 a a (A)  2:3 (B)  1:3 (C)  5:6 (D)  6:5 Solution Ans.  (D) Fb  2gh  gh a2  3gha2 ; Fw   g h  ah  2gh ah  5 g h a 2 [here  h=a];  Fb 3   6   2  2 Fw   5  5/2 Example#5 During  blood  transfusion  the  needle  is  inserted  in  a  vein  where  the  gauge  pressure  is  2000  Pa.  At  what  height must  the  blood  container  be  placed  so  that  blood  may  just  enter  the  vein  ?  [Density  of  whole      blood  =  1.06  × 103  kg  m–3]. (A)  0.192  m (B)  0.182  m (C)  0.172  m (D)  0.162  m Solution Ans.  (A) 2000 Pressure  P  =  hg         h  =  1.06  103  9.8   =  0.192m Example#6 A  liquid  of  density    is  filled  in  a  U-tube,  whose  one  end  is  open  &  at  the  other  end  a  bulb  is  fitted  whose pressure  is  P .  Now  this  tube  is  moved  horizontally  with  acceleration  'a'  as  shown  in  the  figure.  During  motion A it  is  found  that  liquid  in  both  column  is  at  same  level  at  equilibrium.  If  atmospheric  pressure  is  P ,  then  value  of 0 P  is A a PA node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 (A)  a (B)  g (C)  P0–a (D)  P  +  a Solution 0 Ans.  (C) Consider  a  point  B  at  the  horizontal  portion  of  tube,  pressure  from  both  side  should  be  same. PA               P  +  gh  =  P  +  gh  +  a,  P  =  P  –  a 0 A A 0 h h B E 53

JEE-Physics Example#7 The  tube  shown  is  of  non-uniform  cross-section.  The  cross-section  area  at  A  is  half    of  the  cross-section  area  at B,C  and  D.  A  liquid  is  flowing  through  in  steady  state.The  liquid  exerts  on  the  tube- CD Statement  I  :  A  net  force  towards  right. Statement  II  :  A  net  force  towards  left. Statement  III  :  A  net  force  in  some  oblique  direction. Statement  IV  :  Zero  net  force Statement  V:  A  net  clockwise  torque. Statement  VI:  A  net  counter-clockwise  torque. Out  of  these (A)  Only  statement  I  and  V  are  correct (B)  Only  statement  II  and  VI  are  correct (C)  Only  statement  IV  and  VI  are  correct (D)  Only  statement  III  and  VI  are  correct Solution Ans.  (A) The  force  has  been  exerted  by  liquid  on  the  tube  due  to  change  in  momentum  at  the  corners  i.e.,  when  liquid is  taking  turn  from  A  to  B  and  from  B  to  C.As  corss-section  area  at  A  is  half  of  that  of  B  and  C,  so  velocity  of liquid flow at B and C is half  to that of velocity at A. Let velocity  of flow of liquid at A be v and cross section area v at  A  be  S,  the  velocity  of  flow  of  liquid  at  B  and  C  would  be  2 [from  continuity  equation]  and  corss  section  area at  B  and  C  would  be  2S. Sv2 2  Sv2 Sv2  Sv2 2 2 Due to flow of liquid, it is exerting a force per unit time of Sv2 on the tube, where  is the density of liquid, S is cross section areaand v is velocity of flow of liquid. The force exerted by liquid on tube is shown in the figure. Which clearly shows that a net force is acting on tube due to flowing liquid towards right and a clockwise torque sets in Example#8 A  rigid  ring  A  and  a  rigid  thin  disk  B  both  made  of  same  material,  when  gently  placed  on  water,  just  manage to  float  due  to  surface  tension  as  shown  in  the  figure.  Both  the  ring  and  the  disk  have  same  radius.  What  can you  conclude  about  their  masses? A  B  node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 (A)  Both  have  the  same  mass. (B)  Mass  of  the  ring  is  half  of  that  of  the  disk. (C)  Mass  of  the  ring  is  double  to  that  of  the  disk. (D)  More  information  is  needed  to  decide. Solution Ans.  (C)   Ring  have  double  surface  than  that  of  disk. Ring Disk 54 E

JEE-Physics Example#9 A  hole  is  made  at  the  bottom  of  a  large  vessel  open  at  the  top.  If  water  is  filled  to  a  height  h,  it  drains  out completely  in  time  t.  The  time  taken  by  the  water  column  of  height  2h  to  drain  completely  is (A)  2t (B)  2t (C)  22t (D)  4t Solution Ans.  (A)  dh  h a tA 2h Here   dt  A 2g dt g   t  h –A  = (av) = a 2gh   h –1 2dh    t =  a 0 0 Therefore for height 2h   t'   2h     t' = 2 t Example#10 A  piece  of  cork  starts  from  rest  at  the  bottom  of  a  lake  and  floats  up.  Its  velocity  v  is  plotted  against time  t.  Which  of  the  following  best  represents  the  resulting  curve? v vv v (A)    (B)  (C) (D)  t tt t Solution Ans.  (A) As  the  cork  moves  up,  the  force  due  to  buoyancy  remains  constant.  As  its  speed  increases,  the  retarding  force due  to  viscosity  increases,  being  proportional  to  the  speed.  Thus  the  acceleration  gradually  decreases. Example#11 A  flat  plate  moves  normally  with  a  speed  v   towards  a  horizontal  jet  of  water  of  uniform  area  of  cross- 1 section.  The  jet  discharges  water  at  the  rate  of  volume  V  per  second  at  a  speed  of  v .  The  density 2 of  water  is  .  Assume  that  water  splashes  along  the  surface  of  the  plate  at  right  angles  to  the  original motion.  The  magnitude  of  the  force  acting  on  the  plate  due  to  jet  of  water  is  :– node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 (A)  Vv1 (B)   V  (v1 + v2 )2 (C)  V v 2 (D)  V (v +v ) Solution   v2  v1  v2 1 12 Ans.  (D) F  p  m(v1  v2 )  V(v1  +  v2 )     where  m =  V  =  volume/sec t t t Example#12 A tank full of water has a small hole at its bottom. Let t  be the time taken to empty first one third of the tank 1 and t  be the time taken to empty second one third of the tank and t  be the time taken to empty rest of the tank 23 then (A)  t   =  t   =  t (B)  t   >  t   >  t (C)  t   <  t   <  t (D)  t   >  t   <  t 123 123 123 123 Ans.  (C) Solution As  the  height  decreases,  the  rate  of  flow  with  which  the  water  is  coming  out  decreases. E 55

