8.SUBJECT TEST-1-SOLUTIONS

SOLUTIONS TO SUBJECT TEST-01 (PHYSICS) 1. If the normal reaction is always zero, the particle always moves under action of gravitational force. Hence the path of particle (and also the nature of surface) parabolic. At a height h = 5 metre above the projection the speed of the particle is v2 = = = 10 m/s The horizontal component of velocity does not change  v cos = 20 cos 60° cos = or tan =   = tan–1 2. Let v1 and v2 be the velocity of efflux from square and circular hole respectively. S1 and S2 be cross-section areas of square and circular holes. v1 = and v2 = The volume of water coming out of square and circular hole per second is Q1 = v1S1 = L2 ; Q2 = v2S2 = R2  Q1 = Q2  R = . L 3. The free body diagram of the capillary tube is as shown in the figure. Net force F required to hold tube is F = force due to surface tension at cross-section (S1 + S2) + weight of tube. = (2RT + 2RT) + mg = 4RT + mg 4. x = A cos t (as it starts from rest at t = 0) A – a = A cos  (1) A – (a + b) = Acos2 (2) Solving (1) and (2) for A we get A = 2a2 3a – b 5. Frequency of sound reflected by wall f' = v  u f v  Wavelength of sound reflected by wall = v  u  v  u  v f  v  u f 6. V = V – r , but the Centre of mass C can be at any distance r from P. Hence V can be in forward or backward direction or can be zero. Consequently friction can act in backward direction or forward or can be zero. 7. The system acts as a pendulum of length 𝑙 acting under a gravity of g cos . Now , 𝑙1 𝑙2 = AC. 𝑙 = 𝑙. x cos = T = 2 = 2 t dx .dt dt 8. average velocity = 0 t t dx = 0 = t xt  x0 t = Acos  / 6 – 1  / 6 = 3A  – 2 since particle does not change it's direction in the given interval , average speed = 9. B moves in dotted circle of radius 𝑙 with uniform speed. Massless rod 1 can exert force T1 on B only along its length. Hence T1 contributes only to centripetal acceleration of B. The tension in rod 2 , that is , T2 can contribute to both tangential and centripetal acceleration of B. Since tangential acceleration of B is zero. = 3A 2 – 3  A m 1 T1 Hence T2 = 0. m  m C B T2 2 10. Initially the block is at rest under action of force 2T upward and mg downwards. When the block is pulled downwards by x, the spring extends by 2x. Hence tension T increases by 2kx. Thus the net unbalanced force on block of mass m is 4kx.  acceleration of the block is = 4kx m 2  20 sin 37 11. T = 10 3 = 4 × 5 12 = 5 sec. 12 Range = R = 5 × (20 cos 37° + 10) 12 4 5 × (20 × 5 12 + 10) = 26 × 5 = 62.4 m 12. Process AB is isothermal expansion, BC is isobaric compression and in process CA nRT P  P  P2  T. 13. Slope of graph is greater in the liquid state i.e., temperature is rising faster, hence lower heat capacity. The transition from solid to liquid state takes lesser time, hence latent heat is smaller. 14. A B Molecular wt. = 16 M0 Molecular wt. = M0 mass = 2 m0 mass = m0 nA = n0 n 8 B = n0 (A) K.E./atom = f k.T. = 2 f k.T. for both the gases. 2 (B) C rms A  , CrmsB  , (C rms)B = 4 (C rms)A (C) (P)A = (n0 / 8)RT , (P V B) = n0 RT V , (P B) = 8 (PA) (D) nB = 8 nA 15. In the equilibrium position the net force on the partion will be zero. Hence pressure on both sides are same. Hence, (A) is correct. Initially, PV = nRT n1 = P1V1 RT1 = PV RT and n = (2P)(2 V) 2 RT = 4 PV RT  n2 = 4n1 Moles remains conserved. Finally, pressure becomes equal in both parts. Using, P1V1 = n1RT1 P2V2 = n2RT2  P1 = P2 & T1 = T2  V1  n1 = 1  V = 4V V2 n2 4 2 1 3 Also V1 + V2 = 3V  V1 + 4V1 = 3V  V1 = 5 V 12 And V2 = 5 V Hence (B) and (C) are correct. In compartment () : P1' V1 = n1RT1 P  3V   P V  1  5  =  RT  RT     P1' = 5Pv 3v 5 = 3 P Hence (D) is also correct. 16. As V =   = V =  340 340 = 1m first Resonance length  1 R1 = 4 = 4 m = 25 cm  R = 3 = 4  R = 5 = 4 3 m = 75 cm 4 5 m = 125 cm 4 i.e. third resonance does not establish Now H2O is poured,  Minimum length of H O Column to have the resonance = 45 cm  Distance between two succensive nodes =  = 1 m = 50 cm 2 2 & maximum length of H2O column to create resonance i.e. 120 – 25 = 95 cm. 17. Free body diagram of block is as shown in figure. From work-energy theorem Wnet = KE or (40 – 20)s = 40  s = 2m Work done by gravity is – 20 × 2 = – 40 J and work done by tension is 40 × 2 = 80 J 18. In standing waves, particles may have phase differences only 0 or . 