P-3 Allens Made Physics Exercise Solution

JEE-Physics This is a simple L–R circuit, whose time constant the coil in these two time intervals will be opposite to 0.4 each other. the variation of current (i) with time (t) will L  L / R2  2  0.2s be as follows : and steady state current i = E/R = 12/2 = 6A i(A) 02 +1.25 Therefore, if switch S is closed at time t=0, then 0 0.2 0.4 0.6 0.8 t(s) current in the cicuit at any time t will be given by –1.25 i(t) = i (1–e–t/ )L (say) 0 Power dissipated in the coil is P = i2 R = (1.25)2 (1.6)W = 2.5W i(t) = 6 (1–e–t/0.2) = 6(1–e–5t) = i Power is independent of the direction of current through the coil. Therefore, power (P) versus time (t) Therefore, potential drop across L at any time t is: graph for first two cycles will be as follows : V di  L (30e5 t )  (0.4)(30)e 5 t P(watt) L dt  V = 12e–5t volt (b) The steady state current in L or R is i = 6A 2 Now, as soon as the switch is opened, current in R1 is reduced to zero immediately. But in L and R it 2 decreases exponentially. The situation is as follows: i0=6A i 2.5 E L R1 R2 L 0 0.8 t(s) E i0 R1 i= R1 =6A R2 i=0 Total heat obtained in 12,000 cycles will be t=0 S is open H = P.t = (2.5) (12000) (0.4) = 12000 J Steady state condition t=t (c) (d) (e) This heat is used in raising the temperature of the Refer figure (e) : coil and the water. Let  be the final temperature. Time constant of this circuit would be Then H = mwSw( – 30) + m S (  –30) c c L 0.4 L '  R1  R2  (2  2)  0.1s Here m = mass of water = 0.5 kg w S = specific heat of water = 4200 J/kg–K  Current through R at any time t is w 1 m = mass of coil = 0.06 kg i = i0e–t/L' = 6e–t/0.1  i = 6e–10t A c and S = specific heat of coil = 500J/kg–K Direction of current in R is as shown in figure or c 1 Substituting the values, we get clockwise. 1200 = (0.5) (4200) (–30) + (0.06) (500) (–30) 3. (i) Applying Kirchhoff' second law :   = 35.6°C d di d di  iR  L  0   iR  L ....(i) 2. (a) Given R = R = 2, E = 12V dt dt dt dt 1 2 and L=400 mH = 0.4 H. Two parts of the circuit are di d This is the desired relation between i, and . in parallel with the applied battery. So, the upper dt dt circuit can be broken as : (ii) equation (1) can be written as E d = iRdt + Ldi Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 L Integrating we get,  = R. q + Li R1 q =   Li1 ...(ii) RR S R2 E R1 E L Here, = x  x0 0 I0 dx  0I0 n(2) + R2 =  – i x 2x0 2 x 2 SS f (a) (b) So, from Equation (ii) charge flown through the resistance upto time t=T, when current is i , is– Now refer figure (b) : 1 46

JEE-Physics q  1  0 I0 n(2)  Here, q = Ce = 0aCI0n(2) R  2  Li1  00  (iii) This is the case of current decay in an L–R circuit. The corresponding q–t graph is shown in figures. Thus, i  i0et / L ....(iii) q Here, i  i1 , i = i , t = (2T – T) = T and L  L q0 4 01 R T Substituting these values in equation (3), we get : + i T 4 3T T t – 44 LT q0 L  R  n4 5 . After a long time, resistance across an inductor 4 . (i) For a element strip of thickness dx at a distance x becomes zero while resistance across capacitor from left wire, net magnetic field (due to both wires) becomes infinite. Hence, net external resistance, B  0 I  0 I dx R 2 x 2 3a  x R  R net  2 = 3R Current through the batteries, (outwards) 4  2  0I  1  1  i 2E 2  x 3a  x  3R  r1  r2 Given that potential across the 4 Magnetic flux in this strip, 0I  1 1 terminals of cell A is zero. 2  x 3a  x  a dx d = BdS =  E   2E  r1  0  4   E – ir = 0  3 R / r1  r2 1 2a 0Ia 2a  1 1  4   total flux  =   x  3a  x  dx 3 d 2 Solving this equation, we get R  (r – r) a 1 2 a   0Ia n(2)  0 a n (2) 6. Inductive reactance     (I0 sin t ) ...(i) X = L = (50) (2) (35 × 10–3)  11 L Magnitude of induced emf, Impedance d 0aI0n(2) cos t Z= R2  X 2  (11)2  (11)2  11 2   L e   e0 cos t dt Given v = 220V Hence, amplitude of voltage rms where e0  0 a I 0 n (2 ) v0  2 vrms  220 2V   Amplitude of current i = v0  220 2  20A Charge stored in the capacitor, Z 11 2 0 q = Ce = Ce cos t ...(ii)  XL  11  0  R  = tan–1  11  4 Phase difference =tan–1  and current in the loop i  dq = C e sin t...(iii) 0 dt In L–R circuit voltage leads the current. Hence, instantaneous current in the circuit is, Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 i = Ce = 0aI02Cn(2) max 0  i = (20A) sin (t – /4) Corresponding i–t graph is shown in figure. (ii) Magnetic flux passing through the square loop  sin t [From equation (i)] V,I v = 2202 sin t i = 20 sin (t–/4) i.e., * magnetic field passing through the loop is 20 increasing at t=0. Hence, the induced current will O T 9T/8 produce  magnetic field (from Lenz's law). Or the current in the circuit at t=0 will be clockwise –10 2 t (or negative as per the given convention). Therefore, charge on upper plate could be written as, T/8 T/4 T/2 5T/8 q = +q cos t [From equation (ii)] 0 47

JEE-Physics 7 . Out side the solenoid net magnetic field zero. It can be assumed only inside the solenoid and equal to R dB BR 11. induce electric field =   nI. 2 dt 2 0 d  d   d ( 0 n / a 2 ) torque on charge = QBR 2 kˆ induced e = –  (BA) dt 2 dt dt or |e| = (0na2) (I0 cost) E E  (2R ) +Z Resistance of the cylindrical vessel R=  s Ld  Induced current i = e  0Ldna2I0 cos t Q R 2R 8 . This is a problem of L–C oscillations. Charge stored in the capacitor oscillates simple EE harmonically as Q = Q sin (t+ )  0  dL Here, Q0 = max. value of Q = 200 C= 2× 10–4 C  1    dL   dt by dt 0 1 1 = LC (2  103 H )(5.0  106 F ) = 104s–1  QBR2 L Let at t=0, Q=Q then  2  kˆ 0  Change magnetic dipole moment = L dQ Q(t) =Q cos t...(i) I(t)  dt  Q0 sin t ...(ii) 0 dI(t) QBR 2 kˆ and dt = –Q02 cos (t)...(iii) 2 1 2 . Magnitude of induced electric field = (i) Q = 100 C  Q0 1 R dB BR At cos t =  t = = 2 23 2 dt 2 1 1 3 . 4000V VV 200V At cos(t) = 2 1, from equation (iii) : dI  (2.0  10 4 C )(1 0 4 s 1 )2  1  = 104 A/s step up step down dt  2  for step up transformer (ii) Q=200C or Q when cos(t) =1 i.e. t=0, 2... V 10 0  At this time I(t) =–Q0 sin t  I(t) =0 4000 1  V = 40,000 Volt [sin 0°=sin 2=0] for step down transformere (iii) I(t) = –Q0 sin t N1  V 4000   200  Maximum value of I is Q  0 N2 200 200 I = Q0 = (2.0 × 10–4C) (104s–1)  I = 2.0A max max (iv) From energy conservation 1 4 . Current in transmission line Node-6\\E:\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit-9 & 12\\02.EMI & AC.p65 1 L I 2m ax 1 LI2 1 Q2 = Power 600 103  150A 2 2 2 C    Voltage 40,000  Q  LC(I2max  I2 ) I  Imax  1.0A  Resistance of line = 0.4 × 20 = 8 2  Power loss in line = i2R = (150)28= 180 KW percentage of power dissipation in during  Q  (2.0  10 3 )(5.0  10 6 )(22  12 ) 180 103 transmission = 600  103 100  30% Q  3  104 C =1.732 × 10–4C 48



JEE-Physics UNIT # 01 (PART – II) KINEMATICS EXERCISE –I 8. t= 2h t=0 1 A (4  1)ˆi  (2  2)ˆj  (3  3)kˆ 3ˆi  4ˆj g 1 . vˆ = = h 32  42  02 5 t1 B 2h   3ˆi  4ˆj  t= 2  2h 3h v v vˆ = 10  5  2 = 6ˆi  8ˆj g t2 C 20  3  4  20  5  20 2  3h t3 D 2 . Avg. velocity = 20  20  20 = 4 m/s t= g 3 3 . vi  2ˆi Required ratio t : (t – t ) : (t – t ) v f  4 cos 60ˆi  4 sin 60ˆj 121 32    = 1 : 2  1 : 3  2  4 ˆi  4 3 ˆj t=T 22 t=t  2ˆi  2 3ˆj H1 9. h = H – g(t–T)2   h 2 v  v f  v i  2ˆi  2 3ˆj  2ˆi 2 3ˆj  3ˆj  t=0 a 2 2 3ˆj m/s2 dv 1 0 . Velocity after 10 sec is equal to 4 . For v = 0, x = 1,4 and a = v dx 0 + (10) (10) = 100 m/s a x1 = 0 × dv dv Distance covered in 10 sec is equal to so =0 ; a| = 0 × dx =0 dx x=4 1 (10)(10)2 = 50m 2 5.  v 2  v 2 dx dy Now from v2 = u2 + 2as. v x y here v = =2ct ; v = =2bt  v2=(100)2–2(2.5)(2495–400)=25 v=5ms–1 x dt y dt Therefore  4t2 (c2  b2 )  2t (c2  b2 ) v 6.  + 4 × 1) ˆi +(4 + (– 3) × 1) ˆj =7 ˆi + ˆj 11 . It happens when in this time interval velocity v (1) =(3 becomes zero in vertical motion  u | v (1) | = 49  1 = 5 2 m/s  g = 5  u = 5 × 9.8 = 49 m/s 7 . u=0 30s 1 2 . t = t1  t2 ; t = t2  t1 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 AC 2 BC 2 10s 20s  AB = AC – BC x1 x2 x3 C B t2 1 = 1 g  t1  t2 2  1 g  t2  t1 2 t1 x = a(10)2 2  2 2  2 A 12     1 x + x = a(20)2 1 1 22 = gt t 1 2 12 x + x + x = a(30)2  x : x : x = 1 : 3 : 5 1 2 32 123 12 E

JEE-Physics 11 22.    u  j  vR 1 3 . Displacement = 2 [4+2] × 4 – 2 [4 + 3] × 2 vR/M tan  = 12 – 7 = 5 m   vR/M Distance = 12 + 7 = 19 m  vR  vR/M  vM 14. S = S + 10.5    ui  u j vM=u i BA vR tan  t2 23 . For shortest time to cross, velocity should be  10t  10.5 maximum towards north as river velocity does not take any part to cross. 2 t2 = 20t + 21  t2 – 20t – 21 = 0 2 4 . Flag blows in the direction of resultant of VW & VB t = 21 sec Vw = 6m/s 1 5 . When the secant from P to that point becomes the VB tangent at that point VR=2 m/s 1 6 . Two values of velocity (at the same instant) is not possible. VB/R = 4m/s d2x –VW 1 7 . a = dt2 = change in velocity w.r.t. the time     6j  (4i  2j) = 4(i  j)NW For OA  velocity decreases so a is negative VW VB For AB  velocity constant so a is zero. For BC  velocity constant so a is zero.  N-W direction. For CD  velocity increases so a is positive. 2 5 . vmG  (vrm )2  (vrG )2  (20)2  (10)2  10 3 m/s 1 8 . Initially velocity increases downwards (negative) and = 20 after rebound it becomes positive and then speed vrm is decreasing due to acceleration of gravity () 2 0 . Upward area of a-t graph gives the change in vrG = 10 velocity = 20 m/s for acquiring initial velocity, it -VmG=10 3 m/s again changes by same amount in negative direction. 2 6 . The resultant velocity should be in the direction Slope of curve = – 10/4 = – 2.5 of resultant displacement node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 a 10 B 20 vr=5 4 20 t vmr 2.5 60m vm 2  20  time = = 4sec A 2.5 Total time = 4 + 4 =8 sec So time = 60 = 5     v = 13 m/s rm 2  52 2 1 . Initially the speed decreases and then increases. v m E 13

JEE-Physics 3 2 . Time to reach the ground = 2  20 =2 sec 2 7 . 1km 10 vBR  vBRcos u=0, a = 6m/s2 vBRsin vR 20m s = ut So horizontal displacement = 0 + 1 × 6× 4 =12m 1 = vBRcos t 2 1 33. v 2 =u2sin2 – 2g × H ; v 2 =u2cos2 1 = 5 cos y 2 x 4 cos   4   =37° vy v ucos 5 u usin vx 3 H/2 H v = v sin37° = 5 × =3 km/hr  R BR 5 ucos 2 8 . For shortest time then maximum velocity is in the  ucos = 6  v 2  v 2   cos   3 direction of displacement. 7  x y  or   30 2   ˆi  2ˆj  ˆi  ˆj 2ˆi  ˆj 34.   u cos ˆi  (u sin   gt ) ˆj    vQP v v  vx  vy 29. = v usin u 45° tan–11  2 x min 1 ucos tan–1(21 ) Q (2,1) Q 2 u ucos = usin - gt  t= g (sin – cos ) 5 P 35.   aˆi  (b  ct )ˆj (0,0) v So from sine rule 5 = x m in   xm Time to reach maximum height (when ˆj comp. of 90 sin  velocity becomes zero) sin b 2b  b – ct = 0  t = c   Time of flight = c 5  2 sin  cos   12 4 = 5 2   2b range = horizontal velocity× Time of flight = a × 22 55 5 c 3 0 . Time of collision of two boat = 20/2 = 10 sec. 2h node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 3 6 . Time to reach at ground = g As given in question i.e. the time of flight of stone is also equal to 10 sec. so vertical component of du stone initially is 50 m/s and the horizontal component w.r.t. motorboat equals to 2 m/s. Hence   3ˆi  50ˆj h vBG  a1 i  a2 j x In this time horizontal displacement ax ;ay 2h u2  2h 31.   d = u × g  d2 = g a y 14 E

