P-3 Allens Made Physics Exercise Solution

JEE-Physics T  2m 2  3.14  1026 =4.90×10–6sec  qB  1.6  1019  8  10 2 1 7 . v||  =  velocity  parallel  to  B =  v  cos  60° T  2m  2 cos 60 2mk So t2 –  t1 =  2 Pitch  =  v||  ×  T  =  v  cos  60  eB   eB So particle  cover 2r  distance in  +z  direction  in circle So    z-coordinate 2  1 2  9.1  1031  2  103  2  1.6  1019 2  m  v 2  1026  1.28  106 0.1   =  qB   =  1.6  1019  8  102     =  2m B So  coordinates  =  (6.4m,  0,  2m)   B  =  4.7  ×  10–3  T 13. (A)   at  centre  =  (Bst.wire  –  Barc)  ( k ) 18.     4  0 a I  1  1 B B centre 4B AB 4  2 2  / 2         0 I (2 sin 60)  0 I  2  (k )  cos 4 R  3    4  R 60  2 20I  A B          =  C a                     0I  3 3  (k )   6a    1 B vertex 'A '  B AB  BBC  B CD  B DA D           For  charge  particle  -  qvB  =  ma qv0I  3 3  =  0  +  0 I  1   0 I  1   0   =  20 I  6ma    1 4 a  2  4 a  2  4 a   a  (B)        NIA (k )  Bi B centre 2 2 4 8:1 T M B  B vertex 2     I   a2  3  Bj    1 9 . T  3 4 a2     0 Ix Nx2  0 Iy N y 2  ˆi B centre  Rx 4 Ry       =   4 1 4 . (A)  M  B  R  Mg  0 | M  B|| R  Mg|  NIA × B = mgr   I  mg          = – 1.6 × 10–3 ˆi T r B 2  B 2 x y mg x (B)  I  rBx (Bz  has  no  effect  on  torque) y                      EAST Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65 15. Work  done  by  external  agent  in  rotating  the 2 0 . For  B  to  be  zero  at  P,  current  in  B  should  directed conductor  in  one  full  turn =  F2r  =  iLB0.  2r upward Power  =  W.n  =  iB0(2r)n 1 6 . fN;  N  =  mgcos  +  IBcos (i)  B P  0  IA  IB  0  IB  3A   10   2    2  10   11     11        (ii)  B S  B A  B B and   B A  B B   fext  =  (mg  +  IB)  sin    ±  f   0 9.6 2  0 30 2 =  (mg  +  IB) sin   ±  (mg  cos  + IB  cos  )   BS   2 1.6   2 1.2  3 33 B 2  B 2 =   4 2 ×  2 ×  0.1  =  0.75  ±  0.13 A B  ±    fext  =  0.62  N  or  0.88  N =  13  ×  10–7  T 11

JEE-Physics (iii)  Force  per  unit  length  on  the  wire  B1 EXERCISE  –V-A F  =  0 I1I2     =  2  107  9.6  3 1. Magnetic  field  at the centre  of the  coil  is  0i 2 d 2 2. 2R =  2.88  ×  10–6  N/m  0i 0 2i BA 2R BB 2 1 .   (i)  Force  on  electron,  evB  =  F BA ;  BB  2 2R   Hence  1    1.6  ×  10–19  ×  4  ×  105  ×  B  =  3.2  ×    10–20 0  2.5 I Radius  of  the  circular  path  in  magnetic  field, 2  5 2     B  =  5  ×  107  T  =        4A mv p qB qB r=   If  momenta  of  two  charged  particles  is  same  then (ii) For  B  to be zero at R; the position of third wire  0 I  2  107  2.5 =  1m 1 2 B 5 107 r q 22. For  vertical  coil,  0 I1 N 1  BH As  electrons  and  protons  have  same  charges;  so 2 r1 their  radius  of  curvature  will  be  same;  though  their sense  of  rotation  will  be  opposite. 3.49  10 5  2  0.2 3 . Parallel  currents  attract  and  antiparallel  currents  I1  =  100  4   107  0.111A repel each other. If a current is made to pass through the  spring;  the  spring  will  compress  as  due  to parallel  currents;  the  turns  will  attract  each  other. 4. Time  period  of  a  charge  particle  in a  magnetic  field is T  2m qB The time  period  is  independent of  radius  or  speed of  the  charged  particle. For  horizontal  coil;    0 I2N 2  BV  I2  =  0.096  A    0 i1 2 cos  d 2 r2  B  4 r   5. F  i 2 d   i2 2 3 .   For  the  coil B M       = 0i1i2d cos   2 r 6.     F  q v  B ;  The  force  due  to  magnetic  field  is always  perpendicular    i.e. d s  ;  hence  this  force v Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65 can  never  do  the  work  on  a  charged  particle. If  the  coil  is  turned  through  angle  '',  the  restoring For  the  charged  particle  to  pass  undeflected 7.  torque,  MB sin   MB  I through  a  cross  E   and  B ;  the necessary  condition    –  Ia2B  =  ma2   ma2 2     is  F  FE  FB  0   i.e.,  qE  q v  B  0 66     6 IB 2 m m T 6IB   =  0.57  s or  E   v  B       T  =  2      v  B   then E vB If   (where  a  =  side  of  square) E 104 B =   =103  Wb/m2 v 10 12

JEE-Physics 8 . W=Uf–Ut  =  –MBcos600–(–MBcos0) 1 6 . Magnetic induction due to a coil at its centre is along the  axis  of  the  coil.  When  two  coils  are  held MB MB perpendicular  to  each  other,  their  axes  are  also                           =  2 +MB  =  2   MB=2W perpendicular  to  each  other,  hence  the  magnetic induction  will  also  be  perpendicular  to  each  other, Torque  =       so  Bnet  at  their  common  centre  will  be  M B   MB sin 60 B B 2  B 2 1 2  3    2 W   3  3W 22   MB  2   2   0 2 i1   0 2 i 2   0 2 i12 2 B  4 R1    4 R2  4 R  i 2     9 . Inside a bar magnet; magnetic lines of force run from south  to  north  pole. 2 32  42     =  5  ×  10–5T       =  10–7  ×  2  10 2 1 0 . No magnetic  field  is ever present at  any  point  inside thin  walled  current  carrying  tube. 11. Bcentre  of  circular loop  =  0 2ni ...(i) 1 7 . The  time  taken  by  charged  particle  to  complete 4 R 2 m For  a  given  length  L  =  n2R circle will  be   T  qB L 1 8 . Magnetic  needle  kept  in  non-uniform  field R= ...(ii) experiences  both  force  as  well  as  torque. n2 1 9 . In  a  parallel  uniform  electric  and  magnetic  field  , From  equations  (i)  and  (ii),  we  get if  a  charged  particle  is  released  then  it  experiences an electric  force  due  to  which  it moves  in  a  straight B  0 2ni n2  B  n2 line. 4 L So,  Bf n2  Bf  n2Bi Bi  12 2 0 . B solenoid  0 ni 1 3 . When  they  are  carrying  current  in  same  direction, B2  n2i2 100  i / 3 they  attract  each  other  with  a  force  F1  0 2I1 I2 B1 n1i1 200  i l1 4 d B2  B1 6.28  10 2 =1.05  ×  10–2T When  the  direction  of  current  in  one  of  the 6  conductors  is  reversed,  the  force  will  become repulsive  with  a  value 6 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65  F2  0 2 2 I1 I2 ;  F2 2 F1  2 2 1 . Magnetic  field  at  a  point  inside  the  straight  long 4 3d 1  1 ;  F2   3 F1 l1 a 3 conductor  at  a  distances  2 from  its  centre  will  be obtained. 1 4 . Resistance  of  galvanometer current sensitivity 10 a R =   =5 a/2 g  voltage sensitivity 2 As  galvanometer is  to be  converted into  a  voltmeter On  applying  Ampere's  circulated  law,  we  get of  range  1  ×  150  =  150V. So  resistance  to  be  connected  in  series V –  R   = 150 –5  =  10000  –  5  = 9995    a  I  a  2 g 150  2  a 2  2  = Ig  10 –3   1 0  B 2   0   1 13

JEE-Physics B1  0 I ...(i)                   a 2 5 . The  magnetic  field  can  never  produce  a  charge  in 4 a 2a the  speed  of  the  charged  particle,  hence  it  can  never produce  a  change  in  kinetic  energy. Magnetic  field  at  a  point  outside  the  conductor  at Whereas  it  produces  a  change  in  velocity  of  the a  distance  2a  from  the  centre  will  be  obtained  as charged  particle by changing  the  direction of  motion of charged  particle. B 2 2 2a   0 I From  this  we  can  conclude  that  magnetic  field cannot  produce  a  change  in  kinetic  energy  whereas it  can  produce change in  momentum  of the  charged particle. B2  0 I ...(ii) 2 6 . B  0 i  4  107  5  106   T  southward 4 a 2 R 4 On  dividing  equation  (i)  by  (ii),  we  get  B1 1 aA B B2 I I1 30° C 2 2 . As  the  thin  walled  pipe  does  not  enclose  any  net current,  hence  the  net  magnetic  field  at  any  point O inside  the  thin walled  pipe  will  be  zero,  whereas  for the outside points it behaves as a straight long current 27. carrying  conductor. D b B0 =  B1  –  B2  =  0I    0I  6  2a  6  2 2b   2 2 3 . When  charged  particle  goes  undeflected  then   E =  0I (b  a) qE  qv  B  qE  qvB  v  24ab B 3 0 . Net  megnet  field  due  to  both  elements These  two  forces  must  be  opposite  to  each  other   =  2  (dB)  cos   if  and  only  if  v   is  along  E  B =  2  ×  µ0 dI   cos   (here  dI  =  I d ) R   2 R EB  E B     E B =  2  ×  µ0 I d   cos   v  B2 v  B2  2R R BB =  2  ×  µ0 I   Rd  cos D 2R R y /2 µ0 µ0 I l2 B  =  0 2 R  cos d =  2 R 24. A l1 x B     B  B1  B2 O P dB dB sin  2dB cos   Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65  dB   C Let  the point  P  is situated  at  'd'  from  O  outside the dB sin  plane  of  the  paper d d           0 2I1 ˆj ;  B2  0 2I2 ˆi 31.    F  q E  v B  B1 4 d 4 d       q 3ˆi  ˆj  2kˆ  3ˆi  4ˆj  kˆ  ˆi  ˆj  3kˆ   BP 0 2  4 d I1 ˆj  I2ˆi      q 3ˆi  ˆj  2kˆ  3ˆi  4ˆj  kˆ  ˆi  ˆj  3kˆ  e     0 Fy =  11q BP 2 d I12  I22 14

JEE-Physics 3 2 Magnetic  moment  of  elements  ring EXERCISE  –V-B r dr 1 . Current, i = (frequency) (charge) =     (2q)   q   2   R dq dA Magnetic moment R (q,m) (q,m) dM =  di  ×  A  = r2  =  r2 TT  q M = (i) (A)     (R2) = (qR2)   4r2 dr Angular momentum  L = 2I = 2(mR2) =  r  ×   2 M qR 2 q  L  2(mR2 )  2m 2mK 3 3 . r    qB   As  K  &  B  are  constant m 2 . Net magnetic field due to both the wires will be downwards as shown in the figure. So r   q vz y rp  :  rd  :  r  :  :  mp : md : m × x qp qd q I –I B :  : mp : 2mp : 4mp Since, angle between     is 180°. e e 2e v and  B :  : 1 :  2  :  1  r  =  rp  <  rd Therefore, magnetic force   Fm  q(v  B )  0 3. The charged particle will be accelerated parallel (if it 3 4 . Magnetic  field  at  the  centre  due  to  element  ring is a positive charge) or antiparallel (if it is a negative charge) to the electric field, i.e., the charged particle r will move parallel or antiparallel to electric and dq magnetic field. Therefore, net magnetic force on it will be zero and its path will be a straight line. dB  µ0 di  µ0  dq   µ0 ds 4. Total magnetic flux passing through whole of the X- 2r 2r  T  2rT 5. Y palne will be zero, because magnetic lines from a closed loop. So, as many lines will move in –Z Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65   direction same will return to +Z direction from the X-Y plane.        =  µ0  Q  (2 rdr )  µ0 Q dr H  = Magnetic field at M due to PQ + Magnetic field 2r  R 2  2 /  R 2 1 µ0Q R µ0 Q  µ0 Q  1 at M due to QR. But magnetic field at M due to R 2 0 R 2 R R QR = O B dr  R   B   Magnetic field at M due to PQ (or due to current I B in PQ) = H So  graph  1 Now H  = Magnetic field at M due to PQ (current I 2 + magnetic field at M due to QS (current I/2)  + magnetic field at M due to QR  H1  H1  0   3 H1  2 2 2 H1  H2 3 R 15

