P-3 Allens Made Physics Exercise Solution

JEE-Physics Match the column Comprehension # 3 2 2 1 . Both the blocks remains in contact until the spring is 1 . y = A sin t + A sin t cos + A cost sin in compression. In this time system complete half 33 oscillation. By reduced mass concept time period of system A sin t + A3 cos t = A sin  t   2 2  3  (  Extreme mean Extreme ) x=6 2. 1 x=2 x=4 F = 8–2x = –2(x–4) T  2  2 2  2s k 3 . x is positive in II & IV; v is positive if fext > 0  Required time (  )= T  2  1s 22 (IIIIVx Compreshension # 1 2 . v1 2kg 2kg v2 1 . Max. acceleration of 1kg (1kg   ) (0 .6 )(1)(1 0 ) Let velocity of rear 2kg be v1 and front 2kg be v2 = = 6ms–2 then 20 = 2v2 – 2v1  v2–v1=10 (  2kg 1 v12kgv2) Max. acceleration of 2kg (2kg   ) Now by conservation of mechanical energy  (0.4)(3)(10)  4ms2 ( ) 3 1 11 Therefore maximum acceleration of system can be (2)(10)2 = (2)v12 + 2 (2)v22 4 m/s2 2 2 (4m/ s2)  v12+ v22 = 100 But v22 + v12 – 2v1v2 = 100 44 44  2A = 4  A = 2  (k / m )  54 / 6  9 m  v1v2=0  v1=0 as v2  0 1 Comprehension # 4 y 2 . 2A = constant  A  (0,R) v k   2  2 v 1.  T (2R / v) R x Comprehension # 2 a  2 . At t=0, x=0, v=v0 =R 1 . kx R a = R so cos(t + ) = 0 & –sin(t + ) =+1    3 f 2  mR2  1 Comprehension # 5 kx – f = ma, fR =  2    f = 3 kx 1 . As A is at its negative extreme at t=0 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 But x = A cos t (A, t=0) ( the cylinder is starting from x=A) so x–3 = 2 sin (2t + 3/2)  x =3 –2 cos (2t) (x=A)  y B (0,4) kA So f = cos t 3 x A (3,0) 2. 1 1  k2   1 m v 2  1  1  3 mv2 2 . As B is at its equilibrium position and moving towards kA2 = 2 mv2 1  R2  2  2  4 negative extreme at t = 0 2 (Bt=0   ) v 2kA 2 2 (1 0 )(2 )2 40 ms1   so y–4 = 0 2 sin (2t + )  y = 4– 2sin(2t) 3m 3(2) 3 55

JEE-Physics 3 . Distance between A & B (AB)  4.  y = Asint = 2 Asin t T = x2  y2 GHF 2IJK GHF IKJ  (3  2 cos 2t)2  (4  2 sin 2t)2  3 A = Asin 2  = Asin 4 2 T T  9  4 cos2 2t  12 cos 2t  16  4 sin2 2t  16 sin 2t 4   =  T =12s  3 4 2 t   5 5 T3  29  20 cos 2 t  sin 5 . Maximum distance(  )   29  20 sin(2t  37)   =2asin  2  =(2a)(0.9)= 1.8a Maximum distance ( )  6 . Velocity () v =  a2  x2 = 29  20  49 = 7cm v2  a2  x 2 49 a2 16 Minimum distance () v1 2 100 a2 9  = 29  20  9 = 3cm  a 2  x12   a=4.76 cm EXERCISE –IV(A)  length of path (  )=2a = 9.52 cm 1 . The maximum velocity of the particle at the mean 7 . Energy at all the three points are equal position () ()  v = A = A (2n) v1=8 v2=7 v3=4 max x  A = v max 3.14 x+1 2n = 2  3.14  20 = 0.025m If at the instant t = 0, displacement be zero so x+2 displacement equation is (t=0)  1 kx2 1 1 k x 1 2 2 y = Asint =Asin2nt =0.025 sin (40t) m 2 m v 2   m v 2  2  12 1 2 A AA 1 m 64   1 kx2  1 m 49  1 k x  12 2 . S = A cos t – 2 sin t – 2 cos t+ 8 sin t 2 22 2 A 3A 15  2  22 x ...(i) = 2 cos t – 8 sin t 1 2 1 kx2 1 2 1 k x  22 2 m v 1  2  2 m v 3  2 = 5A  4 cos t  3 sin t  5 A cos(t  37) 1 m 64   1 kx2  1 m 16   1 k x  22 8  5 5 8 2 22 2 5A 12  2  2 x ...(ii)  A'= 8 ,  = 37 from (i) & (ii)   =3 & x = 1/3 then, total energy is equal for the maximum kinetic energy 3 . x = Asin (t + ) Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 (  ) At t = 0 x = 1 cm  A sin = 1 ...(i) 1 m 64  1 1  1 m v 2  v max  65 m/s 2 m 9 2 max dx Velocity v = dt = A cos(t + ) 2 At t = 0  = A cos  Acos =1 ...(ii) from (i) & (ii) A2 (sin2 + cos2) =1+1 OR  v   a2  x2  8   a2  x2  A = 2 cm and tan =1   = 4 rad  7   a2  (x  1)2 ,4   a2  (x  2)2 After solving the equation (  ) Vmax= a  65 m / s 56

JEE-Physics 8. As v =  a2  x2 so v1 =n a2  3a2 na 1 2 . (i) At equilibrium position ()    42 dU F=– dx  0  2x–4 = 0  x =2m and v2 = 3 n A2  3a2 2 na = 4 dU (ii) F =– =–(2x–4) = –2(x–2)  A  3a  15 3cm dx 9 . x = 12 sint – 16sin3t  F–x  SHM = 12 sint – 4 (4sin3t) (4 sin3=3 sin – sin3) Here 2 = 2  T  2  2  2 s–1 = 12 sint – 4 (3sint – sin3t) = 12sint – 12sint + 4sin3t=4sin3t 2  motion is SHM with angular frequency 3 So amax = 362 (iii) a = a  2 6  2 3m 2 2 6 1 3 . Frequency of oscillation ( ) 1k 1 600 10 1 f   10. m v 2 =8 × 10–3 2 m1  m2 2 1.5  Hz m 2 ax 1 Let maximum amplitude be A then  m2a2 = 8 × 10–3 (  A) 2 v   A2  x2    4rad/s where x= diference in equilibrium position Therefore equation of SHM (x=   )   (     ) x=0.1 si n  4 t     m1  m2 g  m1g  1 m  4  k k 120 1 1 . (i) Maximum speed of oscillating body and v  0.5  3  1 m/s 1.5 ()  Therefore () 2 vmax = A = A × T  1  2 5 37 5 37 Here A = 1 metre, T = 1.57 s  2  600 6 1  20 A2  1 0 A   cm 2  3.14  1 1 4 . Common velocity after collision be v then by COLM  vmax = 1.57 =4 m/s (ii) Maximum kinetic energy (v  ) () 11 2Mv  Mu  v  u Kmax= m v 2 = × 1 × (4)2=8J 2 2 max 2 (iii) Total energy of particle will be equal to Hence, kinetic energy (  ) maximum kinetic energy. F I(  )  HG JK(iv) Time period of mass suspended by spring 2  1 Mu2 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 1 u = (2M) 2 24 (   It is also the total energy of vibration because the spring is unstretched at this moment, hence if A m is the amplitude, then T = 2 K   so force constant (   )  A b g42m 4  3.14 2  1 F I1 1 Mu b gK= T2 = =16 N/m G JKA2 = Mu2  A = 2K 1.57 2 H K2 4 57

JEE-Physics 1 5 . Additional force ( ) EXERCISE –IV(B) mg g 1 . Let block be displaced by x then displacement in =kA  A= =2.45m springs be (x 44 1 6 . Centre of mass will be at rest as there is no external ) force acting on the system. x ,x ,x and x 4 123 (Suchthat()x=2x + 2x + 2x + 2x     1234 So effective length ( ) Now let restoring force on m be F = kx then (mF=kx)  eff =  m1   2f = k x = k x = k x = k x  m 1 m  11 22 33 44 2 m1 F 4F 4F 4F 4F m1  m2 g     T  2 eff = 2 k k1 k2 k3 k4 g ma2 ma2 1 1 1 1 1   k  4  k1  k2  k 3  k 4  17. T  2 I  2 62  2 2 2a mg 3g m 1 1 1 1 mg  a T  2 k  2 4m  k1  k2  k3  k 4   2  Mg 2 . (i) Let x = = initial compression in spring 0k a a () 2 a 11 COME : k(x + b)2 = k(a–x )2 + (m+M)g (b+a) 20 20 1 8 . T = 2 I where ()  k  2mg mg ba I= m L2  mL2  m L2  1 7 m L2 & = 3L (ii) k  2mg  m  M2  2mg 3   12   4 ba b  a 12 (Distance of centre of mass from hinge) 1 2mg ( )  f  2 b  a m  M   T  2 1 7 m L2  2 17L (iii) Let h = initial height of m over the pan 12(2mg(3L / 4)) 18g (h=m )  1 9 . Moment of inertia about hinged point v = common velocity of (m+M) after collision Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 (  (v=(m+M)) I = mR2 + mR2 = 2mR2 COLM : m 2gh  (m  M)v ....(i) 4R2 4 11 1 R2 2 g 1  2 COME : kx 2 + (m+M)v2 + (m+M)gb= k(x +b)2 mg  1 mg  20 2 20 f 1 I 2 1 2R 2 2mR2 2 M m ab  h   m  ba      58

JEE-Physics 3 . (a) Let the liquid of density 1.5  occupy the left portion 0.2 1 AB and the liquid of density  occupy the right portion  0.2 , cos   1 BC of the tube. The pressure at the lowest point D As tan = 0.2, sin 1.04 1.04 due to the liquid on the left is  dP = Rg [2.5 × 0.2 + 2.5  – 0.5 + 0.5 × 0.2 ] (AB1.5 BC = Rg [ 2.6 ] D) = 2.6 g P1=(R–Rsin) 1.5 g where y = R, the linear displacement. (y= R, )  Restoring force ( )F = 2.6 gAy C This shows that ()Fy  Hence the motion is simple harmonic with force R  constant ()  A k = 2.6 gA Now, total mass of the liquid ()  1.5 D B m  2R A  2R A 1.5p = 5RA 44 4 The pressure due to the liquid on the right is  Time period ()  ()  T  2 m  2 5 rA  k 4  2.6gA P2=(Rsin +Rcos)g + (R–Rcos)1.5 g Since the liquids are in equilibrium ()   1 .9 3 R = 2.47 R seconds. 9.8 P =P or (R–Rsin) 1.5 g = (Rsin+Rcos)g 12 + (R–Rcos)1.5 g Solving, we get () 4 . T sin   I tan=0.2 or  = tan–1 (0.2)    (b) Let the whole liquid be given a small angular displacement  towards right. Then the pressure difference between the right and the left limbs is (2mg m  4 2       2g )  4 dP = P – P d2 2g 2g g 2 21  dt2  4   0    4  2  T  2 g Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 = [Rsin(+) + Rcos(+) g + [R–Rcos() 1.5g – [R–Rsin()] 1.5 g 5 . Given a2 + b2 =  2 +  2 a sR =Rg [2.5 sin()–0.5 cos(] =Rg [2.5 (sin cos + cos sin) 1 2O  x1 y b –0.5 (cos cos–sinsin)] b T x2 tan  = d Q For small  () a sin   ,cos   1  dP = Rg 2 1 1 tan  = 2 P LetOT = x , TQ = x , PT = d , TS = y 12 [ 2.5 sin + 2.5  cos – 0.5 cos + 0.5  sin] 59