JEE-Physics 500 1 OR 3 7 . – 1500= sin37° × t – × 10× t2 ; t = ? 32 Net acceleration of m a 500 3a Distance = 3 cos 37° × t (Horizontal) 4000  a2  (3a)2  10 a x = 3 m 38. d 42. aA=2ms-2 3 aB=1ms-2 A 2 4 B y 1 5 x vP Qv aC C M Here x2 = y2 + d2.  +  +  +  +  = constant 1 2 3 45 dx dy dy  x   dx   x  v .. .. .. .. ..  1  2  3  4  5  0 So 2x dt = 2y dt  dt =  y   dt  =  y  (v) =  a + a + (a – a ) + (–a ) + a = 0 cos  CA AB BC OR  2a + 2a – 2a = 0 CAB Component of velocity along string must be same v  a = a – a = 1 – 2 = –1 ms–2 so vM cos = v  vM = cos  CBA 3 9 . x2 = y2 + d2  Acceleration of C is 1 ms–2 upwards dx dy  2x dt =2y dt v 43. Given  = 2 + 2 x \\\\\\\\\\\\\\\\\\ d  2  2 d d y node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 dy  x   dx  v \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\    d  2  2 2  2 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ dt =  y   dt  = sin  d OR Component of velocity along string must same so at =1  = 12 rad/sec2 v cos = v  v= v v2 M M 4 4 . Centripetal acceleration = R sin  2 2 R1 1 2 R2 v = v = v1 = = 1 2 a R1 R2 v2 4 0 . Net acceleration of load  –    a 14  2 4 5 . = 25 =2acos  2  = 2asin  2   magnitude of acceleration 4 1 . Net tension on M T 2T (3 T )2  T 2  10 T 14  22 80 =2r =  25  × 100  9.9 m/s2 T 20 2T T 4 6 . Given r = m T M  T T Angular velocity after second revolution m   v  50  5 r 20 2 Now from acceleration × Tension = constant 2  2  2 final initial  aM(10T) = a (T)  a = (10)aM = 10a m m E 15

JEE-Physics 25 2  2 4    25 EXERCISE –II 4 32 1. u = u ; u = a cos t 25 20 x 0 y R a t  r  32    15.6 yt x = ut ;  dy = a  cos t dt 0 0 t0 47.  = constant , a = 0 w  x  T y=a sint = asin  u0   2  w  2  rx ,   2, T    T2 2R 22 R R v dv t /  a av   ; a= 2 R 2. = –av2 =  v2 = a  dt inst u t0 So ratio = aav 2 1 v 11  ainst     v u = –at   = – at uv 48.  = 6cm, v = ?, = 2  rad/s. u x t udt   v =  dx 60 30 =  1  aut 0 1  aut t 0 So v      6   cm/s = 2 mm/s u  n (1 ) t 1 n 30 5 au 0 a  x =  a ut = (1 +aut) Difference = 2  cm/s = 22  mm/s 5 1 3. t = x2 + x  1 = (2 x + )v  v  49.  and  remain same but v and a is   2x T proportional to r thus at half the radius, v' = v & a ' = aT 20cm/s 2 2 T2 aT=aR  Acceleration = (  2x)2 v = 2v3 4. x M x B A 5 0 . Let x is the distance of point P from O, the, 7m/s 17m/s from figure O wall v 2 = (7)2 + 2 × a × x ; v = 13 m/s x x P mm tan = or x=htan h (17)2 = (7)2 + 2 × a × 2x h dx d 3m 13  7 =t ;t = 17 13 ; t1  63  = h sec2   a t2  dt dt 12 a 42  d  S  dt   v = h sec2    vB/A 5 . | v A | = 10 m/s node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 So putting values    vC/B h=3,  = 180 – (90 + 45) = 45° vC  vC/B  vB/A  vA  Q we get v  (3 2 )2  0.1 = 0.6 m/s P vA 5 1 . Angular velocity  about centre = 2  12( i)  6  15 (i)   6  351 j  10i 24  24 = 2 × 0.40 = 0.30 rad/sec v = R  15  2 i  351 j 1  4 4 = 0.80 × = 0.40 m/s 2 v2 0.40  0.40  100  72   351 2  3 m / s a=  = 0.32 cm/s2 vC  R 50 4 16 E

JEE-Physics 6 . Time of fall of stone = 2  20 = 2 sec 1 2 . When acceleration is constant the instantaneous 10 velocity is equal to the average velocity in mid of Horizontal displacement of truck in 2 sec the time interval. v a 1 v2  v1  v3  v2  S = 2 × 2 + × 1× 4 . v3 a = t1  t2 t2  t3 . 2 22 22 v2 Length of truck = 6m v1 v2 v2 t 7 . As given 9 =y/6  y  54m 9y  1 3 . <v > = vds 2asds  2 v Average velocity of particle 6  space  ds 3 Displacement 54 B = = = 9m/s ds time 6  <v >  vdt atdt v   vs   time    vt  = 4 : 3 dt dt 2 8 . Distance covered by : train I = (Area of )train I = 200 m 1 4 . x = 40 + 12t – t3. train II = (Area of )train II = 80 m Speed dx = 0 + 12 – 3t2  t = ± 2sec So the seperation = 300 – (200 + 80) = 20 m. dt  x(2) = 40 + 12 × 2 – 23 9.  = (t2 – 4t + 6) ˆi +t2 ˆj ;  = (2t – 4) ˆi +2t ˆj = 64 – 8 = 56 m. r v at t = 0, x(0) = 40  = 2 ( ˆi + ˆj ); when    then  ·  = 0; t = 1s x = x(2) – x(0) = 16 a a v a v v sin v 1 5 . v = v + 9.8 × 0.5 = v + 4.9 BT T T  v – v = 4.9 m/s and 4m v cos BT v 2 – v 2 = 2gs = 2 × 9.8 × 3 = 58.8 BT 2 vT 1 0 . Time to cross 2m is  v sin  ..... 3m t = 0.5s To avoid an accident vB 2 Displacement = 4 + v cos  × v sin  2 8 × =4 + 2 cot  v sin  16 sin   (v + v ) × (v – v ) = 2 × 9.8 × 3 v sin  = BT BT 4 sin   2 cos   v + v = 12 m/s B T node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 v= 16 16. v g m/s2 v g m/s2 2m/s2 min = 1.6 5 m/s 42  22 2m/s2 [ (a cos  + b sin ) has max. value = a2  b2 ] 2h  time of ascent = 11.  = 4 t ˆi + 3 t ˆj ( x = at2 & y = 3/2t2) v g2 v(1)  4ˆi  3ˆj ; v(2)  8ˆi  6ˆj 2h time of descent = g  2  < v > = 12ˆi  9ˆj = ( 6ˆi  4.5ˆj ) m/s ta  82 2   td 12 3 E 17

JEE-Physics 1 7 . Time taken to reach the drop to ground u2 sin 90 100 1 1 2 2 . PQ = R = g = = 10 PQ = 10 9 = 0 + × 10 × (3t)2 10 t4 2 t3 y 9 Q 5 =3t 2 1.8  t 10m/s 6 3 t (x,y) = (10, 6, 0) 1 P 453°7° x (2,0) 1 1.8 x2 = × 10 × (2t)2 = 20t2 = 20  =4m 8 2 9 x= 1 × 10 × (t)2 = 5t2 = 5  1.8 = 1m 2h 2gh 3 29 2 3 . Time to fall = g 2  2R cos   Range = Horizontal velocity × time g 2h x = 2gh × g =2h 1 8 . Time to fall = g cos  so it does not depend on  i.e. the chord position. 1 9 . 3002 = (3t)2 + (4t)2 o vA= 3m/s 2 4 . At maximum height vertical component of velocity 300 × 300 = 25t2 becomes zero. t = 60 300m v2 = u2 + 2as vB= 4m/s Ratio = 32 3 =3:4 42 3 vAsin60° 3 2L 60° 2 0 . For man on trolley 2 vt = L  t = 3 v with respect to ground : vt + 3 = L + 2L 5L For A : 0 = v 2sin260° – 2gh 2 vt 3  For B : A 3 2gh = v 2sin260° = v 2 (3/4) AA 3 2L L 5L L 4L 8gh  2 vt – vt = L – 3 = 3  S = 3 – 3 = 3 v= A3 0 = v 2 – 2gh B 2u sin  vB = 2gh ; vA  2 2 1 . Time of flight 4  g cos 60 ...(i) vB 3  (angle of projection = ) dx Distance travelled by Q on 60° 2 5 . x = 103t ; y = 10t – t2; =103 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 incline in 4 secs is dt 1 3g × 42 = 40 3 dy =0+ × vy  dt =10 – 2t  at t = 5 sec. 22 v becomes zero at maximum height y & the range of particle 'P' is 40 3  y = 10 × 5 – 52 = 25m. 1 3g 26.   t2ˆi  (t3  2t)ˆj ; r = u cos × 4 + × 42 = 40 3   22 v dr   2tˆi  (3 t2  2)ˆj u cos  = 0 ; so  = 90° dt from equation (i) u = 10 m/s  2  a r  d  2ˆi  6 tˆj dt2 E 18

JEE-Physics  d2 x Hence [C] a.v = 4t + 18t3 – 12t = 0 (For ) a = dt2 = 2a. Hence [D]  t = ± 2/3, 0. x(2) = 0 [From (i)]. dy dy 3t2  2 For parallel to x-axis  = 0  = dx dx 2 2 3 3 . x = 2 + 2t + 4t2, y = 4t + 8t2  at t = sec it becomes zero so (c) dx dy 3 v = = 2 + 8t, v = =4+16t   2ˆi  6  2ˆj  2ˆi  12ˆj x dt y dt a (4,4 )  8ˆi  16ˆj 2 7 . Area of the curve gives distance. a = 8; a= 16; a  = constant x y y = 2(2t + 4t2); y = 2(x – 2) ( x = 2 + 2t + 4t2) 28 . Acceleration = Rate of change of velocity i.e. which is the equation of straight line. velocity can be changed by changing its direction, speed or both.  v(t) ED 34.  (3  1  t)ˆi  (0 – 0.5 t )ˆj ...(i) Displacment F C For maximum positive x coordinate when 2 9 . Av. velocity = Av 60° v becomes zero time B x  3 – t = 0  t = 3 sec 3 0 . x = t3 – 3t2 – 9t + 5. x(5) > 0 and x(3) > 0 then  = 4.5ˆi  2.25ˆj . so [A] v = dx/dt = 3t2 – 6t – 9 r (3 )  t = –1, 3 so t = 3 3 5 . [A]  Distance  Displacement Hence particle reverses its direction only once [B] average acc. = change in velocity /time.  Average speed  Average velocity   In interval (t = 3 to t = 6), particle does not reverse a ±0 v ±0 its velocity and also moves in a straight line so distance = displacement. velocity can change by changing its direction [C] Average velocity depends on displacement in 3 1 . Motion A to C  172 = 72 + 2as time interval e.g. circular motion  after one 7m/s 17m/s revolution displacement become zero hence average velocity but instantaneous velocity A BC never becomes zero during motion. [D] In a straight line motion ; there must be B v2 72 +  s  172  72 reversal of the direction of velocity to reach Motion A to  B = 2a  2  = the destination point for making displacement 2 zero and hence instantaneous velocity has to be zero at least once in a time interval. 36.  =  vˆ ; [   speed] v v v (A) v = 289  49 = 13 m/s Velocity may change by changing either speed or direction and by both. B2 7 13 dx t = dt x t0 (B) <v > = = 10 m/s 37. v  x ;  x  [2 x ]4x =t AB 2 4 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 (C) t = 13  7 17  13 t1 63  t  42 1 a ,t = , t2   x =  2  at t = 2  x = 9m 2 a 42 (D) <v > = 13 17 = 15 m/s dv 1 1 BC 2 a = v = x = m/s2 dx 2 x 2 3 2 . x = u(t – 2) + a (t – 2)2 ...(i) at x = 4  v = 2m/s & it increases as x increases so it never becomes negative. dx v = = u + 2a (t – 2) 3 8 . Average velocity dt Displacement Area under v  t curve Therefore v(0) = u – 4a == time interval tim e E 19