JEE-Physics 6 . We can write 9 . The magnetic field at P(a,0,a) due to the loop is    Bkˆ equal  to  the  vector  sum  of  the  magnetic  fields E = E.i  and   B produced by loops ABCDA and AFFBA as shown in the figure.  Velocity of the particle will be along q.  E  direction. Therefore, we can write   A qEˆi Ck v j   P(a,0,a) i E, B v In   and  , A, E and B are positive constants DB while q can be positive or negative. E Now, magnetic force on the particle will be   Fm    B)  q{AqEˆi}  {Bkˆ} AF q(v Mag neti c  f i eld  d u e to loop  A BC DA  wi ll b e along   ˆi           q2AEB(ˆi  kˆ)  and due to loop AFEBA, along  kˆ . Magnitude of Fm  q2 AEB(ˆj) magnetic field due to both the loops will be equal. Therefore direction of resultant magnetic field at P  Since,  Fm is along negative y-axis, no matter what is the sign of charge q. Therefore, all ions will deflect will be  1 (ˆi  kˆ) towards negative y-direction. 2 7 . Ratio of magnetic moment and angular momentum 1 0 . Consider an element of thickness dr at a distance r Mq form  the  centre.  The  number  of  turns  in  this is given by        N b dr L 2m element,  dN   b  a  dr r w h i c h  i s  a  f u n c t i o n  o f  q  a n d  m  o n l y .  T h i s  c a n  b e Magnetic field due to this element  a derived as follows : at the centre of the coil will be-   qr 2   dB  0 (dN)I  0I N dr M = i A = (q f). (r)2 = (q)   2 (r2)  2 . 2r 2 b  a r and L = I= (mr2 ) rb r 2   B  dB  2 0 N I ) n  b  (b  a  a  q ra M 2 q     2m mv mr2 1 1 . Radius of the circle =  Bq L 8 . If the current flows out of the paper, the magnetic or radius  mv if B and q are same. field at points to the right of the wire will be upwards (Radius)  > (Radius)  m v  > m  v and to the left will be downwards as shown in figure. A B  A A B B B  1 2 . Magnetic field at P is  B , perpendicular to OP in the i direction shown in figure. y B So,  B = B sin  ˆi  – B cos ˆj  ;   P(x,y) Here, B =  0I r B 2 r i×  O x B Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65 Now, let us come to the problem. yx Magnetic field at C = 0 sin  =   and cos  =  rr Magnetic field in region BX' will be upwards (+ve)   0I 1 ( yˆi  xj)  0I(yˆi  xˆj) (as r2 = x2 + y2) because all points lying in this region are to the right B 2 r2 2(x2  y2 ) of both the wires.   x x’ 1 3 . Magnetic lines form closed loop. Inside magnet these A are directed from south to north pole. CB Magnetic field in region AC will be upwards (+ve), because points are closer to A, compared to B. 1 4 . If (b–a) > r (r= radius of circular path of particle) The particle cannot enter the region x > b Similarly  magnetic  field  in  region  BC  will  be So, to enter in the region x > b      r> (b – a) downwards  (–ve).  Graph  (b)  satisfies  all  these conditions. mv q(b  a)B   Bq m  (b  a)    v  16

JEE-Physics 1 5 . Electric field can deviate the path of the particle in 20. Ans.  (AC) the shown direction only when it is along negative y-  direction. In the given options  E  is either zero or FBA  0 , a l o n g  x - d i r e c t i o n .  H e n c e ,  i t  i s  t h e  m a g n e t i c  f i e l d because magnetic lines are parallel to this wire. which is really responsible for its curved path.  Options (a) and (c) cannot be accepted as the path FCD  0 , will be helix in that case (when the velocity vector because magnetic lines are antiparallel to this wire. makes an angle other than 0°, 180° or 90° with the  magnetic field, path is a helix) option (d) is wrong FCB  is perpendicular to paper outwards and  FAD  is because in that case component of net force on the perpendicular to paper inwards. These two forces (although calculated by integration) cancel each other particle also comes in  kˆ  direction which is not but produce a torque which tend to rotate the loop in clockwise direction about an axis OO'. acceptable as the particle is moving in x-y plane. Only in option (b) the particle can move in x-y plane. 21. Ans.  (ACD) In option (d) :  Fnet  qE  q(v  B) mV Initial velocity is along x-direction. So, let   vˆi Radius of circular path of charged particle R =  qB v   mV   Fn et qB  qaˆi  q[(vˆi )  (ckˆ  bˆj)] • Particle enters region III if R >     qaˆi  qvcˆj  qvbkˆ • Path length in region II is maximum is I n  o p t i o n  ( b )  : R =  V  =  qB  m Fnet  q(aˆi )  q[(vˆi )  (ckˆ  aˆi )]  qaˆi  qvcˆj T m 16 . Net force on a current carrying loop in uniform • Time spent in region II t =  2  =  qB magnetic field is zero. Hence, the loop cannot translate. So, options (c) and (d) are wrong. From 22. Ans.  (C) Fleming's left hand rule we can see that if magnetic field is perpendicular to paper inwards and current 23. By    direction  of     from  equation    q    B  in the loop is closkwise (as shown) the magnetic force F F v  Fm on each element of the loop is radially outwards, Magnetic  field  is  in  –z  direction or the loops will have a tendency to expand. 30° y i Fm 4m/s x F 30° M,Q 4m/s x=0 x=L  Time  =    /6  M 1 7 . U  M B = – MB cos   QB / M 6QB Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65 Here,  M  = magnetic moment of the loop M 50M  = angle between  M  and  B U is maximum when  = 180° and minimum when   B  = 6Q 10  103  3Q = 0°. So, as  decrease from 180° to 0° its PE also decreases. Comprehension#1 1 . If  B2>  B1,  critical  temperature,  (at  which  resistance 18 . Magnetic  force  does  not  change  the  speed  of charged particle. Hence, v=u. Further magnetic of  semiconductors  abrupty  becomes  zero)  in  case  2 field on the electron in the given condition is along will  be  less  than  compared  to  case  1. negative y-axis in the starting Or it describes a circular path in clockwise direction. Hence, when it 2 . With  increase  in  temperature,  T   is  decreasing. exits from the field, y < 0. C T (0)  =  100  K 19.   C Fm  q(v  B ) T   =  75  K  at  B  =  7.5  T C Hence,  at  B  =  5T,  T   should  line  between  75  K C and  100  K. 17

JEE-Physics Comprehension#2 Match  the  column 4. (p) R dB BR 1 . induce  electric  field  =   2 dt 2 torque  on  charge  =  QBR 2 kˆ E (A) All electric field vector at an angle of 120° Q 2 electric field at centre is zero. E Individually  due to +ve and (–ve) charge   1 +Z  dL (B) Electric potential due to negative charges and    dL   dt positive charges is equal therefore it is zero by  dt 0 (C) On  rotating  the  hexagon  current  in  the  loop becomes  zero   QBR2  kˆ   magnetic  field  at  the  centre  is  zero L 2 E E (D) As  current  in  the  whole  loop  iz  zero  therefore magnetic  moment  is  zero  Change  magnetic  dipole  moment  =  L QBR2 kˆ  +–+–+–+–+ 2 (q) p 2. Magnitude  of  induced  electric  field= R dB = BR (A) Electric  field  will  be  non  zero  due  combination 2 dt 2 of  two  charge  electric  field  is  additive  in  nature Comprehension#3 (B) Electric  potential  due  to  negative  charges  and positive  charges  is  equal  therefore  its  potential 1 . 4000V VV 200V is  zero (C) On  rotating  about  the  axis  current  is  zero magnetic  field  is  zero    [  total  charge  is  zero] (D) Magnetic  moment  due  to  rod  is  zero step up step down V 10  Kq k2 for  step  up  transformer  4000  1 (r)     E   a2 b2   V  =  40,000  Volt for  step  down  transformere N1  V  4000  200 (A) Same  charges  are  spread  at  an  Angle  of  120° Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65 N2 200 200 therefore  electric  field  at  centre  is  zero 2 . Current  in  transmission  line (B) Electric  potential  due  to  negative  charges  and  Power 600 103  150A positive  charges  is  not  equal  therefore  it  is  non-zero Voltage 40,000 (C) Current  in  the  individual  loop  is  non  zero therefore  magnetic  field  is  non  zero (D) As  current  is  non  zero  magnetic  moment  will be  non  zero Resistance  of  line  =  0.4  ×  20  =  8 Power  loss  in  line  =  i2R  =  (150)28=  180  KW  percentage  of  power  dissipation  in  during (s) • transmission  =  180 103 100  30% 600 103 18

JEE-Physics (A) Electric  field  at  the  centre  due  to  symmetrical 42 ML2 distribution  is  zero  2I  = I  =  ML 2   I  =  QS ZZ 3 3 QS (B) Electric  potential  due  to  negative  charges  and   I0 L2 B  3 I0B positive  charges  is  not  equal  therefore  its | | 2 / 3ML2 2M potential  is  non-zero   (C) Current  non-zero I  Magnetic  field  is  non-zero  Angle by which the frame roates in time t is (D) As  current  is  non-zero  magnetic  moment  is 1 3 I0B .(t)2 also  non-zero  =  (t)2   =  2 4M (t) L (i)  = 30° sin   (A) Electric  field  at  M  is  additive  therefore  it  is 2. non-zero 3. R (B) Electric  potential  due  to  negative  charges  and Here, R  mv0 positive  charges  is  equal  therefore  its B0q potential  is  zero (C) Current  is  zero  because  summation  of  charge   sin 30  L 1  B0qL mv0 is  zero mv0   2 mv0   L =  2B0q  Magnetic  field  is  zero B0q (D) As  current  is  zero  magnetic  moment  is  also (ii) In part (i) zero Subjective L 1L sin 30° =      L = R/2 1 . Magnetic moment of the loop,  R 2R M  (iA )kˆ  (I0L2 )kˆ 2.1 Now when L' = 2.1 L   2 R  L' > R Magnetic field, Therefore, deviation of the particle is =180° is as  B (ˆi  ˆj) shown B  (B cos 45)ˆi  (B sin 45)ˆj  2 (i) Torque acting on the loop,  m      ( I0 L2 kˆ)   B (ˆi  ˆj)    v0ˆi  vB  and t  = T/2 =  B0q  M B     vf AB 2   (i) Magnetic field  (B) at the origin = magnetic field I0L2B (ˆj  ˆi )   |   2  I 0 L2 B due to semicircle KLM + Magnetic field due to other     | (ii) Axis of rotating coincides with the torque and   0I (ˆi )  0 I (ˆj) semicircle KNM   B 4R 4R since torque is in ˆj  ˆi  direction or parallel to QS.   0I B Therefore, the loop will rotate about an axis passing  0I ˆi  0I ˆj  ˆi  ˆj through Q and S as shown in the figure. 4R 4R 4R Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65 y  Magnetic force acting on the particle SR   q       q{(v0ˆi )  (ˆi  ˆj)}0I x F  v B 4R PQ   0qv 0 I kˆ F  4R  (ii)  FKLM  FKNM  FKM  | |  Angular acceleration, a   And  FKM  BI(2R )ˆi  = 2BIR ˆi I  F1  F2  2BIRˆi where I = moment of inertia of loop about QS. I  + I  = I  (From theorem of perpendicular axis)    Total force on the loop,  F  F1  F2   F  4BIRˆi QS PR ZZ But I  = I QS PR   Then,  FADC  FAC  or  FADC  ˆi (AC )B 19

JEE-Physics 4 . (i) Given : i =10A, r  = 0.08m and r  = 0.12m. 5 . Let the direction of current in wire PQ is from P to Q 12 S t r a i g h t  p o r t i o n s  i . e . ,  C D  e t c . ,  w i l l  p r o d u c e  z e r o and its magnitude be I. magnetic field at the centre. Rest eight arcs will z  produce the magnetic field at the centre in the same direction i.e., perpendicular to the paper outwards or vertically upwards and its magnitude is B = B  + B o u t e r  a r c s P  Q  y i n n e r  a r c s  1 0i   1 0i    0  ( i )  r1  r2  I    2r2   4   r1r2  R  2  2r1  2  x  S  Substituting the values, we have The magnetic moment of the given loop is :  (107 )(3.14)(10)(0.08  0.12) M  Iabkˆ B T Tor qu e on t he loop due to magnet i c  forc e s  i s : (0.08  0.12)   1  M  B B = 6.54 × 10–5 T (Ver t ically upward or out ward       (Iabkˆ)  (3ˆi  4kˆ)B0ˆi  3IabB0ˆj normal to the paper) Torque of weight of the loop about axis PQ is : ( i i )  F o r c e  o n  A C     a ˆi   ( m g kˆ)  mga ˆj 2  r  F  2 2 Force on circular portions of the circuit i.e., AC etc., due to the wire at the centre will be zero because We see that when the current in the wire PQ is from magnetic field due to the central wire at these arcs  will be tangential (=180°) as shown. P to Q,  1  and  2  are in opposite direction, so they Force on CD can cancel each other and the loop may remain in equilibrium. So, the direction of current I in wire PQ Current in central wire is also i=10A. Magnetic field is from P to Q. Further for equilibrium of the loop: at P due to central wire,  B  0 . i  3IabB  =  mga mg 2 x 1  2  0 I  2 6bB0     Magnetic  force  on  element  dx  due  to  this (ii) Magnetic force on wire RS is :     I B F  0 . i  .dx   0  i 2 dx  I     I bˆj  3ˆi  4kˆ B0   2 x   2  x mag neti c field   dF = (i)   F  IbB 0 3kˆ  4ˆj     (F = iB sin 90°) 6. mg Therefore, net force on CD is- In equilibrium, 2T  = mg  T  =     ....(i) 0 02 x r2 0i2 0.12 dx 0 2 n  3       2 0.08 x 2  2   2  dF x r1         F   i Mag net i c m ome nt,  M =i A  = Q R 2 Substituting the values, BQR 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65 F = (2×10–7) (10)2 n(1.5) or   MB sin 90                                  F= 8.1 × 10–6 N(inwards) 2 Force on wire at the centre Let T  and T  be the tensions in the two strings when Net magnetic field at the centre due to the circuit is 12 in vertical direction and current in the wire in centre is also in vertical direciton. Therefore, net force on magnetic field is switched on (T  > T ). the wire at the centre will be zero. (=180°). Hence, 12 For  translational equilibrium of  ring is  vertical direction,  T  + T  = mg ....(ii) 12 For rotational equilibrium, D BQR 2 (T  – T )  1 22 2 (i) Force acting on the wire at the centre is zero. (ii) Force on arc AC = 0  T1 – T  =  B Q R 2 ....(iii) (iii) Force on segment CD is 8.1 × 10–6 N (inwards). 2 2 20

JEE-Physics Solving equations (ii) and (iii), we have tt mg BQR 2 (iii) = BiNA     dt  BNA  idt T  =   12 2D 00 As T  > T  and maximum values of T  can be BNAQ 12 1 I= BNAQ     3 T0 , we have  3T0  T0  maxBQR2  mg   I 2 2 2D  2 T0  At  maximum  deflection  whole  kinetic  energy (rotational) will be converted into potential energy of D T0 spring.  Hence, 11  max =  BQR 2 I2 =  k2 2 2 max 2qvm m B N A 7 . r  Bq    r  q Substituting the values, we get  max  Q 2I rp  mp q  12 1 9 . B  =  0i   (sin  37  +  sin  53)  4 d r m q p 4 1 2 8 . (i)  = MB = ki (=90°) 0i 7  53°  5    =  12         5   k  MB (NiA )B 4 3x  = NBA ii 37° (ii)  = k. =  BiNA (k = Torsional constant)     70i    k  =  7 53° 37° 4 8 x 4x 2BiNA  k  (as  = /2)  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\01.Magnetic effect of current.p65 21



