SOLUTIONS- Subject Test Part Tesr-1 & 2

SOLUTIONS TO SUBJECT TEST - 1 (ST-1) PHYSICS  Vob   Vob Vob  1.  = 0 1 V   1  V (straight line) ; when V = 0 ;  = 1.  sound  Vob 0 sound  sound 0 and as Vsound  1  0  2 2. 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 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. 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 indi cates the reversal of direction). 5. 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. 6. 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 7. 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 = = 3A 2 – 3  8. 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 9. T = 10 12 3 = 4 × 5 12 = 5 sec. Range = R = 5 × (20 cos 37° + 10) 12 4 5 × (20 × 5 12 + 10) = 26 × 5 = 62.4 m 10. Process AB is isothermal expansion, BC is isobaric compression and in process CA P  nRT P  P2 T. 11. 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. 12. 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 13. 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   5Pv 5 1  5  =  RT  RT  P1' = 3v = 3 P     Hence (D) is also correct. V 340 14. As l =  = 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. 15. In standing waves, particles may have phase differences only 0 or . 16. 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 17. 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 18. 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. 19. 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 20. Area covered by line joining planet and sun in time dt is 1 1 1 dS = 2 x2d ; Areal velocity = dS /dt = 2 x2d /dt = 2 x2 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 1 1 1 A = dS/dt = 2 (2R – r)2  = 2 (2R – r) [(2R – r)  = 2 (2R – r) V B or VB = 2A 2R  r (least speed). (Using values) VB = 40 km/s. SOLUTIONS TO SUBJECT TEST - 2 (ST-2) PHYSICS 1. Independent acceleration of A and B would be a = g (sin –  cos) and a2 = g (sin –  cos) if  >  , a < a , so blocks will be in contact and will have common acceleration a  a  (m1  m2 ) gsin  1m1gcos   2m2gcos (m1  m2 )  g[sin  (1  2 )cos ] 2. r = r = 3 L 2 T1sin30º = T2sin30º + mg T1 = T2 + 2mg 2  3 L  T cos30º + T cos30º = m   1 2     T + T = m2L 2L  T = m 2  2  g  2L  T2 > 0 only when 2 g   m2F 3. The maximum compression in spring for shown situation is 2(m1  m2 )K . Since (m1 + m2)K is same for all situations ; the compression is maximum for spring 3 and least for spring 2. 4. For disc, from torque equation mR2 3 mg R – TR =  2 .... (1) By application of Newton's second law on block we get, T – mg = ma .... (2) a where a = R  (3) 4 g solving a = 3 5. T mg F=3mg mu = 4mV V = Ki = u 4 1 mu2 = 4 2 1 u2 mu2 Kf = 2 × 4m × 16 = 8 = 1 J Loss = 3J. 6. Speed of wave in wire V = = = Maximum time period means minimum frequency ; that means fundamental mode. V f =   V = 2𝑙 = 1 2𝑙 = 35 hz  (f = 35 Hz) and; frequency of first overtone = V = 70 Hz. 𝑙 𝑙 =  7. W = – U = 1 Kx2 = 2 1 AY 𝑙2 2 L 3 V2  2 V1 9. W = Area under the curve = 2 P1V1 P2  2P1 and P1V1 = nRT1 3 . P V Therefore w nRT1 = 2 1 1 P V 1 1 10. nC T = Q  nCT = 6n RT1 dT = 4T1 – T1 = 3T1 n . C . 3T1 = 6nRT1 C = 2 R 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.) 12. If m1  0, S  0 S = Vt m2 (f  0 ) dS𝑙  V    t  m2 f  0 + S𝑙 V  t m2 f  0 0.1  0.1  0.01 102 0.01  1 1 = 10.0 10.0 1.00 102 + 1.00 50 = 8% 13. 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. 14. 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 17. (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. 18. (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. 19. 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 = 1 mv2  1 = 2 1 × 1 × v2  v = 2 m/s. (C) at t = /6 s, x = 2 sin /6 = 1m. (D) KE = 3 2  x = ± 1  1.5 = 2 × mv2  v = m.  x = ± 1m. 20. (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 v  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.