JEE-Physics v 45 . Range = u2 sin 2 25–2t g 5t For  & (90 – ) angles, range will be same so for 55 30° & (90 – 30°)60°, projections both strike at the tt same point. For time of flight, vertical components are responsible 20= 1 25  25  2t  5 t  t = 5, 20 h1  u2 sin2 1 sin2 30 = 1 2 h2 u2 sin2 2 = sin2 60 3 25 y = x2 , y1  1 dy dx 46. ; dt = 2x = 2x v x 4 dt x 3 9 . For returning, the starting point 2 Area of (OAB) = Area of (BCD) 1 × 20× 25 = 1 × t × 4t  t = 5 5  11.2 11 2 2 v = 2 × × 4 (at x = , v = 4) y2 2x  Required time = 25 + 11.2 = 36.2   v = 4m/s ; v 1= 4ˆi  4ˆj ; v =4 2 y 40 . As air drag reduces the vertical component of x velocity so time to reach maximum height will 2 decrease and it will decrease the downward vertical velocity hence time to fall on earth increases. Slope of line 4x – 4y – 1 = 0 is tan 45° = 1 and also the slope of velocity is 1. 41.  Horizontal component of velocity remains constant 4 7 . After t = 1 sec, the speed increases with  v'sin  = v cos  (from figure)  v '  v cot  a = g sin 37° = 6 m/s2  v = g sin37°× 1 = 6 m/s Y  speed = 82  62 = 10m/s (90–) 4 8 . New horizontal range v sin v'cos 1g g 4u2 sin2   = R + × × T2 = R + × 22 4 g2  v'cos v' P v cos = R + 2H ( H= u2 sin2  So from vy = uy + ay t  – v' cos cos2  v 2g ) =v t  cosec = sin–gt–v sin – gt  u2 sin  g 49. h= 2g  u = 12 × 10 × 5 = 10 m/s max 4 3 . As given horizontal velocity = 40m/s u cos  × t = 40; t = 1 sec 25 At t = 1, height = 50 m t = = 1s so no. of balls in one min.  50 = u sin  × 1 – 1/2 × g × 1  u sin  = 55 H 10  Initial vertical component = u sin = 55 m/s As hoop is on same height of the trajectory. = 1× 60 = 60 5 0 . a = – kv + c [k > 0, c > 0] So by symmetry x will be 40 m.  dv =  dt  – 1 n (– kv + c) = t node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 kv  k c u2 sin 2 4900  kv = c – e–kt 4 4 . Range =  480 =  sin 2  ˆi g 980 51. Let acceleration of B aB =a B (90 – ) projection angle has same range. Then acceleration of A w.r.t. B =     15  a B ˆi  15ˆj aA aB P 480m Q This acceleration must be along the inclined plane Time of flight : 15 3 15  2u sin  2u sin(90 – ) so tan 37° = 15  aB 4 15  aB a = –5 B T1= g ; T2= g  = – 5 ˆi  aB 20 E

JEE-Physics 5 2 . (4T) a = (2T) (a ) EXERCISE –III AB TRUE/FALSE aB T 1 .  Acceleration depends on change in velocity not 2  aA = on the velocity. but a = dvB T TT 2 . Velocity and displacement are in same direction. B dt TT 2T t2 t t2 T 11 2T B aB = t  2  aA  2  4 4T 5 10 At t = 2s, aA A 3. S 3rd =S –S = × g × (3)2 – × g(2)2=25m 32 22 a= 2 22 1+1=2 ms–2 4 . Initially packet acquires balloon velocity which is in A upwards direction so it moves upwards for some  time & then in downward. 24 5 . Because all bodies having same acceleration g in 5 3 . For B : dow2 nwards direction. Net acceleration 6 . At highest point, vertical velocity becomes zero and = 52  102  125  5 5 ms–2 total velocity due to horizontal component of velocity & acceleration due to gravity which acts always 54. a + a = 1 vertically downwards. 12 a –a =7 7 . Greatest height 12 a –a =2 u2 u2 31 H= 2g and.....horizontal displacement = g a1 a1  a = 4, 1 a = –3, A C a1+a3  R  2H 2 D 8 . Instantaneous velocity is tangential to the trajectory. a1+a2 B 9 . Trajectory of particle depends on the instantaneous a =6 a1–a3 velocity not on acceleration. 3 Acceleration of D = a + a dv v2 13 R = 4 + 6 = 10 ms–2 downwords 10. a= dt v = speed of particle a = where always t N 5 5 . Block B will again comes to rest if v = v i.e. 3t = (12t)t  t = ½ s acts towards the centre or  to the instantaneous Ac velocity. dv v2 dv v2 v 1 s ds 1 1 . No, because all masses having same acceleration Given   g is in downward direction. dt r ds r  5 6 . node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 ; dv  r v v0 o  n v0   S  v0  eS/r 1 2 . Firstly gravity decreases the speed when particle  v  r v moves upwards and then again increases by same amount in downward direction.  v0  veS / r  v  v0eS / r 1 3 . When the vertical velocity component becomes zero, then the particle is at the top i.e. it has only hori- v2 1 a2s 2s zontal component at that time which never changes 57. tan =   so it is min. at the top. R dv / dt Rav / 2 sR E 21

JEE-Physics FILL IN THE BLANKS  u2 cos2   2 u2  gH  cos   1    60 11 52 1. X= × 2t × at = × t' × at'  t' = 2t 22 As given ucos = 2   u2 sin2   gH  u2 cos2   Total time = 2t+ 2 t = (2 + 2 )t 5 V at u2cos2 = 2 (u2 – gH) 5 aa t' MATCH THE COLUMN X t dx d2x tt dt dt2 1. [A] X = 3t2 + 2 V = = 6t a = aX at' dv [B] V = 8t  a = dt = 8 gx2 [D] For changing the direction 6t – 3t2 = 0 2. y = 3 x – 2  t = 0, 2 sec gx2 Trajectory equation is y = xtan – 2u2 cos2  2 . Slope of v.t. curve gives acceleration (instantaneous) u2cos2 = 1  u2cos260 = 1  u = 2 m/s  at that point  dv a dt 3 . At t = 0, v(0) = 10 m/s; t = 0 ; v(6) = 0 Change v(6) – v(0); v =0 – 10 = –10 m/s 11 V 3 . usin × 1 – × g × 12 = usin × 3 – × g × 32 10 22 0 24 6 2u sin = 40  usin = 20 -10 t usina=20 40  Average acceleration ucos charge in velocity 10 5 Time of flight = 2  u sin  = 4 sec = time = 6 = 3 m/s2 g Average velocity = Displacement 20 1  = time interval 40 sin =  sin =  30° 2 Total displacement = Area of 's (with +ve or –ve) 11 1  h = 20 × 1 – × g × 12 = 15m = 2 × 2 × 10 – 2 × 4 × 10 = –10 m (units) 2 4 . Due to gravity, it acquires vertical velocity and due to  Average velocity = 10 = – 5 m/s horizontal force it acquires horizontal component of 6 3 force and when a velocity having both components then the path of the particle becomes parabolic. a(3) = slope of line which exist at t = 0 I0t = 4 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 a = tan = 10 = –5 2 vy 4 1 v 4 . R = vt   = 1 = 4 radian usin u ucos R vx /2 /2 5. H R 2 ucos v2 = (u sin)2 – 2g × H and v2 = u2cos2 y 2 x 22 E

JEE-Physics  Displacement = 2R sin /2 = 2 sin 2 8 . Because initial vertical velocity component is zero Distance = vt = 4m in both cases. Average velocity = Displacement 9 . Inclined plane, in downwards journey. The time = 2 sin 2 component of gravity is along inclined supports in displacement but not in the other case. Average acceleration = Change in velocity  2  4 sin 2 = 8 sin 2 10 . Maximum height depends on the vertical time 1 component of velocity which is equal for both. 1 1 . Speed is the magnitude of velocity which can't be 5 . Velocity & height of the balloon after 2 sec: negative. v = 0 + 10 × 2 = 20 m/s  1 2 . If the acceleration acts opposite to the velocity then h = 1/2 × 10 × 4 = 20 m the particle is slowing down. Initial velocity of drop particle is equals to the velocity 1 3 . Free fall implies that the particle moves only in of balloon = 20 m presence of gravity.  u = 20 m/s as  g  s After further 2s vs  0 Comprehension#1  height = us  v s × 2 = 20m from initial position 1. Dis tan ce  d / 2   2 Displacement d 2 of balloon  Height from ground = 20 + 2v = 40m 30 m(E) ASSERTION & REASON 30 2 (SW)  u2 sin2 2. 40m(N) For max. range  g  , the projection angle() 1. 10 m(N) should be 45°. 2.node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 b 3 . x = 1, y = 4; x = 2, y = 16 So initial velocity ai + bj tan 45°= a  a = b 1122 3.  Displacement = (x2  x1 )2  (y2  y1 )2 4. Whenever a particle having two  components of 5. velocity then the path of projectile will be parabolic, = 12  122 = 145  12m 6. if particle is projects vertically upwards then the 7. path of projectile will be straight. Comprehension #2 1 . Positive slopes have positive acceleration, negative E Acceleration depends on change in velocity not on velocity. slopes have negative accleration. 2 . Accelerated motion having positive area on v-t If displacement is zero in given time interval then its average velocity also will be zero. e.g. particle graph has concave shape. projects vertically upwards. 3 . Maximum displacement = total area of graph To meet, co-ordinates must be same. So in frame = 20 + 40 + 60 + 80 – 40 = 160 m of one particle, second particle should approach it. 4 . Average speed In air, the relative acceleration is zero. The relative = Dis tan ce  20  40  60  80  40  24 m / s velocity becomes constant which increases distance time 70 7 linearly which time. 5 . Time interval of retardation = 30 to 70. Yes, river velocity does not any help to cross the river in minimum time. 23

JEE-Physics Comprehension # 3 Comprehension # 5 1 . y = 3x – 2x2 1 . If the projection angle is increased, maximum height gx2 will increase. Trajectory equation is y = x tan – 2u2 cos2  2 . Projection angle is 45° & V = 21 m/s, projection g y tan = 3    60 & 2u2 cos2  = 2 speed is V sin 45°=21  V =21× 2 =30m/s 0 0 3 . By the v – t graph the acceleration is y 21 = –10 = –g 2.1  u  5  10 21 4 4 . 28 0 K 5.9 t 2 2.8 39.2 10 H  u2 sin2  , 1 0   3  10 2g  2  2. Max. height 3 m 8 2 10 3. Range of A = u2 sin2 = 10  sin120 = 3 5 . Initial kinetic energy = 1/2 mV 2 g 10 2 0 If mass doubles, then we can sec from (v – t) curve y then velocity becomes half of previous. 1  v0  2 1 / 2 m v 2 2  2  0 2usin 2  10  3  × 2m ×  Hence [B] Time of flight = g = 10 2= 3 2 4. 10 6 . Position of the cable at the max. height point. 5. At the top most point v=ucos= 10 cos60° = 10 (V0 sin 45 )2  V02 2 2g 4g H= 10 Comprehenison # 6 2 1 . In ground frame [A] it is simply a projectile motion. But mg in [B] frame horizontal component of the displacement is zero i.e. in this frame only vertical comp. appear  10  2 which is responsible for the maximum height.  mg = mv2 ; R =  2   10 R  1m 2 . As observer observes that particle moves north-wards. R 10 40 4 VPC N Comprehension #4 45° VP 1 . R = Cv n W E 0 45° observer node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 Putting data from table: 8 = C × 10n VC 31.8 S  31.8 = C × 20n  = 3.9  4 = 2n  n=2 8 3. Frame [D], which is attached with particles itself so the minimum distance is equal to zero. 2 . C depends on the angle of projection. 3 . R = C × v n  8 = C × 10n and 4 . a = 20 m/s2 ; a = 10 m/s2 0 bD 8 R = C × 5n   R = 22 = 2m a = 30 m/s2  bD  Force acting on a body = 10 × 20 = 200N 24 E

JEE-Physics Comprehension#7 EXERCISE –IV A 1 . In vertical direction h = u sin  t  1 gt2 1 . By observation, for equal interval of time the magnitude of slope of line in x-t curve is greatest 2 in interval 3. t2  2u sin  t  2h 0 2 . By observing the graph, position of A (Q) is greater  g  g than position of B (P) i.e. B lives farther than A and – also the slope of x-t curve for A & B gives their velocities v >v .  t + t = 2u sin  12 g .....(i) BA In horizontal direction x  u cos  t  1 at2  t 1 T  2  2 u cos  2x 3. a = a  where a & T are constants  a  a 0 0  t2 – t  0 v t  t  t2   dv = a 1  dt  v = a0 t  2T  2u cos  0  T   t + t = ....(ii) 0 a t 0 3 4  g t1  t2   a t  t2   t  2T  dt From (i) and (ii) = tan–1    dx = 0     0 a  t  t  t 3 4 2 . At maximum height v = 0 t t2 t3  y For a = 0  1– T = 0 t  T = a0  2  6 T   u2 sin2  g  H = = t1  t2 2 T T2 T3  max 8 = a0  2  6 T   a0 T 2g  v dt T3  < v > = 0 T 3 . At maximum range vertical displacement = 0  dt 2u sin  0  t = g . So range R a u cos   2u sin  1 a  2u sin 2 4 . S =u+ (2n–1) by putting the value of n=7 and 9,  g  2  g  = – n2 find the value of u & a, u=7 m/s & a =2 m/s2. = 2u2 sin cos   g  tan  5 . After 3 sec distance covered =1/2 × 2 × 9 = 9m g  a velocity of lift = 2 × 3 = 6 m/s up = 6m/s, a = g  height = (100–9) = 91 m node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65  Time to reach the ground 1 = 91= 6t + × g× t2 t = 3.7 sec 2 Total time taken by object to reach the ground = 3 + 3.7 = 6.7 sec. Time to reach on the ground by lift = 1  2  t2  100  t = 10 sec. 2 So interval = 10 – 6.7 = 3.3 sec 6 . (i) v 2× 52 =3.3 10 t 2.9 52m t 2 × 52× 10 ~32 g 2.1 E 25