JEE-Physics UNIT # 05 THERMAL PHYSICS EXERCISE –I 8 . Let at temperature , volume increases by 2% 1 . Let  = temperature on X–scale corresponding to then according to question ( 50°C on Y–scale 2%)  = Y50°C X 100 = 98 [1+3.3 × 10 (–4)] 375°C -30°C   = 60.4 + 4 = 64.4 °C  50°C FL X Y 9 . L=Lthermal– L =contact force 0  1L= A Y1 (rod – 1) 12 –125°C –70°C  X  (125)  50  (70)  X= 1375 FL 375  (125) 30  (70) 2 L = AY2 (rod – 2)  Y11 = Y22 1 0 . Pressure at the bottom in both arms will be equal 2 . For centigrade and Fahrenheit scale ()  ()   0   0  1  2   t1    t  2 t1  1t2 F  32 C 0 100 1 .1  1  2     C   (140  32) = 60°C  2 212  32 100  0 180 3 . Slope of line AB (AB ) 1 1 . Strain () = C  100  0 100 5   t=    = –12 × 10–6 × (75 – 25) = – 6 × 10–4 F 212  32 180 9 4 . If we take two fixed points as tripe point of water 1 2 . Coefficient of linear expansion of brass is greater and 0 K. Then (t(hanthatofsteel.) 0 K ) TX  0  TY  0 450 TX = 200 TY 9TX = 4TY 1 3 .   0 1  T     0  0T 200 450 0A T  104 100 4 A 2 X  LFP   0B T 106 100 6  B 3 5 . = constant (for all temperature scales) UFP  LFP ( 1 4 . Clearance = R'–R but 2R' = 2R (1+ sT) where ()  R' – R = RsT LFP  lower fixed point ( ) = (6400) (1.2 × 10–5) (30) = 2.3 km Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 UFP  Upper fixed point ( ) X  5  C0  60  5 C  C = 65°C 1 5 . x= A  B  A 1  AT  B 1  BT    A  B 95  5 100  0 95  5 =  AA  BB 100 6  10 5 1 6 . For rod A (A )   =  A(100) 6 . L = 6 × 10–5 = L   = 1  12  106 =5°C 0 For rod B (B )  = 20B(100) 2 7 . Expansions of a metal is same as photographic For rod C (C ) enlargement.  d1 will increase by 0.3% 2 = xA(100) + (30 – x)B(100)             5 4  x 3  d1, 0.3%     & 30–x = 0 3 0 1

JEE-Physics 1 7 .    Expansion in dx = 0 = surrounding's temperature (  )   0 2 1.76  106 x dx 80  60 80  60   1x dx T  100  105  1.2  t  k  2  30  ...(i) 0 = 100 1.76  10 5  x  1.2  10 6   x2  2 and 60  50  k 60  50  30  ...(ii)  2  t  2    0 =3.76 mm  t = 48 sec T2 2 4 .  t  (x22 – x12) For x1 = 0, x2=1 cm 1 8 . d = 0dT     d   aT  bT2  0dT 7  (12–02) For x1 =1cm, x2 = 2cm T1 71    =a b  3 7b T13   t  (22 – 12)    t = 21 hrs 0 2 T22  T12  T23  T13  =  2 aT12  3  0 t3 3 1 9 . For simple pendulum (   )T = 2  T 1  1 25. From Stefan's law ( ):   = AT4    Q 2 2 g  T  1 / 2  T  1cal 4.2J s  m2 cal Assuming clock gives correct time at temperature    5.67  10 8 1  T4  T= 100 K Power = Intensity () 26. 6 1 6 Area 0  24  3600 = 2  0  20 & 24  3600 Power absorbed by the foil = Intensity at foil × Area = 1   4 0  0   0  30C of foil× 2     1.4  10 5 C 1  P  P0 A   A 0 T 4  A 4 d 2  4 d 2  20. Now P'= A 0 (2T )4  A  4P 4 (2 d )2 20°C 27  Q    Q t    40°C A t B    1  20C   K A A 100  70 = K B A 70  35  0 1  1  20S  30 70 (1+20 C)= (1+20S)  1  0  KA = KB KA 1 2  = KB 2 1  2  31 2 1 . r=1+31=2+32  2= 3 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 x dx dx dx 0  28. dRH = KA KA 0 (1  x) 2 2 . Let  = junction temperature ( =   ) Net heat current at junction is zero ()   RH =  dx 1  n(1  x)  dR H  0 KA 0 (1  x)  KA0   200  0 3k(100–)+k(0–)+2k(50–) =0    3 °C = K 1  n (1   0 ) 2 3 . Newton's law of cooling ()  A0 OR  Check dimensionally (   ) t = k[ – 0] 2

JEE-Physics 2 9 . According to Wien's law   3 6 . Heat lost = Heat gained m 1/T  mT. ( = ) As the temperature of body increases, frequency msteam × 540=1100× 1 × (80–15)+20×1× (80 – 15) corresponding to maximum energy in radiation (m)  mass of steam condensed = 0.13 kg increases. Also area under the curve () ((m)  3 7 . Water has high specific heat and due to this it ) absorber more heat in rise of temperature.  E d  T4 (  ) Q 3 8 . When water is cooled to form ice, the energy is 3 0 . t = AT4 = same but Tred < Tgreen released as heat so mass of water decreases. as red Tred =  Tgreen green (see VIBGYOR) (   Area of red star is greater )  () 3 1 . Rate of cooling of water = Rate of cooling of alcohol 3 9 . If intermolecular forces vanish water behaves as gas. (= ) ( )   (250  10)  1  (5)  (200s  10)  5 Number of moles of water = 4.5  103  250 130 67 18  Specific heat of alcohol () ()  s = 0.62 Total volume of water at STP 3 2 . Amount of energy utilised in climbing (STP) ( ) = 22.4 × 250 × 10–3 m3 = 5.6 m3 mgh= 0.28 × 10 × 4.2 4 0 . Heat removed in cooling water from 250C to 00C h= 0.28  10  4.2 = 1.96 × 10–2 m = 1.96 cm (250C  00C     )  60 10 = 100 × 1 × 25=2500 cal Heat removed in converting water into ice at 00C 3 3 . Entire KE gets converted into heat. (0°C  )  ()  = 100 × 80 = 8000 cal Heat removed in cooling ice from KE = ms   10 × 10 × 10 = 2 × 4200 ×  ()    = 0.12°C 00 to –150C = 100 × 0.5 × 10 = 500 cal Total heat removed in1hr 50min Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 3 4 . M= mass of hallstone falling (1 hr 50 min    )  (M=) = 2500 + 8000 + 500=11000 cal m = mass of hallstone melting Heat removed per minute (   ) (m= 11000 = =100 cal/min As Mgh = mL. 110 m gh 10 103 1 41. P P P So   33 103  M L 33 3 5 . HC = ms = ms (1)°C T T T HK = ms  = ms (1)K = ms(1)°C HF = ms  = ms (1°F) = ms(5/9)°C density density  HC = HK > HF increases decreases For equation :  P= RT  Mw  3

JEE-Physics At constant temperature   P 48.   n1CP1  n2CP2 For 1st graph       P At constant temperature. n1Cv1  n2Cv2 For 2nd graph :  P= R T  5R 7R Mw   = 1.5  CP1 = 2 ; CP2 = 2 1 3R 5R At constant P,   CV1 = 2 ; CV2= 2 ; then n1 = n2 T dP 4 9 . vrms  3KT T  m v 2  T  m v 2 For 3rd graph L: dT = constant  P  T rms rms  density  = constant m 3K 4 2 . If temperature is doubled, pressure will also be 5 0 . Change in momentum (  ) R T = 2mv cos(45°) 45° 1 45° doubled as P  M w ()  100% increase   =2 × 3.32 × 10–27 × 103 × 2 4 3 . Volume can't be negative. = 4.7 × 10–24 kg ms–1. ( ) At constant pressure (  ) 5 1 . Here V = aT + b where a,b > 0 V  T or V  (t + 273) nRT nR bb So P =  but  so P2 > P1 aT  b a  b / T T2 T1 3  4 4 . th volume of air at 0°C occupies entire volume at  , 5 2 . PV= nRT  P = Mw RT 4 (0°C )  V1  V2 3 / 4V V  PM w  (105 )(28  103 ) m–3= 1.25 g/litre T1 T2 kg RT 8.3  273 As     = 171°C 273  60 273   4 5 . Ideal gas equation (  ):PV = nRT 53 . U1 = +ve; U2 = 0 So at V = V0; RT1 =  P0  (V0) and at V = 2V0, U3 = –ve  2   U1 > U2 > U3 As volume increases, W = +ve. RT2 =  4 P0  (2V0)  T2 – T1 = 11P0 V0 5 4 . Internal energy and volume depend upon states.  5  10R () 4 6 . Number of moles remain constant 5 5 . PT11 = constant & PV =nRT Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 ()  V T V 12  P1 V1  P2 V2  P1V1  P2V2  V  T12   12  v   T R T1 R T2 RT1 RT2 V T V T n1 + n2 = n1' + n2'  PV  PV  (1.5P )V  (1.5P )V 5 6 . U = 2P0V & W = P0V 273 273 R  273 R  T So Q = W + U = 3P0V = 3P0V0  T = 273 × 3 K = ( 273 × 3 – 273) °C= 546°C 5 7 . When water is heated from 0°C to 4°C, its volume decreases. 0°C 4°C   4 7 . Total translational KE()    33  P V is negative ( ) = 2 nRT = 2 PV Hence Cp – Cv < 0  Cp < Cv 4

JEE-Physics 5 8 . V  T4  V (PV)4 C1  Q1   U  W1   P4V3 = constant  PV3/4 = constant  T   T T    1 (W2 > W1) RR C2  Q2   U W2   C =Cv + 1  x = 3R + 1  3 / 4 = 3R + 4R = 7R  T   T  T  59. feq = f1n1  f2n2  f3n 3  (5n)(3)  (n)(5)  (5n)(6)  50 9. Q = U + W n1  n2  n3 5n  n  5n 11 Q = +ve, as heat is absorbed from the atmosphere 6 0 . U = a + bPV = a+bnRT    U = bnRT = nCvT W=–ve as the volume decrease  Cv = bR  Cp = bR + R ()    Cp  bR  R  b 1 Cv bR b  U = Q – W = +ve – (–ve) =+ve  Internal energy increases. ( ) EXERCISE –II 1 0 . HA = (6 cal/s) × (6 – 2) s HB = (6 cal/s) × (6.5 – 4) s 1 . All dimensions increase on heating. HA 4 8 ()  HB   L21 – 2L1L1  2.5 5 4 4 2. DC2 = L22 –  0 = 2L2 L2  0 = 2L2(2L2) – 2L 1 (1L 1)  1 =42 1 2 . For insulated chambers (  )  4 n1 + n2 = n'1 + n2' 3 . A part of liquid will evaporate immediately sucking (final pressures become equal) latent heat from the bulk of liquid. Hence a part of ()  liquid will freeze.() PV 2P.2V P [3V ]  P '  5P   RTRT RT 3 )  For left chamber () 4 . Qvap = Qfreezing PV = P'V'  5P V '  V ' 3V m.(L) = M(L)   M = m 3 5 L = latent heat of freezing (    ) For right chamber () m = mass of vapour ()  M = mass of freezed   4PV= P'V' = 5P V '  V '  12V 3 5  Fraction of water which freezed 13. P2 = constant ( ) ()  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65  M  M   P = p RT (Ideal gas equation) (  )  m  M m  m 1   M 6 . Mixture may attain intermediate temperature or  P2  P  RT   PT  R  = constant terminal temperatures of fusion or vapourisation.    M   M  ( The graph of the above process on the P–T ) diagram is hyperbola. 7 . Water at 4°C has highest density (P-T) For the above process () (4°C) 8 . Q1 = U+ W1; Q2 = U + W2  P2   P2   P2  P22  P2  P ...(i) Ratio of specific heats         /2 2 ()  1 2 and 5

JEE-Physics P  (vrms )A  MB  MB  1 P1T1 = P2T2  PT = 2 T2  T2 = 2 T ...(ii) (v rms )B MA 16MB 4 1 4 . (i) PV2 = C  TV = C mA If volume expands temperature decreases. No. of mole of A = M A ()  No. of mole of B mB  mA/2  8nA = MB M A /16 V3 (ii) P= KV2  = constant Pressure exerted by a gas in the vessel depends on the number of molecules present inside. T If volume expands, temperature increases ( ()  ) PP 2 0 . Average speed ( ) 1  2  ....  N N(N  1) N  1 vavg =  V V N 2.N 2 (i) PV2 = C (ii) P = KV2 Q = U + W rms speed () Q2 > Q1 as W2 > W1 & U2 > U1 12  22  32  ......  N2 vrms = N 15. Ideal gas equation ( ) P  .R T  N(N  1)(2N  1)  (N  1)(2N  1) M 6N 6 For state A (A): P0  0 R T0 v rms  (N  1)(2N  1)  2 2  2N 1 M v avg   N  1   6 (N  1) 6 For state B (B) 3 2RT 8RT 3RT 3P0 = M R2T0    2 0 2 1 . vP = ;v  M ; v rms  M M 1 6 . Q = U +W  +Q = U + P0(V2 – V1) 2 2 . Fig A : WA = +ve Fig B : WB =+ve Fig C : WC = +ve Fig D : WD = –ve 1 1 1 1  In process Fig–(D), heat is released.  Q = U + P0  2  1   U = Q + P0  1  2  2 3 . Q = U + W and U = nCVT 17. vA  3RT w 3RT v  w2  2 U can be zero if T is zero or Q – W is zero m  3 vB  2m v2 3 (U TQ– W ) Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 1 8 . (vrms)L = (vavg)R 3RT 8RT  M1  3 2 4 . At constant volume, work done by gas is zero. M1 = M 2 M2 8 () 4KT 2 5 . For any process (   ) U = nCVT 1 9 . Average KE per molecule in A & B = In adiabatic process (  )   Q = U + W = 0  U = –W (AB) For any process () (vrms)A = 3RT 3RT M A ; (vrms)B = MB U W Q U P  V  C=   nT  n  T  nT nT nT 6