JEE-Physics From geometry ( )  T2   x  g =  t1 T0  t  2  a  1  g  a  ...(ii) d + y =  sin ( + ) T 2 1   sin and d + 1 cos ( + ) tan  = 1 ( + )  d  12  T = 2 d  2 12  x g g a g ag  Ttotal = 2   a   ga g  a  4x  g g   6 . From energy equation (    a  ga g  a  x ///// F2 x2 8 . (i) F1 x1 mgsin   x mg 1 1  x  2 mv2 + 2 k  a  b  E x1  K 2 /2 x1  2 , f1  2   ,  ma kb2 x 0  T  2 a m   x2 , x2 = , f2 = k + a2 b k  LL    F1 L  F2  mg sin  L 2 2 g ()  7. T =2 (Lift is stationary)  L KL  LKL  mg L 0 L 22 2 T = 2 ga  5 KL2  mg L  K  2mg 1 4 2 5L (Lift is accelerated upward) (ii) If the rod is displaced through an angle , then ( )  ( )  T = 2 L  kL  L  M g  L    M L2   k  Mg 2 –(kL)L–  2 2  2  3  L  ga (Lift is deacelerated upward)   9k     = 3 k 4M 2 m ( )  9 . (i) This system can be reduced to ( Let x = total upward distance travelled    )   (x= )  x1 x  = at2  t= 22 a \\\\\\\\\\\\\\\\\\\\\\\\ keq  for upward accelerated motion M Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 ()  \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\  t T1 t  2  x  g Where   m1m2 (0.1)(0.1)  T0   2  a  1  g  a  ...(i)  = 0.05 kg T =    m1  m2 0.1  0.1  and keq=k1+k2=0.1 + 0.1 =0.2 Nm–1 & for upward decelerated motion  f  1 keq  1 0.2  1 Hz 2  2 0.05  ()  60

JEE-Physics i(nii)oCthoemr spprreinssgio(ninonespringis equal toextensionEXERCISE –V-A )  1 . Time period of a mass loaded spring = 2R    m ()  =2(0.06)  6  50 T  2 m So T  1 ...(i) Total energy of the system (   ) kk E  1 k1 (2R)2  1 k2 (2R)2 Spring constant (k) is inversely proportional to the 2 2 length of the spring, i.e. ((k) )   2 k  1 = k(2R)2 = (0.1)  5  = 4 × 10–5J  (iii) From mechanical energy conservation  k complete spring 1 /    k cut spring 1 n ( )   / n  11 m1v12 +2 m v 2 = E 2 2 2  0.1v2 = 42 × 10–5v=2×10–2ms–1   k  n kcut spring complete spring 10. K2 b Tcut spring k complete spring  1 [From (i) b  T complete spring k cut spring n a a K1  T  T ncut spring complete spring 2. In a simple harmonic oscillator the potential energy is directly proportional to the square of displacement For small angular displacement , net torque towards of the body from the mean position; at the mean mean position is ( position the displacement is zero so the PE is zero ) but speed is maximum; hence KE is maximum.  = (k1a) a + (k2b) b  I = (k a2+ K2b2)  ( 1  1  (mL2 + 3 ML2)  = (k a2 + k2b2)    1 )   = k1a2  k2 b2 2 = k1a2  k2 b2 3. The time period of the swing is ( ) L2 (m  M ) L2 (m  M ) 33  Hence frequency ( )  T  2 eff g 1 k1a2  k2b2 Where leff is the distance from point of suspension f = 2 = 2 L2 (m  M ) to the centre of mass of child. As the child stands Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 3 up; the leff decrease hence T decreases. (leff  leff T  4. T  2 M 5T Mm and  2 k3 k On dividing ()5T  M  m 3T M 25  M m  m 16 9 M M  9 61

JEE-Physics 5 . As maximum value of (  ) 1 1 . Natural frequency of oscillator = 0 A sin   B cos  = A2  B2 ()  so amplitude of the particle (   )  Frequency of the applied force =  ()  = 4 12  12  4 2 Net force acting on oscillator at a displacement x (x)  6 . A harmonic oscillator crosses the mean position with =m(02–2)x ...(i) ...(ii) maximum speed hence kinetic energy is maximum Given that ( ) F  cost at mean position (i.e., x=0) From eqs. (i) and (ii) we get ((i) (ii) ) (m(02–2) x  cost ...(iii)  ) Also, x =Acost ...(iv) Total energy of the harmonic oscillator is a constant. From eqs. (iii) and (iv), we get PE is maximum at the extreme position. 1 ( m(02–2)Acost  cost A   m 02  2  ) 1 2 . Initial acceleration ( )  7 . Maximum velocity ( ) F 150.20 = 10 ms–2 = vmax = A m 0.3 Given ( )  13. Ans. (4) (vmax)1=(vmax)2 1A=2A2 A1 2 k2 m k2 1  cos 2t 1 1  A 2  1    k1 y = sin2 t = =  cos2t m k1 2 22  motion is SHM with time period 8 . The time period of a simple pendulum of density ()  when held in a surrounding of density  is = 2  2  (   ) 1 0 0 t   3   14. y1=0.1sin  tmedium  tair    dy1 1 0 0 t  v1  dt  0.1  100 cos  3  4  10 3 4 3 3 3   t water  t0  t0 =2t0  10 cos 1 0 0 t    4  1 1 3   3  10 3 y2=0.1 cos100t m1  v2  0.1  sin 100t  0.1 cos 100t   2  9 . T = 2   T2  kk Phase difference between v1 and v2 11 11 (v1 v2   )  t2  k1 & t2  k2  t2 + t2  k1 + k2 1 0 0 t   100t   1 2 1 2 3  2  6     Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 11 1  T2 t2 t2 T2 But k1 + k2 =    1 + 2 = k d2x 1 5 . dt2  x  0 1 0 . The total energy of a harmonic oscillator is a constant On comparing the above equation with the equation and it is expressed as (  of SHM(        ) )  E  1 m2 (Amplitude)2 d2x  2 x  0 2=   T  2 2 dt2  E is independen to x instantaneous displacement. (Ex) 62

JEE-Physics 1 6 . The expression of time period of a simple pendulum 2 1 . W.Df = Kf+Uf – Ki –Ui is ()  11  – f.x= kx2 + 0 – mv2 – 0  T  2 eff 22 –15x = 5000 x2 – 16 = 0 g eff  5000x2 + 15x – 16 = 0  x = 5.5 cm Where leff is the distance between point of suspension 4m/s and centre of gravity of bob. As the hole is suddenly unplugged, leff first increases then decrease because 2 of shifting of CM due to which the time period first Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 increases and then decreases to the original value. x \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ (leff  leff  1   k1  k2 22. m 1 7 . Maximum velocity of a particle during SHM is The original frequency of oscillation () ()  vmax 2  A  4.4  T 7  103 T = 0.01s 1 k1  k2 f 2 m  75  On increasing the k1 and k2 by 4 times, then f' 1 8 . KE =  100  TE becomes (k1k2)  1 m 2 A2  x2   3 1 m 2 A 2   x  A 1 4 k1  k2   2f 2 4 2  2 f  2 m Now from x = Asint we have A  A sin t   2 t    t  1 s 2 3 . Mass = m, amplitude =a, frequency =  2  T  6 6 (=m, =a, =  1 2 a2 m  (KE)av =  2m 2 a2 4 2  1 9 .  x = 2 × 10–2cost  v = (2 × 10–2) () (–sint) = –2 × 10–2 sint Speed will be maximum if    2 4 . a = –2x , v =  A 2  x2 a2T2 + 42v2 = 4x2T2 + 422(A2 – x2) 1 t =  t = = 0.5s 4 2 22 = T 2 2x2T2 + 422(A2 – x2) 20. x = x cos  t   = 422A2 = constant ( ) 0  4  v= dx = –x0 s i n  t    aT 2xT = –2T = constant ( ) dt  4   xx dv   1k 1k  a = dt  4  2 5 . n1 = 2 M ..... (i) and n2 = 2 ..... (ii) = – x 0 2 c o s t  Mm = x02 cos    t    according to conservation of linear momentum  4  ()   Mv1 = (M + m)v2 M1A11 = (M + m)A22  4  From equation (i) & (ii)  x02cos   t  =Acos(t + ) A1  M  m  2  M  m  M A2  M  1  M  Mm =  .  M m M  3  A = x02 and   4 4 63

JEE-Physics 26. X = A sin t EXERCISE –V-B 1 X = X + A sin (t + ) 2 0 X – X = X + A sin (t + ) – A sin t 2 1 0 1 . U(x) = k|x|3 Become X2 – X1 max =X +A [U] [ML2 T 2 ] 0 [x3 ] [L3 ]   [k]    [ML1 T 2 ] |A sin (t + ) – A sin t) | = A   = 3 max m Now, time period may depend on 2 7 . By using T = 2 Ag () Where m = 3d and A = 2 T  (mass)x (amplitude)y (k)z 3d d [M0L0T] = [M]x[L]y [ML–1 T–2]z = [Mx+z Ly–zT–2z] T = 2 2g T = 2 g Equating the powers, we get 28. k11 = k22 = k. k1 = 5k ()  2 –2z =1 or z = –1/2 y – z = 0 or y = z = – 1/2 2 9 . Equation of damped simple pendulum 1 Hence, T  (amplitude)–1/2  T (a)–1/2  T  () a d2x d2x g 2.  = 2 1 = 2  dt2 = –bv + g sin   dt2 = –bv +  x = 0 1 2 3 bt By solving above equation x  A 0e 2 sin (t + ) A0 2  k1 1= 2  2= 1 so  = 3 3 At t = , A = 2b k 3 0 . FBD of piston at equilibrium Force constant k  13 k = k (FBD) 1 2 length of spring 3 . U (x )  k(1  e x2 )  Patm A + mg = P0A ...(i) It is an exponentially increasing graph of potential energy (U) with x2. Therefore, U versus x graph will be as shown. FBD of piston when piston is pushed down a distance ( (U)x2 x(x Ux FBD) )  U K d2x ...(ii) X Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 Patm + mg – (P0 +dP) A = m dt2 From the graph it is clear that at origin.Potential Process is adiabatic (  )  energy U is minimum (therefore, kinetic energy will be maximum) and force acting on the particle is also zero  PV   C  dP  PdV because V ( Using 1, 2, 3 we get f 1 A 2 P0  2 M V0 )  64

JEE-Physics dU mg sin  mg sin  F = = –(slope of U–x graph)=0. dx Therefore, origin is the stable equilibrium position. Hence, particle will oscillate simple harmonically sin  mg cos about x=0 for small displacements. Therefore, correct   mg option is (d). (a), (b) and (c) options are wrong due to mg following reasons : bob is () (Net forceonthe F = mg cos  (figure b) x=0 net (d) (a,(b) (c) or ) Net acceleration of the bob is () dU g = g cos  (a) At equilibrium positoin F = =0 i.e, slope of eff LL dx U–x graph should be zero and from the graph we can T = 2 geff  T  2 g cos  see that slope is zero at x=0 and x=±. Now among these equilibriums stable equilibrium position is that where U is minimum (Here x=0). 5 . In SHM, velocity of particle also oscillates simple Unstable equilibrium position is that where U is harmonically. Speed is more near the mean position maximum (Here none). and less near the extreme positions. Therefore, the Neutral equilibrium position is that where U is time taken for the particle to go from O to A/2 will constant (Here x =± ). Therefore, option (a) is wrong. be less than the time taken to go it from A/2 to A, (F=dU =0 U–x or T < T . 12 dx ( x=0x=±   0A/2 U(x=0) A/2A UT< T .)  1 2 U(x=± 6 . Potential energy is minimum (in this case zero) at ). (a) mean position (x=0) and maximum at extreme (b) For any finite non–zero value of x, force is directed positions (x= ±A). towards the origin because origin is in stable ((x=0 )  equilibrium position. Therefore, option (b) is incorrect. (x(x=±A)) At timet=0, x=A.Hence, PE should be maximum. Therefore, graph I is correct. Further is graph III, PE is Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 (b)) minimum at x=0. Hence, this is also correct. (c) At origin, potential energy is minimum, hence kinetic (t=0 x=A energy will be maximum. Therefore, option (c) is also IIIIx=0 wrong. ()  (c)) 4 . Free body diagram of bob of the pendulum with 7 . Block Q oscillates but does not slip on P. It means that respect to the accelerating frame of reference is as acceleration is same for Q and P both. There is a force follows: (FBoffriction between the two blocks while the horizontal plane is frictinless. The spring is connected to upper ) block. The (P–Q) system oscillates with angular 65