JEE-Physics 2.1 = 32 – 2.9 29  14  14 + 3.3  17 a ; t= +1 t 2.1 2 3t 1 12. 0 1 u=0 2 (ii) Height= 52  [ 32 + 2.9] × 14 = 293.8 –1 7 . Deaceleration of train , (i) Area under a – t curve the change in velocity 1 v2  u2 20  20 a=  = 100 km/hr2 u = 1 × 1 + × 1 × 1; u – u = 1.5 m/s 2s 2  2 2 20 20 1 u2  1.5 m / s (u = 0) 0 Time to reach platform =  hr 100 5 1 upto 3 sec : u = 1.5 – × 1× 1 =1m/s  Total distance travelled by the bird 2 1 u – u = 1m/s  u3 = 1m/s (u = 0) = vt = 60 × =12km 3 0 0 5 0 10 1 3 . VA=10m/s 12 m/s 8 . t = t – 0.6 = 4 = 2.5 v 150 m Distance travelled to stop 10m/s 4 t 10 12 0.6 3.1 v v 2 2 25 36 5t 6t 1 Car A Car B Stopping distance = 0.6 × 10 + × 2.5 × 10 Total distance = 25 + 36 = 61 m covers by both car 2  Remaining distance = 150 – 61 = 89 m 6 + 12.5 m= 18.5 m 9 . (i) Height = upward area under v–t curve = 20m v 2× 45 =3sec 1 4 . Let v = v – (–3v) = 4v 20 10 ground AB 20 20 t 100 M v 2 25 TRAIN A 60 M 3v TRAIN B (ii) Total time of flight = 2 + 3 = 5sec 1 0 . Total time = 1.5 + 3.5 = 5s time = 160 = 4 sec 4v v  10m / s 15 2× 61.25= 12.25 velocity of train v = 10 m/s A v(m/s) 11.25 10 vB = 3 × v = 30 m/s t 1.5 61.25 1 5 . Direction of flag = Resultant direction of the wind node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 velocity and the opposite of boat velocity 1 1 . From given situation :     72 (i  j)  51j vW vB 2 60  20 4000 = 36 2i  (36 2  51)j  36 2i EAST  (i) a = = = 160 km/hr2 avg 1.00 – 0.75 25 1 1 6 . For A : 30t = S/2 = 60 (2 – t )  t = 4/3 hr (ii) Area = × [20 + 60] × 0.25 1 11 (Here S is the total distance and t is time up to 2 1 which A's speed is 30 km/hr) 25 = 40 × = 10km 100 26 E

JEE-Physics 1  30  4  20. u sin × 1 – 1 g(1)2 = u sin × 3 – 1 × g × (3)2 ×  3  2 2 For B : a × 22 = × 2 = S 2 2u sin = 40  usin  = 20m/s  a = 40 km/hr2 u2 sin2  20  20 2g 20 (i) (a) v = 40t= 30  t = 0.75 hr Max. height=   20m B (b) v = 40t = 60  t = 1.5 hr B (ii) There is no overtaking. R3 21. Vertical displacement of particle = 1 7 . Relative velocity of A w.r to B, 2 aa 2 60° 60° V time = =  30° AB v  v cos  v (1  cos ) R3 n 60° 2 dd 18. t= vB = 600s, drift=v × vB d vw w vB vR 1m Time for this = 2R 3 3R 120 = v × 600s; v = 2 g w w 5 sec g d vw vB t = = 750 d vR  uˆi gtˆj uˆi 3R ˆj  uˆi  3Rg ˆj 2 vw2 v(t)     g  v B  g  vw  2 4  vw 2 9 22. 780 = u sin  × 6 + 1 × g × 36  vB  2 1  =  = 5  v B  25 780 – 180 = u sin  × 6 usin  = 600 =100 m/sec vw 3 1/5 1 6  vB   m / sec vB 5  3/5 3 i.e. food package dropped before 10 secs 1000 = u × 10  u = 100 m/s d1 g  (16 )2 d  600   200m VB = 600 = 3 h= 2 = 1280 m.  ˆi  ˆj V 23. Bomber v v(0)  19.  v cos v sin 53° 37°  cos ˆi  g t )ˆj t v(t) 800m  v (v sin   0.6v g  v(t)  v2 cos2   (v sin   gt)2  (2v sin   gt) ˆj  v(t)  v(0) 2 < v(t) > = = v cos  ˆi + 0.6 v 2 (0.6v)2  20 = +   800 ....(i) 2 g  2g  g According to question (v cos )2  (v sin   gt)2 (i) By solving equation (i), we get v = 100 m/s. (ii) Maximum height : node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 = (v cos )2   2v sin   gt  2 (0 .6 v )2 (0.6  100)2  2  = 800 + = 800 + = 980m 2g 20  2v sin   gt 2 (iii) horizontal distance v2cos2 + (v sin – gt)2 = v2cos2 +  2  = Horizontal velocity × time of flight = 100 cos 37° × 20 = 1600m gt 3gt (iv) horizontal component vsin – gt = – vsin +  = 2 v sin v = u = 100 cos 37° = 80 m/s 22 HH 4  v sin  vv = uv – 10 × 20 = 100 sec 37° – 200 t  3  g  = 140 m/s  v = 80ˆi – 140ˆj ,  = 802  1402 strike v E 27

JEE-Physics H 28. v = 2t2; a = dv = 4t  a (1) = 4 2 4 . distance covered by free falling body T T dt 2 v2 (2  12 )2 H  1 gt2 H a=  =4 22 t NR 1 ; g 2 2 (4)2  42  a= a T  a N  32 a4 2 H/2 29. (v + v) t = 2R, (0.7 + 1.5) t = 2 × 22 ×5 A B 7 t = 2  22  5  10  100 sec = 14.3 sec 7  2.2 7 vsin v H/2 Acceleration of B = v 2 1.52 = 0.45 m/s2 B   vcos  R5 30. a = ar ; r = 2r ;  = 2t2   = 1 t t2 In same time, projectile also travel vertical distance 31. (a)   0  1   t2  t = 2sec 22 H H H 1H (b) v = 0 +  2 =  m/s , then  v sin  g 2 2 2 2 2g v sin   gH ...(i) also   v cos  H v cos    g O (1,0) A SOA= × 1m g; ...(ii) a=  m/s2=dv H 2 dt From equation (i) and (ii) tan   H v2 sin2   v2 cos2   gH  2 g  H  2  3 2 . r = 2.5 m, a = 25 m/s2 v  gH 1  H2  t net 3 m/s2 (a) Radial acceleration = 25 cos = 25 × 2 d 10 3 v2 1/2 25.  10 2 cos 45  10 10 2 sin 45  3 (b) 25 2  25  v  125 4  m/s d = 20 × 1 = 20 m. 26. Here aB (3T) = (aA) (2T) aA = 3 aB 1 2 (c) Tangential acceleration = 25 sin = 25 × m/s2 2 33. According to 2T 28 1 72v2  t2  R  t  5R 2T   6v 2 25R Using R  vt  1 72v2  25 2R 2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65  2  2 5 R 36v2 aAB = aA – aB = 3 a0 – a0 = a0 72v2 v 2 2 a T  25R B R  v5R  R  11  A E 6v 6      27. at  6ˆi    R    2ˆj    3 kˆ rad/s2 Angular velocity : = v + t v   8ˆj        R ˆi    2 kˆ rad/s  v  72v2  5 R  v 12v  17v ar  v   R 25 R 2 6v R  5R 5R 289v2 Angular acceleration  = 2R = 25R

JEE-Physics EXERCISE –IV B ˆi  2ˆj Velocity of second ship = u  5 1. VB T = 18, VB T = 6; VB  20 =3 tan   2u / 5  2  10 5 VB  20 VB  20 VB  20 20 km/h 10  u 10 5 10 5 A2 A1 5  A B 10 1 VBT 20 2 (i) t =  sec , minimum distance = 10 km  V + 20 = 3V – 60 vB  40km / h B B 4. –25 m/s After 5 sec height of balloon = 25 × 5 = 125 m  T = 6 VB  20  6  60 = 9 min (i) Minimum speed VB 40 1 20  80 125 = u  252  (u – 25)2 = 2500; 2 [10 + 30] ×  3  2. (i) Area = + 10 × 2 4 0  2g 3 u – 25 = 50; u = 75 m/s 1 20 + [10 + 30] × = 4000 2 (ii) u = 2 × 75 = 150 m/s 125 = (150 – 25) t – 5t2 v 125 = 125t – 5t2  t2 – 25t + 25 = 0 30m/s 5. v = v – v = v – (–v ) = v + v 12 1 2 1 2 1 2 240 = 80 + t a1 a2 3 3 max  10m/s v1  v2 2 2 a1  a2 80 t     max 3 20 240 sec 20 240–  3 a = –a – a = –(a + a ) 12 1 2 12 400 800 400 = 4000 6. Let t = time of accelerated motion of the  2400   helipcopter. 3 3  Distance travelled by helicopter 400  800  1200 = Distance travelled by sound = 1600 3 3 = 800  1 ;  = 1 1 80 1600 2 6  × 3 × t2 = 320 (30 – t)   t = sec 23 Final velocity of helicopter node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 (ii) Dist. travelled = 10  24 0  3 80  = 800 m 80   1 / 6  v = u + at = 0 + 3 × = 80 m/s 3 3 . It's velocity is 10ˆi 7. V = (4 + 2) ˆi  3ˆj , V = (–3 + 2) ˆi  4ˆj A B (–10, 10 ) (10t–10, 10 ) River P Q 2km/hr O 53 5 37° 4 4 53° 3 100 100 Time to cross the river t = ; t =  displacement after time 't' = 10ˆi × t A3 B 4 E 29

JEE-Physics 100 100 11. u = 10 3 m/s Y-axis Drift = 3 × 6 = 200 m ; Drift = –1 × 4 P Q Remaining distance = 300 – 200 ; 25 m 100 100 100 100 h g g cos30° (t ) =  ; t =  g cos30° 30° 60° total A 3 8 B4 6 O 800  300 1100 600  400 1000 t =  ;t =  A 24 24 B 24 24 tA  165 sec tB  150 sec 8 . From figure (a) 120° V B (i) v(t) = (u – g cos 30°t) ˆi – g sin t ˆj 2d 60° 60° d From given situation time to cross = 3 V u – g cos 30° t = 0 V t = 2 sec d V 60° Minimum time t = v (ii) Velocity u = 0, a = g cos30° = g 60°  Ratio = 2 3 xx 2 A g  vx = 0 + 2 × 2 = 10 m/s (iii) Distance PO = 9. v Car 3= vC  v = 10 3 m/s 10 3 cos 90  t  1  g sin 30  22 tan 60° = v H  10 C 2 PO = 10 m  h = 10 sin 30° = 5 m Normal u sin602 vCar (iv) Maximum height = h + 2g  3 2 =5+ 10 3  2  = 16.25 m 30°30° 20 vH (v) Distance PQ u = 10 3 m/s Y-axis PQ vH/C OQ =  2 10 3 u2 sin 2 1600  3 2g cos30 10 m O 10 3 1 0 . Range (OA) =   80 3 OQ = 10 3 g 10  2 10  80  80  3  PQ = PO2  O2 h  80 3  tan 60  2  v2 cos 60 Time to strike  vcos 60°× t = 80 3  2 = 102  10 3 = 20 m node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 80 3  2 10 3 (u sin )2              t= = 1 v 1 v 1 2 . For stone : 2h = & h = (u sin )t – gt2 2g 2 160 3 480  9 2402  38400 h9 3  40h  20h 2 v v v2  t = 10  t = 0.8h = 10 20h v2 – 1600 – 18v = 0 v  18  324  6400 h 2 u v(constant)  h Pole h  v  50m / s Horizontal displacement : vt = u cos  t 2 30 E

JEE-Physics  v( 2  1) 20h  u cos   2 20h EXERCISE –V-A 10 10 1 . Kinetic energy of a projectile at the highest point v2 = Ecos2() where E is the kinetic energy of  projection,  is the angle of projection. u cos  2  1  1 2 E  2   1 3 . u cos t = D ....(i) Ehighest point = E ( c os 4 5 ° )2= E 1 ....(ii) 2 u sin  t – gt2 = –H u2 sin 2 10 m/s 2 2. R = g 30°  t 2u sin   u2 sin2   2gH  D g u cos  102 sin 60 R= 10 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65(u sin )2D2 tan2  h  10m2g 4(H  D tan ) 10mR  10  3  5 3  8.66 m  H = h + H = H + D2 tan2  2 max 4(H  D tan ) 3 . Both horizontal direction speed is same u h v cos = v0  cos= 1   = 60°  0 2 2 4 . When a body is projected at an angles  and 90–; the ranges for both angles are equal and the H corresponding time of flights for the two ranges are t1 and t2. D R  2u2 sin cos   1 g  2u sin   2u sin 90   g 2  g   g  11 1 1 4 . s = ut + at2  a = (u sin )t – gt2  2 gt1t2  R  t1t2 22 u=2ag 5 . Khighest point = [K ]Point of projection cos2  a a KH  K cos 60  KH  K 4 2a t= u sin   u2 sin2   2ag 6.   K (yˆi  xˆj ) ; vx = Ky; dx v = Ky g dt 2 u2 sin2   2ag dy t = g similarly = Kx For horizontal motion : 2a = u cos  × t dt u cos   2 u2 sin2   2ag dy x  2a = g   = 60° Hence dx = y  y dy = x dx, by integrating y2 = x2 + c. u2 u 2 R 2 Rmax= g ; max 7. Area = r2= g2 t  2a  2a 2 a u2 u2  u cos 2 ag  1 g 8 . Hmax = 2g =10 m and Rmax= g =20 m 9 . u = 5 and tan = 2 2 gx2 so by y = x tan – 2u2 (1 + tan2)  y = 2x – 10x2 (1+4)  y = 2x – 5x2 25 E 31