19. Since the angular speed and angular acceleration of each moving point on the rod is same, acceleration of each moving point on the rod is in same direction, that is, parallel. Hence statement-1 is false. 20. For a disc rolling without slipping on a horizontal rough surface with uniform angular velocity, the acceleration of lowest point of disc is directed vertically upwards and is not zero( Due to translation part of rolling, acceleration of lowest point is zero. Due to rotational part of rolling, the tangential acceleration of lowest point is zero and centripetal acceleration is non-zero and upwards). Hence statement 1 is false. 21. Both the stones cannot meet (collide) because their horizontal component of velocities are different. Hence statement I is false. 22. From statement-2, if the component of relative velocity normal to line of impact is non-zero, they shall not have same velocity after collision. Hence statement-2 is correct explanation of statement-1. 23. Dimensional analysis gives the unit of the proportionality constant but provides no information of i ts numerical value. Hence statement-1 is false. 24. An isothermal process is an example to show the possibility "for both the pressure and volume of a monoatomic ideal gas to change simultaneously without causing the internal energy of the gas to change". Hence Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. 25. Speed of wave in a medium is property of the medium. 26. With increase in temperature speed of sound in air shall increase and wavelength of sound will increase.Hence statement-1 is false. 27. (A) If velocity of block A is zero, from conservation of momentum, speed of block B is 2u. Then K.E. of block B = 1m(2u)2 = 2mu2 is greater than net mechanical energy of system. Since this is not possible, 2 velocity of A can never be zero. (B) Since initial velocity of B is zero, it shall be zero for many other instants of time. (C) Since momentum of system is non-zero, K.E. of system cannot be zero. Also KE of system is minimum at maximum extension of spring. (D) The potential energy of spring shall be zero whenever it comes to natural length. Also P.E. of spring is maximum at maximum extension of spring. 28. (A) If resultant force is zero, Psystem will be constant. → (B) If resultant torque is zero, Lsystem will be constant. (C) If external forces are absent, both Psystem and Lsystem will be constant. (D) If no non conservative force acts, total mechanical energy of system will be constant. 29. KE = 1 mv2 = TE  v = = 2 m/s max 2 max max amplitude A = vmax  = 2m. x = A sin t = 2 sint v = 2 cost = (A) v = m/s  x = ± m. (B) KE =  x = ± 1 mv2  1 = 2 m. 1 × 1 × v2  v = 2 m/s. (C) at t = /6 s, x = 2 sin /6 = 1m. (D) KE = 3 2 1  1.5 = 2 × mv2  v =  x = ± 1m. 30. (A) If P = 2V2 , from ideal gas equation we get 2V3 = nRT  with increase in volume (i) Temperature increases implies dU = +ve (ii) dW = +ve Hence dQ = dU + dW = +ve (B) If PV2 = constant, from ideal gas equation we get VT = K (constant) Hence with increase in volume, temperature decreases Now dQ = dU + PdV = nCvdT – PK K T2 dT [ dV = – T 2 dT] = nC dT – PV dT = n(C v T v – R) dT  with increase in temperature dT = +ve and since Cv > R for monoatomic gas. Hence dQ = +ve as temperature is increased (C) dQ = nC dT = nCv dT + PdV  n (C + 2R) dT = nC dT + PdV  2nRdT = PdV  dV dT = +ve Hence with increase in temperature volume increases and vice versa.  