JEE-Physics For Q = 0, C = 0 (adiabatic process)  Heat absorbed (  ) () 9 11 2 6 . Slope of isothermal process (   ) = RT0 + RT0 = RT0 2 2 P P m1 =  V V 3 2 . CP – CV = R; M (0.20 – 0.15) = 2 [M = molar mass] 2 Slope of adiabatic process (   )   M = 0.05 = 40 g P P CP m2 = V   V = – m1 = C v m1 CP 0.2 42  1   CV 0.15 3 f 2 7 . WAB = (2V0 – V0)P0 = P0V0  f = degrees of freedom (  )=6 [Isobaric process] [ ] WBC = n R (2 T0 )n V0 = 2P0V0.n2 3 3 . For constant pressure process Q 2 V0 (Q ) [Isothermal process] [ ] = nCPT = 30nCP 3 For constant volume process Q 2 P0 V0  P0 V0 2P0 V0.n2 (Q)  Q AB   5 5  WBC 4 n 2 n  7  = nCVT = C P T  T = 42°C = 42K 2 8 . PV = C; n P + nV = nC 34. Wadiabatic = nR (T1  T2 )  6R  nP = – nV + nC  y = mx + c  1  m = –  = – [2.10  2.38]  1.4  1  R (T1  T2 )  6R  T2 = (T – 4)K (1.30  1.10) 5/3 1  The gas is diatomic () 2 9 . WOBC = WODA 35. Ans. (C) UCA = nCV(TA – TC)  Net work done ()  3 = WOBC – WODC = 0 = 1× R× (300 – 450) = ( –225 R) 2 3 0 . Final pressure ( )  kx0 UAB = nCV(TB – TA) 3 S = 1 × 2 R × (600 – 300) = +450 R Workdone by gas = P.E. stored in spring = 1/2kx02 UBC = nCv(TC – TB) (=)   3 = 1 × R × (450 – 600) = – 225R Change in internal energy  2 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 U = |–W|= 1/2 k x 2 0 As gas expands, T is negative. (T)  UABCA = UCA + UAB + UBC = 0 3 1 . Work done = Area of ABC with V–axis 36. A T (V,P) V0 (=V -ABC)  P0 B P = P0(2V0 –V0) + 0 = P0V0 = nRT0 = RT0 V0 V 2V0(Let) Change in internal energy = nCVT P0 2 ( ) 2V0 V = 1  3 R  4 T0  T0   9 R T0 Let V0 = initial volume = 2V0 = final volume 2 2 7

JEE-Physics (V0 = =2V0 =  41. Ans. (A) V = volume of any state For body A (A ) P = AT4 (V= P  100    3004 A then P0  P0 / 2  P  P0 / 2 2 V0  V 2V0  V For body B (B)  nRT = PV = V  3P0  P0  P  (T4 ) = (1–0.5 – 0.3) × [ × 3004]  2 2 V0 V A   = 0.2 × 100 = 20 W/m2  T–V curve is a parabola with vertex above. 4 2 . From Stefan's law of cooling : (T-V)  () Hence temperature first increases then decreases. ()  3 7 . Work done in process 1–3 is greater than that in eA(T4 – T04) = ms   dT  process 1–2. While change in internal energy is same  dt  for both processes  Q2 > Q1.  1 × 5.8 × 10–8 ×  × (0.08)2 × (5004 – 3004) (1-31–2 4.2 ×   dT     dT  =0.067° C/s  =10×90×  dt   dt  ) 4 3 . For surface areas to be same 3 8 . Intensity in first case () () a 2 ( 1) A T14 Ssphere= Scube  4 R 2  6a2  3 P1  4 d12 R I1 = 4 R 2 1 Volume ratio () Intensity in second case ()  I2  P2  A T24 Vsphere  4 R 3 2R 6 Vcube 3  ( 1) 4 R 2 4 d 2 2 2 a3 a  A T14 A T24  d2  T2  2  (Mass of water in sphere) > (Mass of water in cube) d1  T1  Given I1 = I2  4 d 2  4 d 2  (>)  1 2 Energy host by radiation depends on the surface 39. For sphere ( )kA   dT   P area. Hence initial rate of energy loss by the two  dr  area equal. But mass of water inside the sphere is greater, hence it will cool slowly. dr.P T r2 dr.P     k (dT)  dT (  A  k  4 r2    0 rr1 )   kT  P  r2  r1  kT 4 4 . For same rate of heat transfer the body having higher Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 4  r1 r2   t = (r2 – r1)  P 4R2 conductivity will have lower temperature difference. If cylinder with higher conductivity is connected with 4 0 . For black body   hot reservoir first then the function temperature Tb, will be closer to hot reservoir temperature. 0T = 0 T '  T1 = 2 T 2 (  P AT4 1 Tb    P'T' = (16P) (2T) = 32PT ) P AT '4 16 8

JEE-Physics 4 5 . Newton's law of cooling implies that rate of cooling is 5R proportional to temperature difference if the 4 9 . For the cube, net resistance = 6 temperature difference between body and () surrounding is small. ((Where R = thermal resistance of each side) (R= ) ) H  100  0 Then, 5R / 6   d  = tan2  (2 – 0) and For side A (A  dt  2  d tan1 (1 – 0)  tan 2   2  0  H  100  A  A = 60°C  dt  1 = tan 1  1  0  3R 4 6 . From newton's law of cooling. 5 0 . Heat current flow rate is uniform everywhere. () ( )  A(T4 – T04) = ms   dT   dt  5 1 . Heat lost (  )  .4r2[(T0 + T)4 – T04] =  4 r3c   dT  = A (T4 – T04) = –4r2(T04 – T4) 3  dt   d(mL )  L dv  L 4r2 dr  12T03  dT dt dt dt   Pr c  (T – T0) =  dt dr  –4r2 (T04 – T4) = L4r2 dt t T dT    radius decreases with time K(T – T0)dt = – dT  k dt   0 T1 (T  T0 ) ()  T = T0 + (T1 – T0)e–kt where K = 12T03 rc 47. KA   dT       A   dT    5 2 . TP = 50°C; TQ = 45°C  dx   T   dx   Heat will flow from P to Q. Q Q (  PQ)  T2 dT   5 3 . For same power of radiation Q2 T dT Q x ()      PA = PB = PC  eAATA4 = eBA TB4 = eCATC4 T1 11 T  A dx ....(i) &  T  A 0 dx ...(ii) & ATA = BTB =CTC (eA : eB : eC = 1 : 2 : 4 ) 0 T1  TA .TC  TB or e A TA e C TC  e B TB & A .C  B  n  T1      L n  T1      x  T2    T   Q   Q    A and A   T2  x / L  T1  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65  T = T1 48. H  T1  T2 and 2H = T1  T2 R 54. R  KA 1 KA  KA ;R' 1  3R R' L /2  L/2 4L KA  4 R R' 2 3L / 4 KA L/4 (where R & R' are thermal resistances). (RR' )  HI  T  1.2W A BA C R C B R  1 R' 1 L 3L kA k 'A  kA kA 3L 3L 7k HII = T  4 T = 4  1.2  1.6W  k' = (k' = cond. of ADB wire). 3R / 4 3 R 3 3 9

JEE-Physics  KA(4  0) 3KA(0  4)  100  0  55. Q    x = 7.5 m  LL  10  x x 61. Fig A : 20  4 ...(i) ...(ii) –4°C  kA  kA  ICE X 3K Fig A : 20 = Q = (100 – ) kA  kA  t 0°C  L L  WATER (10–X) K +4°C 4/2 Equation (i) (ii) 1 = 2 / t t = 1 min. 5 7 . Rate of cooling ( )  = ms   d  = 4AT03T  TA  TB TA  TB k1 .k2  dt  LL k1  k2 62. Q    k3  L   4A × T03 (50 – 20) = 10 k3A k1A k2A and 4A × T03(35 – 20) = ms   d = m s  0.2  T2  T1 KA (T2  T1 )  KA (T2  T1 )   dt   60  x  4x 3x  x  63.  =   f Q  ms = 60  15 10  = 1500 J/°C KA 2K.A 0.2  30  then f = 1/3 5 8 . For a grey body  6 4 . For an elemental spherical shells,  + r + t = 1 if  = 0.4, r = 0.6 ()  then t = 1–0.4 – 0.6 = 0.  dT   r2 dr T2    dr  The body is opaque ()   Q Q = K4r2  r2  4K dT r1 T1   r2  r1  = 4K(T1–T2)     r1.r2  Q  r1 r2   r2  r1  Q 5 9 . Wien's displacement law.(  )  1T1 = 2T2 = b = 2.8 × 10–3 km 6 5 . P = P0 – aV2  3000 × 1 = T2 × (2)  3 × 10–3 km  1 = 1 m & [2– 1] = 9 m From ideal gas equation (  )   2 = 10 m  T2 = 300 K PV = nRT  RT = (P0 – aV2)V (n=1) 6 0 . (i)  = K1A (  25)  K 2A (25  ) T=  P0 V  aV 3   dT  0   P0  3aV2   R  dV  R  Q t1 t2  KA(  25)  K A (25  )   = –5°C  V P0 and d2T 6aV 2 3 3a dV 2   R (< 0) –25°C Q +25°C (P0  aV 2 )V  2P0 P0 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 R 3R 3a  Tmax = t1=2 t2=3 K1 K2 6 6 . Let x= percentage of water solidified then heat lost = Heat gained (ii)  = K1A (  25)  K 2A (25  ) (x= Q t1 t2 = )  x × 3.36 × 105 = (100 – x) × 21 × 105 2A(  25) 3A (25  )  x 100  86.2% 2 3      = 0°C 1.16 10

JEE-Physics 6 7 . Q1 + 36 = Q2 Q1 7 2 . At 30° true length is given by (30° )  KA (100  )  36  KA (  4) LL Q2 = SR (1+zinc T)   = 76°C =100 (1+26 × 10–6× 30) =100.078 cm At 0°, True length is given by (0° ) 6 8 . He and Ne are monatomic gas.   SR 1  zinc T 00.078 (He Ne) = 1  glass T  1  8  106  30 Vf =100.054 cm Vi 69. Q = W = nRT n 7 3 . Tensile Stress ( ) Q 1500  = (Ysteel)   YS T T  nRnVf / Vi  0.5  25 / 3  n3     b  S  1500  360K = 200 × 109 (0.8 × 10–5) (200) = 3.2 × 108 Nm–2 0.5  25 / 3  1 = 0.32 GNm–2 7 0 . Area under the curve is equal to number of molecules 74. = V  V0   3 t of the gas sample. A0 A0 (75. Requiredheat Available heat ) 10 g ice (00C) 5 g steam (1000C) 1 800 cal  2700 cal Hence N = 2 a V0  aV0 = 2N 10 g water (00C) 5 g water (1000C)  1000 cal 1 V0  a  10 g water (1000)  1  2 So available heat is more than required heat therefore final temperature will be 1000C. Vavg = N 0 vN (V )dV  N 0 C. V0 V  dV  3 V0  Vavg 2 (   100°C ) V0 3 Mass of vapour condensed (   )  1  1 v0 V2  a  V02 800  1000 10 N N 0  V0 V  2 = =g V 2N ( V )dV 540 3 0  V2  dV  rms  Vrms  1 Total mass of water (   )  V0 2 10 40 1 = 10+ = =13 g 3 3 3 Area under the curve from 0.5 V0 to V0 is 3/4 of Total mass of steam (   ) total area (0.5 V0V0 3/4 )  10 5 2 = 5 – 3 = 3 = 13 g Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 7 1 . Length ()   0 1  T  =  (1+20) 76. At normal temperature (  )  0 Area ()  f5 Cv= R= R A=A0 (1+T) = 6  2 (1+40) 2 2 0 At any temperature (  )  Volume () V=V0 (1+T) = 3 (1+3T)=03 (1+60)  f  1 f 0  2 2 R=R Density   Cp – Cv= – = 1 0  0 from process ( ) T = k1V2  T 1  60 and ideal gas equation (  )PV = nRT 11

JEE-Physics we have PV–1 = constant   x = –1 EXERCISE –III R RR  C = Cv + =Cv + =Cv + 1x 1 1 2 Tr ue/False At normal temperature (  )  1. v = 3RT rms M w (False) 5R C= 2 R+ 2 = 3R 3RT 3R(2T) 3RT 7 7 . PV2 = constant  PV–2  P  2 V 2. v = M w , v 'rms  Mw /2 2 Mw =2v (False) PV rms rms  Bulk modulus ( ) 3. C – C = R  C > C (Ans  True) p v p v K= P  2P 4 . Energy radiated per second = AT4  V V () Q1  1  2  4000  4 1  Q2  4   2000   16  1   (1 6 )  1 (Ans  F) As PV = nRT So K  V2 and K  T2 5 . (PE) > (PE) (Ans – True) rared compressed 7 8 . CPmix = n1Cp1  n2Cp2  n3Cp3 6 . Equal volume at NTP contains equal molecules. n1  n2  n3 (NTP) (4 )  7 R   2  5 R   1(4R) 16  2   2 7 . Higher temperature means higher internal energy   R ()  4 2 1 7 R3 R Match the column 79. C= CV + = R+ <0 1x 2 1x 1 . When A & B are mixed ( AB) 5 3x 100  1  x < 0  1 < x < 1.67 ms (T – 20) = (2m) s (40 – T)  T = 3 = 33.3°C When A & C are mixed ( AC) ms(T – 20) = (3m) s (60–T)  T = 50°C When B & C are mixed ( BC) (2m) s (T –40) = (3m) s (60 –T)  T = 52°C When A, B & C are mixed ( A,BC) ms (T –20) + (2m)s(T–40) = (3m) s (60–T)  T=46.67°C 2 . Isobaric process   Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65  P = constant () Isothermal process    T = constant   U=0U=constant () Isoentropy process   Q  S = =0  Q=0 T No heat exchange ( ) Isochoric process    V = constant ()  dW = PdV = 0 12