JEE-Physics EXERCISE –V-B Subjective 1.(i) u is the relative velocity of the particle with respect Single Choice to the box. total displacement 2 1. v = = = 2m/s Y X av total time 1 gx=gsin gcos u g 2 . v2 = 2gh [ it is parabola] P  ux=ucos Q and direction of speed (velocity) changes. uyg=suinsin 10  t  10 3. a = – 11 at maximum speed a =0 u is the relative velocity of particle with respect to x 10 t  10  t = 11 sec the box in x-direction. u is the relative veloicty with 11 y 1 respect to the box in y-direction. Since there is no Area under the curve = 2 × 11 × 10 = 55 velocity of the box in the y-direction, therefore this is the vertical velocity of the particle with respect to ground also. aa Y-direction motion 4 . S = u + (2n–1) = (2n–1) (Taking relative terms w.r.t. box) n2 2 aa uy = + u sin ; ay =– g cos S = x + (2n+1) = (2n+1) 2(n+1) 2 11 s = ut + at2  0 = (u sin ) t – g cos × t2 Sn 2n  1  Sn1 = 2n  1 22 v 2u sin  v0  t = 0 or t = g cos  5. v   v0  x  v0 X-direction motion   x0  (taking relative terms w.r.t box) a   v0 n  v0   v0  x0 x 1  x0   x0  x     a ux = +u cos & s = ut + at2 2 –v02 a    v0 2 x  v 2 x0 2u sin  u2 sin 2  x0  0 x0 a = 0  s = u cos × g cos  = g cos  x x MCQ's (ii) For the observer (on ground) to see the horizontal displacement to be zero, the distance travelled by 1. x = a cospt ; y = bsinpt;   a cos pt ˆi  b sin pt ˆj  2u sin  r the box in time  g cos   should be equal to the  sin2pt + cos2pt =1 range of the particle. Let the speed of the box at node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 the time of projection of particle be u. Then for the x2 y2 motion of box with respect to ground.   =1 (ellipse) a2 b2  ap sin ptˆi  bp cos(pt)ˆj ; v =  = apˆi u = –v, s = vt + 1 at2, a =–g sin v t 2p x 2 x  u2 sin 2  2u sin   1  2u sin   2 g cos   g cos   2  g cos    ap2 ptˆi  bp2 sin ptˆj ;  = bp2ˆj sx    v  g sin  a 2p  a = t  On solving we get v  u cos(  ) av 0 cos    p2 a cos ptˆi  b sin p tˆj   p2  a r 32 E

JEE-Physics 2 . Let 't' be the time after which the stone hits the 3 . (a) From the diagram B object and  be the angle which the velocity vector  VT  VBT makes an VB A u makes with horizontal. According to question, we have following three conditions. angle of 45° with the x-axis. Vx=ucos (b) Using sine rule 45° 45° V ucos VB = VT O 1.25m gt-usin=|Vy| sin135 sin15 u |Vx=Vy| 0 ucos  V = 2 m/s B Integer Type questions (i) Vertical displacement of stone is 1.25 m. 1 . With respect to train : 1.25 = (u sin) t – 1 gt2 where g=10 m/s2 Time of flight : 2vy 2 5 3  3 2 T= g   (u sin) t = 1.25 + 5t2 ...(i) 10 By using s = ut + 1 at2 2 (ii) Horizontal displacement of stone we have 1.15 = 5T – 1 aT2  a =5 m/s2 = 3 + displacement of object A. 2 Therefore ( u cos) t = 3 + 1 at2 2 where a= 1.5 m/s2  (ucos) t =3 + 0.75t2...(ii) (iii) Horizontal component of velocity (of stone) = vertical component (because velocity vector is inclined) at 45° with horizontal). Therefore (ucos) = gt– (usin) ...(iii) The right hand side is written gt–usin because the stone is in its downward motion. Therefore, gt > u sin. In upward motion u singt. Multiplying equation (iii) with t we can write, node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 1 & 2\\02 Kinematics.p65 (u cos) t + (usin) t = 10t2 ...(iv) Now (iv)-(ii)-(i) gives 4.25 t2–4.25 = 0 or t = 1 s Substituting t = 1s in (i) and (ii) we get u sin = 6.25 m/s  u = 6.25 m/s and u cos = 3.75 m/s y  u = 3.75 m/s therefore   u xˆi  u yˆj x u    u  3.75ˆi  6.25ˆj m/s E 33



JEE-Physics 17.  3 2 =  3 m 2 720 2 MI  about  AB  is  =  0  +  0  +  m  ×   2 4 2 5 .  =  60   =  24   0  =  24  –    ×    8      =  3   sin60°= 23 24   =I  =    ×  3  =  72 60° B 2 6 . Rod  rotates  about  its  one  end  in  a  horizontal  plane A ()  Mg 5L ML2 5g   = I   =  ×      26 3 4L I 2  m2 m 2  27 .   Book    does  not  rotate  so  for  rotational 1 8 . MI  =    =  12 6 equilibrium  the  net  torque  becomes  zero. (  4  10 –8  1 0 –2 2 )     man 0 1 9 . I  .7  ×  10–24  ×  10–3  ×    ×  2   2  w e ig h t +    =  1.7  ×  10–24  ×  16  10 –20 ×  2  ×  10–3  =  13.6  ×  10–47      =  –     =  –   W  b anticlockw ise  man   2  4 w e ig h t 2 2 2 8 . For angular acceleration () = I 2 0 .   =  T   =     =  2  rad/sec.  20  ×  0.2  =  0.2  ×      =  20  rad/sec2 K.E. =  1  I2  I =  2K.E.   I  K.E.     =  0  +  t  =  0  +  20  ×  5  =  100  rad/sec 2 42 2 9 .   Density  of  steel  >  Density  of  wood 2 1 . MI  is  more  when  mass  is  for  away (>  () MI  about O >  MI  about O I   =  dm ×  (2a)2  =  M  ×  4a2         [in  fig(a)]                [in  fig(b)] rods     =   0 =  I0 '   I   =  dm a2  =  Ma2 I 0 ' I0 ring 2 2 . From  perpendicular  axis  theorem 3 0 . Angular  acceleration    =   ( I   =  (10  +  9)  ×  0.3  –  12  ×  0.05  =  5100   y=x     =  10–3  rad/sec2 II 3 1 . Centre of mass of the rod ()   =  C.M.  from  'O'    = 1 L M2 M2   0 L2   I + I =   I =   2  0 (0x dx) x   12 24 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65 M2 M2   C.M.  =  2 L   from  O 3   MI  of  two  rods  =  2×  = 24 12 Moment  of  inertia  about  O  is 2 3 . I  =  x2  –  2x  +  99 (O) MI  about  axis  pass  through  centre  of  mass  will  be dI   =   dm r2 =  L (0 x dx) x2 = 0 L4 4 minimum  when  (     I =   0 )  O dI X            = 0  2x – 2 = 0   x=1  Angular acceleration   dx 2 4 . Plane  of  disc  is  (X  –  Z) weight  0 L2  g  2 L 4g  =   2  3 =  =   I  =  I  +  I  40  =  30  +  I    I =  10 I 0 L4 3L y x z z z 4 21

JEE-Physics 50. a  =  g sin    K2     a  =  time;  K2   =  2 EXERCISE  –II R2 R2 5 1  K2 R2  M 2  M 2      M 2   1 0 M 2   For  solid  sphere  which  is  minimum  in  all  of  them. 1. I  =  2  ×   3  3   I  =  3   So  a  is  maximum  and  time  to  reach  bottom  is minimum.  (      2.  When  ball  moves  towards  ends  of  the  tube  MI aincreases ( ) ) 11 51. As  given  (KE)   =  (KE)   =  I2  =  Mv2  R translational 2 2 (  For  system   fext   &   ext   are  zero) 1 1 K2   Angular  momentum  &  linear  momentum =  2   MK22  = MR22;  R2   =  1   Hence  ring remains  constant 2 () 1 1 MR2 v2 5 2 . mgh  = Mv2 + 2 ×  R2   I    =  I   I   >  I        <   22 11 2 2  21 21 2v2  v2 3v2 4 gh gh  =   =  gh    v  =   P P 44 3 2 2I m2 6P 3. P  =  I  or       m 1 1 MR2 v2 3v2 2 53. Mv2  + 2 ×  R 2 =  mgh   h  = 4 g 22 12 1 12 v2 t      1  m  54. Mv2  + ×  MR2  ×  =  Mgh So   2 / 12P 2 2 5 R2 6P m v2 1 + 2 =  gh  v  =  10gh 4 . The  prism  will  be  topple  when  the  torque  due  to  F  5  7 becomes  greater  than  the  torque  of  weight  about the  front  corner  of  the  prism. 2 (F  5 5 . Angular  velocity  of  rod  in  vertical  position ( ) —4  /2 ) 3a a mg              F ×  2 = mg ×  2  F =  3 1 ML2 2  MgL ;    =  3g 5 . As rod does not slip on the disc, so the kinetic energy 23 2 L of the rod is ( upper  part  of  rod  rotates  through  an  angle    its  ) L centre  of  mass  will  rise  (1  –  cos) 4 1 I2 1  ML2  M R 2  2                             =  2 2   =  ×   12  ( L(1  – cos)) I  –  MI  of  rod  about  axis  passing  through  centre  of 4 disc  &   perpendicular to  the  plane. node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65 From  energy  conservation  (   ) (I–  1  M  L / 22  2 )  =  2  2  3  3g + 1 MgL= 1 MgL  cos Max height ( ) H =  (v sin 45)2  v2 2 4 4 2g 4g 6.    60o v v cos45° 56.   v1  =  v2        v1 x x v2 H x – 45° x   v1  =  v2x  +  v1x    ;  x  =  v1 Angular  momentum  ( ) v1  v2 mv v2 mv3 =  mv  cos45°  ×    H  =  ×    =  2 4g 4 2g 23

JEE-Physics 1 9 . Sphere  is  on  verge  of  toppling  when  line  of  action 11 = mv2(1+1)+ of weight passes through edge. (    2 4 . KE v=r m 2v )system2 2  2v 1 2m v v R – h (m+2m)2v2  + m(2v)2  =  6mv2  2v 2 v=r R –h R 2 5 . As  we  displace  to  the  right,  lower  point  of  rod  is cos =  R   h = R – R cos  having  tendency  to  move  towards  left.  So  friction mg is  in  direction  right  so  center  of  mass  will  move right. 2 0 . In  figure  (a)  µN +  N   =  mg  ...  (i) ( 1  2 mg 3  N   =  µN ,  N   =    f  =  mg  1 22 1  µ2 a 10 2 1 . In  figure  (b)  N =0,  N   =  mg...(ii) µN1 )  12 f  =  µN   =  mg fa 9 N1 2 6 . Frictional  force always acts in  such  a way to prevent     fb 10 b2 µN2 mg the  sliding  or  slipping.(       3 2 2 .  For purely rolling  () ) v = R  2 7 . As the disc comes to rest, ()  v So   0  =  v0–gt     t  =  v0 v = R        P a g So  at  point  'P'  the  resultant  velocity  makes  45° Also 0 =    – t    t =  0 with  horizontal  'x'  axis  and  it  is  also  the  angle 0 between  'v'  and  'a'.  ( P    x45° v0  0 v0  g Thus  g    0  va) 2 3 . For  motion  from  A  to  B  from  energy  considerations v0 g mr  I mr2 / 2 1   r0   mr2  mr2  mr2  mr2  (AB )  2 mgh 1 mv2 1 1 v2 2gh 29. Angular  momentum   = r  2 2 2  3 p                or   As  v   increases  when  falls  down  then    also  and  so p Now,  for  motion  from  B  to  C angular  momentum  is  also  increases.  As  particle ( BC) falls  down  the  length  of  position  vector  decrease node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65                mgh  1 I2  1 m  2gh  so  MI  about  'O'  also  decreases. 2 2 2  3  (v  p   mgh  mgh  1 I2   1 I2  mgh ...(i)  2 32 2 6 O thus  remaining  translational  KE  ()    =  v   r =  r mgh 5mgh KE =mgh  –   3 0 . Force  diagram  of  plank  and  sphere t 66 () hence  required  ratio  (  )  m2 KEt  5mgh  6  5  c  f P  KEr 6 mgh m1 F P m1 F r m2 f F–f = m1a 25

JEE-Physics 5 . (A) When  insects  move  from  A  to  B      moment  of Comprehension  #  1 inertia  increases  and  when  if  move  from  B  to 1 . Case  I Case  II C  moment  of  inertia  decreases  Ndt  mv1  0  Ndt  mv2 A R R 2  Ndt  I2 B 4 Ndt  I1  0 as  2  >  1    KE2  >  KE1 C 2 . Point  of  percussion  at  which  (fr  =  0) (AB(fr = 0)  BC I )    =  MR   h0  =  0.4R (B) (Nmoo metonrtuqmue  ofa  tchtes   poanrt iclae  rpemaratiinc lec onasntagnutlarComprF(heo hr> e hnh 0s> i ohn0   friction  act  in  forward  direction    )   #  2 )  1. f  =  mg(R)  =  I =  (0.1)5  ×  10  ×  1                  r I  =  constant  () 2     first  decreases  &  then  increases       =  MR2      =  2.5  rad/sec2 5 (  )   =  0  –  t Fr (C) Remain  constant  (  )  0  =  40  –(2.5)  t    t  =  16  sec L2 Comprehension  #  3 (D) KI  =       first  decreases  &  then  increases 2I                      ()1 . By  applying  conservation of  angular  momentum about  O. 6 . For translatory motion ( )  (O)                         F–  fr  =  macm...(i) v For  rotatory  motion       \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\            FR + frR = I ...(ii) By  solving  equation  (i)  and  (ii) 2 2 mR2v ' MvR+ mR2=mv'R+  MVR 5 5R 2FR2 2F  1   a cm   acm  =   I  mR2 m  K2  o v 1 R2     \\\\\\ O\\\\\\\\\\\\\\\\\\\\\\\\ K2 2 mR2  7 mv ' R  v' =   2 R (A)  For  R 2 minimum  acm  is  maximum  in  sphere 55 7 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65       (KR22 2. By  applying  angular  momentum  conservation  law acm )  () K2 F mvR  =  7 m v R 2   v' 5v v (B) For  R 2    acm  = m that  is  fr  =  0 5 R 7 o' \\\\\\\\\\\\\\\\\\\\\\ K2 2 6F F Change  in  angular  momentum  gives  angular (C)  For  R 2  3    acm  = 5m that  is  fr  =  5 impulse ()  K2 1 4F (D)  For      acm  = 3m 2  5v  2 R2 2   R Ndt  I  0 = 5 mR2  7R   7 mvR 27