JEE-Physics Example#13 A closed cylinder of length '' containing a liquid of variable density (x) =0 (1+x).  Find the net force exerted b y  t he li qui d on t he a xi s of rotat i on. (Take t he c y li nder  to b e massle ss a nd A =  c ross sec t i onal area  of c y li nder )  A (A)  0 A2 2 1  1  (B)  0 A2 2 1  2  (C)  0 A2 2 1           (D) 0 A2 2 1  4   2 3  2 3  2  2 3 Solution Ans.  (A)  F  2 xAdx  (1 x ) x dx 0 A2 2 1 1  x dx  0  2 3 20 A 0    dm  =  Adx;  dF  =  (dm)2  x        Example#14 The  graph  shows  the  extension  of  a  wire  of  length  1m  suspended  from  the  top  of  a    roof  at  one  end  and  with a  load  W  connected  to  the  other  end.  If  the  cross  sectional  area  of  the  wire  is  1  mm2,  then  the  Young's  modulus of  the  material  of  the  wire  is W(N) 100  (10mm) 80 1 2345 60 40 20 (A)  2  ×  1011  Nm–2 (B)  2  ×  1010  Nm–2 1 (D)  None  of  these Solution (C)  ×  1011  Nm–2 Ans.  (B) 2 Y  F/A  W  W  YA  slope   Y   (slope)  1  40  20   2  1010 Nm 2  /  A    A 1 0 6  (2  1)  103  Example#15 Extension The  diagram  shows  a  force-extension  graph  for  a  rubber  band. Consider  the  following  statements node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 I.  It  will  be  easier  to  compress  this  rubber  than  expand  it. II.  Rubber  does  not  return  to  its  original  length  after  it  is  stretched. III.  The  rubber  band  will  get  heated  if  it  is  stretched  and  released. Which  of  these  can  deduced  from  the  graph- Force (A)  III  only (B)  II  and  III (C)  I  and  III (D)  I  only Solution Ans.  (A) Area  of  hysterisis  loop  gives  the  energy  loss  in  the  process  of  stretching  and  unstretching  of  rubber  band  and this  loss  will  appear  in  the  form  of  heating. 56 E

JEE-Physics Example#16 A  solid  right  cylinder  of  length    stands  upright  at  rest  on  the  bottom  of  a  large  tub  filled  with  water up  to  height  h  as  shown  in  the  figure-I.  Density  of  material  of  the  cylinder  equals  to  that  of  water. Now  the  cylinder  is  pulled  slowly  out  of  water  with  the  help  of  a  thin  light  inextensible  thread  as  shown in  figure-II.  Find  the  work  done  by  the  tension  force  develop  in  the  thread. h   (A)  mgh Fig u re -I  F ig u re -II  (D)  mg(0.5+h) Ans.  (C) (B)  mg (C)  0.5  mg Solution Work  done  by  tension  force  =  work  done  by  gravity  =  0.5  mg Example#17 Ans.  (A,C) n  drops  of  a  liquid,  each  with  surface  energy  E,  joining  to  form  a  single  drop (A)  some  energy  will  be  released  in  the  process (B)  some  energy  will  be  absorbed  in  the  process (C)  the  energy  released  or  absorbed  will  be  E(n–n2/3) (D)  the  energy  released  or  absorbed  will  be  nE  (22/3–1) Solution E=T  4r2 4 R 3  n  4 r3 n  R3 R  =  n1/3r 3 3 r3 Surface  energy  of  big  drop  E'=T4R2  =  T4n2/3r2  =  En2/3 Energy  released  =  nE–E'  =  nE–n2/3E  =  E(n–n2/3) node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 Example#18 A  sphere  is  dropped  into  a  viscous  liquid  of  viscosity    from  some  height.  If  the  density  of  material  and  liquid  are   and    respectively  (  >  )  then  which  of  the  following  is/are  incorrect. (A)  The  acceleration  of  the  sphere  just  after  entering  the  liquid  is  g        (B)  Time  taken  to  attain  terminal  speed  t    0 Ans.  (A,B,C) (C)  At  terminal  speed,  the  viscous  force  is  maximum (D)  At  terminal  speed,  the  net  force  acting  on  the  sphere  is  zero Solution Acceleration  will  be  less  than  g     .  Time  will  depend  on  density  .  Viscous  force  may  be  maximum  or    minimum  both  are  possible E 57