JEE-Physics frequency . The spring is stretched by A. Then kx = k2x2 x + x = A  x + k1x1  A 11 1 2 1 k2 (QP QP k2A (P–Q)Ampx1litudke1ko2fkp2 oin=tAPwixll1 = k1  k2  ext. in k . be the max. 1 A) 1 (Pk)   kk So amplitude of point P is k 2A  .   k2 m  m 2m (P ) k1  Maximum acceleration in SHM = 2A L x 12. x =  ( 2 kA (i) F = – k   L   a=  2  m 2m Now consider the lower block. 1 = – k   L  × L   2  2 ()  x L Let the maximum force of friction = f 2 = – k   2  × L m  2 ()  kA kA  f = ma  f = m ×  f =  kL2 m mm 2m m 2 L2     2   12 net  k  m L2 2 8. y = Kt2  d2y  2K  a = 2m/s2 (as K = 1 m/s2) 1/ 2 y dt2   k6      k6  f = 1 6k   m    m  2 m T1  2  g and T = 2 g  ay MCQ 2 1 . From superposition principle :  T12  g  ay 10  2  6 () T22 g  5 y=y +y +y 10 123 11 y1 = a sin t + a sin (t + 45°) + a sin (t +90°) 9 . kx2  (4k)y2   = a[sin t + sin(t + 90°)] + a sin (t + 45°) 22 x2 = 2a sin (t + 45°) cos 45° + a sin (t + 45°)  ( 2  1)a sin(t  45) = A sin (t + 45°) 1 0 . Displacement equation from graph Therefore, resultant motion is simple harmonic of ()  amplitude. A = ( 2  1) aand which differ in phase by 45° relative to the first. X = sin   t    2  2    4 T8 4  (A= ( 2  1) a  ) Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 Acceleration ()a = 2 sin   t –  4 Energy in SHM  (amplitude)2 16  A2   a   E resul tan t  E sin gle ( 2  1)2  32 2 4 32 At t  s, a = – cm/s2 3 32   = 1 1 . In series spring force remain same; if extension in k E 3 2 2 E sin gle 1 resultant and k are x and x respectively. 212kk2 . ForA = –B and C = 2B 12   x x  X = B cos 2t + B sin 2t  2B sin 2t  4  12 66

JEE-Physics This is equation of SHM of amplitude 2 B (mm 12 0 ( 2 B  If A = B and C = 2B, then X = B + B sin 2t This is also equation of SHM about the point X=B.   Function oscillates between X=0 and X=2B with  a(mplitudeBx=. B  Velocity of centre of mass = v x=0x=2BB)   0 Paragraph  Location/x–coordinate of centre of mass at time 1 . For linear motion of disc ()  t=v t 0 (tx-=vt)  0 Fnet = Ma = –2kx + f m1x1  m2x2  x = m1  m2 where f = frictional force (f=  )   m1[v0t  A (1  cos t)]  m2x2 For rolling motion ()   MR 2    Ma  Ma Fext  v t= m1  m2  2   2  2 2 0 fR = –   R  f = – = –  (m + m) vt = m [v t–A (1– cos t)] + mx 1 2 0 10 22 Fext 4kx  mvt + mvt = m v t–m A(1– cos t)] + mx 2 3 10 20 10 1 22 Therefore Fext = –2kx – = –  m x = m v t + m A(1– cos t) 22 20 1  x = vt + m1A (1– cos t) ...(i) 2 0 2 . Total energy of system (   ) m2 E= 1 Mv 2  1  1 +2× 1 kx2 = 3 Mv2 + kx2 (ii) To express  in terms of A. 2  2 2 4 0 (A) 0 dE = 0  3 M (2 v ) dv  2k   dx   0  x = vt – A(1– cos t)  dx1 = v – A sint dt 4 dt  dt  1 0 dt 0   dv   4k  x  0   = 4k  d2 x1  A2 cos t ....(ii) dt  3M  3M dt2 3 . Using energy conservation law x is displacement of m at time t. 11 ( )  (t m x ) 11 1 Mv02 1  1 1  d2x1 = acceleration of m at time t. 2 2 2 dt2 1 =2× k x 2 1 (t m )1  MR2  When the spring attains its natural length  , then 2kx1 – fmax = Ma & fmaxR =  2   0 acceleration is zero and (x – x ) =  21 0 3 M g ( But fmax = µMg  x1 = 2K 0 (x –x ) =  ) 0 21 . x2 – x = 0 Put x from (i) 1 2 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 3 1  92M2g2   Mv2 =Kx 2=    v = µg 3M  m1A  4 0 1 K 4 0 K  v0t  m2 (1  cos t)  [v0t  A (1  cos t)]  0   SUBJECTIVE   =  m1  A(1– cos t) 0  m2  1 1 . (i) Two masses m and m are connected by a spring 12 When d2 x1 =0, cos t = 0 from (ii). of length  . The spring is in compressed position. It is dt2 0 held in this position by a string. When the string snaps,  =  m1  the spring force is brought into operation. The spring 0  m2  1 A. force is an internal force w.r.t. masses–spring system. No external force is applied on the system. The velocity of centre of mass will not change. 67

JEE-Physics 2 . A sphere of radius R is half sumerged in a liquid of 3 . A small body of mass attached to one end of a density . vertically hanging spring performs SHM. (R) ( m For equilibrium of sphere () ) Angular frequency ( )= O x Amplitude ()= a R Under SHM, velocity ()  v =  a2  y2 Weight of sphere = Upthrust of liquid on sphere. After detaching from spring, net downward (= ) acceleration of the block = g. (     =g) V  Height attained by the block = h  Vg = ()g (=h)  2 where = denisty of sphere(=) v2 2 (a2  y2 )  ...(i)  h = y + 2g  h  y  2g  = 2 For h to be maximum (h) From this position, the sphere is slightly pushed down. Upthrust of liquid on the sphere will increase and it will dh act as the restoring force. dy =0, y = y*. ( 22y * Restoringforce () ) ddhy=1 + 2 2g ( 2 y *)  0 =1– 2g F = Upthrust due to extra–immession 2y * g  g =1  y*  2 ()  Since a2 > g (given)  F = –(extra volume immersed) × g g ( F = –()×g)  a > 2  a > y*. y*  (mass of sphere m) × acc(a) = –R2 gx from mean position < a ( <a) F = –R2x × g 4 3g x  R3 ×  × a = –R2gx  a = – 3 4R g y*  . Hence 2 a=– 3g  2 x, [from (i)] 4R 3g Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65  x 2R  a is proportional to x.(x  )  Hence the motion is simple harmonic. ()  Frequency of oscillation (   1 a 1 3g = . 2 x 2 2R 68

JEE-Physics UNIT # 08 CURRENT ELECTRICITY EXERCISE –I 1 . j  Current density n  Charge density 9. j = –nev v d1 = j d n1e j n1  1  n2  4n1 v d2 = n2e , n2 4 v d1  n2  4n1 = 4 : 1 v d2 n1 n1 I 2 . j  A  nevd 4I 16I This is balanced wheat stone bridge d2  nev ...(i) d2  nev ' ...(ii) From maximum power transfer theorem Internal resistance = External resistance 4I v  From equation (i) & (ii) 16I  v   v' = 4v = 3. vd  Ane i As A  so vdvP > v 3R  6R Q 4 = 4 = 2R R = 2  3R  6R 2envA 3 10. V2 V P Initially, I  R 2R 4 . i = nev A; I   (nevA)  nevA d4 2  2  Power across P= P= R X Y  4 R  R  L L L2 L2 L2 d 5.    A L AL V m 2V 4V2 2V2 Finally, I = 3R , Power PX  9R , Py  Pz  9R d,  same for all as the material is same for all. 25 9 1 Hence P increases, P decreases. R1 : R : R = 1 : 3 : 5 = 125 : 15 : 1 xy 2 3 Alternative method : Brightness  i2R when S is closed current drawn 6 . R  L L  L2  R  L2 from battery increases because R decreases. i.e. AL V eq current in X increases. So brightness of X increases and current in Y decreases. So brightness of Y decreases. 7 .  i2R  S  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 R XX eq Y Y  Balanced Wheatstone Bridge I2R  V 2 2 R  R  r)2 11. P= =  R = (R  As 1 1  7  36 So R = 36  7  85 9 12 36 7 AB 7 7  is constant and (R+ r) increases rapidly Then P  (R+r) P 8. P = i2R  10 = i2 5  i2  10  2  i = 5 30 60 2V 20 12. 2 i4  i5  P4   i 2 4, P= (i2)5 2  2  5 1 V = IR 2 = (I)(20)  I  10 A P4  1  P4  P5 10 P5 5 5 =2 , P= cal/s 4 5 47

JEE-Physics   676 S R1  A1 , R2  A2 From (i) & (ii)  13. As A < A so R > R S 625 12 12 In series H = I2Rt H  R ; H > H (676) (625) = S2  S = 650 1 2 V2 1 22. E=  V    & E=  V 2   '   '   In parallel H= t H  ; H < H    3  3 /  2 R R 12 14. 1 V =  + i(r)  12.5=  + 2 (1)  =12 V (As the battery is a storage battery it is getting 23. Potential gradient ( )x= E   9r  10r   L  charged) According to question () ()  E  E  9r      5L 2  10r   L  9 1 5 . The correct answer is R = 0 4V 0.8 2 4 . Potential gradient ( ) I1 16. R=0.8  x   5   4.5   1.5 Vm -1 0.8  0.5  4.5   3  I2 4V 0.8 1.6 I – 0.8 I = 4 ...(i) 3 12 Here (x) (AC) =3  AC   2m 1.6 I – 0.8 I =4 ...(ii) 1.5 21 2 5 . Potential gradient ( ) from eq. ( ) I = I =5 12 x=  8 12   4  2 Vm 1   16  voltage difference across any of the battery. )  0.2 I Vb Effective emf of E and E Va 1V 12 V –1 + 0.2 × 5 – V = 0 (E E ) ab 12 V – V = 0 Volt E2  E2 ab E = E  r2 r1  1 volt 1 7 . V = IR  0.2 = I (20) 1 / r1  1 / r2 2 I = 0.01A (through the galvanometer) Balancing length AN = 1 1  1 m  25cm  2   2  4 g I G = (i – i )S  (0.01) (20) = (10 – 0.01)S gg  S = 0.020  18. R = V  G  910  10 V  90 V2 V2A r2 v ig  10 3 R   26. P    V  same V = 10 No. of divisions 10  100   = 0.1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 12 27. (25W– 220V) 1 9 . 20 + R =  R=100 V12 2202 0.1 R1 R1  25 P= , = 1936 1 12 20. I   0 . If i=0, 42 (100W–220V) potential difference is equal of EMF of cell. = 12V V22 2202 P= R2 , R2  = 484   2 100 21. P  Q  P  S ...(i) In Series (I same) S 625 Q 625 Q P P 676 H=I2Rt, H  R so if R >R then H > H S 676   12 1 2  Q S ...(ii) R is likely to fuse 1 48

JEE-Physics V2 V2A 33. I = 4V = 0.2 A 2 8 . P  R  L ...(i) wire 0.4  50 Potential difference across voltmeter, V2A 10V2A () P     L  9L ...(ii) V = Ir – 2  L  10   2 sint = 0.2 × 50 x –2  2 cost = 10 V 10 10 2V 10 from eq. (i) & (ii) P   9 P 4V V x P 10 P  P  1 0V (let)  9   100   100  100  11.11% 4V P P9  V = 20  (cost) cm/s 2 9 . In parallel combination the equivalent resistance is less than the two individual resistance connected and in series combination equivalent resistance is 1,2,3 1',2',3' more than the two individual components. 34. O 4',5',6' O 7',8' 4,5,6  7,8 Points1, 2, 3.........8 are of same potential and ) 1', 2', 3'.........8' are of same potential. 3V 2 8V 1 Er B (1, 2, 3.........8       2 10V 1', 2', 3'.........8' ) 30. A BA  3R 3V R= eq 8 3 5 . Total length of wire ()  E1  E2 10  8 = 90 + 90 = 180 m ; r1 r2 2 2 E= 1 1  1 1  1 volt and Total resistance of wire ()  = 180/5 = 12  r1 r2 22 nE n  1.4 r= r1r2 =1 . Therefore A B As I = R  nr  0.25 = 12  5  n  2  n = 4.7 r1  r2  Total number of cells required = 5  2V 31. Ans. (A) () 3 2 . Given circuit can be reduced to  6 6 4 12 V Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 9 V 9 9 36. A 6 6 B A 6 6 B 9 1  A   B Req=10 A 1 A 4V 4V Reading of ammeter (  )  4 =  1A 3 1 Reading of voltmeter (  ) = 3 × 1 = 3V 49