JEE-Physics 3 . Let R = R then R = R = 2R as R =  7 . Initially rate of heat flow will be maximum at A and minimum at B as there is no temperature difference BC AB AC kA across section B. 100  TB  TB  0  T = 67.7°C (A B  2R R B B )   Q    Q  and  Q   2  Q     t    3  t  dQ t AB BC t AB AC In steady state dt will be same. 4. For(A) (dQ)  12 2 dt as PV = mNv2 = E so in P = E, 3 rms 3 3 dQ dT In steady state  KA = same E is transtational kinetic energy of unit volume. dt dx (E)  dT  dT  1  dT  dt  dx  A  dx  For (B): &  0    In U = 3RT, U is not internal energy of one mole as will be maximum at B & minimum at A for monoatomic gas U = 3/2 nRT (U (BA)  ) For (C): 8. Q As Q = nCdT and dT = In W= P(V – V ); w is work done in isobaric process. nC fi (w) Therefore molar heat constant C is the determining For (D) : factor for rate of change of temperature of a gas as In U = nCvT heat is supplied to it. It is minimum for isochoric process U is change in internal energy for every process. a monoatomic gas  C V  3 R  , resulting in greatest  2  (U ) of 5 . From given V–T graph we cannot tell the nature of  dT  gas ( V-T) slope  Q  i.e. curve 1. slope of V–T graph (V-T   )=nR (C,  P From graph  nR    nR    n    n  C V  3 R    P  A  P  B  P   P  2  A B  Cannot say anything about n A & PA dQT n B PB 1) Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 RT 6 . Isothermal bulk modulus = P = For isobaric process of monoatomic gas and isochoric process of diatomic gas, their heat capacities are same ()V 5RT  5 R  , therefore both are represented by curve 2. Adiabalic bulk modulus = P= 3V  2  () For isobaric process of diatomic gas 7 that is CP  2 R Slope of PV graph in isothermal process P RT represented by curve 3.Q axis represent isothermal V V2 (PV =– =– process and T axis represent adiabatic process. Slope of P–V graph in adiabatic process ( (PV)=– P   5P 52R    V 3V 13

JEE-Physics 2 No heat current flows through rod CD CP  7 R (CD ) 2  3 RR 2R Q T So B E B E  2R/3 )  BE R/2 R/2 R 4 V0  V 2 4 V0  15 V02 = For (B) V0 PdV  2VdV =  9 . 15 units W R 2R/3 R/2 13R/6 F A BE V0 F A From PV = nRT, 2V2 = nRT For (C)   2 V22  V12  nR T   nRT = 30V 2 Total heat current form A to F, 0 (AF) U  nC V T  nR T  30 V02 = 30 12 = 30 5  I = 100  0  600  1  1 2 13 13R 7/5 1 R = 75 units 6 Q = 75 + 15 = 90 units Let temperature of B be T then Molar heat capacity (  ):  B R 5R (BT ) B R5 C= C + = 2 R+ 1  1 = 2 R  2  3R V 1x I  100  TB 600 700 C R   T = 13 13R B 25 For (D) = 3 × 3 = 25 units As heat current is inversely proportional to heat resistance.  1 0 . P = M W RT (  )  For (A) : So heat current in BD (B D ) For AB P  V  T  V2  T  –2 =  2R  I  2 For BC V = constant   = constant   2R  3I For CA P = constant  T = constant R For (B) For AB P  T   = constant  TB  TD  2I 2  600  R/2 3 3  13R  For BC T = constant  P  TD  700 200 500 C For CA P = constant  T = constant   For (C) 13 13 13 For AB P = constant  T = constant Comprehension Based Questions For BC T = constant  P Comprehension#1 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 For CA V = constant   = constant For (D) 1 , 2 Let heat is supplied at a rate of k cal/min then For AB  T  P T2 1 For BC T = constant  P For A  = constant P  (k cal/min)  1 k × 1 = m(0.5) 11 k1 × 4 = m(80) k1 × 2 = m(1) (2)  k = 20 m  1= 40°C, 2 = 40°C 1 C Therefore initial temperature = –40°C 11. For (A) R R ()   R RR R/2 Final temperature ( ) = +40°C Ry  KYA  2KXA  2 AB EF R/2 R/2 D 14

JEE-Physics Comprehension#2 0 0 2  1   L T  T 1 1 . P  V  PV–1 = constant   x = –1 1   L T 2 L  T    R 5R R  C = C+   =3R v 2 2 2 1  Temperature= T+ 2. Q = nCT = n(3R)T = n(3R)(3T) = 9nRT = 9P V 11 L Comprehension#5 Comprehension#3 1 . On increasing temperature (giving heat), U increases. 1 . Since P is constant = 1 atm, heat added will cause Now r increases for A while decreases for B. temperature rise.From the phase diagram, A will avg sublime while B will first melt and then boils. ( ,U A (P=1 atm, A   B)  r Br  avg avg 2 . From the phase diagram, at 2 atm & 220 K, A is gas 2 . The equilibrium remain unchanged but average distance increases. & B is solid.(2atm220 KA B)  () 3. We have ravg   1.0003  0.9 9 9 9  r0  2  Comprehension#4 1 . Let M be the mass of solid (M) (from equilibrium position) ( )  1M  r  T    2  105 / K r0  Volume displaced ( ) = 2  s M Comprehension#6 Thrust force ()=L  g  Mg 2s  L = 2S 2. Ans. (D) 2 . If Liq > solid Also L Vg = Mg Q = 1  4200  2 J / sec = 420W t 20 As T , L, so V displaced  T, L, V  Fraction of solid submerged should increased. d 2 1 dt 20 400 () 3.  K  TS  TL    K 40  K  3. If fraction of solid submerged doesn't change, then 40 d t  1 t 400 ()60 20  K dt ; n 20   40 60 0  0 V0 1  3T   constant   liq  3S  n  20    1 t  t  277 s 1   Liq T  40  400 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 4 . If h doesn't change (h)Com pr e h e ns io n# 7 V = Ah  L=2S 5 . If volume change in solid is zero. 1. As Q = nACvAT = nBCvBT  nC = nC A vA B vB () Let at T', solid sinks (T) But volume is constant ( )  So  PAVA = nART & PBVB = nBRT  0 V g  VS g Initially   nA  PA 2.5 5  C vB 5 5 / 2R 2 nB PB   C vA   1.5 3 3 3 / 2R Finally () TVg = VSg= 0  Gas B is diatomic & gas A is monoatomic 2 (B A) 15

JEE-Physics 2. As nA = 5 125 5  60  C Pmix  1.56 3 nB so M A  3  M B   mix C Vmix  5 M = 4M  Gas A = Ar , gas B = O 4 . Internal energy ( ) He = 100 J B A 2 3 . Number of molecules in A (A) Internal energy (  ) H = 200 J while mixing, they don't interact 125 () =nN = 40 N A  3.125N A Internal energy of mix (  )  AA 125 3 = (100 + 200) J = 300 J 4 . U=nC T=   2  300  2812.5 cal Comprehension#10 V 40 2 Comprehension#8 1. Q1  T1  Q1   T1  Q 2   300  (80)  87.9kcal 1 . At temperature above 4°C, temperature of water Q2 T2  T2   273  above is less as compared to below as water is heated by radiation of longer wavelength. (4°C     87.9 80  2.W=Q1 – Q = – = 7.9 kcal = 33.18 kJ 2  ) 3 .  T2 273   10.1 T1  T2 27 2 . At temperature below 4°C, temperature above is less as compared to below & thus water remain is it is C o mp r eh ens i o n# 11 due to higher volume at the upper surface. (4°C 1.Change is entropy  S =Q  )  T  Unit of entropy ()= JK–1 3 . Ldm  T  kA , dm  Adx 2 . When milk is heated, its entropy increases as it is dt x irreversible process. (  LAdx  TkA x2  Tk  t )   L dt x 2 Thickness, x  t1/2 3 . After a long time disorder is increased. 4 . Transition of ice starts from the top & decreases below (    )  to the bottom. () Comprehe n s ion#1 2 Comprehension#9 1 . Stress developed at junction are same 1. M mix  n1M1  n2M2  5 4 2 2  24 g ()  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 n1  n2 7 7 Y111T = Y222T  Y111 = Y222 53 2 . As cross sectional area is same & equal and opposite  2R   5R force acting on both rods. 2. C Vmix  2 2  fmi x R  fmix  3.57 2 ( 7 ) 53 75 So F/A= same.  2R   5R  2R   5R 3 . C V mix  2 2 , C Pmix  2 2 3 . Let shifting in junction be x towards right then 7 7 16

JEE-Physics (x) EXERCISE –IV A     1 1 T  x ,     22 T  x 1. h2 =L 2 – 21 = constant    1    2 24 2 1  h  2L1L1 =0 2 2 4    Y2      2L L2 – But Y1       1 2  12  4L2 (L22) =L1(L1 1) 12 (Y11  Y22 )T  4L222  L211 So x = Y12  Y21 2 . Here: Comprehension#13  R  d   = L(1+ 1) ...(i)  2  1. Q = 0  UBC =–W =–400 J  d  d/2 BC BC  2  d/2 R   = L (1+2)...(ii)  2 . For complete cycle ( ) Q =W R  Q + Q + Q = W + W + W d 1   1  2  T  AB BC CA AB BC CA   2    700 + 0 +(–100) = 700 + 400 + W  R   CA  W = –500 J (1  2 )T CA 3. = Wnet 600 600 Since  1  2  T  1 Q input  100   100   85.71%  2  700 7 dd  Comprehension#14 = (1  )T (1  2 )(t2  t1 ) 1 . 450 = m(0.5) (150)  m = 6g 2 . L = Q  800  450  350  175 cal/g Temperature difference 3 . (i) Thermal current = RH m6 63  (= R H ) Q 200 10 3. s= m T  6(240  150)  27 cal/g°C b gwhere 1 = K1A + K2A = K1  K2 a2 RH   a Comprehension#15 = (60 + 40)(3 × 10–2) = 3 WK–1 T1  T2  Thermal current ( ) T1 1.   0.5  T1  560K = 80 × 3 = 240 W (ii) Ratio of thermal currents (  ) Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 ' = 0.7 = T1  T2  T1 ' 280 2800 HCu KCu 60 T1 T1  = = = 1.5 HAl K Al 40 3  T' – T= 1120 1 1 3 4 . In steady state rate of flow of heat in the whole system will be same. 2. max = T1  T2  100  600  300  100  50% T1 600 ( )  KA(200  1) = 2KA (1  2 ) = 1.5KA(2  18)    T2 1 T2 1 1000  200 –  = 2 – 2 & 2 – 2 = 1.5 – 27 T1 6  100 3 3 112 12 2 3. =1–  '=1– T1   T2  K  1 = 116°C, 2 = 74°C 17

JEE-Physics 5. Q = KA 1  2  t  mL  m × 335 × 103 9 . By using Wien's displacement law  (   )  = 0.01 × 0.54  45  6  60  60 m = 0.261 kg b 2.89  10 3 T= m = 1.5  10 6 =1927 K 5  102 Therefore mass of ice left in the box after 6 hours (6)10.(i)PA = P B = (4 – 0.261) kg = 3.739 kg  e AA T4 = e  AT4 A A B BB HFG KJI GFH IJK 1/4 1 eA 0.01 4 6. T = eB TA = 5802=1934K B 0.81 K1 K2 K3 10°C +20°C (ii) According to Wein's displacement law          F I5802 Power required ( ) HG JKATA= T  B= AB = 3A B 1934 B  = B K eq A T1  T2 Also  – = 1 m  – 3 =1m   = 1.5 m  BA B B   A T1  T2   1 2 . Let m = mass of steam required per hour =  1 2 3     (m=)  K1 K2 K3    Heat needed (  )   = (10 × 1000 kg) × (1cal/g) × (80–20)  = = 60 × 104 cal/hour A T1  T2 1  2  3 Heat supplied (   ) k1 k2 k3 = m × 1 × (150–100) + m × 540 + m × 1× (100 – 90) = 1 3 7  3 0  = 9000 W Heat needed = Heat supplied  2.5  1  25   10 2 ( =)  0.125 1.5 1   60 × 104 = 600 m = 1000 gm = 1 kg Q KA 20  10 2KA 10   1 3 . Heat needed to bring ice to freezing point 7 . = same So = t  ()   2 =10 = 5°C = 10 × 0.2 × 20 + 200 × 0.5 × 20 = 40+2000 = 2040 cal 8 . (i) Temperature gradient ( ) 2040 Time taken to reach 0°C = 100 s = 20.4 s = T1  T2 100  0  = =100°C/m (0°C) 1 (ii) Steady state temperature of element dx : Heat needed to melt ice dx ()  –20 29.50 T = 100 (1–x) =2040 + 200 × 80 =18040 cal20.4 Heat absorbed by the element to reach steady state 180.4 240 ()   temp time dQ = (dm)s T = dx s T  0 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65  dQ = 20 [100 (1–x)] dx Total heat absorbed by the rod ()  1 Time taken to melt ice ()  18 Q =  dQ  2000  1  x dx =1000 J 0

JEE-Physics 18040 1 7 . (i) At triple point (temperature = –56.6°C) & = = 180.4 sec pressure = 5.11 atm), the solid, the liquid & the vapour phases of CO co–exist. 100 2 Heat taken in 4 min (4)   (   =–56.6°C)  = 5.11 = 100 × 4 × 60 = 24,000 cal ,CO  Let  2  = final temperature then (  ) ) heat lost = heat gained (  =  ) (ii) Fusion point & boiling point both decrease on 18040 + 10 × 0.2 ×  + 200 × 1 × (–0) = 24,000 decreasing pressure. 24000 18040 ()    = 202 = 29.50°C (iii) Critical temperature is 31.1°C and critical 1 4 . When A & B are mixed ( AB)  pressure is 73.0 atm. On temperature higher than critical temperature it can't be liquified. 12°C 19°C 28°C A B C ( 31.1 °C   73 .0  ) (iv) (A) Vapour (B) Solid (C) liquid (A)(B)(C) 16°C 23°C =? 1 8 . No. of moles (initially) = No. of moles (finally) mS 12 + m S 19 = m(S + SB)16 3S = 4S ...(i) () = A B A B A 76  V0  76  V0  PV0  PV0 When B & C are mixed ( BC)  273 273 273 335 mS 19 + mS 28 = m(S + SC)23 4SB = 5S ...(ii)  P = 83.83 cm of Hg B C B C 1 9 . P\" = P' + 5 cos 60° = (P' + 2.5) cm of Hg when A & C are mixed ( AC)  mS 12 + mS 28 = m(SA+SC)   = 20.3°C AC 1 5 . Heat lost ( ) = 3  1 m v 2  PP P’ 46cm 4  2  5cm 60° Heat gained ( ) = ms(327–27) + mL P” 44.5cm  As heat lost = Heat gained (  =  ) For constant temperature process : So 3  1 m v 2  = ms × 300 + m × L () 4  2   46  44.5  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 3 P ×  2  = P' × 46 = (P' + 2.5) (44.5)  v2 = 0.03 × 1000 × 4.2 × 300 + 6× 1000× 4.2 44.5  2.5 8  P' = 1.5 & P = 75.4 cm of Hg  v = 12.96 m/s 1 6 . Heat gained = Heat lost (  = ) 2 0 . Let m = mass of neon gas then (m=)  100gm m gm n=  m  28  m  from PV = nRT water steam  20 40  24°C 10°C 90°C  m 28  m   20 40  100× 1× (90–24) = m× 540+m×1× (100–90)  105 × 0.2 =  × 8.314 × 300  m = 12 g m = 4.074 g; m = 23.926 g Ne Argon 19