JEE-Physics Ans.  (A,B) Example#19 Some  pieces  of  impurity  (density  =)  is  embedded  in  ice.  This  ice  is  floating  in water  (density  =  w).  When  ice  melts,  level  of  water  will (A)  fall  if  >  w (B)  remain  unchanged,  if    <  w (C)  fall  if  <  w (D)  rise  if  >   w Solution Level  will  fall  if  initially  the  impurity  pieces  were  floating  along  will  ice  and  later  it  sinks. Level  will  remain  unchaged  if  initially  they  were  floating  and  later  also  they  keep  floating. Example#20 A  glass  full  of  water  is  placed  on  a  weighing  scale,  which  reads  10  N.  A  coin with  weight  1  N  is  gently  released  into  the  water.  At  first,  the  coin  accelerates as  it  falls  and  about  halfway  down  the  glass,  the  coin  reaches  terminal  velocity. Eventually,  the coin  rests  on  the  bottom  of  the  glass.  Acceleration  due  to  gravity is  10  m/s2.  The  scale  reads (A)  10.5  N  when  the  coin  accelerates  at  5  m/s2. (B)  11.5  N,  when  the  coin  decelerates  at  5  m/s2. (C)  11  N,  when  the  coin  moves  with  terminal  velocity. 10.00 N (D)  11  N,  when  the  coin  rests  on  the  bottom. Solution Ans.  (ACD) When  coin  moves  with  terminal  velocity  or  rests  on  the  bottom    Reading  =  10  +1  =  11  N When  coin  moves  with  acceleration  of  5  m/s2 Reading  =  10  +  1  (½)  =  10.5  N Example#21  to  23 Velocity  of  efflux  in  Torricelli's  theorem  is  given  by  v  2gh ,  here  h  is  the  height  of  hole  from  the  top  surface, after  that,  motion  of  liquid  can  be  treated  as  projectile  motion. 2m(i)         (ii)  node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p6530° O 2 1 . Liquid  is  filled  in  a  vessel  of  square  base  (2m  ×  2m)  upto  a  height  of  2m  as  shown  in  figure  (i).  In  figure  (ii)  the vessel  is  tilted  from  horizontal  at  30°.  What  is  the  velocity  of  efflux  in  this  case.  Liquid  does  not  spills  out? (A)  3.29  m/s (B)  4.96  m/s (C)  5.67  cm (D)  2.68  m/s 2 2 . What  is  its  time  of  fall  of  liquid  on  the  ground? (A)  1s (B)  1s (C)  1s (D)  2s 2 3 5 2 3 . At  what  distance  from  point  O,  will  be  liquid  strike  on  the  ground? (A)  5.24  m (B)  6.27  m (C)  4.93  m (D)  3.95  m 58 E

JEE-Physics Solution 21. Ans.  (B) The  volume  of  liquid  should  remain  unchanged.  Hence,  2  ×  2  ×  2  =  1 x  x  2   2            x  1.42m 2 3  Now  h  =  x  sin  60°  =  1.23  m       v  2gh  2  10  1.23  4.96m / s 22. Ans.  (C) 2m v  H = 2  sin  30° =  1m         t  2H  1 s g5 30° H O AB 23. Ans.  (D) OA  2 cos 30  3m  AB  vt  4.96 s          OB = 3.95 m 5 Example#24  to  26 A  cylindrical  container  of  length  L  is  full  to  the  brim  with  a  liquid  which  has  mass  density  .  It  is  placed  on  a weight-scale;  the  scale  reading  is  w.  A  light  ball  which  would  float  on  the  liquid  if  allowed  to  do  so,  of  volume  V and  mass  m  is  pushed  gently  down  and  held  beneath  the  surface  of  the  liquid  with  a  rigid  rod  of  negligible volume  as  shown  on  the  left. rigid rod L 2 4 . What  is  the  mass  M  of  liquid  which  overflowed  while  the  ball  was  being  pushed  into  the  liquid? (A) V (B)  m (C)  m  –  V (D)  none  of  these 2 5 . What  is  the  reading  of  the  scale  when  the  ball  is  fully  immersed (A) w –Vg (B)  w (C)  w  +  mg  –  Vg (D)  none  of  these 2 6 . If  instead  of  being  pushed  down  by  a  rod,  the  ball  is  held  in  place  by  a  thin  string  attached  to  the  bottom  of  the container  as  shown  on  the  right.  What  is  the  tension  T  in  the  string? (A)  (V–m)g (B)  Vg (C)  mg (D)  none  of  these node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 24. Ans.  (A) Ball  will  displace  V  volume  of  liquid,  whose  mass  in  V. 25. Ans.  (C) Weight  mg  is  entered  while  weight  Vg  of  liquid  is  overflowed. 26. Ans.  (A) F  F  = T + mg  Vg = T  + mg  T = (V – m)g T mg 59 E

JEE-Physics Example#28 A  light  rod  of  uniform  cross-section  of  10–4  m2  is  shown  in  figure.  The  rod  consists  of  three  different  materials. Assume  that  the  string  connecting  the  block  and  rod  does  not  elongate. A 15 cm 20 cm 10 cm YA = 2.5 ×  1010 N/m2 B YB = 4 ×  1010 N/m2 C YC = 1 ×  1010 N/m2 D E 10kg Column  I  (Points)   Column  II    (Displacements  in  m) (A) B (P) 4 (B) C (Q) 9 (C) D (R) 1 2 (D) E (S) 1 5 (T) 2 4 Solution Ans.(A)  P  (B)  Q  (C)  T  (D)  T Y  F/A          F               B 100 0.1  4  106 m  4m   /  YA  2.5  1010  104  c  B  100 0.2  5  106 m  5 m   c  9m 4  1010  104   D  C  100 0.15  15  106 m  15m   D  2 4 m 1  1010  104  Displacement  of  E  =  Displacement  of  D  =  24  m Example#29 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 An  ideal  liquid  is  flowing  in  a  tube  as  shown  in  figure.  Area  of  cross-section  at  points  A,  B  and  C  are  A ,  A   and 12 A   respectively  (A   >  A   >  A ).  v ,  v ,  >  v   are  the  velocities  at  the  points  A,  B  and  C  respectively. 3 1 3 2 12 3 h1 h2 h3 A B C Column  I (P) Column  II (Q) (A) h  is (R) Less  than  h 1 (S) 3 (T) (B) h  is More  than  h 2 60 3 (C) v   is Less    than  v 1 3 (D) v   is More  than  v 2 3 The  maximum  value  amongst  the  three  velocities. E