JEE-Physics Comprehension  #  8 a0 Comprehension  #  10 1 , 2 For  translatory  equilibrium Fm 1 . Torque  needed  to  lift  the  block  =  mgR ()   fr () F  –  f   =  ma ....(i) Torque  due  to  wrench  on  =  FL  as  L  increases,  force r0 decreases fr m For  rotatory  motion (=FLL  a1 ) () fr  R  =  I....(ii) 2 . By  applying  work  energy  theorem For plank ( ) f   =  ma ...(iii)   KE  =  w +w r1 B  wrench By  solving  equation  (i)  (ii)  and  (iii)  F  fr   2 fr  fr 0  =  –mg(1)  +  w    w  =  mg  m   m  m wrench wrench a  –  R  =  a 0 1     work  done  and  power  will  remain  same ()   fr  =  F  ;  ap  =    F 4 4m f 3. Comprehension  #  9 L 1 . Normal  of  the  cube  shift  towards  the  point  'A'  to balance  torque  of  friction. Comprehension  #  11 (sAafmteer  vsoelmocei tyti mdeu eb  ototh  m  bolomcekn taunmd   disc    move  with conservation.  As A given  in  question  50%  of  total  kinetic  energy  of system  is  lost. 2 . For  toppling  before  translation  torque  of  F  and  f r ( about  centre  of  mass.(      F  fr  )  50%) net  =  f  a  fr  a   =  r  mga  =  0.2mga  2   2  and  torque  of  normal  about  centre  of  mass (mg a/2 = 0.5  mga)12 12 1 m r 2 2   mv2  1  1 m r 2 2 ...(i) 2 0  2 2 Torque  of  normal  >  Torque  of  mg  and  fr By  impulse  equation   due  to  friction. (>mg fr)  ()    Body  will  translate  before  toppling 1 1 2 2   mr2  m g rt ...(ii)    () 2mr 0 3 . Torque of fr and F is (frF)  Block  will  achive  velocity  v  in  time  t node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65                       =  rmga  =  0.6  mga (t  v Torque  of  normal  ( ) v so t  g ...(iii) mga So  by  solving  equation   (i),  (ii),(iii) =  2   0.5  mg; N 1 . v  r0 F 4 fr.F  >  N      2 .  By eqn (iii)  Time  r0 4 g fr 3. By  equation (ii)  &  (iii)  0r  r  2v   0 2   body  will  topple  before  translate    ()  So  change  in  angular  momentum 29

JEE-Physics 5. (i)  In  this  problem  we  will  write  K  for  the  angular LL / 2 g L g 2g momentum  because  L  has  been  used  for  length  of / 4 22 4 = 22   i.e.    =  L   x   (sin t)  / 2   the rod. ( K 0 L)  Substituting  this  value  in       = 12v 7L M 2g 12v 7 V L = 7L   i.e.  v= 2 10  1.8 =3.5  m/s. 12  6 . System  is  free  to  rotate  but  not  free  to  translate.            A OC BA OC B During  collision,  net  torque  on  the  system  (rod  A  + rod  B  +  mass  m)  about  point  P  is  zero. L/2 L/4 L/4 Just before collision Just after collision Therefore,  angular  momentum  of  system  before collision=  Angular  momentum  of  system  just  after Angular  momentum  of  the  system  (rod  +  insect) collision. (About P). ( about  the  centre  of  the rod  O  will  remain  conserved  just  before  collision  and  after  collision  i.e.,  K   =  K . if (  O     P A+B)  ) K = K P  if L  ML2  L  2  P 4   4      Mv  I   12  M   L 7 A   M v 4 48  ML2    i.e.,    12 v 7 L (ii) Since  the  weight  of  the  insect  will  exert  a  torque  the angular  momentum  of  the  system  will  not  be B conserved.  Let  at  any  time  t,  the  insect  be  at  a distance  x  from  o  on  the  rod  and  by  then  the  rod mv has  rotated  through  on  angle  ''.  Then  angular momentum  of  the  system  will  be (Let  be  the angular  velocity  of  system  just  after 0x)t H c( o el lr ies ,i o   In  , =   t hmLei o=nm Lef nt  omf vi(n2er)t i=aI  of system about P)  ML2 2  dJ dx (I=P)  12  dt dt J  =   M x   =2Mx   and    =  Mgx  cos  =  Mgx  cost      [  =  t] =  m  (2)2  +  m (2/3)  +  m  2      2  A B    2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65 12   dx  g  Given  :    =  0.6m,  m  =  0.05kg So   Mgx  cost  =  2Mx dt   dx =   2  cost  dt According  to  given  condition  i.e.  for                     m   =  0.01 kg  and  m   =  0.02  kg AB )  Substituting  the  values,  we  get    ( )                                        I  =  0.09  kg–m2 L  Therefore,  from  Equation  (i)  ( (i) ) x = L/4, t = 0 and for x =  2 , t =  2  [as t = 2 ], 2mv (2)(0.05)(v)(0.6)   I     =  0.67  v...(ii) the  above  equation  becomes  ( )   0.09 L g Now,  after  collision,  mechanical  energy  will  be   2 cos t dt 2 conserved. ( )            2 dx =  0 L 4 31

JEE-Physics 9. Given  mass  of  disc  m  =  2kg  and  radius  R  =0.1  m 10.   = I and  =    = rf  sin r F (m = 2kg R=0.1m) for  maximum  torque  the  angle  between  r  and  f (i) FBD of any one disc is (FBD ) must  be  90°  i.e.  perpendicular  to  the  heavy  door. Frictional  force  on  the  disc  should  be  in  forward (rf 90°   direction.  Let  a   be  the  linear  acceleration  of  COM ) 0 of  disc  and    the  angular  acceleration  about  its COM. Then, (11.Moment  of  inertia  about  an  axis  passing  through a0intersection  point  (      ))  ff  1 2 1 a   =   ....(i) 0 m2  =  mr2  ×  4  =  2  ×   4  ×  4  = 2  f R 2f 2f 2kg 2kg F = 12N 2   I  1 mR 2  mR  2  0.1  10f ...(ii) 2 1/2m 3 F =12 3 2 Since,  there  is  no  slipping  between  disc  and  truck. Therefore.  Acceleration  of  point  P  =  Acceleration 2kg 2kg 1 =6 of  force  about  this  point  =  12  ×  of point Q (Torque  2 P=Q ) ()    a  +  R    =  a   f +  (0.1)  (10f)  =  a   Angular  acceleration    0   2  = =6× 2 = 12rad/s2   3 f  a  f  2a  2  9.0 N     f  =  6N 2 3 3 I Since,  this  force  is  acting  in  positive  x  –  direction. 1 2 . For  translational  equilibrium  T   +  T   =  2mg 12 (x- ) ()  Therefore,  in  vector  from    (6i )N T1 T2 f ()  0.5m 20cm=0.2m mg mg 0.8m 1z 2 For  rotatory equilibrium  ()    xy Take  torque  about  extreme  left  point P Q ()  f f =  mg[0.5  +  0.8]  =  T   ×  1 2     (6i )N T  =  mg  ×  1.3    T  =  2mg  –T   =  0.7  mg  r f f 2 1 2 (ii)  Here,    (for  both  the  discs)     Ratio  T1  0.7mg  7 T2 1.3mg 13       0.1j  0.1k and      0.1j  0.1k rP r1 rQ r2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65 Therefore,  frictional  torque  on  disk  1  about  point 13. Let  the  mass  of  sphere  be  'm'  &  the  linear O (centre of mass). (O1 acceleration  of  sphere  be  =  a )  (ma)      f  (0.1j  0.1k )  (6i)N  m F 1 r1 R       (0 .6 k  0.6j)   0.6(k  j) N–m f   1  (0.6)2  (0.6)2  0.85 N–m For  translatory  motion  ( )   and  1      0.6(j  k ) F  +  f  =  ma ...(i) 2 r2 f Similarly,     For  rotatory  motion  (for  no  slipping)     =  I    and             1  2 =  0.85  N–m 33

JEE-Physics 2 1 . We  have FBD  as shown  (FBD  EXERCISE  –IVB 1.(i) As the cylinder rolls without slipping about an axis passing through C.M, hence mechanical energy of the cylinder will be conserved i.e. ( So  cylinder  stop  at  when  translatory  and  rotatory  both  stops.  So  constant  acceleration  is  there  )  ( R ) RR olling R a  g sin   fr   (1)       (2) m Side view  After break of contact  fr  mg cos  a = g sin–g cos So  we  have  equation v=u+at  (  as  v=0)  ( U  + K E )   = ( U +  K E ) 2 4 1 4 =(g sin–g cos) t  11 g sin   g cos   mgR + 0 = mgR cos +  I2 +  mv2 By  =  I 22 v1 but  =    and I =  mR2 . R2 mR2 1  1 m R 2   Torque  by  friction   =fr R   fr R  2 Therefore   mgR = mgR cos +   2  2 mg cos= 1 mR   2g cos  v2 1 v2 4 2R             R 2 +  mv2  R 2 = g (1 – cos) ...(i) 2 3 We  know  a=  R  g sin   g cos   2g cos  (as 30) When  the  cylinder  leaves  the  contact,  normal     1  so  t  4  t=  1.2  sec reaction () 3 3 g 3  2 1  3 3  N = 0 and  = c. mv2 v2 ...(ii) Hence mg cosc =  R  R = g cos c 2 2 . Rod  rotates  about  the  C.M.  let  the  point  'P'  at  a 4 distance  'x'  from  C.M.  having  a  zero  velocity. (i) & (ii)  = g (1 – cos) = g c os c (3 x P)  4 4        cos c =  7  =c = cos–1  7  So       x×  P 2  2m/s 4  For zero velocity () At the time it leaves the contact cos  = cos c =  7 x (ii) ()      x  –  2  =  0 On substituting it in equation (i), P x  = 2 = 2 =  m 4 gR 1  4 4  2  3 7  gR node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65         v =  7 1 (iii) At the moment, when the cylinder leaves the contact x  =   m           (down to  the centre of mass) ()       () 4 gR v =  7 Therefore rotational kinetic energy KR = 1 I2 ()  2   = 1  1 m R 2  v2 = 1 mv2 1 m  4 g R  =K = mgR ...(iii) 2  2  R2 4 =4  7  7 R 35

JEE-Physics    (ii)     Velocit y of CM () =MV–m v Initial momentum = Final momentum M m (=)  Vcm  m1v1  m2v2 0 = mvr –I       I=mvr...(i) m1  m2 By energy conservation () Velocity of C.M remains constant as the 1 kx2  1 mv2  1 I2 ...(ii) 22 2 () Value of Moment of i ner t ia about poi nt O is  fext  = 0 for system (O)    (iii) From the angular momentum conservation 25 0 .6 2 25 0.3 2   0.6 2 0 2  2   2  () I11 = I22 I    2   2  .3 18 12 18   12  M m 2   V  v  61 M  m    So I =  75  . So by equation (i) We have I11 ( )  2 =  I2 2 = M m 20 61 11 61 11 15 75   9  v  0.15   75   9  v  100 (M  m) Vv  2     5 11 v     55 v 2 =   61  4 244  20    Put this value in equation (ii)   (iv) Work done by couple () 76.5 =  1  61   5 112 v2  1  11  v2 2 75  61  4  2 9 = change in rotational  kinetic energy ( )   76.5  1 61 25 121 v2  1  11  v2   2 9 1 1 2 75 61  61 16      =  I222  2 I112 2  121 11  96  61 18  1 Mm     76.5 v2        =  M m (V + v)2  1  0   2178  64416  2  76.5 = v2   105408  (By putting I1, I2, 1 & 2) we get the answer 6 . (i) To calculate mass M  we have to balance torque 76.5  v2  66594 about point O. 105408 (MO)v2 = 121.08  v  11 m/s 7.   Power dL  d dw  d =  dt dL ×    =  2  m r 2 2 ×   =  =  node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65 dt dt dt dt So  25  30  M  15  2  30  ;   M =  11 Kg 18 9 dL dr ( L = mr2 ;  = = 2 mr ) Tor q u e d u e to ro d  A B a nd  C D i s  zero. dt dt (ABCD ) dL dr = 2 mr × r= 2 mr22  [  = V = r (ii) When thread is burnt then energy stored in spring is distributed in term of velocity of block and to rotate dt dt P o w e r  =  2 m r 2 3 the fream. (  ) By angular momentum conservation. ()  37