dQ = dU + dW = +ve (D) dQ = nC dT = nCv dT + PdV or n (Cv – 2R)dT = nCrdT + PdV dV or – 2nRdT = PdV  dT = –ve  with increase in volume temperature decreases. Also dQ = n(Cv – 2R)dT with increase in temperature dT = +ve but Cv < 2R for monoatomic gas. Therefore dQ = –ve with increase in temperature. SOLUTIONS TO SUBJECT TEST – 02 (PHYSICS) 1. When the container is at rest with respect to the Earth, there is pressure on the walls due to the weight of the water. The pressure results from the contact force between the water and the container. In free fall, both the water and the container have acceleration of g and the contact force is zero, so removing part of a wall by making a hole produces no outward flow. (Note that some of the water is in contact with the air, which is not accelerating, so there is still atmospheric pressure on the water.) 1 M g h  1 MV2  2.    = MST  1 10  50  1  25 2 2  = 56.25 × T 2  2  T = 500  625 56.25  2 = 10°C. 3. The extension developed in the string due to small values of '' is : x = h sin   h Torque about 'O' :  = (Mg sin ) L + (kx)h or,   mg L + kh2 = (mgL + kh2)   Also;  = I  = mL2  (2) From (1) and (2) : mL2  = (mg L + kh2) 1 or  = 2  gL  kh2   L  m  Now : T = 2 = 2   = 1 T 1 = 2L d 4. dm.2R = 2T sin 2 Rd 2R = 2T d 2  2R2 = T  v = Also speed of string is R = 2R2 = R  The velocity of disturbance w.r.t. ground = R + R = 2R. 5. During 1st collision perpendicular component of V, V becomes e times, while IInd component VII remains unchanged and similarly for second collision. The end result is that both VII and V becomes e times their initial value and hence V" = – eV (the ( – ) sign indicates the reversal of direction). 6. Disc (B) moves in a circle of 3R radius. Distance by which its centre has moved = 2 (3R) = 6R 6R Angle turned by it = 2R = 3 radian.  Vob   Vob Vob  7.  = 0 1 V   1  V (straight line) ; when V = 0 ;  = 1.  sound  0 sound sound 0 and as Vob Vsound  1   0  2 4 8. At x = 0 the phase difference should be . hence y2 = a cos (t + kx + 3 ) Alternate solution y2 = a cos t + kx + 0)             t  3 0    0 3  y = y1 + y2 = a cos t – kx + 3 ) + a cos t + kx + 0 ) = 2a cos    × cos kx  2   2      y = 0 at x = 0 for any t  kx     0 3 2  = 2 at x = 0   = 4 3 . Hence y2 = a cos t + kx + 4 ) 3 9. From given graphs : 3  3  3 ax = 4 t and ay =   t  1  4   v = 8 t2 + C At t = 0 : v = – 3  C = – 3 3  v = 8 t2 – 3  3 t2    3 dt  .... (1)  3 t2  t   Similarly ; dy =   8 4 dt  .... (2) As dw = F. ds = F.(dx ˆi  dy ˆj) W 4  3 ˆ  3  ˆ  3 2  ˆ  3 2  ˆ  dw    4 t i   4 t  1 j .  8 t  3  i   t 8  t  4  j dt 0 0          ∴ W = 10 J Alternate Solution : Area of the graph ; ax dt = 6 = V(x) f  (3)  V = 3. and ay dt = –10 = V(y) f  (4)  V = – 6. Now work done =  KE = 10 J 10. The force F1 causes extension in rod. F1 F2 causes compression in left half of rod and an equal extension in right half of rod. Hence F2 does not effectively change length of the rod. F2 C 11. S = (100.0)(10.0)(700.0)  (1.0)(3.0 103 ) 80.0  10.0 3.0  1.0 = 3.5 × 103 J/kg°C (According to addn. and multiplication rule of S.F.) Vt 12. If m1  0, S  0 S = m2 (f  0 ) dS𝑙  V    t  m2 f  0 0.1  0.1  0.01 102 0.01  1 1 S𝑙 V  t m2 + f  0 = 10.0 10.0 1.00 102 + 1.00 50 = 8% 13. As the temperature increases, heat loss to surrounding increases. After some time the rate at which heat is lost becomes equal to rate at which heat is supplied and an equilibrium or steady state is achieved. Hence temperature becomes constant after some time. 14. The equation of wave moving in negative x-direction, assuming origin of position at x = 2 and origin of time (i.e. initial time) at t = 1 sec. y = 0.1 sin (4t + 8x) Shifting the origin of position to left by 2m, that is, to x = 0. Also shifting the origin of time backwards by 1 sec, that is to t = 0 sec. y = 0.1 sin [(4t – 1) + 8(x – 2)] 15. As given the particle at x = 2 is at mean position at t = 1 sec.  its velocity v = A = 4 × 0.1 = 0.4  m/s. 16. Time period of oscillation T = 2  2  1 sec.  4 2 Hence at t = 1.125 sec, that is, at T seconds after t = 1 second, the particle is at rest at extreme 4 position. Hence instantaneous power at x = 2 at t = 1.125 sec is zero. 17. First, let us check upto what value of F, both blocks move together. Till friction becomes limiting, they will be moving