JEE-Physics Solution Ans.  (A)  Q  (B)  P  (C)  R  (D)  ST By  using  A v   =  A v     A   <  A   <  A     v   >  v   >  v .  Now  by  using  P  +  ½  v2+  gh  =  constant 11 22 231 231 We  have  P   <  P   <  P     so  h   >  h   and  h   <  h 231 13 23 Example#30 A  square  plate  of  1m  side  moves  parallel  to  a  second  plate  with  velocity  4  m/s.  A  thin  layer  of  water  exist between  plates.  If  the  viscous  force  is  2  N  and  the  coefficient  of  viscosity  is  0.01  poise  then  find  the  distance between  the  plates    in  mm. Solution Ans.  2 Av Av 0.01  10 1  12  4 F d   2m d F 0.002 Example#31 A horizontal oriented tube AB of length 5 m rotates with a constant angular velocity 0.5 rad/s about a stationary vertical axis OO' passing through the end A. The tube is filled with ideal fluid. The end of the tube is open, the closed end B has a very small orifice. The velocity with which the liquid comes out from the hole (in m/s) is O'  AB  2m 5m O Solution Ans.  2 Apply  Bernoullie  theorem  between  two  ends P0 5 x2 dx  P0  1 v 2  2 52  32   1 v 2    (0.5)2  ×  16  =  v2   v  =  2m/s 2 2 2  3 node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65 Example#32 A bal l of densit y 0 fal ls from re st from a poi nt P onto t he sur face of a liquid of densit y  i n t ime T. It enter s the liqu id, stops, move s u p, a nd  retur ns to P i n a total t ime 3T. Neglect v iscos it y, s ur face tensi on a nd  splas h i ng. Fi nd  P t=3T the ratio of  0 . t=0 Solution                                                      Ans. 3  It strike the surface of liquid with velocity v  = gT t=T t=2T 1 In water (liquid) its time of flight  T =  2v1 2gT     3  0  1 g   1 g 0  0 3T t= 2 E 61

JEE-Physics Example#33 A  cylinder  fitted  with  piston  as  shown  in  figure.  The  cylinder  is  filled        with  water  and  is  taken  to  a  place  where there  is  no  gravity.  Mass  of  the  piston  is  50  kg.  The  system  is  accelerated  with  acceleration  0.5  m/sec2  in positive  x–direction.  Find  the  force  exerted  by  fluid  on  the  surface  AB  of  the  cylinder  in  decanewton.  Take  area of  cross–section  of  cylinder  to  be  0.01  m2  and  neglect  atmospheric  pressure  (1  decanewton  =10N) AD y x Ans.  5 B 5m C Solution Force  due  to  piston  =  50  ×  0.5  =  25  N Force  due  to  fluid  =  (ah)  A=A  (1000  ×  5  ×  0.5)  =  2500  Pa  ×  0.01  m2  =  25  N Force  on  the  surface  AB  =  50N  =  5  decanewton Example#34 A  container  filled  with  air  under  pressure  P   contains  a  soap  bubble  of  radius  R.  The  air  pressure  has  been 0 5R reduced  to  half  isothermally  and  the  new  radius  of  the  bubble  becomes .  If  the  surface  tension  of  the  soap 4 water  solution  is  S,  P   is  found  to  be  12nS .  Find  the  value  of  n. 0R Solution Ans.  8  4S  4   P0 4S  4  4  5R  3 4S  P0 16S  125 96S  R   3   2 5R  3  4   P0  R  2 5R  64  P0  R 3        P0 R Example  #35 Water (density )  is flowing through  the uniform  tube of  cross-sectional area A with a  constant P speed  v  as  shown  in  the  figure.  Find  the  magnitude  of  force  exerted  by  the  water  on  the v 60° curved  corner  of  the  tube  is  (neglect  viscous  forces) Solution A node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-4\\Fluid Mechanics\\English\\Theory.p65  3 vy | P x | mv sin 60  mv v 2  mv  x    | P net |  3 Px2  Py2   9  3  mv | P y |  mv  mv  4 4  22   3  dm  . v  =  3 A v2      Since, dm  A (v dt)  dm  A  v | P net |  3 mv    | F net |   dt   dt 62 E

JEE-Physics RAY OPTICS INTRODUCTION The branch of Physics called optics deals with the behavior of light and other electromagnetic waves. Under many circumstances, the wavelength of light is negligible compared with the dimensions of the device as in the case of ordinary mirrors and lenses. A light beam can then be treated as a ray whose propagation is governed by simple geometric rules. The part of optics that deals with such phenomenon is known as geometric optics. PROPAGATION OF LIGHT Light travels along straight line in a medium or in vacuum. The path of light changes only when there is an object in its path or where the medium changes. We call this rectilinear (straight–line) propagation of light. Light that starts from a point A and passes through another point B in the same medium actually passes through all the points on the straight line AB. Such a straight line path of light is called a ray of light. Light rays start from each point of a source and travel along straight lines till they fall on an object or a surface separating two media (mediums). A bundle of light rays is called a beam of light.  Apart from vacuum and gases, light can travel through some liquids and solids. A medium in which light can travel freely over large distances is called a transparent medium. Water, glycerine, glass and clear plastics are transparent. A medium in which light cannot travel is called opaque. Wood, metals, bricks, etc., are opaque. In materials like oil, light can travel some distance, but its intensity reduces rapidly. Such materials are called translucent. REFLECTION OF LIGHT When light rays strike the boundary of two media such as air and glass, a part of light is bounced back into the same medium. This is called Reflection of light. (i) Regular / Specular reflection : When the reflection takes place from a perfect plane surface then after reflection rays remain parallel. It is called Regular reflection. (ii) Diffused reflection When the surface is rough, light is reflected from the surface from bits of its plane surfaces in irregular directions. This is called diffused reflection. This process enables us to see an object from any position. NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 LAWS OF REFLECTION incident rayplane normal to mirrorreflected ray  Incident ray, reflected ray and normal lies in the same plane. ir  The angle of reflection is equal to the angle of incident i.e.  i = r. O mirror reflecting surface normal normal normal ir en reflected ray incident ray reflected ray i r\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\incident rayreflected ray incident ray r O  \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ OO In vector form rˆ = eˆ  2 eˆ.nˆ nˆ E 1