JEE-Physics EXERCISE –II dq 5. I   2 16t dt 1 . Free–electron density and the total current passing Power : P = I2R = (2 – 16t)2R through wire does not depend on 'n'. 1 (nHeat produced=  Pdt  8 (4  256 t2  64 t)Rdt )   0 2. E eq  E1r2  E 2 r1 2 3 1 4  2 =  256 t3  64t2  R 1/8 = R joules r1  r2  7  4 t  3 6  34 2   0  1V 3 6 . It is the concept of potentiometer. 2 () DB 2V 4 B i1 7 . By applying node analysis at point b r= 3 4  12 ; i = 2/7 1 V R1 V/2 R1 V/4 eq 3 4 7 2  12 A A 13 ab R2 R2 R2 7 B 00 0 V VV V V  2 2 4 2  0  R1 1  R1 R1 R2 R2 2 V > V =2 1 ; = 2 BD  13  V –V – 13 V DB O 2R/3 From Figure 1 : RR 4 7R 2 8.  R 2R  V + 4i –2–V =0; – 2 + 4i = 0 9. 3 B 1 D 13 AR B R 66 11R i = A; V = 3 – 3 × R= 13 G 13 AB 18 21 6 19 For wheat stone Bridge condition is R1  R3 VG  V , V = 1+1 × =V = V R2 R4 13 H 13 H 13 (R1  R3)   A  2   A R2 R4 3. 28V   4 =  5 Therefore null point is independent of the battery voltage. ()   B    B  1 0 . V = E – ir  V = –ri + E Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 R = 14  I = 2A; V = iR = 7 volt Slope of graph 'V' and 'i' gives 'r' intercept of graph eq AB 'V' and 'i' gives E tan  = y  r . 4 . Both '4' and '6' resistors are short circuited x therefore R of the circuit in 2 is 10 A. ('V' 'i' 'r''V' 'i'  eq ('4' '6'   2   tan y  Req= x 10 A.) E  =  r .) Power ()= VI = 200 watt 1 1 . V = E + ir and in charging current flows from positive terminal to negative terminal. Potential difference across both 'A' and 'B' = 0 (V = E + ir   ('A' 'B' =0) ) 50

JEE-Physics 1 2 . Slope of 'V' vs 'i' graph give internal resistance 50 × 10–6 × 100 = 5 × 10–3 × (R) R  1 r=5 For voltmeter I (R + G) = V ('V'  'i' r=5  g 50 A (R + G) = 10V   R + G = 200 k  R  200k Intercept gives the value of e.m.f. E = 10 volt (E10 volt  18. 20 20 1 20 5 i=   A ) min R min 200 10 Maximum current is ( )i = E  2A 20 20 2 75 B max r i=   Amp G R max 250 25 max 13. If n batteries are in series than the circuit can be made 1 nE nr Potential = iR =  75  7.5V 10 as i  nE  E min PM nr r i.e. independent of n. 2  75  6V 25 Across potentiometer V = i R  max PM (n i nEE nr r 1 9 . If e.m.f of c is greater than the e.m.f. of the 'D' nE nr I =0 n)  r So r does not play any role of zero deflection in g a lva n o m e te r. 1 4 . If n batteries are in parallel than the circuit can be  C 'D'  made as i  nE (n   Ir=0 r r   i nE )  r E r/n i i is directly proportional to n . n 15. In parallel combination current gets divided 20. 30V 3/4 30V therefore parallel combination supports i = i + i 3 30V 1/4 12 is 20A in series current remain same therefore the series combination supports i = 10A. 1 ( 1/4 i=i+i Both 30V are in parallel 12 20A 30 1 i  3 i  0  i = 30 A 44 i=10A ) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 1 6 . As power in 2 is maximum when the current in it is maximum. Current in it will maximum when the 2 1 . Assume DE  R1 EC  R2 value of R is minimum   R = 0 (2eqVR1 =+V R = 1  2 R R=0) BE eq Means balance wheat stone bridge Heat =i2RT(36)(2) = 72 W R1 1 1 7 . For Ammeter I G = (I – I ) R P  R ; 1  R1  2 gg Q S 1 R2 Ig G R R2  R1 1 1  R1 I – Ig R1 2 51

JEE-Physics R1 1  R1  2  R 2  2R1 26. 1 Between A and B R 2  2R1 1  0 R1 = –1 + 2 R2  2  2 ( 9 + 0.9) × 10 × 10–3 = (I–10mA) × 0.1 1  990 mA = I – 10mA  I = 1000 mA = 1A CE  R2  2  2 2 2 7 . When S open ( S )  22 ED R1 2 1 Assume resistance of AB = R (AB =R) 22. B i/3 Ci Resistance of wire per unit length. i/6 ()  R A i/3 2i/3 x= L D i/2 i/6 i/3 F H G i/6 i i/3 i/6 i/6 E So current in FC=0 23. 12  1A  V =1=1 × X  X=1 E x I= X  Y r R 12 Now in AC When Y shorted I = ERL 1r   6 10=12 – Ir  10  12  12  r RL2 E = 12V 1  r When S closed  10+10 r = 12 + 12 – 12r 2  10r =2  r = 0.2  E1  E2  E1 E R 5L 5E 5 12  5V 24. r1  r2  R r1  R ;(E + E) (r + R) < E (r +r +R) V=     1 2 1 11 2 1 R L 12 12 12 E R + E R + E R < E r + E R; R(E + E ) < E r 5 1 2 1 12 1 12 12  10  6 – I r = 5  6 r 5  r=2  On solving we get E r > E (R + r ) 1   12 2 1 2 5 . If all were in series all of them would have being Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 getting discharged. But since, 2 are in opposite V R   slope  polarity, they will be getting charged. 28. I (            ) V = E+iR getting charged V  (nE  4) So R  i P = I2R R nR R P as (nE–4) as 4 batteries will be cancelled out P  I2 = E+  nE  4  R , = E +  E  4  = 2 1  2  E  n R   n  n  52

JEE-Physics 2 9 . Rearranged circuit between A & B is : EXERCISE –III (A  B )  Match the column 1 . For potentiometer short circuit = x  A B  1 A 37 B ()  x Depends only on primary circuit (due to symmetry) (x )  Total resistance of circuit ()( A ) E  x    if secondary circuit remain same 11 72 9  =  =3.i= =3A 33 3 (B) R  x  1 if secondary circuit remain same (C) Heat produced in cell ()  2. = I2 r = (3)2 × 2 6W S.C = 1 if x remain same  3  = S.C=   x  Current in resistance connected directly between 1 A & B  7  3  7  1.4A After closing the switch net resistance decreases 15 5 therefore there will be increases in the current. After closing the switch V2 becomes zero hence (A  B  V = V1.  7  3  7  1.4A ) ( 15 5 S2  V= V1) 30. A dx B S 2r 4r V x dx dx RR rx2  r2 (1  x)2 v1 v2 r =r +rx = r(1 + x)  dR = x x A R1   dx   1  1 1  , After short circuiting current in the resistance 0 rx2 (1  x )2 r 2    becomes zero therefore power become zero. R2  1 dx   1    (  r 2 (1  x )2 r 2 1  1  1  ) For null point R1 10  R1  R2 Comprehension–1 R2  1 . Power through fuse ()  10 P = I2R = h × 2r 1 1  1  1  3  2  h = heat energy lost per unit area per unit time 1 1 2 2 1 ()  3 + 3 = 4  = 1 m I = current. Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 3  2r  I2  h  r3  I  r3/2 3 1 . VP – VQ = 2(2R)  4 R = 24 – (2R)I1 r 2 RR  I1   r1  3 /2  4  3 / 2 8  I2   r2   1  1 PAQ 2A 1.5R 24V 0.5R    RS I1 R 3R V2 TU 20kw  2000= I2 V 2 . P = VI 20  IR = 12 – 2 R, E – I (4R) = 4R, I + I = 2  V = 200 volt  V < 200 volt 2 12 1 3 . At maximum power delivery R = r, so  = 50%  E = 20R – 48 () 53

JEE-Physics Comprehension–2  1 E1 1 1 E1 1 . As potential of 1, 2 and 3 are same potential  I = E+ R1 &I =  E - R1  R1  2 2  R1 R2  2 difference across them 'zero'. 1 (1,23  1 = 0.3  R1 = 20 R1 6 )  1 1 1 0.3 R= and +=  2 40 R1 R2 4 o 23 E1 Now as R1 = 0.3  E1 = 0.3 × 20 = 6V 2 . As 1, 2 and 3 are having same potential therefore we can draw it. Comprehension–5 (1,2 3)  1 . In balancing condition, current in the circuit should be zero which happens at =20 cm according to graph. O 1,2,3 ( = 20 cm ) R = R/3 ; R = R/3; R = R/3 2. At balance point    20 01 02 03 V   6  1.2V 100 100 3. As point 1,2,3 are equipotential V = I R 12 ()  (1,23)   V = 0 therefore I = 0 for R , R ,R 3. At  = 0, applying kirchhoff's 2nd law in the circuit 12 23 31 containing cell, = IR Comprehension–3 (= 0  ,= IR) 1 . Current is maximum when resistance in the circuit is minimum. i.e. when S1,S3,S5 are closed because where I is the current at = 0, &  is the emf of the then all resistances will be shortcircuited I = V0 . max R (cell. R=   1.2  30 I 40  103 S1,S3, S5 (=0 I  I = V0 .  R   1.2  30 ) max R = 40  103 2 . After regular closing of switches, total resistance I decreases gradually. Comprehension–6 (1.V=E+ir 50 12 0.04 = 12 + (0.04) (50) )  = 12 + 2  14 V 3 . P = V02 , P = V02 so P1  7 2. Ans. (A) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 1 R 2 37 P2 37 R Loss in power ( )  7 Comprehension–4 = i2r = (50)2(.04) = 100 W E1  E2 , I + I = E2 E2 E1  E2 3. Ans. (C) R1 1 2 R2 R2 - R1 1. I=  I = Total input ()  – Loss in power ()  1 2 I2 R1 I1+I2 = Useful power ( ),  Input power   14 (50)  700 w E2 I1 R2 Loss in power ()   100 w, E1 Rate of conversion (  )= 600 watt 54

JEE-Physics EXERCISE –IV A 5 . By applying perpendicular Axis– Symmetry () 1. I  40 = 10 9.6 60V 16  2.5 A 4 0.4 8 R= R  R2  A B AB 7  R1  R2  I1  I 20V 6 . By applying perpendicular axis symmetry. Points 6 lying on the line 'AD' have same potential therefore I1   48  2.5  2A  I2 = 0.5 Resistance between AB and CD can be removed  60  R= 9 AB  V = 0.5(7) = 3.5 volt 'AD'  ABCD  2 . V = IR 12 A=12V R= 9A VAB = 1V 1A AB VBC = 2V 12V 1 VCD = 3V 0V B B=11V 6V  AB 6V C =9V C  D D=6V 3 . By symmetric path method Points E, F and B, C 7 . (i) When switch S is open (S ) are Equipotential  R = 1 AD E, F B,C ) 36V A 4A 4A 6  2 D ab 6  1 1 1 2 B 2 1 C 4 . By perpendicular Axis symmetry all points 1, 2, 3 V –V = (36–6×4) – (36–3×4) = –12V ab are at same potential therefore junction on this (ii) Total current through circuit = 36V  9A 22 4 line can be redrawn as R = R . ()  AB 35 (1, 2 , 3  36V R=22 R .) 3A 6A AB 35 6  1 a b Therefore I = 3A Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 I 6  AB 2 8 . (i) Chemical energy consumed = 3 watt ( ) 3 (ii) Rate of energy dissipation = i2R = 0.4 watt ( ) 2r/3 + r/2 + 2r/3 (iii) Rate of energy dissipation in resistor A B  A 2r B () = (E – ir) = 2.6 watt 2r/3 + r/2 + 2r/3 (iv) The output energy to the source = 2.6 watt () 55

