14.COMBINED TEST - 2-SOLUTIONS

SOLUTIONS TO COMBINED TEST - 2 (PAPER - 1)  2a  a  mg  a  1. KA ×  3 2  =  2      3mg A = K 2. The FBD of blocks is as shown From Newton's second law 4mg – 2T cos = 4 mA (1) and T – mg = ma (2) 4 cos = 5 and from constraint we get a = A cos (3) Solving equation (1), (2) and (3) we get acceleration of block of mass 4m, a = 5g downwards. 11 3. Let the density of water be , then the force by escaping liquid on container = 2Sgh Vg  2Sh   g  acceleration of container a = =   V  V  Now  = Sh  a = V Shg V 4. As the temperature of water is increased from 2°C to 3°C the density of water increases (remember anamolous behaviour of water), also the volume of sphere increases. Therefore bouyant force on sphere due to water shall increase. 5. Apparent frequency of S1 and S2 heard by observer is f1 = v f v  u and f2 = v f v  u  Beat = f – f2 = 2uv f v2  u2 6. When the particle crosses D, its speed is half the maximum speed.  v = or vmax 2 vmax R  R or x = 2 R  Distance BD = 2x = 3 R 7. (B) The two are in series. KA = 3KB i.e. thermal resistance of A (say RA) is one third that of B (say RB) i.e. RB =3 RA  Thermal current = 36  9 4RA RA  9   Temperature difference across 'A' =   .R = 9°C Ans. A  A 8. (Moderate) Velocity of A and velocity of B relative to A are as shown.  VB  VBA  VA  V = and direction of = 5 VB is away from VA . 9. For block to be in equilibrium : F = N + mg cos 37° f = mg sin 37° FBD : For F to be minimum, For no sliding f < µN  mg sin 37° < 1 [F – mg cos 37°] 2  2mg sin 37º + mg cos 37º < F  F  2mg × 3 5 + mg × 4  F  2mg i.e. F > 200 N 5 10. 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 11. The mechanical strain  𝑙 MCQ = 𝑙 =  T = 1.21 × 10–5 × 20 = 2.42 × 10–5 The tension in wire  𝑙 = T = Y 𝑙 A = 2 × 10 × 2.42 × 10 × 10 = 48.4 N 11 –5 –6  speed of wave in wire V = Since the wire is plucked at 𝑙 4 = from one end = 22 m/s The wire shall oscillate in 1st overtone (for minimum number of loops)  = 𝑙 = 1m 12. V Now V = f  or f =  (After collision) v = vA + vB (1) = 22 Hz. 1 = vB – v A 2 v vB – vA = v (2) 2 v = 3v B 4 vA = v 4 m  3v – 0  3mv B =  4  =  4 1 1  v 2  3v 2  3 loss in K.E. = mv2 = m  4      4 mv2. 13. (AB) x = 2 ab2 c3 2      16 lnx = lna + 2lnb –3lnc 1 dx = x 1 da + a 2 db – b 3dc × 100 c dx × 100 = x da × 100 + a 2db × 100 – b 3dc × 100 c = 1 + 2 × 3 – 3 (–2) Maximum error =  13 % 14. W = – U = 1 Kx2 = 2 1 AY 𝑙2 2 L 15. 10 = 2t t = 5 H = ut + 1 at2 = 5 × 5 + 1 2 2 (–10) 52 = 25 – 125 = 100 m Integer Answer Type 16. x = 4 Sol. N = mg f = ma As f must be static friction (No slip condition) f  N  ma   mg or mao  mg  mA2  mg      = 2  T  T  2 42A   gT2 17. 