JEE-Physics 2 1 . No. of moles withdrawn () )  m 1 =n –n = So MV 2 = nC T 20 v 12 M  m= M(n – n) = M  P1 V1  P2 V2   V= 2nC v T  2C v T = 36 ms–1 1 2  R T1 RT2  0 M Mw 15  105  30  103 11  105  30  103  2 5 . For gas trapped in the tube (  )   32  300  8.314  290  8.314  =0.139 kg PV = PV = PV 11 22 33  (76–h) A =(76–hcos60°) A'= 76 × A 2 2 . (i) Dotted lines correspond to ideal gas  66 × 40 = 71 × ' = 76 × \"  = 37.2 cm & \" = 34.7 cm () (ii) T1 > T2 (On high temp. real gas behaves as ideal gas) ( 60° h )  h (iii) PV  m .R  10 3 × 8.314 = 0.26 J/K  T M 32  103 h n R TA  2R  300  300R P PA 2 R R5 R 3 2 6 . VA = V0 = 23. = 0.4  C =   C = C – R = 2R A B P v P CP 0.4 2 (i) Atomicity = Monatomic, V= n R TB 2R  400  400R  4 D C B PB  3 V0 V0 V (= 2 Degree of freedom = 3   =3)  V = nRTC  2R  400 = 800 R = 8 V C PC 1 30 (ii) Cv = 3   CP 5 n R TD  2R  300 R CV  PD 1 V= = 600 R = 2V 2 3 D 0 (iii) Mean gram–molecular kinetic energy For cyclic process () ( ) 3 U = 0, Q = W = R× 300 = 450 R W =P (V –V )  2  4 V0  V0   105  2V0  105 2 AB A B A  3  3 2 4 . (i) Let n = number of collisions per second per unit area, (n = W=P V n2 =2 × 105 × 4 V0 n2  8V0 n2 × 105 BC B B 33 ) Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 change in momentum (  ) W= P (V –V ) = 1 0 5  2 V0  8 V0    2V0  105 CD CD C  3  3 = 2mv  Pressure exerted on wall ( ) W = –P V . n2 = – 105 × 2 × n2 DA DD n (2mv) = = P W = W + W + W + W = 1152 J 0 AB BC CD DA  n  2  32  10 3 3  8.3  300  105 (i)  Q = 1152 J (ii) W = 1152 J (iii) U = 0 6.02  1023 0.032  PV 2 7 . (i) P V =nRT and P V = nRT 11 1 22 2  n = 1.95 × 1027 2 (ii) If vessel is suddenly stopped then KE will utilized in 1 increase in temperature. (      V 20

JEE-Physics  PV > PV 22 11 2 8 . (i) T = PA VA  5  104  10  120.3K  nRT > nRT1 T > T A nR  2000  2 2 1  4  For same volume, P V = nRT  8.314 11  T < T & P V = nRT 2 12 2  P < P T = PB VB = 2T = 240.6° K P(104N/m2) 1 2 B nR A (ii) P V =P V =2PV=P2V  T = T PC VC 10 B C 22 11 1 2 nR TC = = 2TB = 481.3° K 5A D For state 3: 10 20m3 V P Let V = 3V/2 2P 2 T = PD VD = T = 240.6°K 3 3 D nR B P = 3P/2 then 3 3P 3V 9 P1 (ii) No. we can not predict the direction of reaction. PV =  PV V 2V 33 2 2 4 () V  T > T & T > T (iii) Process ABC : 3 2 3 1 W = PV = 10 × 104 × ( 20 – 10) = 106 J (iii) P = mT + C  P C m P T 2  2000  3R  U = nCVT =  4   2  × (T – T ) CA  P1  C   P2  C  1  T1   T2    = 2.25 × 106 J Q = 3.25 × 106 J T Process ADC  From ideal gas equation W = 5 × 104 × (20 –10) = 0.5 × 106 J () U = nC (T –T ) = 2.25 × 106 J V CA  Q = 2.75 × 106J P1 V1  P2 V2  P1V1  T1   P1  C  2 9 . (i) W < W (Area under PV–graph gives work) T1 T2 P2 V2 T2  P2  C  AC ABC (PV)  C (ii) U =1 0J  V1  1  P1  A V2  C < 1 V < V Q = 200 J P 1  P2  1 2 AC 15 10 B C 1 W = × (6–2) × (15 + 5) AC 2 V1  C V2  C v 4  20 5A T1 T2 = =40 J 246 V (iv) V = mT–C ;  2 1 2 T Q = U+ W 200 = (U – 10) + 40 C then P V = nRT  UC = 200 + 10 – 40 = 210 – 40 = 170 J 11 1 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65P1 V1  nRT1 (iii) Q = UAB + W= (U – U) + W P2 V2 nRT2 AB AB B A A B 4× 105 8× 105 N/m2 N/m2PV=nRT2= (20– 10) + 0= 10 J 22 P 3 0 . (i) Work done by gas  P1  T1  V2   V1  C   V2  () P2 T2 V1  V1   V2   C P1  C 1 1 0.2 0.5m3 V P2  1  V1   C = (4+8) × 105 × (0.5–0.2) 1  V2  1  P1 > P2 2 = 1.8 × 105 J (ii) Increase in internal energy ( ) 21

JEE-Physics U = nC T = n R ( T2  T1 )  P2 V2  P1 V1 P V = P V     VB    PA 3 V  1  1 A A BB  VA  PB  2  (8  0.5  4  0.2)  105 = 4.8 × 105 J For Process C  D (CD ) 5 1 PV= PV  3 CC DD (iii) Amount of heat supplied ( )   VC    VB   3  PD 3  VD   VA   PC P Q =U + W = 6.6 × 105 J   P= (iv) Molar specific heat of the gas 2 D 2C ()  At end points A and D (A D  Q 6.6  105  R  PA  PD  3PC   3PC   T = 500 K = n T = (P2 V2  P1V1 ) = 17.1 J/mole–K TA TD  2  1000 TD D 3 2 . (i) For adiabatic process ( ) 3 1 . Given that PA T V –1 = T .V –1 T = 1000 K 11 22 A n=1 B  300 V5/3–1 = T2.(2V)5/3–1 T = 189 K 2 C 2 D V (ii) Change in internal energy ( ) P =  3  P,  = 5/3 B A 3 U = nCVT= n R(T – T) 2 2 1 1  2  2 /5  3  P =  3  P,  0 . 8 5 3 C A = 2 × × 8.314 × (189–300) = –2767 J For process AB (AB ) 2 (iii) Work done by gas () PA1 TA  PB1 TB P1 V1  P2 V2 n R ( T1  T2 )  1  1 = = = 2767 J 1  3  2 / 5  2  TB = TA  PA   =1000 ×  PB  3 3 . (i) P (ii) A P0 = 1000 × 0.85 = 850 k For Process B  C (BC ) P0/2 C B V0 2V0 V PB  PC  T = T  PC  = 850 × 1 TB TC C B  PB  = 425 K 2 (iii) Work done by gas () (i) W= nR (TA  TB ) VB Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 AB  1 VA W = nR(T ) n 3RT n2 AB A = A  1  8.314  (1000  850) = 1870.2 J W = P0 (V – 2V ) 5  1  3 BC 2 0 0 (ii) QBC = UBC + WBC= nCVT + 0 =– P0 V0  n R TA 3 2 2 =– 2 RTA 33 W =0 = n. R(T – T )= 1 × × 8.314 (425 – 850) CA 2C B 2 WABCA = RT (3 × 0.693 – 1.5) = 0.58 RT A A = –5300.175 J  QABCA = W= 0.58 RT ABCA A (iii) For process A  B (AB ) 22

JEE-Physics 3 4 . For polytropic process ( )  3 6 . Number of moles ()  T (5.66V)n–1  2  PV 1.6  106 0.0083 16 T V n–1 = T V n–1  TVn–1= = 5.66n–1 11 22 2 n= RT = 8.3300 = 3 Taking log both sides ()Heatis supplied at cosntant volume n2 = (n–1) n 5.66 (  2 so Q= nCVT  n = 1.4 = 1 + f  f = 5 Q 2.49  104 (i)  Degrees of freedom (  )=5  T = T + nC V = 300 + (ii) Work done by gas () 2 1 16  3R  3   2  = 300 + 375 = 675 K = – P1 V1  P2 V2  PV  P2 (5.66 V )  12.3PV P2 T2   1 1.4 1 As V = constant So P1 = T1 PV PV  1 1.4  675 22 11  Where =  P = P  5.66  2  P=   (1.6 × 106) = 3.6 × 106 Nm–2 2 3 0 0 3 5 . (i) P 37. PV = constant  (105) (6)5/3=(P ) (2)5/3 2 A B (TA = TC)  P2 =(105)(3)5/3 Nm–2 C P1 V1  P2 V2  1 W= V0 2V0 VV (ii) Process AB (AB):        105 6  103  105  35 / 3 2  103 VA  VB =  972J TA TB 5 1 3 38. n A C PA  n B C PB  = n A C VA mix  n BC VB T= VB T = 2T = 600 K B AA VA  5   7  Process BC  BC): 19 1  2 R   nB  2 R   13 =  3   5   n =2 mole  2   2  B 1 T V –1 = T V –1 R  nB R BB CC  V = 80 2 Litre =113 L 3 9 . (i) For a cyclic process () C Q + Q + Q + Q = W + W +W +W Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 For end states A & C A C 1 2 3 4 1 234 PA .VA  PC VC = nR  P = 0.44 × 105 N/m2  W4 = Q1 + Q2 + Q3+ Q4 – W1 – W2 – W3 TA TC C = 5960 – 5585 – 2980 + 3645 – 2200 + 825+1100 (iii) Work done (  ) = 765 J W = P (2V –V )= nRT = 600 R; W (ii)   Wnet  W1  W2  W3  W4 AB A 0 0 A BC Q given Q1  Q4 = nR (TB  TC )  3 nR(600–300) = 900R 2200  825 1100  765  1 2 = 5960  3645 = 0.1082 23

JEE-Physics 4 0 . PT = constant  P2V = constant  PV1/2 = constant   4 2 . PV PV 103 10 3 For this process C = C + R = C + 2R = nRT 1 V V  n= RT = 25 / 33 25 1 1/2 5 P  C = 37.35 –2 (8.314) = 20.722 = R B V 2 f5  R= R  f=5 2 2 AC 4 1 . The maximum temperature of the gas will be during V process BC. For process AB (AB) (BC ) W = 0, Q = U = nC V T Process BC can be represented by straight line, y = mx + C 1  3  297  25   2  2 (BCy=mx + C = R (300–3) = = 148.5 J ) For process BC (BC) So P = mV + C Putting point B & C gives  1   25 3P = 2mV + C ...(i) P = 6 mV + C ...(ii) Q = 0, W = nR (T –T ) =  25   3  [300–3]  1 P 12  5 1 So subtracting 2P =–4mV So m = –  3  2V = 148.5 J and U  W  148.5 J For process CA (CA) U  0 W = nRT n  VA  = – 6.9 J and Q = W = – 6.9 J  VC  Thermal efficiency ( ) P Wnet 148.5  6.9 From (ii) P =  .6V + C  C = 4P = Q =sup plied 148.5 =0.954 2V Hence we get equation as (  ) y    P  x  4P ...(iii) 4 3 . Slope ()  2V   P mP where y is pressure and x is volume of gas. =– V = – tan 37°  V = tan 37° Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 (yx)  V   3   4105   3  3  m =  P   4  =  2  105   4  = 2 = 1.5 Putting y from above. Now we have x y = nRT    P x2  4Px = nRT ....(iv)  V For maximum temperature (   ) 4 4 . U  nC V T , Q = nC P T and dT  2P x  4Px  25 2500 dx  2V W = nRT = (1)  3  (100) = J 0   = 0 3 Hence x = 4V Putting in (iii) U  n CP  R  T  nCPT  nRT We get nRT = 2P(4V) = 8PV max 8PV  25 500 = 1000 – (1)  3  (100) = 3 J So T =  x=8 max nR 24

JEE-Physics 4 5 . P A + Mg = P A T1=300K T2=500K EXERCISE –IV B 12 n=1 n=1 Extra force needed P1,2V P'1,2V 1 . (i) For the right chamber () ( ) P2,V P'2,V  2 4 3 P0  1  = P2A  P1A  Mg n=1 n=1  32  P 1–T  = .T2 = P2A  P1A  P2 A  P1A 00 = P2  A – (P1)A  2 4 3 P0  .V2  32  T = 9 and P0 .V0  2 4 T0 T0  9   n R T  nRT  A 5000  4 T0  =  V 2V  =N 3 9 32 8 ()  V = 4  243 V0  27 V0 47. New length of gas column 2 h h 9h For the left chamber () 2 + 16 16 = = P0 V0 P1 .V1 243 T0 T1 32 P0  where P = 1 New volume of gas ()  =  9 h  A = 9 8 46  16 V V1  2 V0  27 V0  27 V0 16 0 P = P + kx  T = T0   243 P0   4 6 V0  9  23 0 1 P0 V0  32   2 7  T 16 0 As PV = constant so P V 1.5 = P 9 V0  1.5 = 12.94 T 00  16  0 (ii) Work done by the gas in the right chamber : () 64 37  9   P= P = P + kx  kx = P  4  27 0 0 27 0 1  R T0  T0 P1 V1  P2 V2  nR (T1  T2 )  h  h   37  1  1 5  1  3 But kx = (3700)  16  so (3700)  16  =  27  (105) 35   × R × T = –1.875 RT Now T V –1 = T2 V2 1 2 40 0 11  9  0.5 2 . Final volume of chamber ( )   16   (273) (V )0.5 = (T ) V0 = V + Ax = 3.2 × 10–3 m3 0 2 0 T2 = 4 Final pressure in chamber (  )   × 273 = 364 K Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 3 kx = P + = 2 × 105N/m2 0A From ideal gas equation (  )  P1 V1  P2 V2 T1 T2 T = (P V )  T1  = 800 K 2 22  P1 V1  Work done by gas () 0 .1  Kx   A   P0  Adx = 120 J 0 25