JEE-Physics 25 25 25 25 I 25 ('2R'  'R 'R=2R+ R  xx 9 . 10V 5V 20V 30V (2R  R x )(R ) 15V 30V 45V 55V 25V 'R'R= (R  2R  Rx )  Rx ) eq 5  By solving above equation R = 3  1 R 10 x 5 11 3A 3A 9A 5A A0 1 4 . In loop ABCDEA 20–4i–2i –10=0 Taking point 'A' as reference potential and its 1 potential to be '0' : I = 20 A   2i–i =–5  i =–2i+5 1 1 ('A' G 2 ii1i2 F D ) 10V 10V i2 i1 i 2 2 2 2 Power supplied by 20 V cell A E 10V  20 V  10V 10V C = –20 × 1 = –20 W B 1 0 . By applying node Analysis 2 ()  In loop ABCDFA 20–4i–10–2i =0  i =–2i+5 2 2 r2 V2 In loop ABCDFGA 20–4i–10–4(i–i –i ) = 0 12 x –V2 0 Put the values of i & i 12  10–4i–4 (1+2i–5+2i–5) =0 r1 V1 V1  10–4i–20i+40=0  i = 25 A 12 x  V2 x  V1 1 1 V2 r1  V1r2 r2  r1 0  x     r1 r2 1 5 . Circuit can be redrawn as   r2 r1 5A 2 D x  V2r1  V1r2 1 11 A B 5A r1  r2 , req = r1  r2 10A 2 C 20 10 3 11. P = P  2 R1  2 R2 30 1 2 (R1  r)2 (R 2  r)2 3 V R= 2 ; I  = 20 A eq R2  r  R2 Req R1  r R1 , r  R1R 2 Current In I = I + I ; I = 15 A CD AC B CD i0  i0t dt  T0 T0 1 2 . By taking 'O' as a reference potential as current   through '4' is zero there should be no potential 1 6 . (i) i  t  i0 ; dq  i0dt drop across it  i0 T0 i0 T0 2 2 ('O' '4' Q   i0 T0  )  t Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 4R 6 4 6 (ii) i = i 1  T0  2A 2A 0 2 1  t 6V 4V 10 (iii) Heat = i2Rdt [ i = i   ] 10V 0 T0 00 0 i 2 T0 2 i 2 0 0 0 Value of 'R' for this condition = 1    t2  i 2 d t  tdt T02 0 T0 'R'  1 3 . '2R' and 'R ' are in series therefore R = 2R + R  i 2 T03  i 2 T0  i 2 T0 xx 0 3 0 0 and it is in parallel with T02 'R'R = (2R  R x )(R )  Rx Heat  i 2 T0 eq (R  2R  R x ) 0 3 56

JEE-Physics 1 7 . Submission of current at the Node 'X' is 2 2 . Power developed in it is maximum when external (Node 'X' ) resistance = internal resistance. 2 x 4 (      =  10 2/3  0 ) 2 4 nr 9n2 10–x 324 / n  R   4  n  12 324 2 3 . Applying KVL x 10  x  0  3  2x 10  0  2 2 4  2  I1 I1+I2 7V 3  15x – 20 – 60= 0 x  80 I2 15 1V Current = V  10  3  = 1 A R 15 2  1 8 . Potential difference across  voltmeter is same as 7 = 2I + 3 (I + I ), 1= 2I + 3 (I + I ) that of 200 1 12 2 12 200 I1= 2A, I2= – 1A  100 200 Power supplied by E1 = a = E1I1 = 14W V1   200  10  20 V Power supplied by E = b = E I = –1W  300  3 10 2 2 2 1 9 . 5 – ir = 4  i = 1A Therefore a + b = 14–1 = 13W 1×R=4VR=4 2 4 . Heat developed will be maximum for the resistor 20. R1  40  4 ...(i); R1(R2  10)  1 '4' because (P.D.) will be maximum for the branch R2 60 6 R2 10 containing '5' and '4' ('4'  R R + 10R = R × 10...(ii) '5' '4'       12 12 By solving (i) and (ii) R1  10 ; R = 5 )  2 3 I A'9V  B'3V 2 1 . (i) Current due to primary circuit 25. A B 4 I1 () I 4 i  1Ep 10 D' 2 C'  R pm  r 10 = 1 Amp By applying K.V.L  V= 1R  V = 9volt V – 4I – 9 – I – 3 – 4I = V PM A1 B Potential gradient ( )=9 16 = +8I + I + 12 1 12 8I + I = 4V ....(i) 1 9 By applying K.V.L. in loop A'B'C'D'A'  12  Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 ( ) = 4.5   = 6m – 9– I + 2 (I – I ) = 0 1 1 11 (ii) i Ep 10 1 –3I + 2I = 9 ...(ii)  A 1 R PM  r  R ext 9  1  10 2 By solving (i) and (ii) Current in 2 resistance is 3.5A. iR 1 V = PM = × 9R = 4.5 volt 2  R 1 r  R R r2  J.d A 0 R  J0 2  rdr dr  0     26.(i) I    Potential gradient x = V  4.5  J0  2rdr R L 12 0 S.C = x   4.5   8  3V R2 R2   2R 2  J0A 1  12   J02    J0  6   3  2 3   3 V = E – ir, iR = 3  i  2 R R 0  r  J02 r2dr J0 2R 2  2J0A  R  R 3 3 J (ii) J0 0 3  2 rdr   = 4.5 –  2  r = 3  r = 1  57

JEE-Physics 2 7 . Potential gradient = x = 0.2  =2Volt 3 . By applying nodal analysis at note 'B' and 'C'. E = x 5 ('B'  'C' ) 21 30 ,10M I 2R VB 1.5 = (0.2) D 100 R 1 r RO  = 7.5m 1 (a) i 2 , V = ir  x  12 G RB R R 2R 35 70 1 100–V RR x =S.C. =1.5 Volt A 1 RC R  12 1  15  1  15  7  8.75m V 100  2V 100  V  0  0 70 10 12 2R R R (b) S.C.=  R E r1  R =  1 .5  5  1.25 300 400     6   7V – 300 = 0  V =  and V = B 7 C 7 S.C. = x I= VB  VC  100 2 R 7R BC 1.25 = 0.2 ( )   = 6.25m 22 EXERCISE –IV B I  200  300  500  5  100  7R 7R 7R 7  R  i.e. times the length of any side. 1 . 4 . By path symmetry potential of points A,B,C,D is same. (A,B,C,D ) By symmetry D and F are at same potential and C and E . And by symmetry C and E are at same potential. So we can removed DF and CE (D F C E  CE R eq = ra   1  Req  ra (  2) 4  2 8 DF  CE ) 5. I  I0 sin  2t   T  As dq I so Q  T/2 I0 sin  2 t dt  2 I0 T dt  T  2 0 A  +x Total heat generated 3 3 T I2Rdt T I20 R sin2  2 t dt 2.   /3 Req=   –x  0  0  T B x 2–x C I 2 R T 1 4 t  I 2 R T   0 0  cos T  dt  0 2 2 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 1 13 1 2   Q RT Q 2 2 R  R eq (  x)  (5  x)   2T   2       8T  (5  x)  3(  x)(5  x)  2(  x)  (  x)(5  x) 6. R  A 1 52  x  2  x  152  3x2  12x  R L 0e x / L xdx  dR  Req (  x)(5  x) 0 0A R= (  x)(5  x)  dR eq 0 R  0 L 1  1  0L (e  1) eq 212  12x  3x2 dx A e  Ae R (max) = 3 I V  V0 A  e eq  R 0 L  e  1  11 58

JEE-Physics 7 . Current with both switches opened is - (i) divided by (ii) () 10  R   100 50  R  50R  9 100 10  R   10R 50  R 30 V 1.5 1   i Req 450 300 After closing the switch (), 10  R  500  15R   9 100  11R 50  R  30 V +V =V 12 13 After solving R = 233.3  +V = 100 SG 99 1  0.99 32 2 1 10. R= S  G ; RA  300 A 100 9 2 7 i2 R V2   6 6 3(2+ r + 0.99) = 12 7 300 1.5 volt  2 + r + 0.99 = 4 i  r = 1.01  V2 1800 12 r By kirchoffs first law ()  i =7  1  1 ; iR = iR A 1800 300 1800 2 2 11 1 2 R = 1800  3  =600 I (R+G) = V; I S = (I–I )G; I (S +G) = 4IG 8 . (0.01) G = 0.1(R + R + R ) 12 3 g g gg By solving the above equation we get the answers. G = 10 (R + R +R )...(i) 1 23 0.01 25 EXERCISE –V-A G R1 R2 R3 1 . In order to convert an ammeter into a voltmeter, 0.1 one has to connect a high resistance in series with it. (0.01)G = 1(R + R )...(ii) 12  0.01 R3 25 R1 R2  1A 3 . The emf of the standard cell E 100 (0.01)G = 10R ...(iii) 1      E100 0.01 R3 R2 The emf of the secondary cell e 30 25      e30 R1 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 By solving Equation (i), (ii) and (iii) R = 0.0278 ; R= 0.25; R = 2.5 E  100  e  30E 1 2 3 e 30 100 9. 4 . Ig=1A; G=0.81; I=10A 10A 1A 0.81 9A Assume 1 division have x ampere When r=10 E S  10  10R / 10  R  9x ... (i) When r=50 90  Ig  1   Ig  S   0.81  0.09 E S   I  G ; 9 90  10  50R / 50  R   30x ...(ii) 59

JEE-Physics 5 . On redrawing the circuit between A and B we get 8 . Let resistances be R1 and R2 A B  R1 R2  A R1R2 I  3V =1.5A 3V 3 3 then S = R1 + R2 and P = R1  R2 2 B 3  n  S  R1  R 2 2  R1  R1  2 P R1R2 R2 R2 2  R2  2 =  R1  3 R1   4  n min  4 R1 3 3 A  B 1 r12 9. Given that   4 & r1 2  A1  r22 4 2 3 r2 3 A2 9 3V In parallel : I1R2=I2R2 3V hence I1  R2  2  A1 34 1   6 . For a given volume, the resistance of the wire is I2 R1 A 2 l1 4 9 3 expressed as  2 x y Volume 10. R  R  2 R2  2  2 R2  R1 R1    R1  4 3 So, the change in resistance of wire will be 300% 20cm 80cm x y x1 y 300%   A D 20 80 y 4 2 4x 7 . 6V 2 C 1.5 6 Now On redrawing the diagram, we get I = =4A 1.5 6 a 100-a 2 1.5 B AC  4x y 2   a  50 cm Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 1.5 a 100  a 6V 1 2 Voltage across R=2V R =2V 3 Hence, voltage across 500=10V 1.5 1.5 5 00 =10V 500 G  12V R 2V 6V 6V 10 1 Current through 500= 500  50 A 60

JEE-Physics 500=101A B B  A  A   A AB AB AA B B AA 500 50 As 500 and R are in series value of  A 500 R   A BdiaB 2 1  22  2 R  VR  2  100 B Adia A 2 = 2 IR 1 / 50 1 8 . Given that 1 3 . E,R1 E,R2 R =100°C 100 RT°C = 200 T=? R100=R0[1+(100)] ...(i) RRT]...(iii) R On dividing eq. (2) by eq. (1), we get 2E R T  1  T R100 1  100 Current in the circuit (  )R1  R2  R On solving, we get T=400°C potential difference across cell with R2 resistance PR S1 R2  Q S2 2E 19.  E  IR2  E  R1  R2  R  R2 But potential difference = 0 2E Under balanced condition  E  R1  R2  R  R2  R = R2 – R1 ()  1 4 . Current supplied by the source to the external P R  P  R S1  S2  resistance  Q S1S2 Q S1S2  S1  S2 E I= Rr If () r>>R; I= E 20. 10 C r A which will be constant(   ) 5 10 C 20 B A 10 1 5 . The internal resistance of a cell DB  r  e    l1  =  240  1 2=2 5V  v T  1 R  l2  1 R  120 Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 On redrawing the circuit, we get ( ) 1 6 . Kirchoff's first law is based on law of conservation C of charge.Kirchoff's second law is based on law of 10 20 conservation of energy. A 10 B  5 10  D  1 7 . Specific resistance (B)A ; diameter dB=2dA 5V (B)A ; d B=2dA It is a balanced Whetstone bridge having Reff as B ? for Resistance 1  A B 30 15 Resistance Reff = 45  10 A 61