1 3 . Coin  will  leave  the  contact  if  N  =  0 JEE-Physics g D F' C so  mg  =  m(2A)   A  2 O 14. I  I  I  I  Iabt.A due to A due to B due to C due to D Am B A EB m 19. f mm Mgsin DC   Iabt.A  0  m l cos 452  m l cos 452  2 The  force equation   (  ) 2l Mgsin–f=Ma ...(i) Iabt.A  ml2 ml2  2ml2  3ml2 The torque equation  (  )  2  fR=I ..(ii) 2 for pure rolling motion ()   1 5 . kˆ F F a Ia Ia  ˆi  ˆj   ;  fR  ; f  R2 r RR Position  vector  of  O  w.r.  to  given  point   (O  From  equations  (i)  and  (iii),  we  get Torque about P (P r F Ia  Ma  Mg sin   a M  I   Mg sin   R2 R2         ˆi  ˆj  F kˆ  F ˆj  ˆi  F ˆi  ˆj 1 I  g sin  MR2    I  Mg sin   Ma   a  1  MR2 16. Conserving  angular  momentum,  we  get 2 0 . A  central  force  cannot  apply  the  torque  about  the ()  centre,  hence  the  angular  momentum  of  the  body  ' under  the  central  force  will  be  a  constant. (  m m )  M 2 2 . 1 I2  mgh   where    I= m2 23 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65  m R 2  mR 2  MR 2  MR 2    1 m 2 2  mgh    h =  2 2 2 3 6g     m m  2 3 .  Angular momentum ( )   2M    L  r p 17. The  moment  of  inertia  of  a  uniform  square  lamina where    u cos t ˆi   u sin  t  1 gt2  ˆj about  any  axis  passing through  its centre  and  in  the r  2  plane  of  the  lamina  is  same  hence  IAC=IEF  ˆi t )ˆj] (pm[u cos  (u s in   g   =  – 1 mgv0t2  cos kˆ  IAC=IEF) L  r p 2 39

JEE-Physics ( N        A   Bv F G L  A =  and B = 0. f1 mg2L 4 . From  the  theorem   In  this  condition  (    N  =  0  =  fr  (about  P) the  block  will  topple  when  ( )   y L mg v=R F  >  mg    FL  >  (mg)  2 F  >  2 M Therefore,  the  minimum  force  required  to  topple O x the  block  is  F  mg 2   ...(i) () L0  LCM  M(r  v) We  may  write  : Angular momentum about O= Angular momentum 7 . Mass of the ring (M = L about CM  + Angular momentum of  CM about Let  R  be  the  radius  of  the  ring,  then origin. (O= L (R)L =2R    R  =  2 ) Moment  of  inertia  about  an  axis  passing  through 13 O and parallel to XX' will be (XX'   L0  =I  +  MRv  2 MR2  +  MR(R)  2 MR2 O)  Note  that  in  this  case  both  the  terms  in  Equation  (i) 1   I0=  2 MR2 i.e.  L CM and  M(r  v ) have  the  same  direction  . That  is  why  we  have  used          L0  =  I  +  MRv We  will  use    L0  =I    ~  MRv Therefore,  moment  of  inertia  about  XX'  (from if  they  are  in  opposite  direction  as  shown  in  figure. parallel  axis  theorem)  will  be  given  by  (XX'  Y   13 V IXX' =  2 MR2  +  MR2  =  2 MR2 Substituting  values  of  M  and  R OX 3  L2  3L3 5. Net  external  torque  on  the  system  is  zero. IXX' =  2 (L )  4 2   8 2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65 Therefore,  angular  momentum  is  conserved.  Force 8 . Mass  of  the  whole  disc  =  4  M acting  on  the  system  are  only  conservative. Therefore,  total  mechanical  energy  of  the  system ()  is also conserved. (M(oment of inertia of the  disc about the g)iven axis  1(4M)R2  =  2MR2 2 )  Moment  of  inertia  of  quarter  section  of  the  disc 6 . At  the  critical  condition,  normal  reaction  N  will  pass     () through point P. ( 11 NP)  (2MR2)  =  MR2 42 41

JEE-Physics 1 6 . Iremaining=  Iwhole  –  Iremoved MCQ        i.e.    1.  A L dL A L 1 1 R 2  2R 2  dt  2  3   3     I  =  (9M)(R)2  –  2 m   1m  ....(i) dL  This  relation  implies  that  is  perpendicular  to dt 9M  R  2   . (  dL,A   R 2  3  A L L Here,  m    M both   and  dt Substituting  in  Equation  (1),  we  have    I  =  4MR2  Therefore  option  (a)  is  correct. 21 23 For  (B) 1 7 . MR2  =  Mr2  +  Mr2   MR2  =   Mr2    52 52 L A L Aˆ Component  of     in  the  direction  of      r  2 ( A L )  15 R  d   dAˆ dL As   L  Aˆ   L    Aˆ  0 dt dt dt  1 mv2  +  1  v 2 3v2  1 mR2 18. 2 2 I   R   mg  4g      I  =  2 The  component  of  L  in  the  direction  of  A does   Body  is  disc. not change with time. ( A L  )  1 9 . Condition  for  no  toppling For  (C)  ()  Here,  L.L  L2  Differentiating  w.r.t.  time,  we  get W h () N  2    >  µN  2        L. dL  dL .L  2L dL    2L. dL  2L dL 2 2 dt dt dt dt dt h  > µ   3  > µ µ <  3     but  as  the  coefficient  of  friction  is  greater  than  1 But  since,  L  dL    L. dL  0 block  will  topple  at  some  angle  (   dt dt 1)Therefore,  from  Equation  (i)  dL  0 dt  wori th   m taimgnei.t u(de of LL i.e. L does not cLhange 20. (A) dp   =  F  as  F  =  0   dp  =  0, ) ext ext dt change  in  momentum  is  zero 2 . In case of pure rolling ()  () (B) Kinetic  energy  of  particle  is  scaler  quantity f  mg sin    (upwards) f  sin  therefore.  It may  change  for  a system,  if  external mR2 force  is  zero 1 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65 I (  Therefore,  as    decreases  force  of  friction  will  also ) decrease.()  (C) F =  0    does  not  indicate  that  torque  is  zero ext 3 . On  smooth  part  BC,  due  to  zero  torque,  angular therefore  there  may  be  change  in  angular m(Feoxtm =e0ntum.trvreeamlonacsiiltanyts i oacnnodan l shteaknnicnte.e  tWtihche   i rotationa l  kinetic  energy le  movin g  from  B  to  C energy  converts  into )  gravitational  potential  energy. (D) Internal  force  on  the  system  may  decrease  due (BC to  the polential  energy  (     BC )           ) 43

JEE-Physics  m R 2  R  2   5R  2 15mR2 (Acceleration  of  plank =  acceleration  of  top  point   4    4  8  4  m   m  I  of cylinder (=) From  conservation  of  mechanical  energy, a  =  F  f1 ...(ii) 1 ()  m2 Decrease  in  potential  energy  =  Gain  in  kinetic a  =  f1  f2 ...(iii) 2 energy    m1   3mgR  =  1 15mR2  2    16g   (f1  f2 )R 2  8  5R I Therefore, linear speed  of particle at its  lowest point (I  =  moment  of  inertia  of  cylinder  above  CM) () (I= a2=2a2 v   5R    5R 16g 5gR =  (f1  f2 )R a2  4  4 5R  v  1 2 m1R 2 2 . We  can  choose  any  arbitrary  directions  of  frictional   2(f1  f2 ) ...(iv) forces  at  different  contacts. (m1R ) a   =  R  = 2(f1  f2 ) ...(v) 2 m2 a1 m1 F (Acceleration  of  bottom  most  point  of  cylinder  =0) f1 () In  the  final  answer  the  negative  values  will  show  the opposite  directions. (a) Solving  Equation  (i),  (ii),  (iii)  and  (v), ( 8F )  we  get  a1  3m1  8m2   Let f   =  friction  between  plank  and  cylinder 4F R a2 1   a2  3m1  8m2 and         (     )  (b) f1  3m1F ;  f   =  m1F f =  friction  between  cylinder  and  ground 3m1  8m2 2 3m1  8m2 2 Since  all  quantities  are  positive.          (     )  a   =  acceleration of  plank  (  ) 1 a   =  acceleration  of  centre  of  mass  of  cylinder 2      () and 3 . (i)  Let  just  after  collision,  velocity  of  CM  of  rod  is  v   =  angular  acceleration  of  cylinder  about  its  CM. and  angular  velocity  about  CM  is  .      ()(v node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65 Directions  of  f   and  f   are  as  shown  here ) 12 Applying  following  three  laws  : (f  f ) 12 m v0 L m 2  CM CM v x Since,  there  is  no  slipping  anywhere L After collision 2  Before collision   a   =  2a   ...(i) 12 45

JEE-Physics Solving  Equation  (i)  and  (ii)  for  r, Now,  let  F   be  the  force  applied  by  the  hinge we  get    r  =  0.4  m  and  r  =  0.1  m x But  at  r  =  0.4m,    comes  out  to  be  negative along x–axis. Then, ( x- F) x (–0.5  rad/s)  which  is  not  acceptable. F F  +  F  =  ( 3 m)a     F   +  F   =  ( 3 m)   (r=0.4 m,  rad/s )  x x 4 m x (i) r  =  distance  of  CM  from  AB  =  0.1  m 3F   F   +  F  =  F   F   =  – (ii) Substituting r = 0.1m in Equation (i), we get =1 rad/s  i.e.,  the  angular velocity  with  which  sheet x 4 x4 comes  back  after  the  first  impact  is  1  rad/s. Now,  let  F   be  the  force  applied  by  the  hinge y along y–axis. (y-    (r = 0.1m  (i) =1 rad/s  F) y 1 rad/s Then, )  F   = centripetal  force    F  = 3 m2 yy 6. Angular  momentum  of  the  system  about  point  O  will (iii) Since,  the  sheet  returns  with  same  angular remain conserved. (O  velocity  of  1  rad/s,  the  sheet  will  never  come ) to rest. ( 1 L   =  L if rad/s   L2  M L2   3mv )  3   mvL = I =  m   =  L 3m  M   5 . (i) The  distance  of  centre  of  mass  (CM)  of  the 7 . For  rolling  without  slipping,  we  have f  ()  system  about  point  A  will  be:    r  3 Ma      (A )  Mgsin  Therefore,  the  magnitude  of  horizontal  force  a=R exerted by the hinge on the body is ( )   Mg sin   f  R  fR   Mg sin  f  2f M  M M A  1 M R 2   2  y    3/2 x Mg sin    f =  3 CM Therefore,  linear  acceleration  of  cylinder, F C () B F  =  centripetal  force    F  =  (3m)  r2 a  =  Mg sin   f  2 g sin  M 3    2 8 . by  angular  momentum  conservation   F  =  (3m)  3  node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\02-Rotational Motion.p65       F   =  3 m2   (ii) Angular  acceleration  of  system  about  point  A is  (A )  A   3     3F IA F  2 4m 2 m 2 initially finally Now,  acceleration  of  CM  along  x–axis  is (x-) Idisc  0  =  Idisc  2Iring    50 0.42  a  =  r  =      3F      a   =  F  10  50 0.42 2 2  6.25  0.22  x  3   4m x 4m   2 2     =  8  rad/sec. 47



JEE-Physics UNIT # 06 (PART – I) SIMPLE HARMONIC MOTION EXERCISE –I 4. = A  5 x A sin t= 2  t = or 1 . 2 = 2   =   f      1 Hz 6 6 2 2 2  Phase difference ()= 5    2 or 120° 66 3 2 . -A OA OR -A O A A/2 A cos =A     from phaser ( ) A/2 A/2 A/2 1 cos   A  2 = 60° phase difference ()2 = 120° cos   A/2 1    A2 3  2t  Total phase difference between them 5 . x = a sin t = a sin  T  ()  2    5   T   3  8   2  T     a T    a sin  4   2 At t = 8 , x = a sin 3 . 10 5 OR 0 t=0 t=T t=T/8 t=3T/4  t=T/4  From figure ( ) t=T/2 maximum amplitude ( ) A = 10 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65  =t;   2 T       Position of particle at t = 0, T8 4 4 (t=0)  x =5 x x a let equation of SHM is ()  As cos   so a2 x = Asin(t+) At t = 0, x = 5 2 2 5 = 10 sin  = /6 and   T  2   Thus, equation of SHM x = 10 sin  t   6. x = a sin t, x = a sin(t + )  6  1 2 Greatest distance ( )  () =|x –x | =2a sin   3a  sin   3 2 1 max 22 24 47

JEE-Physics Now according to question ()x =x 1 0 . x = 2sint & 12 y = 2sin  t    2 sin t  2 cos t  a sin t = a sin (t + )  4     –t = t +   t = 2      a7  x2  y2  2xy  2  x = a sin   = a cos 2  1 2 4 which represent oblique ellipse ()  7 . Minimum phase difference between two position 1 1 .  x = A sin t ()  x = A sin  & x + x = A sin(2) 1 1 2  2 2  A 4 5   x = &x +x =A = 53° + 37° = 90° 4   1 2 12 53° T 8 4 37° 3 A x1  1 3  x2 = A– 2 . x2 2 1 Therefore (  )= T  20 OR 4 4  Time taken = 5s Suppose amplitude be A and distance traveled in 1 OR sec be x and in 2 sec be x . 12 -5 -3 45 (A 1sx2s  1 x) 2   2  2   34 =t  2 1   -A t=0 A T 20 10 84 x1 x2 5 5  80 2  1  Therefore    4  cos   cos   x1 6 2 4A  from figure 1 + 2 = 53° + 37° = 90° or 2  x1  A  =t     t t = 5 sec & x + x =A 2 12 -A 4 A 2 10  2 1 x1  1 A  2  . Therefore x2 2 1 8 . x = a sin (t + )  x = at t = 1s, x=0 = a sin(t+)    = – 2  v = a  cos (t +) 1 2 . Maximum possible average velocity will be around 1 mean position.  At t=2s, 4 =a cos(2+) ()  1  a  2  co s()  a co s     a 1  a  3 Average velocity in time (   )  4  6  3  3  3  2  2 ab  T 2 A / 2 4 2A 9.  4 T/4 T Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 x=Acost  a =Acos and a+b=Acos2 OR  a2  M.P. A  a+b=A[2cos2–1] =A 2  A 2  1 T/8 T/8 -A 2a2 3T/4 xx T/4   A  a  b  A2+(a+b)A –2a2 = 0  A a  b  a2  b2  2ab  8a2 T/2 A  2 Note : Please correct the answer of the question 48