JEE-Physics Change in internal energy   Process C  D Q =0 Process D  A CD  U = nC T v Q = nCvT DA  U   P1 V1   3 R  T = 600 J 5 R  (T – T)  R T1   2  =1  2  A D  Heat Supplied (  )=120 + 600 = 720 J = –108313.753 J  Efficiency  3. work output  Qin  Q out = Heat input Q in    Q out   100%  1  108313.753   100% 1 Q in  185156.937    = temperature of disc (  )  = 41.50% instead of V /V =2  = constant temperature ( ) CD Note : please read V /V =½ 0 CD Heat input to disc ()=KA(0  ) in the question. L 5. A 5°C D C 95°C Heat utilised by disc ( )= ms d 95°C h h2 dt B h1 5°C (where s = specific heat of disc) Pressure at the bottom of A – B limb : (s=)  350 KA t (A-B)  dt  d   ms d  KA(0  ) P + 95gh1 = P + 5gh ...(i) dt L 0 B 300 0   msL 0 Pressure at the bottom of C–D limb : msL  0  300 (C-D) KA  0  350  t  n  t = 166.32 sec P +  gh = P +  × g × h (P = P )...(ii) B 95 05 2B C 4 . Let V =V P 2 B0 (Let)P0 Solving we get,  = 2 × 10–4 °C–1  = × 10–4 °C–1 Process A  B 3 B C 6 . Heat flow for three sections will be same. D T V –1 = T V–1 () AA B T = 909 K 27°C 1 2 0°C B Process B  C A V0 8V0 16V0 V 1st Air Space 2nd Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 PANE OUT SIDE VB  VC  T= 7272K PANE TB TC C Room Process C  D : 27  1  1  2  2  0 L L L T V –1 = T V –1  T = 5511.15 K   CC DD D Heat flow (  )  KA 1st Pane  KA  A.S.  KA  2nd Pane Process A  B Q =0  Q = 26.48°C ; Q = 0.52°C Process B  C AB 1 2 Q = nC T Heat flow rate ( ) BC P 7 d 27  1 27  26.48 = 1 × R (T – T )   dt  L   0.01  2 CB  KA  1st Pane  0.8  1 = 185156.937J = 41.6 watt 26

JEE-Physics v0 v0 9 . Work done by gas () 2 7. xx 1   P0 Adx   kx.dx 00 Let () kx2  50 J =P Ax + n = no. of moles in vessel–1 1 02 (n =1 )  1 k n = no. of moles in vessel–2  50 = 105 × 4 × 10–3× 0.1 + (0.01) 2 (n =2 )  2 1 P = initial pressure in both vessels Spring constant  k = 2000 N/m 1 (P =) Heat supplied (  ) 1 Q = U + W = nCVT + W P = final pressure in both vessels. 2 = n R T 2  8.314  50  50 = 1295 J (P =)    1 +W =  5  1 1  3 Initially ( 1 0 . In free expansion, temperature remains constant volume of vessel–1 = volume of vessel–2 () –1–2) n1RT0  n2RT0 n =n V 3V P1 P1 12 Finally, V1  n1RT1 & V2  n 2 .R T2   n1R  T2 P2 P2  P2  Initial temperature () T = P0 V0 V1  T1  275 & V + V = V + A V2 T2 271 1 2 0 0 nR Displacement of mercury droplets After compression ( )  ()  T (4V)–1 = T(V)–1  T = 2T 00 Change in internal energy () L  V1  V2  4  V0  A   0.26m R A 546  A  U= nCV.T = n    1  8 . For each state T  n R (2 T0  T0 )  1 P () 3P0 D(3T0) C (2  105 )  (103 ) 2  (1.5  1) = 400 J n  P0 V0  P0 .(7V0 / 2) 3 P0A(7 V0 / 2) 3P0 V0  n R T0  P0 V0 T0 TB P0 TCT0 TB 7T0  1  1  2D Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 1 1 . For equilibrium ()  V0 7V0 V 7 21 5 T = T; T = T B 20 C 20 T = 3T . m1 m2 D0 h1 h0  Heat absorbed ( ) h0 = (nC . T)A  B + (nCvT)BC h2=0 P = n4R 5 T0  + n 3R(7T ) = 31.nR T = 31P V m1g +gh0= m2g + g h  2  0 0 00 A1 A2 0 Work done   =(3P – P ) (V – 7 21 V) m1  2kg, m2  1kg  A1  A 2 20 0 0 0 = –5P V 00 27

JEE-Physics For final equilibrium ()  P0 A  P  A2 m1g  m2g 258R TR ...(ii) ...(iii) A2 A1 For the CO gas (CO    )  22  m block will fall down (m ) P0 A  P  A3 1 1 For constant temperature forcess and pressure being 430R TR constant ()  V =V final  258 1  430 1 initial  1  1 h A + h A = h A + h A  h = 30 cm & h = 0  645 2.5 &  645 1.5 01 02 11 22 1 2  3 2 1 2 . Let h = empty space over Hg–column   +  +  = 3   (1 + 2.5 + 1.5)= 3 i 123 1 (h =Hg)  1   = 0.6  and  = 2.5  = 1.5 For constant temperature process 1 2 1      = 1.5 × 0.6 = 0.9 3 () For the entire system  PV = PV  4 × h = 5h U1 + U2 + U3 = 0 11 22 1 2 where h – h = 1  h = 4 & h = 5  n1R (T  T1 )  n1R (T  T2 )  n3R (T  T3 )  0 1 2 2 1 1 1 (2 1) (3 1) True Faulty  P  0.6  P0  (P  1.5  P0)  (P  0.9  P0)  0 Reading Reading  7 1  5 1  7  1  5   3   5 73 69 75 70 74 x (i) Total length of tube ( ) 13  P  12 P0 = 69 + 5 = 74 cm or 70+4=74 cm 1 4 . For compartment C (C ) (ii) When faulty barometer reads 69.5 cm.  4 V0   27 (69.5cm)  9   P  8 P0 PV = P 00 P V = P V ; 4h = 4.5 h; h = 4.44 P0.T /1– = P.T/1– 11 22 1 0  True reading = 69.5 + 4.44 = 73.94 (iii) When the barometer reads 74 cm P T –3 =  27 P0   T 3  T 3 0 0  8  =T 20 (74 cm) For compartment A (A ) PV = P2V2 4 × 5 = (74 –x)2  x = 69.528 cm 11 27 7 P= P Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 1 3 .  = 372°C;  = –15°C;  = 157°C  CO2  5 1 23 A 80 P0 V0  P1 V1 RT0 RT1 For the H gas (H )  27 P0  2 V0  4 V0  R T0 21 22 8  9  P0 V0 T P1 V1  P2 V2  P0 A  P  A1  T =  =T = 40 R T1 R T2 645R TR 1 1 R ...(i) For the He gas (He)  For compartment B (B ) 28

JEE-Physics P0 V0  P1 V1  P0 V0  P1 V0 21 1 5 . For the process () T0 T1 T0  4 P0  21   P1  4  T0 5 V nC T = n RT = Ua v 2 a V (i) Final pressure in A (A) 27 P  nRT = PV = 2 V P  2 a 5a 5 V 80    W  PdV  4 a V Vf Final pressure in B (B) 21 P Vi 5 40  U = a V Vf  100 Final pressure in C (C) 3P Vi 20 (ii) Final temperature in A (A)= 21 T0 4  W = 5 U = 80 J 4 (i) W = 80J Final temperature in B (B   )= 21 T0 Q = U + W = 100 J + 80J = 180 J 4 Final temperature in C (C  )=3 T0 (ii) C = Q U W U  4 U 5 2   nT nT nT (iii) Heat supplied by heater = (U + W) all chambers (= (U+ W) ) = 9  U   9  5 R  9 R 5  nT  5 2 2 Q = (U + W ) + (U + W ) + 0 11 22 = (U – W ) + (U + 0) 13 2 = U + U – W 1 23 = n1 R T  n 2 R T   n3RT  1 1 2 1   3  1  17 17 = P V + P V + P V = 18 P V 2 00 2 00 00 00 (iv) Work done by gas in chamber B = 0 (B) Work done by gas in chamber C (C)  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 nRT = –U    1 = –P V 00 Work done by gas in chamber A (A) = (–) W = –(P V ) = P V chamber 00 00 (v) Heat flowing across piston –I (-I)  17 = U2 = PV 2 00 29

JEE-Physics EXERCISE –V-A n1  n2  n1  n2  mix  1 1  1  2  1 6. 1 . The temperature of gas molecules depends on the average kinetic energy associated with the disorderly Number of moles in 16 g He =4 motion (i.e., random motion) of the gas molecules. (16 g, He =4) The orderly kinetic energy of the molecules of the Number of moles in 16 g O2 = 0.5 gas container will increase in the lorry, whereas disorderly kinetic energy will still remain the same; (16 g, O2 =0.5) hence the temperature of the gas molecules will 57 remain unchanged.  He  3  O2  5 (Onreplacing nHenO2 ;  He,  O2     mix  1.62 7. We have, molar heat capacity = molar mass × )  specific heat capacity per unit mass ( = x 2. v rms  3RT  ) M  C = 28 C (for nitrogen) and C = 28 C pp VV Where T is the temperature of the gas molecules Now C –C = R or 28C – 28 C =R pV pv in kelvin and M is the molecular mass of the gas C –C = R M  p v (T 28 ) 8. U = U + U2 (n +n ) CT = n C T +n C T 1 12 v 1 v1 2 v2 TH2  TO 2 TH2 320    T  vH2  vO2  M H2   P1 V1  P2 V2 T1T2 MO2 2 32 P1 V1 T2  P2 V2 T2  TH2  20K 7 9. v  RT  v1   1M 2  4 5 3. n1  n2  n1  n2 M v2 2M1 5  32  1 1 1 2 1 3  1 1  1  1 460 21 200  1 5 7 5  1  v2  25  8  v2  460 m/s / 3 1 / 21   2  3  5  4    3  24 10. Ans. (4) 1 2 2 2 16 Energy of the diatomic gas 4 . Given P  T3 ()   PV = RT 5 nRT  5 PV  5  8 104   = 5 × 104 J Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65  P  (PV)3  P3V3P  P2V3 = constant 2 22 4  PV3/2 = constant    CP 3 1 1 . Q = ms = 0.1 × 4184 × 20 = 8.4 kJ CV  2 5. Monoatomic ( ) n 1= 1 5  1 2 . n1C v1 T1  n2C v2 T2  n 3C v3 T3 1  3 7 =(n1 + n2 + n3) CVmix T 2  5 Diatomic () n2=1 n1 T1  n2 T2  n3 T3 n1  n2  n3 n1  n2 n1 n2 As C V1  C V2 so T =  mix  1 1 1 2 1     mix  3 2 30

JEE-Physics 1 3 . Specific heat at low temperature is 1 5 . All reversible engines work for different values of temperature of source and sink hence the ()  efficiencies of all such engines are different.The incorrect statement is all reversible cycles have same  T  3 efficiency.  400  Cp  3 2 (  Q   m.c.dT  )  4 100  32  T 3 dT 20 1000  400  16. Ans. (3) When water is cooled to form ice, the energy is  32  1  T4  released as heat so mass of water decreases. 10 (400)3  4  (   32  1 204  44  )   (400)3 4 10  32  1  160000  256  17. 1 T2 1 T2  0  T2 0 or T =   (400)3 4 T1 T1 10 = 0.002 kJ Which is not possible ()  = T2 = Q2 1 8 . Heat can't flow from the body at lower temperature T1  T2 W to body at higher temperature is a consequences of  20  0.02 W = 0.028 kJ second law of thermodynamics. (  300  20 W  4 0.002  )  W = 0.0148 kJ 300  4 W 1 9 . The instantaneous thermodynamic state of matter 14 . This is free expansion of the gas in which is denoted by pressure, volume and temperature. temperature will remain constant i.e. on the other side temperature will T. (  )  ( T)  2 0 . Tsource  627C  627  273  900K P,T Vacuum Tsin k  27C  27  273  300K Efficiency () Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 v/2 v/2   1  Tsink  1  300  1  1 Tsource 900 3 Number of moles ( ) n1 P V / 2 K T   2  Output  Work P 'V Finally number of moles n2  KT 3 Input Heat Input ()  2  3 × Heat Input = Work n1 = n2 2 PV P1V P' P  × 3×106 × 4.2 = Work  2 2 KT 3  Work = 8.4 × 106 J 31