JEE-Physics C 30 For parallel combination () 20 15 10 11 1 A 10 B  5 10 R R1 R2 D  11  1  R eq R 0 (1  1 t) R 0 (1  2 t) 5V 5V 1 11  The current delivered by the source is  R0 R 0 (1  1 t) R 0 (1  2 t)  2 1  pt I  V  5  0.5A 2 (1 + pt)–1 = (1 + 1t)–1 + (1 + 2t)–1 R 10 using binomial expansion (  ) 2 1 . Let the resistance of the wire at 0°C is R0 also let the temperature coefficient of resistance is . 1  2 2 0°C R02 – 2pt = 1 – 1t + 1 – 2t p =  28.   R  2 R= R50=R0[1+(50–0)]....(i) A Similarly R100=R0[1+(100–0)] ...(ii) R R 2 On dividing equation (ii) by equation (i), we get   = 2[0.1] = 0.2% increase. RR  R100 1  100 6  1  100 29. R = R1 + R2 + R3 + R4  R  5 100  5 R 50  1  50 ; 5 1  50 100  6  300  5  500  1  200 R = R1 + R2 + R3 + R4 = 20   1 / C For combination R 100  20 100  5% 200 R 400 3 0 . i = 0.2 A,  = 4 × 10–7 -m, A = 8 × 10–7 m2 1 x  i  0.02  4  107 = 0.1 V/m On replacing   200 / C in equation (i), we get A 8  107 1  1  1  1 3 1 . Due to greater heating as H = I2R 200  4  5=R0 5 0 5  R0 25W get fused. 5   H=I2R  4   5  R0  R0  4 2 5W  32. 22. 55 R  R  55  8  220  20 80 2 6 2 4 . Choosing A as origin,(A ) 240 I Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 E  j   2r 2 VC  VB I a b 1 dr I  1 1 2 5 .   a   r2  2 a  b 2 a 6 240 VB  VC   I 1  a 1 b   2  a   60 2 7 . For series combination ( ) (1 2 0 )2 R bulb  60  240 S  1R 01  2R 02 R 01  R 02 120 V1  246  240  117.07 R01 = R02 = R0 (given) S  1  2 2 62

JEE-Physics (1 2 0 ) 2 3 . Current I can be independent of R6 only when R1, R heater  240  60 R , R , R and R form a balanced wheatstone’s 234 6 120 bridge. Therefore, R1  R3  R1R4 = R2R3 V2  54  48  106.6 R2 R4 So change in voltage = V1 – V2  10.4 Volt  R1, R2, R3, R4 R6      3 3 . To increase the range of ammeter, resistance should  I  R6     bpearadleleclr)esaosetdota(Sl oresaidstdainticoenatol sahmumntetceorndneeccrteeadseisn. RR12RR 3  R1R4 = R2R3 4 ( ) 3E2 () 4 . In the first case t = msT..(i) R  V2   H  R t  When length of the wire is doubled, resistance and mass both are doubled. Therefore, in the second case ( EXERCISE –V-B NE2 .t = (2m)sT ..(ii) Single Choice 2R 1 . Net resistance of the circuit is 9. Dividing eq. (ii) by (i), we get  current drawn from the battery, N2 =2  N2 = 36  N=6 18 9 5 . The circuit can be redrawn as follows   2R 2R 2R 2R 9 2R i = 9 =1A = current through 3 resistor rr r P QP r Q 3 A 2 C 2 2R 2R 2R 2R 2R 2R 9V 1A i2 i4 4R 8 8 4 2r P i1 i2 i3 2Rr 4R 2 B 2 D 2 R+r QP Q Potential difference between A and B is A B  V2 V2 V2 V2 VA – VB = 9–(3+2) = 4V = 8i1 6. P= so, R=  R = & R =R = RP 1 100 23 60  i = 0.5 A  i = 1 – i = 0.5 A 1 2 1 2502 Similarly, potential difference between C and D Now, W1 = R1  R 2 2  R1 and Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 CD  VC – VD = (VA–VB) – i2(2+2)= 4 – 4i2=4 – 4 W= 2 5 0 2 and 2502 (0.5)=2V= 8i3  i3 = 0.25 A 2 R1  R 2 2  R2 W3  R 3 Therefore, i =i –i = 0.5 – 0.25  i = 0.25 A 423 4 W :W :W = 15:25 :64  W< W < W 123 1 2 . As there is no change in the reading of galvanometer 2 3 with switch S open or closed. It implies that bridge 7 . Ammeter is always connected in series and voltmeter in parallel. ( is balanced. Current through S is zero and S  S8.Theratio  AC will remain unchanged. CB  AC  I =I, I=I. CB R GP Q 63

JEE-Physics 9 . P=i2R Current is same, so PR. 2 R 2R 2 In the first case it is 3r, in second case it is (2/3)r, in 1 6 . G G r 3r III case it is & in IV case the net resistance is 32 P=i2R PR.  R > 2  100 – x > x 3r, (2/3)r,  Applying PR  QS r IV 3r   2 x ..(i) R  x  20 ...(ii) We have  3 2 R 100  x 2 80  x R < R <R <R  P < P <P < P Solving eq. (i) and (ii) we get R=3 III II IV 1 III II IV 1 1 7 . Given circuits can be reduced to 10. 54 3 RPQ = 11 r, RQR = r and RPR = r 11 11  R is maximum 1 3V 1  3V PQ 2 32= = 32= P1 = 1 9W P2 12 18W 1 1 . BC, CD and BA are known resistance. The unknown resistance is connected between A and D. 2 3V BC, CD BA A D  P3 = 32= 9W  2 2 1 2 . Vab = ig.G=(i–ig)S  i = 1  G  ig V2 S  1 8 . P  and 100W > 60W > (i-ig) R S V2 V2 V2 1 11 40W  R100      R 60 R 40 R100 R 60 R 40 [Note : Although ( ) 100 = 60 +40 so at i a ig G b room tempeature () Substituting the values, we get i=100.1 mA V2 V2 V2 1 11      1 3 . W=0. Therefore, from first law of thermodynamics, R100 R 60 R 40 R100 R 60 R 40 (Applicable W=0.   Only at room temperature) U= Q = i2Rt= (1)2 (100) (5× 60) J = 30 kJ ()] 1 4 . Current in the respective loop will remain confined Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 in the loop itself. Therefore, current through 2 1 9 . resistance = 0. Current always flow in closed path.  2  1 5 . H = I2Rt I  same So H  R  , same. 20. R  L L   independent of L R = r2 A   H = 4H Lt t BC AB So H  R  1 r2 64

Multiple Choice 7.5 amp JEE-Physics v 15 V 2 K R p  1.5  6  2 . The rheostat is as shown in figure. Battery should 1 . I= R e q 24V 7.5  be connected between A and B and the load between C and B AB CB 9 V 1.2 K I = 240  60 = 7.5 mA 32 8 (A) Current ()I is 7.5 mA (B) Voltage drop across RL is 9 volt RL =9 volt (C) P1 = V12 × R2 225 1.2  1.6 P2 R1 = 2  81 v 2 2 (D) After interchanging the two resistor R and R 12 RR  12 V = 2.4  7 = 3.5 mA 3. I = R eq (48) P1 = V12 RL  v1  2 21 V P2 R L (v2 )2  v2    24V 4 . Slide wire bridge is most sensitive when the 3 V resistance of all the four arms of bridge is same.  9 2 =9  3  Hence, B is the most accurate answer. Assertion - Reason  1. Ans. D B Subjective Problems  1 . (i) There are no positive and negative terminals on the galvanometer because only zero deflection is needed.  5. (ii) Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-07 & 08\\03-Current Electricity.p65 G J1   2 2 R and J2    2 R as J1  2.25 x  R J2    R  2   1 / 2  12 A JB CD 42 42 (iii) AJ = 60 cm  BJ = 40 cm so R  2  2.25 1  2R 2  R  4 If no deflection is taking place. Then, the Wheatstone’s bridge is said to be balanced,   Hence, X  RBJ X  40 2  x= 8 12 R AJ  12 60 3 65



JEE-Physics UNIT # 04 (PART–I) GRAVITATION EXERCISE –I GM G  4 R 3 4 R2 3 G  R G 1 1 6.67  10 11 6. g   R2 g = .2 2 .04 3  1.67  109 1. F1  F2    g  R  g  R  g  3g g 3R 2M F12 7. GM g mg 10 g '  (R  h)2  49 ; w' = 49  49 = 0.20 N 1 M F13 M Apparent weight of the rotating satellite is zero 3 because satellite is in free fall state.  Fnet  F1 ˆi   F2 ˆj  F ˆi  ˆj  1.67  10 9 ˆi  ˆj ( ) 1C 2. F  rm ; F  rm 8. t 2h t'  2h 6 2h  6 sec g =1 sec; g'  g This force will provide the required centripetal force ( ) 9 . g' = g – 2r cos 60 g' = g – 2R cos2 60 Therefore g' = 0, g = 2R cos2 60 m2 r = C 2  C 60° rm ; m r m 1 T  2  T  r m 1/2 m 2r  60° 3. At P : g = GM G81M 0 r = R cos 60° x2  (D  x)2 2 R R D – x = 9x; 10x = D MP 81M 4g , t   2  g R x D–x 4g x D 9D from the earth 1 0 . Acceleration of small body w.r.t. earth=g–(–2g) =3g from the Moon and Now from second equation of motion 10 10 (=g–(–2g) =3g 4. g  GM  )  r2 R is reduced to R/2 and the mass of the mars H= 1 (3g)t2 t 2H 2 3g becomes 10 times (R  R/2     10) OR node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 44 X CM  2mx1  m(x2 )  2mH  0  2H gmars  10 gearth and Wmars= 10 Wearth = 80N. 2m  m 2m  m 3 1 2h  g  2h H  1 gt2  t  2H R  g R 32 3g 5. g '  g  ; h  h 1 ; g '  g 1  d  1 1 . Gravitational field inside the shell is zero. But the 1=2 R2 R  force on the man due to the point mass at the R centre is ( ) g '  d  h g decreases by 0.5% gRR 48

JEE-Physics GMm GMm Gm GM ( FNew  3R 2 ; Fold = R 2 V1  md ( M m ) ; V2 = Md M m) Change in force   2 GMm G m( M  m ) ; V2  G M( M  m) 3R2 V1  d d 1 2 . By applying conservation of energy G 2 V  ( M  m) ()  d KEi + PEi = KEf + PEf 1 8 . There will be no buoyant force on the moon. 1 mv2  GMem  0  GMem 2R 2R (Eventually balloon bursts) (    )  1 mv2  GMem  1  1 2GM 2G2M  2 2GM 2 R 2 R ; v'e = 19. ve  R/2 R 1 mv2  GMem  u  GMe ve =2 (11.2 km/sec) = 22.4 km/sec 2 2R R 13. PEi =  GMem  mgR ; PEf =  GMem   mgR 2 0 . To escape from the earth total energy of the body R 2R 2 should be zero KE+ PE = 0 (  ) mgR 1 mv2  GMm  0  KEmin = m gR e Increase in PE is 2 5R 5 2 1 4 . Centre of gravity of the two particles. 2 1 . There is no atmosphere on the moon. () () X CG  W1 X 1  W2X 2  (0)(0)  (mg)(R ) R 2 2 . K.E. =  1 GM1M2 W1  W2 0  mg 2r r= 2R for the first and r = 8R for the IInd The centre of mass of the two particle system is at K.E1   1 8R   4 :1 K .E 2  2R 1  ()   GM1M2 , P.E1 R P.E2 = 4 : 1 M(R )  m(0) R Similarly P.E. is  2M 2 X CM   Put the ratio of K.E  2 P.E 15. Ig GM V   GM , 2 3 . Relative angular velocity when the particle are  R2 , R moving in same direction is      V=IgR=6× 8 × 106 = 4.8 × 107  0 k 1+ 2 (1 + 2) t = 2 dV dx 16.  dV   Ig.dx ;  v   x3 2 rad / sec  24 6 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 r  2 = ; 1 = 0–V =  1 2   V = k  k When the particles are moving in the same direction 2x r  2r2 V  2x2 then angular velocity becomes 17. () 'Emq'uiisli(bmriumpositionofthe neutral pointfrommass)Bysubstituting(1–1a2)nd(2 1in– 2) t = 2 get equation we  m 1  2  24 hrs =  m  M  d 2 4 . At points A, B and C, total energy is negative. V1  Gm1 V2  Gm2 (A,B C) r1 ; r2 49