JEE-Physics 2 T  there is no loss of energy ( )  =t = T8 4 dE 8 8kxv x cos   x A dt  0  3 kxv + 2mva =0  3  2mva cos   A  4A 2  x= 4kx 4kx 4k   Average velocity ( ) a   2 x    3m 3m 3m total displacement 2x 2A / 2 4 2A T  2  2 3m = =   4k total time T/4 T/4 T 1 3 . Time period ()=4(1+1) = 8 1 8 . Let the natural length of the spring =  0 vv () A Mean B 2s From figure ( ) Position 1 4 . Maximum KE (  )= 2× 5 = 10J 0 Total energy ( )= 15 + 10 = 25J   1 5 . For maximum displacement  F=4N F=5N () F=9N 1 Mg mg Mg(x) = k(2x)2  x   x  x0 2 2k 2k 16. Both the spring are in series (  )   k (–0)...(i) 4= K(2K ) 2K 5 = k (–0)...(ii)  K= = eq K  2K 3  9 = k (–0) ...(iii) Time period () T = 2 K eq iii  i 5 k    eq. iii  ii  4  k      m1m2 Here () = m where = m1  m2 2 m. 3 3m   5  4 2 2K 4K  T = 2 = 2 OR 19. m k 2k v keq m m f 1 k f1  m2 m m  f2 m1 m f1  M  m  1/2 1 m 1/2 f2  M  M  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 xx     Total extension ( )=2x  By energy conservation (  )  20. T = 2 geff ...(i) a0 0 E  1 K eq 2x 2  1 mv2  1 mv2 T0  2  T 2 2 2 2 g  a ...(ii) mg Now E  1 2k 4x2  1 mv2  1 mv2  4 kx2  mv2 g a  4  a  3g (upwards) 23 2 2 3 g dE  4 k 2x dx  m 2v dv  dt 3 dt dt 49

JEE-Physics 2 2   2 25. 2 1 . 1  2 =  and  3  They will be in phase if ()   (1–2) t = 0, 2, 4...  mg mgcos 2 2  t  1  2     2  6 sec 3 22. mgsin =–mgsin  I=-mgsin  m(2+k2)=–mg(sin~)         g   2     g g  2  2g   k2 T  2 where    so T  2 2  k2 T  2  2 2  k2 2  k2 g  2   g  2  T  1  2 1 g 23. f f   g 2  T2   2   4    k 2  0 f1    n 2 2 1  n  1  2 f2 2    n      1   n  1 1 2 2 6 . For weightlessness ( )  2 4 . Center of mass 2m of a system is at a distance from mg = m2a  g = (2f)2 (0.5) peg P is  and moment of inertia of the system 2g  2f = 2 g  f = 2 22 2 m 2 (2mP 2 7 . Ans. (C) is 3 m Time period for spring block system is T  2  2m2 ) k 2 2 3 does not effected. (  T  2 m )   k P /2 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 C.M.  28. Ans. (C) mgsin mg mgcos U(x) = ax2 + bx4 U F = – =–2ax – 4bx3  –2ax for small x x  2a   r  F  mg sin  So m2 = 2a    22 m  sin    for small  2 9 . at x  3 A  I  mg sin  2 22 2 m 2      3g  2    mg 1  3  1 3 2 2 2 2 2 m2  A 2  4 A 2  8 m2 A 2 KE = =  3g  T  2  2 2 2 KE is increased by an amount of 1 m2 A 2 . Let now 2 2  3g 2 50

JEE-Physics amplitude be A then total KE (2k)(k) 2 1 k=  k eff 2k  k 3 ( 1 m2 A 2 A1 For 2nd condition 2 m 1 1 k )  T = 2 k  f  T  2 m KE = 1 m2 A2  1 m2 A2 f1  k/2 3 18 2  5 1 m2  3  f2 2k / 3 2 8 2  4  = m2A 2  A 2  A 2  A1  2A 1 3 8 . In an artificial satellite ( )   111 1 g =0  T =  3 0 . Here mv2 = kx2  m2(a2–x2)= m2x2 eff 222 2 T/2  x  a 4 2 2cm  2a sin t dt  22 2 2  3 9 . <acceleration( )>  0 T /2  dt 3 1 . Total energy (  ) 0 1 a2 4 0 . Required time (  TT T E= m2a2  E  T2 )=   2 4 12 6 I 2  k2 4 1 . KE at centre ()  3 2 . T  2 mg  2 g = 1 m2 A 2  = 1 m42f2 A 2 3 3 . x = 3sin 2t + 4cos 2t = 5 sin(2t+) 22  a = 5, v = a = (5)(2) =10 KE at distance x (x ) max = 1 m42f2 A 2  x2  3T 2 3 4 . For (A) : at t= , particle at extreme position Difference ()= 1 m  42 f2 x2 = 22f2x2m 4 () 2  a = –2x  F  0 For (B) at t = T/2, particle at mean position 4 2 . From the graph, equation of acceleration can be (t=T/2 written as () v = A(maximum) () a = – a cos t For (C) : at t=T, particle at mean position max (t=T)  velocity can be written is () a = –2x =0 v = – v sin t. For (D) : at t =T/2, particle at mean position max (t=T/2) 11 mv2 KE = mv2= max sin2 t so x = 0  U  1 kx2  0 2 2 2 Hence the graph is as shown in A (A )  Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 35. sin  13   sin 2    sin   1 4 3 . a=82–42x = –42(x–2)  =2  6  6  6 2 Here a=0 so mean position at x =2 a (a= 0 x=2    ) Now x = a sin(t) = 2 Let x = Asin(t + ) 3 6 . T    2       As particle is at rest at x=–2 (extreme position) T 2 8 2 2 and Ampliutde = 4 as particle start from extreme position. Therefore 3 7 . For 1st condition k k= ( x=–2        eff 2 =4)  x –2 = –4cos2t  x = 2–4 cos2t 51

JEE-Physics 44. x = A (1+ cos 2 t) sin (21t) k k  1 0 m , 2  4m 2 2 = A [sin 21t + cos 2 t + sin 21t] 3. 1  0 2 A [sin 11 (1–2) t = 0, 2, 4,.... t=0 = 0 21t + sin (2 (1 + 2)t – sin (2(1–2)t)] 2 2 Required ratio ( )=1: (1–2) : (1 +2)  t  2 4  2=4 (1–2) 1  2 45. y  sin t  3 cos t = 2 sin  t     1  2   k  2 N / m 3   2 To breaks off mg (mg) 4 . When cylindrical block is partially immersed mg = m2min A  g  22min    g () 2 F = mg  3Ayg = A(60× 10–2) g moment it occurs first after t = 0 B (t=0) y = 20 cm  Maximum amplitude ( )= 20 cm 2  2 sin  t1    t1    t1     2 40cm  3  6 6 6 g 20cm EXERCISE –II Restoring force when it is slightly depressed by an amount of x. 1 . v = A cos t, a =–2A sint (x)  v 2  a  2  A    2 A  F= – (Vg) = – (Ag)x   1  Straight line in v2 and a2  T  2 m Ah  2 h  2 A 3g 3g (v2a2) Ag  2 60  102 2 s = 3  9.8 7 2 . Initially finally kx FB k(x+ y) FB 5 . At equilibrium mg ()= kx /2 (/2+ y) 0 k Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 mg (0.2)(10) m mg mg  x0  k  200 = 0.01 m = 1 cm kx + F = mg Amplitude of SHM ()=1cm B Frequency ()= 1 k1 200 kx = mg – Vg   5Hz kx = mg – Ag ...(i) 2 m 2 0.2 2 Let cylinder be displaced through y then restoring 6 . The position of momentary rest in S.H.M. is extreme force, (y position where velocity of particle is zero. )  ( f = –[k(x+y) + F – mg] )  net B f =    ky  A    y  g   a= g N kx  2  mg  a = g/2 mg net A a= 0 Motion fnet = – [ky + Agy] ma = – [ky + Agy] Position A/2 a = g/2 A/2 a= g    k  Ag  f  1 k  Ag A m 2 m 52

JEE-Physics As the block loses contact with the plank at this From energy conservation ( )  position i.e. normal force becomes zero, it has to be the upper extreme where acceleration of the block 11  v2 1  v  2 1 m will be g downwards. mv2= m  2  2  2  2 2k  m  kx 2  xm  v 22 m ( 10.Let small angular displacement of cylinder be  then g ) restoring torque 10 ()  2A = g  2 = 0.4 = 25   = 5 rad/s 3 )T=2 2 I = –k(R)R where I= MR2 5 2 Therefore period ( = s Acceleration in S.H.M. is given by a = 2x  d2  kR 2   0  d2  2k   0  2  2k dt2 3 MR2 dt2 3M 3m ()  2 From the figure we can see that,At lower extreme, acceleration is g upwards (      1 1 . As  < so T = 2 L g) g  N – mg = ma  N = m (a +g) = 2mg At halfway up, acceleration is g/2 downwards 1 2 . mgh  1 (M  m)v2  1 kx2 22 (g/2) g1 mgh  1 kx2 x  2mgh 1 / 2  mg – N = ma  N = m (g – ) = mg 2   k  2 2 At halfway down acceleration is g/2 upwards (g/2 ) 13. f  1 k g3 2 m  M  N – mg = ma  N = m (g + ) = mg 22 7. x =3 sin (100 t), y = 4 sin(100t)  4 x 1 4 . y = 10sin (t + y 3 Maximum KE (  )= 1 m2A2  Motion of particle will be on a straight line with 2 slope 4/3.  64  1 m2 A2  1 m2 A 2  x2  100 2 2 (4/3 )  64A2= (A2–x2)100  x = 0.6 A As r  x2  y2 = 5 sin (100 t) 1 m 2 A2  x2   1 m2 A2  x  so motion of particle will be SHM with amplitude 5. 22 (5)  A2 – 0.25A2 = A2x   x = 0.75 means 75% of energy r 8.  xi  yi  (A cos t)i  (2 A cos t)j 1 5 . Maximum speed =( ) v  A  v0 Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65  x = A cos t, y = 2 A cos t  y = 2x 0 0  The motion of particle is on a straight line, periodic and simple harmonic So equation of motion x = v0 sin 0t 0 () ()  1 6 . x = 3 sin100t + 8cos250t 9 . At maximum compression v A  v B & kinetic energy = 3sin(100t)+ 4(1+cos100t) of A–B system will be minimum ( =4 + 5sin(100t + 37°)    A-B  )  Amplitude ()= 5 vA vB Maximum displacement ( )  v1 = 4 + 5 = 9 cm so vA = vB = 2  KAB = mv2 4 53

JEE-Physics 1 7 . Velocity of 3kg block just before collision 1 9 .  a =–2y and at t = T, y is maximum so acceleration (3kg)  is maximum at t = T. (a =–2y t = T  y  t=T) =  a2  x2   k  (a2  x2 )  m  3T 3T Also y=0 at t = 4 , so force is zero at t = 4   900 (22  12 )  30 m / s (t = 3T  y=0)  3  4 Velocity (of combinedmassesimmediatelyafter t)he (AttT2,v=0=PE=osc)illation energy collision  (3)(30) = 10 m/s 36 EXERCISE –III New angular frequency (  )  Fill in the blanks ' = k  900  10 11 1 m9 1 . mv2 = m2(a2–x2) = 0.5 and m2x2 = 0.4 22 2 Therefore ( )v' = ' a '2  x2  a2  x2  5 a  3 m= 0.06 m  10 = 10 a '2  12  a   2m x2 4 50 1 8 . From energy conservation ( )  18 2 . nT1 = (n+1)T2 = 18  n=9  T2= 10 =1.8s 3 . k1xL = k2(1–x)L = kL 1 k1  1 k f0  f2 f1  2 m1 2  x(1  x) x(1  x)m 2 (2n 1 )  2 t 0  4   4 . vdt 4 sin dt 8 dt  1) <v> =  2 (2 n 1)  (2n m/s 11 dt k12 = mg (1 + x) + kx 2 2 1 21 0 12 2mg 15m2g2 5. 2a = g  a  g  k 1  k2 2  0 5mg 4mg 6 . v12 = 2(a2–x12), v22 = 2(a2–x22)  1  k    k 4mg  v 2  v 2 & a v 2 x 2  v 2 x12  If  > k the lower disk will bounce up. 1 2 1 2 2 x 2  x12 v 12  v 2 2 2 ( > 4 m g  )7 . x1= A sint, x2=A sin (t + ) Node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 5 & 6\\02.SHM.p65 k x2 – x1 = A(cos –1)sin t + A sin  cos t Maximum value of x2–x1 Now If   2mg then maximum normal reaction k from ground on lower disk (    2mg  = A 2 (cos   1)2  A2 sin2   2A sin    A  2  k )   5  5 N = 3mg + k(x + ) = 6 mg   or   = or 0 26 6 33 54