JEE-Physics GOLDEN KEY POINTS • Rectilinear propagation of light :In a homogeneous transparent medium light travels in straight line. • When a ray is incident normally on a boundary after reflection it retraces its path. i = 0, r = 0 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\i=0, r=0No light C \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ C i = 0, r = 0 Virtual image plane mirror concave mirror convex mirror • The frequency, wavelength and speed does not change on reflection. • Eye is mostly sensitive for yellow colour and least sensitive for violet and red colour. Due to this reason : • Commercial vehicle's are painted with yellow colour. • Sodium lamps (yellow colour) are used in road lights. REFLECTION FROM PLANE MIRROR Plane mirror is the perpendicular bisector of the line joining object and image.  The image formed by a plane mirror suffers lateral–inversion, i.e., in the image formed by a plane mirror left is turned into right and vice–versa with respect to object. left right right left front object image back mirror When a watch placed in front of a plane mirror then watch is Its object and its time is object time and image of watch observed by 01:45 a person standing in front of mirror then time seen by person. (i) Object Time = AH Image Time = 12 – AH Its NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 (ii) Object Time = AH BM 10:15 Image Time = 11 – 60' – AH BM (iii) Object Time = AH BMCS Image Time = 11 – 59' – 60\" – AHBMCS  A plane mirror behaves like a window to virtual world. A AB  Plane mirror Real Space B Virtual Space 2 E

JEE-Physics  To see complete image in a plane mirror the minimum length of plane mirror should be half the height of a person. From figure. HNM and ENM are congruent EN = HN 1  MD = EN = HE 2 Similarly EN' M' and LN' M' are congruent 11 Length of the mirror MM' = MD + M'D = 2 HE + 2 EL 11 = (HE + EL) = HL 22  Minimum of length of mirror is just half of the person. • This result does not depend on position of eye (height of the eye from ground). • This result is independent of distance of person in front of mirror.  Deviation for a single mirror ir  = 180 – (i + r);  i =  r;  = 180 – 2i    A E Total deviation produced by the combination of two plane mirrors which are \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\M2 inclined at an angle  from each other. = 1 + 2 = 180 – 2 + 180 – 2 = 360 – 2 ( + ) ...(i) incident2 reflected C From QAB, + 90 – + 90 –  = 180  = +  ...(ii) Putting the value of  in (i) from (ii),  = 360 – 2 B normal O mirror mirror  90–   90– Q \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ M1 1 A\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ If there are two plane mirror inclined to each other at an angle  the number of image (n) of a point object formed are determined as follows. (a) If 360  m is even then number of images n = m – 1 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\  bisector object (b) If 360  m is odd. There will be two case.   (i) When object is not on bisector, then number of images n = m (ii) When object is at bisector, then number of images n = m – 1 NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 (c) If 360  m is a fraction, and the object is placed symmetrically then no. of images n =nearest even integer  S.No.  in m  360 Number of images formed if degree  object is placed asymmetrically symmetrically 1. 0    2. 30 12 11 11 3. 45 8 7 7 4. 60 6 5 5 5. 72 5 5 4 6. 75 4.8 – 4 7. 90 4 3 3 8. 112.5 3.2 – 4 9. 120 3 3 2 3

JEE-Physics  If the object is placed between two plane mirrors then images are formed due to multiple reflections. At each reflection, a part of light energy is absorbed. Therefore, distant images get fainter.  Keeping the mirror fixed if the incident ray is rotated by some angle, the reflected ray is also rotated by the same angle but in opposite sense. (See Fig. 1) (Fig. 1) (Fig. 2)  Keeping the incident ray fixed, if the mirror is rotated by some angle, then the reflected ray rotates by double the angle in the same sense. (See Fig. 2)     v on  v in , v op = v ip vop vip though speed of object and image are the same v = component of velocity of object along parallel to mirror. op v von in v = component of velocity of object along normal to mirror. on v = component of velocity of image along parallel to mirror. ip Stationary Plane Mirror v = component of velocity of image along normal to mirror. in        If mirror is moving v ip  v op and v im n   v om n vop vmp vmn    vip     v in  vmn = – v on  v mn  v in  2 v mn  v on  v mn = component of velocity of mirror along normal. von  vin v op = component of velocity of object along mirror.  v on = component of velocity of object along normal  NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 v ip = component of velocity of image along mirror.  v in = component of velocity of image along normal. Example Find the velocity of the image. Solution  = (–10 cos 37°) ˆi = 8ˆi and  (10 sin 37°) ˆj = 6ˆj v ox voy =       v iy v oy v ix  2 v mx  v ox = 2 (–2i) – 8ˆi = 4ˆi and   6ˆj 4E