JEE-Physics Internal energy and entropy are state functions and not path functions.  2 1 . 26.  n1C v1 T1  n 2C v2 T2  n1C v1  n 2 C v2 Tf A(ssumeT >T)Tfn1Cn1vC1 Tv11  n 2 C v2 T2  n2 C v2 22. 12 12 1 5 R T0 1  3 R  7 T0  6 R T0 3  2 2  3  4R 2 T0 Heat given = Heat taken   1 5 R 1 3 R 22 1C v1 (T1–T)= 2Cv2 (T–T2) 2 7 . Work done in adiabatic process Here C v1 = C v2 &  = P1 V1 , = P2 V2 (    ) 1 T1 2 T2 W  R T1 – T2     1  R T2 – T1  2 3 . The first law of thermodynamics does not introduce  –1 W the concept of entropy. + 103  8 .3 7  = 1+0.40 = 1.40 14 6  (=1 103 )  The gas must be diatomic 2 4 . Heat supplied Q = Area FBCEF (  ) (Q=FBCEF )  2 8 . Let  is the efficiency of heat engine and  is the T2 = 2T0 B corresponding coefficient of performance of a refrigerator working between the same temperature. T1 = T0 C (  A  FE The relation between  and  is S1 =S0 S2 =2S0 =T (S –S ) + 1 (T –T )(S –S ) (   ) 21 2 121 2 1 Output  =  –1=10–1=9 Also  = Input Work done W = Area ABC (W=ABC)  1 Energy absorbed from the =2 (T –T )(S –S ) reservoir at lower temperature 2 121 = 1 Work done on the system 2 (T2 – T1 )(S2 – S1 ) W       1 (  ) Q T1 (S2 – S1 )  2 (T2 – T1 )(S2 – S1 )  T2 – T1  2 T0 – T0  1 Energy absorbed Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 T2  T1 2 T0  T0 3  =9 = 10J So, energy absorbed ( )= 90 J 2 5 . The internal energy of a system is a state function, i.e., change in internal energy only depends on the 2 9 . As a thermodynamic system is taken from state i to state f; then the internal energy of the system initial and the final position and not on the path remains the same irrespective of the path followed chosen. ( i f  )  (Q –W)iaf = (Q–W)ibf  50 – 20 = 36 – W Hence U1 = U2  W = 6 cal 32

JEE-Physics 3 0 . WAB = Work done in isobaric process 35. 1  T2  1  T2  5 ... (1) T1 6 T1 6 () and 1  T2  62 1  T2  62 2 ... (2) = RT = 2R × (500 – 300) = 400R T1 3 T1 3 Work done on the gas = – 400R By solving equation (1) and (2) () T1 = 372 K and T2 = 310 K 3 1 . WDA = Work done in isothermal process 3 6 . V =V0(3) ()  P1 4 = 2.303 RT log P2 = (3.14)(10)3 [3 × 23 × 10–6] [100–0] 3 = 28.9 cc 1  105 37. Strain  2  1  t and Stress = 2.303 × 2R × 300 log 2  105 1 Y Strain = 2.303 × 2R × 300 × (– 0.3010) = – 414R Stress = Y  t Word done on the gas (   ) 3 8 . W = Area bounded by curve = P0V0 = – (– 414 R) = 414 R ()  3 2 . WABCDA = WAB + WBC + WCD + WDA 33 2 QAB = nCVT = n × 2 R × T = 2 P0V0 = R(500–300)+2.303R (500)log 5 2 QBC = nCP T = n × 2 R × T = 5 P0V0 1 Total heat supplied (   )  + R(300–500) + 2.303 R(300) log 2 = 276R 3 13 = 2 P0V0 + 5P0V0 = 2 P0V0 Work done on the gas (  ) W P0 V0 = – 276 R  = Q × 100 = 13P0V0 × 100 = 15.4% 3 3 . TB = T1, TC = T2,  = 1.4 40. = 1  T2  × 40 = 1– T2 T2 = 300 K VB = V, VC = 32 V T1  100 500 100 TB VB –1  TC VC –1 P A T1 B 60 300  –1 Again 1 T1  T1 = 750 K 100 TC T2  VB  TB  T1   VC  D T2 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 C 2T0 4T0 V 2P0 =  1   –1  1  32  4 T2 1 3 P0 T0 2T0 =1– T1 =1– = = 0.75 4 4 V0 2V0 3 4 . 1 Mv2  f RT and heat supplied = nC v (2 T0  T0 )  nC p (4 T0  2 T0 ) 22   1  2  T    1 Mv2  n.3 R T0  D5R (2T0 )  13 n R T0  13 P0 V0 2 2 2 2 f 2R 33

JEE-Physics 4 1 . Amount of heat required by a body of any mass of  = K undergo a unity change in temperature is known as T2  T1 = T2  T1 A heat capacity or thermal capacity of the substance. R 3x ( 1 Comparing with the given result f = 3 )  () 4 2 . Black board paint is more close to a black body. 4 7 . Rate of flow of heat (   )  ()  T   T1 – T2  = KA  x  = K ( 4 r r )  r2 – r1  1 2 4 3 . Infrared radiations are detected by pyrometer. () to Stefan's law ()  4 8 . According 4 4 . The power radiated by a sphere of radius R at Earth temperature T is r0 (T R) Sun R P = T4(4R2) Where   = emissivity of the material of sphere. r ( = )  = Stefan's constant ( ) T = Absolute temperature ( )  Power radiated by Sun ( ) R = Radius of the sphere (  )  =(4R2)T4 P1 T14 R 2  4000  4  1  2 1  16 Intensity of the sun received by the earth P2 1  2000   4  42 16     24   1 () T14 R 2 2 Power of the sun = 4r2 4 5 . According to Stefan's law, power radiated by a perfectly black body is (P= AT4;P=4R2T4) I44Rr22T4  R 2 T 4 r2 P2  R  2  T2  4  Radiant power incident on the earth = I r02 P1  R   T1   2  P2=64P1 () 1  where r02 is the projection of the earth's area x 4x 4 6 . T2 2K T1  receiving the energy from the sun.(r02    x  R2T4 r02 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 R1=Resistance of left part = P  r2 KA (R1 = 4 9 . Let the temperature of the interface be T0. 4x 2x (T0) R2=Resistance of right part =  T1 l1 T0 l2 T2 2KA KA (R2 =)  K1 K2 Total Resistance ( )  R1  R2 x 2x 3x Thermal current (I) ( )    KA KA KA Temperature difference = Thermal current ( ) Thermal resitance 34

JEE-Physics Thermal resistance ( )= 1  EXERCISE –V-B K A 1 . Average rotational KE (  )  T1  T0  T0  T2 1  / K1A  / K2A = 2 × kT 1 2 2  T0  K 12 T1  K 2  T2 K12 1 (for diatomic gas) ()  K 2  1 2 . Initial conditions P V = n RT, P V = n RT 1 12 2 5 0 . In steady state, temperature decreases linearly along Final condition (P1–P)2V = n RT the bar. ( 1 (P –1.5P) 2V = n RT 22 ) n1RT  P 2V () ddQt   d   dx  i.e. KA n2RT  1.5P  n1 1 2  mA  2 2V n2   mB 3 (  ) = d dt 1.5 3 51. Rate of cooling – = k( – 0) d d 3. For A : Q = nCPT =(nC ) (30)   0  0 P = –kdt      kdt For B : Q = nCVT = nCVT=nCP(30)  n( – 0) = –k t + C  correct answer is (2)  CP  7  T  30  C V  =30 × 5 =42 K stress 5 2 . Y = strain  stress = Y × strain 4 . At 2880 K : F  Y(T)  F = YST b 2.88  106 nm  K S m  T = = 1000 nm 2880K Therefore force by one part on other part () E U2 = 2F = 2SYT 5 3 . By Newton's Law of cooling T = TS + (TH – TS)e–kt  (nm) 500 1000 1500 ()  Therefore U > U & U > U TS  Temperature of surrounding (  )  21 2 3 TH  Temperature of body at t = 0 5.  v = RT sound MW (t=0    ) K  constant ( ) T  Temperature of body at time t  vN2   N2  MHe = 7 1 43 3 v He M N2  He    ( t) 5 28 1 5 5 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65  n1C V1  n 2C V2 2   5 R   4  3 R  11R n1  n2  2   2  6 So graph is 0 6. C Vmix = = = 24 t 11R  U = (n + n ) C T = (6)   T = 11RT 1 2 V mix 6 7 . T1V11  T2 V21 but V = AL T1 =  V2  5 / 31 =  L2 2/3 So T2  V1   L 1  35

JEE-Physics 8 . 1 T1  2 T2  3 T3  b so T1  T3  T2 1 1 . As P = constant 9 . The temperature of ice will first increase from –10°C so PV  nRT  V nR 1   = to 0°C. (–10°C 0°C) VT PV T Heat supplied in this process will be: Q = mS (10) 1i 1 2 . v sound  RT v1 m2 M W so v2 = m1 ()   m=mass of ice (  )  90°C S = specific heat of ice ( ) 0°C i 13. Then, ice starts melting ()   T Temperature during melting will remain constant (0°C) 90°C ((0°C) ) kA 90  T   kA 90  T   kA T  0 Heat supplied in the process will be Q = mL 2    ()   90 – T + 90 – T = T  3T = 180  T = 60°C L = latent heat of melting (  ) 1 4 . If dW = 0, dQ < 0 then dU < 0 Now the temperature of water will increase from 0°C The temperature will decrease ( ) to 100°C. (0°C  100°C)  1 5 . As  mono   dia so 2  monoatomic & 1  diatomic Heat supplied will be (  )Q = mS (100) 3w 1 6 . For complete cycle ( ) where Sw = Specific heat of water (S = ) U  0 so W = Q = 5 w  WAB  WBC  WCA =5 But W =0& Finally water at 100°C will be converted into steam BC at 100°C and during this process temperature again remains constant. Temperature versus heat supplied graph will be as follows W = 10 (2–1) = 10  W = 5 – 10 = – 5J AB CA 1 7 . At constant temperature PV = constant Temp. ( PV=) Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 100°C dV V    dV / dP  1 0°C So PdV + VdP = 0  dP =– P V P –10°C Q1 (Q1+Q2) (Q1+Q2+Q3) 1 8 . Black body radiates maximum number of wavelength Heat supplied and maximum energy if all other conditions (e.g., temperature surface area etc.) are same. So, when the temperature of black body becomes equal to the (100 °C100°C ,treamdipaetera,tmuraeximofumtheenfeurgrnyaacned, the black body will it will be brighted of all.Initially it will absorb all the radiant energy incident )  on it, so it is the darkest one. (  P 2 W >W >W  1 213  10. 3   V   V1 V2 )  36

JEE-Physics 19. Ans. (C) 2 6 . Power radiated ( ) For same temperature difference less time is taken Q = eAT4 and m T  b for x, this means e > e . (  x  So (300) T = (400) T = (500) T xy 123 e>e)   3T = 4T = 5T xy 123 According to Kirchoff’s law ( )a >a and A : A : A = 4 : 16 : 36 = 1: 4: 9 xy 123 14 9 is maximum. Q :Q :Q = : :  Q 2 0 . 1     a T  2S T A B C 81 256 625 B 1 1 s 1 s 2 7 . Net heat absorbed by water 160 Js1  2 = a  1  2 = a  s ()  =1000 –160 =840 J/s 21. P A PA Q msT  tt CB BC 2  4200  77  27 1000 Js1 T V 840 = t 2 2 . 2kg ice (-20°C) 5kg water (20°C)  t = 500 s = 8 min 20 s 20 kcal 00 kcal 2 8 . Heat transfer in warming of glass of bulb due to 2kg ice (0°C) 5kg water (0°C) filament is through radiation. 160 kcal 2 kg water (0°C) (  )  Final temperature ( )0°C The temper of sun is () 29. welding arc ature hi n turn a higher than that of Amount of ice melted whic greater than tungsten 100  20 filament. = = 1 kg ( 80  Final mass of water ()  )  = 5 +1 = 6 kg 2 3 . Temperature of liquid oxygen will first increase in the 3 0 . 1 calorie is the heat required to raise the temp. of 1g same phase. Then, phase change (liquid to gas) will of water from 14.5 to 15.5°C at 760 mm of Hg. take place. During which temperature will remain (760 mm14.515.5°C  1g constant. After that temperature of oxygen in gaseous 1°C) state will further increase. (V 32. V VT ) PT2 = constant & PV = nRT Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 P V T V 3 P3  V  T3  V = 3 T  VT = T   V 2 4 . P1  W < 0 & P > P  RT  P1  1M 2  4 3 1 M P2 2M1 3 33. P= V V1 V2  1  4  2  8 2 3 3 9 k 2 A   T1  T2  25.q L = 1  3 4 . For two rectangular blocks q L= kA T1  T2  q2 1 2  = ()  2 4 q1 37

JEE-Physics  MCQ’s R1  kA  2R ; R2 = R 2kT 8kT 3kT 2kA ;v  In configuration 1 (1 )  1. v = m ,v = m  vP  v  vrms P m rms Equivalent thermal resistance = 3R Average KE of a molecule ()  (   =3R) 33 = kT = mv 2 In configuration 2 (2 ); 2 4P Equivalent thermal resistance = 2 2 . There is a decrease in volume during melting of an R ice slab. Therefore negative work is done by the ice– water system on to the surrounding  W =–ve 3 ( (   =2 R ) )  3 Heat is absorbed during melting Rate of heat flow (   )  ()  T T  Q = +ve  U  Q  W = +ve Q1  3R t1 and Q2  2R t2 3  T t1  3 T t2  t2  2 t1  2 sec. d 3R 2R 9 3 . Radius of curvature ( ) R = 1  2  T 3 5 . Process FG is isothermal (FG  ) 4 . C – C = R Always constant ( ) pv so work done (   )=nRT ln  Pi  Cp  Pf  C v =  decreases with atomicity ()  32P0  (C + C ) and C .C depends on degree of freedom pv pv = 32 P V ln  P0  = 160 PV ln2. therefore it will be more for diatomic gas 00 00 [(C + C )  C .C  pv pv Process GE is isobaric (GE  )  ] So work done (   )  = P|V| = P |(V – V )| 5 . (i) For isothermal process curve should be hyperbola. 0G E () = P |(32V – V )| 0 00 (ii) Work done & change in internal energy are both = 31 P V negative. (  00 Process FH is adiabatic (FH  )  so (32P )V 5/3 = (P )V 5/3  V = 8V  )  00 0H H 0 (iii) For higher pressure volume is increasing and lower Since process FH is adiabatic so Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\01.Thermal Physics.p65 pressure volume is decreasing. (    (FH   )   PH VH  PF VF ) 8 1 6 . (A) From 0 to 100 k the major part of graph lies  P0 8 V0  32P0 V0 )  36P0 V0  5  1  3 Process GH is isobaric so work done in linear region and very small part in non-linear region, therefore to a reasonable approximation ( GH     )  between 0 K – 100 K, graph of C vs T is linear. = P |(32V – 8V )|= 24 P V (0 100K 00 0 00    0 K–100K 38