together. Using the FBDs F 10 kg block will not slip over the 15 kg block till acceleration of 15 kg block becomes maximum as it is created only by friction force exerted by 10 kg block on it a1 > a2(max) F  f 10 f = 15 for limiting condition as f maximum is 60 N. F = 100 N. Therefore for F = 80 N, both will move together. Their combined acceleration, by applying NLM using both as system F = 25a 80 a = 25 = 3.2 m/s2 18. If F = 120 N, then there will be slipping, so using FBDs of both (friction will be 60 N) For 10 kg block 120 – 60 = 10 a  a = 6 m/s2 For 15 kg block 60 = 15a  a = 4 m/s2 19. In case 80 N force is applied vertically, then For 10 kg block 80 – 60 = 10a f a = 2 m/s2 For 15 kg block in horizontal direction. F F – f = 15a a = 4/3 m/s2, towards left. Sol.20 to 22. FBD of rod and cylinder is as shown. Net torque on rod about hinge 'O' = 0  N × L = mg × L or 2 Net torque on cylinder about its centre C is zero.  f R = f R or Net torque on cylinder about hinge O is zero.  N × L = N × L + mgL or N2 = 23. (i) (ii) Work done in process AB 2RT0  3T  w AB =  3  1  1   0   w = – 8RT0 QAB = du + w = ncv T – 8 RT0 QAB = – 16 RT0 (ii) for process B  C, WBC = 0 dU = nc T = 2 3 R . 8T0 = 8RT v 2 3 QBC = 8RT0 and C  A (isothermal process) W = 2R (3T ) 𝑙n Vf 0     27  V i  i  total work done WCA = 18 RT0 𝑙n 3 and QCA = 18 RT0 𝑙n3 W = WAB + WBC + WCA = –8RT0 + 0 + 18RT0 𝑙n 3 Wnet = RT0 (18 𝑙n 3 – 8) and Qnet = QBc + QCA = – 8RT0 + 18RT0 𝑙n 3 Qnet = RT0 (18 𝑙n 3 – 8) 24. From the frame of wedge The component of velocity of ball normal to wedge is u = 10 – 10 sin30° = 5 m/s.  Time of flight = 2u gcos 30 = 2 sec. Hence the particle strikes the wedge after t = 2 sec.  330  10  25. Frequency appearing before collision = 300   330  20 34 = 31 × 300    Conservation of momentum  mu + 2mv = 2m × 20 – m × 10 u + 2v = 30 u  v  1  u – v = 15 30 2 Solving u = 20, v = 5  330  20  Frequency appearing after collision = 300  330  5  = 310  300 ;   325 fi 1105 So their ratio = fa = 961 Ans 26. v = 𝑙 1 = t = t 1 for vertical motion 1 = g t2 2 x = 0.25 (i.e. level goes down from 0.81m to 0.25m. Using equation of continuity) dx – . dt = . 4 × 10–4 on solving t = 1000 s 27. Since e = 1  Final normal component of velocity = v cos 37º 5 . As the angle of rebound is equal to the angle before impact. Therefore, both normal & tangential components of velocities must change by the same factor.  Tangential velocity after impact becomes Let the time of impact be t. v sin 37º 5 .  v cos 37º  mv cos 37º  5 6mv cos 37º N =   = t 5t where N is the normal force imparted on the ball by the wall. 6 mv cos 37º Frictional force = N = 5  t v sin 37º   v sin37º  m v sin 37º 5  mv sin37º 5  6 mv cos 37º Also frictional force = t  t = 5 t 2  = tan370   = 2 . 3  1  1  x  x = 5 3 3 4 2 2 10 28. Since the spool rolls without slipping a = R (1) The F.B.D. of spool is shown in figure. R for linear motion  T – f = Ma r a and for rotational motion M T MR2  a  f.R – T.r = 2  R  f   For maximum value of T, f = Mg 3MgR on solving T = R  2r 3  (0.4)  3 10  0.2 Using values T = 0.2  0.2 = 18N 29. As block goes down by distance x, water comes up by distance y. As both are measured from initial level of water, compression in the spring is x but the block is in depth (x + y) in water. Applying conservation of volume 0.2x = (1m2 – 0.2 m2).y x = 4y  y = x 4 Thus total depth of block in water = Free body diagram in equilibrium : Fb = (0.2)(5x/4) (1000) (10) For equilibrium : mg = kx + fB y x x + × = 5x 4 4 1800 = 2000 x + (0.2)  5x  (1000) (10)    4   18 = 20x + 25x  x = 18 45 m = 40 cm 30. Area covered by line joining planet and sun in time dt is 1 1 1 dS = x2d ; Areal velocity = dS /dt = 2 x2d /dt = 2 x2 2 where x = distance between planet and sun and  = angular speed of planet about sun. From Kepler's second law areal velocity of planet is constant. At farthest position A = dS/dt = 1 (2R – r)2  = 2 1 (2R – r) [(2R – r)  = 2 1 (2R – r) V 2 B or VB = 2A 2R  r (least speed). (Using values) VB = 40 km/s.