JEE-Physics EXERCISE –II Gm2 Tcos  6 . T sin  = 2 ; T cos  = mg  GMm tan  = Gm ;   tan 1  Gm  Gm2 Tsin 1 . Net force towards the centre  m2(9R)= (9R )2 g2  g2  2 mg  8 . Gravitational field and the electrostatic field both  2  GM  T  2  27  2 R are conservation in nature (    729R 3  g )  2. V = –Eg.dr ()10. Both field and the potential inside the shell is non zero Because field is uniform (  2 = –Eg.20  E   1 ; V = + 1 [4]  2 ) 10 10 5 work done in taking a 5 kg body to height 4 m 1 1 . Case I : = m (change in gravitational potential) (5 kg4m Ui + Ki = Uf + Kg GM em  1 mv2   GMem  0 = m) R2 R  h1 = 5 2  2 J G M e m 1 2GM R   GMem  5  R 2 R2 3 R  h1  m e 3 . when r < r1, gravitational intensity is equal to 0 ( r < r1)   1  1 1 h1 = R M2 R 3R  2 R  h1 M1 Case II : r1 Ui + Ki = Uf + Kg r2  GMem  1 mv2   GMem  0 R2 R  h2 GM1 when r > r1, gravitational intensity is equal to r2 G M e m 1 2GMe GMem R 2 R2 R  h2 when r > r2, gravitational intensity is equal to  m R   G(M1  M2 ) ( r > r1      r2  1  1  1 R 2R R  h2  h2 = R GM1  r> r2     Case III : r2 Ui + Ki = Uf + Kg G(M1  M2 ) r2   GMem  1 m 4GM e R   GMem R 2 R2 3 R  h3 k 4.  dV   E.dr ,  dV   r dr node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 11 1     v = –k log r + c at r = r0; v = v0 R 3R R  h3 h3 = 2R  v0 = –k log r0 + c  c = v0 + k log r0 By substituting the value c from equation 1 2 . By applying work energy theorem change in K.E. (c) = work done by all the forces   v= k log  r0  + V0    r  K.E. = Wg – Wfr ; Wg > Wfr  dU d therefore KEf increases due to the torque of the 5. F   dr (ax  6 y ) ; Fx = –a and Fy = b air resistance its angular momentum decreases dr therefore A,C (       a2  b2 KEf)  F  aˆi  bˆj  acceleration= m 50

JEE-Physics 1 4 . Fnet = force due for sphere + force due for cavity GMr (+) 1 8 . Gravitational field intensity F = R 3  GMm  R  0 GMm  mg () R 3  2   2R2 2 Inside the sphere (  )  (F1  r1, F2  r2) 1 5 . Pressing force by the particle on the wall of tunnel F1  r1 of r1 < R & r2 < R is and acceleration is mgsin F2 r2 ( mgsin) Gravitational field intensity (  ) Pressing force ( ) I 1 (Out side the sphere())  r2 GMx R GM = mgcos   R3 2x 2R2 r22 F1 r12 Pressing force is independent from 'x' thus it is  F2  if r1 > R and r2 > R constant (x ) 19. Gravitational potential ( ) V  GM R GMx x2  R2  GM 4x2  R2 (B) Gravitational field at the point x from the centre gsin = R 3 4 of the coil is (x) x2 2R3 GMx  (R2  x2 )3 /2  2 0 . Gravitational potential due to hemisphere at the centre is V because distance of each mass particle from the centre O is R. If the distance between the point and mass is changed potential will also change R (V x is increases from to R, thus acceleration R 2 R R  x)  increases ( )  2 1 6 . Motion of m (m  ): 2 1 . Acceleration of the particle from the centre of the earth is directly proportional to the distance from m CM 2m the centre ( 2r/3 r/3 )  node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65  2r  Gm 2m  2 r3  a = GMx  a  x  a = 2x m2  3  =  T=  2 R3 r2  3Gm Particle will perform oscillatory motion.  T  r 3 / 2 and T  m1/2 () 1 7 . Due to symmetry the gravitational field at the origin is zero. The equipotential line will take the shape 2 2 . By energy conservation(   )  of a circle in yz plane. Ki + Ui = Kf + Uf ( 1 R 2  11GMm 0  GMm 2 2  8R yz) 2R  K   51

JEE-Physics GMm 11  1   1 K  R2  EXERCISE –III R  8 2  2  4  True / False 7GMm KR2 1 . True  2 . False, total energy must be negative. 8R 8 7GMm ( )  K  R3 3 . True, two (negative) masses attract each other. 3GMm (      )   2 3 . P.E. of the system is equal to U i   2R () Fill in the blanks work done (  ) 1 . COME : U1 + K1 = U2 + K2   3GMm 0 + 0=– 3GMm 1 mv2  v = 3GM 3 + 2 = 2 ve  U   U f  U i  U i   2R 2R R 24. m v1 d v2 M GM r 2. vorbital = Total energy of mass M will become zero, it will be escape (M  Kinetic energy after firing(    ) ) 1 m 2v0 2 = 10GMm =  2  R K+U=0 2 1 Mv2  Gm1m2  Gm2m2  0 3 . COME : U1 + K1 = U2 + K2 2 dd 1 MV2  G M 2  M  M2  v v 2 d m m 1 GMm 1 1 GM  0 + 0 = – r + 2 mv2 + 2 mv2  v = r V 4G M1  M2  d GM 4 . geff = R 2 –2R = g – R = 0  g 10 1 = 6400 103 = 800 rad/s R = 1.25 × 10–3 rad/s 5 . Kepler's third law is the consequence of conservation of angular momentum. ()  node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 6 . Area enclosed by earth's orbit ()   L  T = 4.4  1015 × (365 × 86400)  2m  2 = 6.94 × 1022 m2 7 . COME : K1 + U1 = K2 + U2  1 1 m  2GM    GMm 0 – GMm  h=R 2  2  R   R R h 52

JEE-Physics Match the Column (C) At C and D, gravitational field and potential remains same 1. ( CD r )  2r (D) As one moves from D to A, field decreases ( DA)   GM  2 mvr= 2 L 4. (A) Kinetic energy in gravitational field increases For (A):L'=mv'r'= m  2r  (2r) = if the total work done by all forces is positive. For (B) : Area of earth covered by satellite signal (  ) increases. (       Potential energy in gravitational field increases )  (B) as work done by gravitational force is For (C) : Potential energy ( ) U'  GMm U GMm GMm negative. (  and    2r 2 2r r ) For (D): Kinetic energy (  ) (C) For mechanical energy in a gravitational field 1K to increase, work done by external force K' = mv'2 =  K' < K should be non–zero. 22 (  va ) rP ra 2 . COAM : mvara = mvprp Comprehension Based questions vP (A) At perigee () Comprehension#1 rP < ra  vP > va (r) 1 . By applying conservation of angular momentum (B) Distance from sun at the position of perigee ()  decreases (q) (mv0R cos  = m v (R + h) ) v  v0R cos  R  1 Rh  R  GMm h  v0 cos  > v (C) Potential energy at perigee UP = – rP 2 . By applying conservation of energy GMm () node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 (UP= –rP 1 GMem  1 mv2  GMem 2 mv02 – r2 (R  h) rPUP  Solving above equation () (D) Angular momentum remains same (p) ()  h  v 2 sin2  0 3 . (A) Potential at A Potential at B 2g (AB) Alternate : As height increases gravitational force decreases and hence the acceleration. Therefore (B) We can not compare about gravitational field at A and at B height will be more than H  v 2 sin2  0 (AB 2g ) 53

JEE-Physics Comprehension # 2 EXERCISE –IV(A) 1. G2Mm GM  F  (R  Rx )2  a  2R2 1 . F1 = F42  F41  F41 h  1 at2 4hR2 hR2 G8 G4 2  2 2 G  = 4L2 + 2L2 (2cos 45) = 2 GM  t2  t 2 GM L2 GM GMh F2  = G8  G2 2 cos 45 2 . V at surface = 2as  2 2R2 h  R = F24  F21  F21 4 L2 2 L2 R R2 R2 = G 2  2  F1  2 L2 F2 If a = 0, t1 =  ; but a >0 ; t< v GMh GMh 2 . mg – T = ma .... (1) G (2 M )m  1 mv2 GMm GMm T + mg' = ma .... (2) 0 2   3. COME  By addiing (1) and (2) (2R  h) R 2R  v  GM a  g  g' T  m(g  g)  mg ' mg' R ; T 22 T Comprehension # 3 T  m  g g '  m   g 1  2h  mg 1 . As the distance of the star is doubled the potential  2  2 g R    energy becomes half of the initial and the velocity  1 mg 2 mg GMm of the particle will become 2 times of its initial T   2R R R3 value because K.E. = 1/2 P.E. GMm ( 3.F1=4R2   F2 =force due to whole sphere – force due to  1 KE=1/2P.E.) cavity 2 (F2=–)  2 . Its kinetic energy GMm GMm 7GMm  F2  7 F2 = 4R 2  18R 2  36R 2 F1 9 4 . For the line 4y = 3 x + 9 4dy = 3dx; 4dy – 3 dx = 0....(i) For work in the region, dW = E.dxˆi  dyˆj = 3ˆi  4ˆj.dxˆi  dyˆj = 3dx–4dy (from equation (i)) = 0 5 . Total mass of earth node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 M  4   R 3 1  4   R 3  R3  2 3  2  3  8  M  4 R 3 1  72  24 Acceleration due to gravity at earth's surface = GM  4 G R 1  72  R2 24 54

JEE-Physics R 9 . No. of pairs for P.E. can be calculated by using Acceleration due to gravity at depth 2 from the ( ) 4  R  2  n(n  1)  2  = nC2= 2  28 G  3  1  4 G 1 R   6   surface  R 2 where n is the no. of mass particles.  2  (n)  Now according to question out of 28 pairs :– 4GR 1  72   4GR1  1  7 12  sides of cube 24 6 2 3 12  face diagonal 6 . Speed at surface (   )= v (let) 12  body diagonal 10. M x m a M,R U  GM1M2 for the point Mass d u=0 COME RR RR g=0 d U  GmdM U   GMm dx x x  GMm  0 = GMm  1 mv2 GMm a dx GMm n    a 2R R 2  a x   a  U     GM 11. GMm Ui =  GMm v = R ...(i) Uf   R (1  n) R Inside the shell () g= 0 dis tan ce 2R R3  U = Uf – Ui =  GMm  1 n  1  R 1    t= = =2 speed v GM By applying energy conservation GM G(5) ( )  7. V1 =  G (R 2  x2 )1 / 2 (16  9 )1 / 2 1 mv2 GMm 1 2GM 1 GM 2  R 1  1  n  ; v = R 1  1  n  V2  (R 2  x2 )1 / 2   3 3 4 G(5) G5 1 3 . Net torque on the comet is zero then the angular  (9  27)1 / 2  6 momentum is conserved. ( ) G work done = m[V2 –V1]  6 = 1.11 × 10–11 Joule 1 4 . (i) Orbital velocity ( ) node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 8 . Potential at centre MM ()   GM 1 2GM v0 = = r2 R M M GM  4GM   4 2GM  =  r r   r = 2R = R + h  h = R = 6400 km Potential energy of the system (ii) COME : GMm =  GMm  1 mv2  0 ( ) 2R R 2 4GM2 2GM2 5.41GM2  v= GM gR = 7.9184 m/s     2 R   55

JEE-Physics EXERCISE –IV(B) 4R2  4H2  8RH  9R2 16R2 16RH  0 4H2 – 8RH – 3R2 = 0 GM 8R  64R2  48R2 1 . Orbital velocity ( ) v0 = r1 = After impulse (  ) v1= kv0 8 COAM : mv1r1 = mv2r2....(i) = R±R R2  3 R2 4 COME : GMm k  1 m v 2  r1 +  2 0  R = R± 7 = GMm +  1 m v 2  ...(ii) r2  2 2  2 4 . Point P, where field is zero Solving equation (i) and (ii) r2  k r1 2  k (P,)  16M M 2a a Pu x 2 . Let r = distance of apogee ( ) 6GM GMm G 16M  m COAM : 5R R = vr ...(iv)  10a  x2  x2  x = 8a COME GMm 1  6GM  COME : – G 16M  m  GMm  1 mv2 :– +m   R 2 5R 2a 8a 2 GMm 1 mv2 ....(ii) G 16M  m GMm 3 5GM =– + =  v= 2 r2 8a 2a a 3R 8GM 5 . COAM : mv1 (2R) = mv2(2R)  v1 = 2v2...(i)  r = 2 and v = 15R COME : GMm + 1 m v 2 = GMm  1 m v 2 ...(ii) Orbital speed at r (r  ) 2R 2 1 4R 2 2 GM 2 GM 2GM = r = 3R  v1 = 3R  Increase in speed (  ) v 2 1  Radius of curvature at perigee = g1 2 8 = GM    () R  3 15  2GM 4R2 8R  RP =  3. v1 H  R  3 GM R 3R GM 3 2 2R 1  3R  Gm 6 . s = 2 g1t02 ...(i)  R  H  8R v1  v 2  GM  v2  2GM 1 ...(ii) s g1 s g2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 0 2R s = 2 g2(t0–n)2 u0, t0 u0+u, t0n R  H ...(iii) u0 = g1t0 ...(iv) u0 + u = g2 (t0–n) GM 9R2 GM 2GM   R  H2 R  u2 2R 8R R h After solving we get g1g2 =  n  9R2 4R 7 . Loss of total energy ()   –1= = |TE|final – |TE|initial = Ct 4 R  H2 R  H –1 = 9R2 16R2 16RH  Ct = GM sM e 1  1 2  R e R S  4 R  H 2 56