JEE-Physics REAL AND VIRTUAL SPACES A mirror, plane or spherical divides the space into two ; (a) Real space, a side where the reflected rays exist. (b) Virtual space is on the other side where the reflected rays do not exist. Real Virtual \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\Real VirtualVirtual Real\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ Space Space \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\space spacespace space OBJECT Object is decided by incident rays only. The point object is that point from which the incident rays actually diverge (Real object) or towards which the incident rays appear to converge (virtual object). Point Point Real Object (Object) Point (Real) Virtual Object IMAGE Image is decided by reflected or refracted rays only. The point image is that point at which the refracted / reflected rays reflected from the mirror, actually converge (real image) or from which the refracted /reflected rays appear to diverge (virtual image). I \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ OI \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ O (Virtual image) (Real image) SPHERICAL (CURVED) MIRROR Curved mirror is part of a hollow sphere. If reflection takes place from the inner surface then the mirror is called concave and if its outer surface acts as reflector it is convex. NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 M' M' \\\\ M' r \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ i \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ principal P P axis C CF P CF spherical M spherical M M surface mirror concave mirror convex mirror E5

JEE-Physics DEFINITIONS FOR THIN SPHERICAL MIRRORS (i) Pole is any point on the reflecting surface of the mirror. For convenience we take it as the midpoint P of the mirror (as shown). (ii) Principal–section is any section of the mirror such as MM' passing through the pole is called principal– section. (iii) Centre of curvature is the centre C of the sphere of which the mirror is a part. (iv) Radius of curvature is the radius R of the sphere of which the mirror is a part. (v) Principal–axis is the line CP, joining the pole and centre of curvature of the mirror. (vi) Principal– focus is an image point F on principal axis for which object is at infinity. parallel to axis \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\M rM \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ i normal parallel C F to axis centre of focus P FC curvature P focus centre of curvature M' M' (vii) Focal–length is the distance PF between pole P and focus F along principal axis. (viii) Aperture, in reference to a mirror, means the effective diameter of the light reflecting area of the mirror. (ix) Focal Plane is the plane passing through focus and perpendicular to principal axis. ///////////////////////////////////////// ///////////////////////////////////////////// AA C CFP FP Focal B Focal B plane plane (x) Paraxial Rays Those rays which make small angle with normal at point of incidence and hence are close to principal axis.  \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ n //////////////////////////////////////////////////////// n ( is very small)  ////////////////////////////////////////// NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 P C /////////////////////////////////////////////////// C (xi) Marginal rays : n ( is large) Rays having a large angle of incidence  CP 6E

JEE-Physics SIGN– CONVENTION //////////////////////////////////////////////////////////////////ylight positive ////////////////////////////////////////////////////////////////// light negative positive positive P FC C FP negative positive x positive negative positive negative positive negative O negative • Along principal axis, distances are measured from the pole ( pole is taken as the origin). • Distance in the direction of light are taken to be positive while opposite to be negative. • The distances above principal axis are taken to be positive while below it negative. • Whenever and wherever possible the ray of light is taken to travel from left to right. RULES FOR IMAGE FORM ATION (FOR PAR A XIAL R AYS ONLY) (These rules are based on the laws of reflection i = r)  A ray parallel to principal axis after reflection  A ray passing through or directed towards from the mirror passes or appears to pass focus, after reflection from the mirror, through its focus (by definition of focus). becomes parallel to the principal axis. //////////////////////////////////////////////////////////////////M parallel //////////////////////////////////////////////////////M P f = –ve M' F P CF M' NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65  A ray passing through or directed towards centre  Incident and reflected rays at the pole of curvature, after reflection from the mirror, of a mirror are symmetrical retraces its path (as for it  i =0 and so r =0). about the principal axis i = r. M M //////////////////////////////////////////////////////////// ///////////////////////////////////////////////////// C F P F i dcireenctteerdotof cwuarrvdasture focus M' rP M' E7

JEE-Physics RELATIONS FOR SPHERICAL MIRRORS  //////////////////////////////////////////////////////////////////M Relation between f and R for the spherical mirror D A  For Marginal rays In ABC, AB = BC C 2 P B AC = CD + DA = 2BCcos  R = 2BCcos  BC  R and BP = PC – BBCP  R  R R 2 cos 2 cos  M' Note : B is not the focus ; it is just a point where a marginal ray after reflection meets.  For paraxial rays (parallel to principal axis) ( small so sin , cos 1, tan ) . Hence BC = R R and BP = 22 R Thus, point B is the midpoint of PC (i.e. radius of curvature) and is defined as FOCUS so BP = f = 2 (Definition : Paraxial rays parallel to the principal axis after reflection from the mirror meet the principal axis at focus) R  BP is the focal length (f) f= 2 ////////////////////////////////////////////////////////////////////////  Paraxial rays (not parallel to principal axis) Such rays after reflection meet at a point in the F focal plane (F'), such that C P FF ' FF ' F' FP f  tan       FF ' = f Relation between u,v and f for curved mirror An object is placed at a distance u from the pole of a mirror and its image is formed at a distance v (from the pole) If angle is very small :   MP ,   MP ,   MP  \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\M NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 uR v P from CMO,   from CMI,  =  +    = –  O CI so we can write  =  +   = –   –  = –  2 =  +  2 11 111 R  v  u  f  u  v Sign convention for object/image for spherical mirrors Real object u – ve Real image v – ve Virtual object u + ve Virtual image v + ve 8E