JEE-Physics dv GM v gR2 GM R  x GM v dh R3   R  h2 vdv   R  x 2 dh R  h2 v0  8 . ;  =  v2  v 2  1 h   5 1  R or 0 0  R  h 0 x =  2   gR2 2 v  v 2  gR2  1  1  GM 0  R  R  1 2 . (i) Orbital velocity of each particle v0 = r 2 h () v 2  v2  2gh 0 v 2  v2  2gRh 1 h  0  R  Masxeipmauramtion orbit-1 R  h ; orbit-2 m2 9 . m1 Astronaut satellite M For satellite : GMm1 –T = m12R ...(i) (ii) Maximum separation before collision = 2 r R2 () GMm2 (iii) Relative velocity before collision R  r2 ()  For astronaut : +T = m22 (R+r)...(ii) 2GM vRel = v12  v 2 = r 2 Dividing eqn(i) by (ii) gm1  T  m1R2 1 3 . Average velocity of satellite P, gm2R2 T m2 (R  r)2 (P)  R  r 2 vP  1 GM R R3  m1R  m2 R  r  WP = 2R 2R 2R r   gm1m2   r   2  = T R Average velocity of satellite Q, R R3   m  (Q)   m   T= m2g   r  R  r2  2  0 vQ 1 GM R 1 3R 3R 3R WQ = = = m2g 1  r   1  2r   = 3m2gr Least time when P and Q will be in same vertical line R  R   R ( PQ ) 3 100 10  64 2 2 2R 3 / 2 6 6 = = 0.03 N  = 1 2 = GM 2 2  3 3  6400 1000 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 1 0 . The area on earth surface in which satellite can 1 4 . (i) At equator ()  not send message ( GMm m2R = R 2 )R = 4R2 [1–cos] P 2 G 4 R 3 3  x2  R2   T = = R 3 . 3  T = G = 4R 2 1    x 11. g below surface (  ) g= GM 3 R 3 (R–x) (ii) T = 6.63  1011  3000 = 1.9 hr GM g above surface (    )g= R  x 2 57

JEE-Physics 1 5 . (i) Necessary centripetal force = Gravitational force 18. F  GMmx  m2 x R3 (= GM2 r3 Particle will perform a oscillation  M2r = 4r2  T = 4  GM with angular speed 2  GM R3 (iii) COME : KE + PE = 0  1 mv2 2GMm =0  v= 4GM ;  r  d (2 GM ) 2  r  2  R3 r 1 6 . Let x = distance of the particle from the surface R3  R3 Ti  2 GM  t1  2 GM Acceleration, (x= ) If acceleration is constant ( )   vdv = GM R  x dx R3 GM  1 at2 1 GM v x 1 x dx g  R2 ; S 2 R  R2 t2 R  vdv   g  2  ve 0 2R3 2R 3 t1  GM GM t2 22  t2   t2  ; v = v 2  x2   dx 1 9 . Relative velocity when satellite revolving anticlockwise e  2g  x  2R  dt ()  t0 dx (1 + 2) t = 2   4  2  t  2  ; t  24  x2   3 24  17   dt =  2g R  x  2R  0R If it moves in same direction (     0 dx )  g 3R 2  x  R 2   4  2 t = 2  3 24  t = R R Let x – R = 3R sin  dx = 3R cos d  30 24   t = 2  t = 1.6 hrs. 24 15 R R x  R R d  g sin 1  3R  0 t   g 2 0 . By applying conservation of linear momentum  t= R sin 1  1 () g  3  m(v) – m(v) = 2mv'; v' = 0 1 7 . Range of throw is = 10m Initially, energy of a satellite 'A' and 'B' is (AB) u2 EA   GMem ; EB   GMem g = 10  u2 = 100  u = 10 m/s 2R 2R ve  2GM ve  Me  Rp Total energy ( ): EA  EB   GMem R  vep Re Mp R node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 After collision velocity of satellite becomes zero then  Se   Re 2 the K.E. = 0, therefore total mechanical energy  Sp   R p  ve  (S = density) becomes (KE=0 vp )GMe 2M 11.2 Re 1 10  10 Rp 2  Rp  11.2  R   2  Re 10  6.4  106  = 40.42 km 11.2  2 58

EXERCISE –V(A) JEE-Physics OR Initial position 2. Required KE = GMm  mgR R 3 . Energy required (   ) = Uf – Ui GMm GMm   GMm   1  1 Just before collision =  R  3 2  3R 2R = G M m   2  3  = GMm  System centre of mass situation   R  6  6R   Fsystem  0  4 .  T  R3/2 M 0  5M 12R  M d  5M d  3R    d  7.5 R M  5M M  5M R 3/2  R   T2  T1  2   5h 43/2  40h 1 5. Force on M block (M)F= GM5m mv2 m 144R2 x 6 . F  R  x x mv2 GMm R  x 2  R  x 9R m RF 2R 5m GM GM R2 FF v = R  x = R2 R  x gR2  gR2 1/2  = R  x =  R  x  12R F 5GM GMm GMm a =  8 . U  U f  U i   2R  R m 144R2 GMm mgR Force on 5 m block (5m ) = 2R 2 GM5m  9. n 1 F = 144R2 GMm GM 2 rn r n1  T    r2 GM5m GM F   m2r    a = 144R2 5m  144R2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 Relative acceleration ( )  11. gh  g 1  2h ; gd  g 1  d   R  R  5GM GM 6GM GM a = 144R2    144R2 144R2 24R2  2h   d  gh = gd  g  R   g  R   2h = d 9R = 1  GM  t2 2 24R2 18R  24R2 1 2 . Work done = Uf – Ui = Ui= 0 – (–0) t2 = Ui  GM1m2  6.67 104  100  10  101 GM R 10 102 1 5GM 18R  24R2  7.5R s   2 144R2 GM 6.67  1011  100  10  103  10  10 2  6.67  10 10 59

JEE-Physics 1 3 . Electronic charge is independent from g, then the EXERCISE –V(B) ratio will be equal for 1. (g 1 ) Single Choice GM M 1 . Force on satellite is always towards earth, therefore, 14. ve  R ve  R acceleration of satellite S is always direccted ve  e  Rp  Me  Re / 10 towards centre of the earth. Net torque of this Re Mp Re 10Me gravitational force F about centre of earth is zero. v 1 Therefore, angular momentum (both in magnitude e and direction) of S about centre of earth is constant 11 1 throughout. Since the force F is conservative in  10 ; v1e  110 m/s nature, therefore mechanical energy of statellite v 1 remains constant. Speed of S is maximum when it e 16. gh  g  g  1  h 2 9 is nearest to earth and minimum when it is farthest. 9 R 1 h  2  (S R     F h S F 1+ =3  h = 2R  S   R )  r 17. m 4m x G  m G  4m r 2. T2  R3; with Re = 6400 km, x2 = (r – x)2  x = 3 Potential at point the gravitational field is zero T2  6400  3 between the masses. (2 4 )2  36000    T  1.7 hr (Forspysatellite R is slightly greater than )  (RR e) 3Gm 3  G  4m Re  TS > T  TS = 2hr V =– – r 2r 3Gm 9GM 3 . Figure shows a binary star system. = – [1 + 2] = – rr 18. Gm2 = mV2 = V  Gm m m MB RB RA MA 4R2 R 4R The gravitational force of attraction between the 1 9 . PEi + KEi = PEf + KEf –mgR + KEi = 0 + 0 stars will provide the necessary centripetal forces. node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 KEi = +mgR = 1000 × 10 × 6.4 × 106 In this case angular velocity of both stars is the Work done = 6.4 × 1010 J same. Therefore time period remains the same. (           )     2  T  60

JEE-Physics V 4. M M L L Total energy of m is conserved for escape velocity K .E + P.E = K.E + P.E ff ii 0+ 0 = 1 mv2  2  GMm   v  4GM  2 GM 2  L  LL Subjective 1 . Total energy at A = Total energy at B (A=B) (KE)A + (PE)A = (PE)B B VR e 100 100 A 99R 100 R 1 m  2GM   G M m 3 R 2   99R 2    GMm 2 R    100   R h  2R 3 On solving we get h = 99.5 R node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\03-Gravitation.p65 61

JEE-Physics UNIT # 02 (PART – I) NEWTON LAWS OF MOTION & FRICTION EXERCISE –I 9 . Just after release T = 0 due to non–impulsive nature of spring. So acceleration of both blocks will be g 1. Force on m = Force on m  a = m2a2 1 2 1 m1 10 . Case (i) : F = 2T 11 2 . ma = mg – T a1  4mg  2mg  2mg  g F1 min max 6m 6m 3 TT 75 mg g = mg – mg =  a = g 100 4 min 4  T – 2mg = 2m × 1 3 3 . For BC = 0, a = 2g  g  0  20 ms2  T = 8mg  F = 16mg 2 5 1 4 4 8 1 1 3 3 For BC = 2m, a = (2  1)g  3g  30 ms2 2 5 1 8 8 Case (ii) : F2 T F = 2T 22 4 . Impulse = Change in momentum 4mg  2mg g T a=  2 6m 3 = m(v – v) = 0.1  0  4 = –0.2 kg ms–1  4mg – T = 4m × g 2 1  2  2 3 8mg 16mg T=  F = 2T = 5 . Impulse =  Fdt 23 2 2 3 I  Impulse = 0.25 × 1 = 0.25 II 1 2mg  mg 2mg  mg g  Impulse = × 2 × 0.3 = 0.30 2 11. a = g; a = = 1m 2 3m 3 1 mg  mg  mg g III  Impulse = 2 × 1 × 1 = 0.50 a = = ; a >a >a 3 2m 2 1 3 2 1 a1 IV  Impulse = 2 × 1 × 1 = 0.50 MT 1 2 . T = M  a ...(i) a/2 2 1 m 1 T 6 . sin = x  tan = x2  1 20– 2 = 2× a ...(ii) For body Ncos = mg N 1 Ma a a=0 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 Nsin = ma  20    2a (constant T velocity) 22 2 ma & T  1  g  T  20N 2 2 g mg 20–10 = 2a  a = 5 m/s2  a = gtan= x2  1 8. Acceleration of particle =  m1  m2  (g  a) 5M  2  5 20  10  4  m1  m2  4 20 – M  5   2 1 (g  a)  g a= g M = 8 kg  2  1  2 2 34 E

JEE-Physics 1 3 . Acceleration 1 8 . Block starts sliding when kt = µmg 0 Net force 3  250  (100)g sin  µs,µk m F = kt = Total mass = 100 750  260 so for t  t, a = 0 = = 4.9 ms–2 0 100 F  µkmg kt m m – µk and for t  t, a = = 0 1 4 . (A) – Pulling force on bricks = 2F 31 (B) – Pulling force on bricks = F 20   0.4  20  (C) – Pulling force on bricks = F 1 9 . a2 = 2 2 (D) – Pulling force on pulley = F/2 2 T = 10 3 4  T = 0 = 5 3 2 2 1 5 . For pulley C, 60° 2T 31 10   0.5  10  a= 2 2 m1g  g 1 Acceleration of m = m1 1 1 Acceleration of m = m2g  g = 5 3  2.5 ; a < a 2 m2 1 2 OR N As 2 < 1 so block will move separately. 1 6 . FBD of block w.r.t. wedge 30° mg mg 30° Acceleration of block w.r.t wedge 3 2 0 . fmax = µN =  4  (80) = 60 mg 3  m g  1   3 1  N 30 2  2   2  f = = g m 100sin 100cos = 80 1 at2 , 1 = 1 3 1  gt2 Now from S = ut + 2 2  2   100 t= 4 = 0.74 s Total force exerted by plane  3 1g = f2  N2 1 7 .  +  +  +  = constant = 302  802 along OB 1234 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\03 NLM.p65 x1 1 2 B 2 1 . N  15 3 ; f = 15  Total Force A 4 3 x2 C Nf ..1+ ..2+ ..3+ ..4= 0 15 x + x + x + x – x = 0  2x + x = 0 30 30× 3 1 1 2 2 1 2 1 2 30° But acceleration of C = g downward ( Tension in string is zero as A is massless)  Acceleration of A = 2g upwards  = 15 3 2  152 = 30N E 35