M AG N I F I C AT I O N /////////////////////////////////////////////////////////////////////// JEE-Physics Transverse or lateral magnification /////////////////////////////////////////////////// M Linear magnification m  height of image  hi ///////////////////////////////////////////////////object A FP height of object ho ho M' ABP and A'B'P are similar so hi  v  hi v C B' ho u ho  B u image hi A' Magnification m   v ; m   v  f  fv  hi u u fu f ho erect-image inverted-image O (positive m) (negative m) P I P O I If one dimensional object is placed perpendicular to the principal axis then linear magnification is called transverse or lateral magnification. m hi v  ho u Magnification Image Magnification Image |m|> 1 enlarged |m| < 1 diminished m<0 inverted m> 0 erect  Longitudinal magnification  If one dimensional object is placed with its length along the principal axis then linear magnification is called E longitudinal magnification. Longitudinal magnification: mL  length of image  v2  v1 image object \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ length of object u2  u1 v1 v2 u2 u1 :mL dv For small objects only   du NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 differentiation of 11 1 gives us dv du   dv   v 2 so mL  dv   v 2  m2 hi= mho  0du    vu f v2 u2 du  u   u  Superficial magnification wi=mwo If two dimensional object placed with its plane perpendicular to principal axis its magnification is known as superficial magnification Linear magnification m  hi  w i ho ho wo wo h = mh , w = mw and A = h × w i oi o obj o o Area of image : A = h × w = mh × mw = m2 A obj image i i o o Superficial magnification ms  area of im age  (ma)  (mb)  m2 area of object (a  b) 9

JEE-Physics IMAGE FORMATION BY SPHERICAL MIRRORS (ii) Object : Placed in between infinity and C Concave mirror Image : real, inverted, diminished in between C and F (i) Object : Placed at infinity Image : real, inverted, diminished at F m  1 & m< 0 m  1 & m<0 u= M ///////////////////////////////////////////////////////////////// ho F /////////////////////////////////////////////////////M C C P F P hi M' hi M' (iv) Object : Placed in between F and C Image : real, inverted, enlarged beyond C (iii) Object : Placed at C Image : real, inverted, equal at C m  1 & m<0 (m = – 1) /////////////////////////////////////////////////////M /////////////////////////////////////////////////////M h0 C h0 F P CF hi O M' P hi M' (v) Object : Placed at F (vi) Object : Placed between F and P Image : real, inverted, very large Image : virtual, erect, enlarged and (assumed) at infinity (m<<–1) behind the mirror (m > + 1) /////////////////////////////////////////////////////M M//////////////////////////////////////////////////////////////// FP hi C h0 P F M' M' NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 For concave mirror M agnification m  1 & m < 0 Object Im age m 1 & m<0 – F m = –1 ––C C–F m 1 & m<0 C C m < < -1 C–F ––C m >>1 Just before F towards C – Just after F towards P  10 E

JEE-Physics Convex mirror Image is always virtual and erect, whatever be the position of the object and m is always positive. M M Object placed Object placed I at infinity in front of mirror /////////////////////////////////////////////////// /////////////////////////////////////////////////// OP IF C P FC Virtual, erect, diminished at F virtual, erect and M' (m<+1), between P and F very small (m<<+1) M' GOLDEN KEY POINTS • Differences in real & virtual image for spherical mirror Real Image Virtual Image (i) Inverted w.r.t. object (i) Erect w.r.t. object (ii) Can be obtained on screen (ii) Can not be obtained on screen (iii) Its magnification is negative (iii) Its magnification is positive (iv) Forms in front of mirror (iv) Forms behind the mirror • For real extended object, if the image formed by a single mirror is erect it is always virtual (i.e.,m is + ve) and in this situation if the size of image is : Smaller than object the Equal to object the Larger than object the mirror is convex mirror is plane mirror is concave m<+1 m=+1 m>+1 M M // /// / / / /// / /// / /// // / // // / /// /// / // / / / // ///M / // // / / // /// / // /// // /// /// // / /// / // // / / // / // OP F OP I F PI I O M' M' M' Example The focal length of a concave mirror is 30cm. Find the position of the object in front of the mirror, so that the image is three times the size of the object. Solution As the object is in front of the mirror it is real and for real object the magnified image formed by concave mirror NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 can be inverted (i.e.,real) or erect (i.e.,virtual), so there are two possibilities. /////////////////////////////////////////////////M (a) If the image is inverted (i.e., real) ///////////////////////////////////////////////// O m  f  3  30  u = I CF P fu 30  u – 40 cm 40cm Object must be at a distance of 40 cm in 120cm M' front of the mirror (in between C and F). (b) If the image is erect (i.e., virtual) M I m  f  3  30  u =  20 cm f  u 30  u O Object must be at a distance of 20 cm in CP front of the mirror (in between F and P). F 20cm 60cm 30cm M' E 11

JEE-Physics Example f A thin rod of length is placed along the principal axis of a concave mirror of focal length f such that its image 3 which is real and elongated, just touches the rod. What is magnification ? Solution Image is real and enlarged, the object must be between C and F. One end A' of the image coincides with the end A of rod itself. 5 / //// /// / // / // / /// /// //// // // / / /// / // /// /// M 2 P 111 f M' So v = u , vA + vA = –f i.e., v = u = – 2f A A AA A BF so it clear that the end A is at C.   the length of rod is f B' A' C f 5 f 3 3 3 f5 2f  Distance of the other end B from P is u = 2f   f B 33 if the distance of image of end B from P is v then 111 vB 5 f B      vB 5 f 2 f 3 1 51 f  the length of the image | v B | | v A | 2 f  2f  2 f and magnification m vB  vA  2  3 uB  uA 1 f 2 3 Negative sign implies that image is inverted with respect to object and so it is real. Example A concave mirror of focal length 10 cm and convex mirror of focal length 15 cm are placed facing each other 40 cm apart. A point object is placed between the mirror on their common axis and 15 cm from the concave mirror. Find the position of image produced by the reflection first at concave mirror and then at convex mirror. Solution For M mirror O act as a object, let its image is I then, f=+15 f=–10 11 //////////////////////////////////////////////////////////////////// //////////////////////////////////////////////////////////////////// u= – 15 cm, f = –10 cm   1 1  1  v = – 30 cm NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65  v 15 10 Image I will act as a object for mirror M its distance from mirror M . I2 I1 O 1 22 15cm 15cm u = –(40 – 30) cm = –10 cm 1 1 11 1 1 1  v1 6 cm       so v1 u1 f v1 10 15 M2 M1 So final image I is formed at a distance 6 cm behind the convex mirror and is virtual. 2 Example The sun subtends an angle  radians at the pole of a concave mirror of focal length f. What is the diameter of the image of the sun formed by the mirror. Solution 1 C Since the sun is at large distance very distant, u is very large and so 0 B' u F  11 1 1 1  v = –f A'  ////////////////// vu f  v  /////////////////////// f The image of sun will be formed at the focus and will be real, inverted and diminished P Arc A'B'  = d A'B' = height of image and  = =  d = f Radius FP f 12 E

JEE-Physics VELOCITY OF IMAGE OF MOVING OBJECT (SPHERICAL MIRROR) ///// // ///// ///// //// /// /// /// // /// //// //// ///////////////M j ( a ) Velocity component along axis (Longitudinal velocity) O i When an object is coming from infinite towards the focus of concave mirror  11 1 1 dv 1 du  v2   m 2  M' v  u  f   v2 dt  u2 dt  0  v ix   u2 vox v ox v ix  dv  velocity of image along principal-axis; v ox  du  velocity of object along principal-axis  dt dt  ( b ) Velocity component perpendicular to axis (Transverse velocity) m= hI v = f  h =  f f u  h0 h0 = fu I    u  dhI    dt  velocity of image r to principal-axis   m v oy   dhI   f  dh0  f h0 du ;   m2h0   ˆj  dho dt  f  u  dt (f  u)2 dt v iy  f v ox   dt r to principal-axis   velocity of object Note : Here principal axis has been taken to be along x–axis. POWER OF A MIRROR The power of a mirror is defined as P   1 100  f(m) f(cm) NEWTON'S FORMULA ///////// /// // // / // / // // // / // / // / // /// /// // / ////M P In case if spherical mirrors if object distance (x ) and image distance (x ) 12 are measured from focus instead of pole, u = –(f+x ) and v = –(f+x ), u 12 11 1 1 1 1 I OF by     vu f (f  x2 ) (f  x1 ) f on solving x1 x2  f2 This is Newton's formula. M' GOLDEN KEY POINTS • Convex mirrors gives erect, virtual and diminished image. In convex mirror the field of view is increased as compared to plane mirror. It is used as rear–view mirror in vehicles. • Concave mirrors give enlarged, erect and virtual image, so these are used by dentists for examining teeth. Due to their converging property concave mirrors are also used as reflectors in automobile head lights and search lights NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 /////////////////////////////////////////////// C field of view • As focal length of a spherical mirror f=R/2 depends only on the radius of mirror and is independent of wavelength of light and refractive index of medium so the focal length of a spherical mirror in air or water and for red or blue light is same. E 13

JEE-Physics REFRACTION Refraction is the phenomenon in which direction of propagation of light changes at the boundary when it passes from one medium to the other. In case of refraction frequency does not change.  Laws of Refraction (i) Incident ray, refracted ray and normal always lie in the same plane . en i In vector form (eˆ  nˆ).ˆr  0 1 r 2 (ii) The product of refractive index and sine of angle of incidence at a point r in a medium is constant.  sin i =  sin r (Snell's law) 12 In vector form 1 eˆ × nˆ = 2 rˆ× nˆ Absolute refractive index c It is defined as the ratio of speed of light in free space 'c' to that in a given medium v.  or n = v Denser is the medium, lesser will be the speed of light and so greater will be the refractive index,  v < v , G > W glass water Relative refractive index When light passes from one medium to the other, the refractive index of v1  medium 2 relative to 1 is written as 1 2 and is defined as v2  1 2  2  (c / v2 )  v1 1 (c / v1 ) v2  Bending of light ray incident ray normal R According to Snell's law, 1 sin i = 2 sin r air i D (i) If light passes from rarer to denser medium 1 = R and 2 = D r refracted ray so that sin i  D  1 i > r incident raywater sin r R NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 In passing from rarer to denser medium, the ray bends towards the normal. (ii) If light passes from denser to rarer medium 1= D and 2 = R normal refracted ray rare medium r sin i  R  1 i < r R sin r D water D i In passing from denser to rarer medium, the ray bends away from the normal. denser medium 14 E

JEE-Physics APPARENT DEPTH AND NORMAL SHIFT If a point object in denser medium is observed from rarer medium and boundary is plane, then from Snell's law we have D sin i = R sin r...(i) A B If the rays OA and OB are close enough to reach the eye. R p r pp i sin i  tan i = dac and sin r  tan r = dap here d = actual depth, d = apparent depth r ac ap I So that equation (i) becomes D p p  dac  D  1 D i   R dap d R 2 d ac ap  dac O object  (If R = 1, D = ) then d ap so d < d ...(ii) ap ac The distance between object and its image, called normal shift (x)  d ap dac   dac  1  1   x  dac   d ac 1  x=d –d   ;    ...(iii) If d = d then x  d 1   ac ap   ac    dap   image dap dac image dac object object Object in a rarer medium is seen from a denser medium R I image shift apparent dap x dac  1  R  1  1 height O B dap 2 D  dAC i actual i dac height d=  d i.e., d > d AB ap ac ap ac A high flying object appears to be higher than in reality. D denser NODE6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Unit No-11\\Ray-Optics\\Eng\\01. Ray theory-Part1.p65 medium x = d – d  x = [– 1] d ap ac ac LATER AL SHIFT The perpendicular distance between incident and emergent ray is known as lateral shift. Lateral shift d = BC and t = thickness of slab N In BOC BC d d = OB sin(i – r) ...(i) A i : sin(i  r) = = OB  O OB r i–r cos r = OD t t D In OBD : =  OB  ...(ii) C OB OB cos r t d From (i) and (ii) d  sin(i  r) DB cos r E 15