05 Sol. (T4 – T 4) . (6a2) t = (d . a3) s.T  t = das T 6.(T 4  T 4 ) = 4.8  103  0.9  2.0 103  5 6  6  108  (4004  2004 ) = 5000 s. X = 5 18. 5 Sol. From energy conservation,  3 GM  1   2 R  m0 =  m V2 2 V = = V = 5 × 104 m/sec. 19. 3 Sol. Restoring torque = mgx x = CP cos  = CP [a is very small] x = R  mgR = l = mL2  12  12 gR    L2  T  2   2𝑙 d𝑙 20. As ; d𝑙 =  𝑙 dT  𝑙 𝑙 T = aT dT 0 T2  𝑙n4 1/ 2 𝑙n2 = a 2  T =  a  = 2 SOLUTIONS TO COMBINED TEST - 2 (PAPER - 2) 1. Speed of wave in wire V = T A = = Maximum time period means minimum frequency ; that means fundamental mode. V f =   V = 2𝑙 = 1 2𝑙 = 35 hz 𝑙 =   (f = 35 Hz) and; frequency of first overtone = V = 70 Hz. 𝑙 2. Frequency of horn directly heard by observer v  v0 f c Frequency of echo = v f v  vc Frequency of echo of horn as heard by observer. v f. v  v0    v  vc  v  Frquency of Beats :  1  1  2vc (v  v0 ) f = (v + v0) f v  v v  v  = (v2  v2 ) 3. Area under the curve is equal to number of molecules of the gas sample. Hence N = 1 . a . V 2 0   aV0 = 2N V0  a  2 V V = 1 v N(V)dV = 1 C . .V dV = V  avg 2 = avg N 0 N 0  v0  3 0 V0 3  v0  a  2 V 1 2 r ms = 1 V 2 N(V)dV N 0 = 1 V 2  N 0  V0 .V  dV = V0  2 rms V0 = Area under the curve from 0.5 V0 to V0 is 3 of total area. 4 4. Given v = x & t = 0 v = 1, x = 1 a  dv  dx dt dt & v. dv  x dx  a = v = x v v  v.dv   x.dx 1 1 v 2 v  x 2 v  v = x dx  x dt v dx v  x   dt  𝑙nx  t 1 0 x  et 5. Be equation of trajectory (y = x tan  y  x tan 1 x    R  4  3 tan 1 3     ...(i)  R    3  4 tan 1 4  ...(ii)  R    dividing (i) & (ii) 1 3  4 3  R   4   3    3 4 1 4   16 1 R   91     7  64  27 R R R  37 7 6. T = 2kx 5kx = mg mg 110 1 x = 5k = 5 10 = 5 (ii) (1) (g) (x) = 1 (1)(v)2  1 (1)(2v)2 x  1 (10)(0.2)2  1 (10)(0.4)2 2 2 2 2 V2 2 = (1 + 4) + (0.4 + 1.6) 2 V = 7. VB Clearly , V = 6 m/s  By string constraint, • • • • x1 x2  x3  x 4  0  –6 + VB + VB + VB = 0  VB = 2 m/s  Vc  VD C D Also, 2  VB VC VD  VD = – 2 m/s  V = 2 m/s  Thus   6  2  8 m / s VB   4 m / s   4 m / s   6  2  4 m / s 8. (AB) Sol. Using trajectory equation 10  20  20 10 = 20 tan (90 – ) – cot2  – 4 cot + 3 = 0 cot  = 1, 3 2  400  cos2(90  )  1  so any angle between, tan–1  3  and tan–1 (1)   1 tan = 3 , 1 9. From conservation of momentum mv = mv' cos30° + mv' cos30°  v' = v  2cos 30 10. Loss in kinetic energy 1 1  v 2 1 = mv2 – 2 × m    mv 2 2 2  3  6 11.  = 4m and f = 500 Hz.  V = f = 200 m/s  V =  T =  v2 = (0.1) × (200)2 = 400 N 12. Since integral number of waves shall cross a point is 5 seconds, therefore power transmitted in 5 seconds is =

× 5 = 22 f2 A2  v × 5 = 2 × 2 × (50)2 × (2 × 10–3)2 × (0.01) × 200 × 5 = 2 5 13. K.E. = 1 2 axis2 = 1 [m𝑙2sin2 + m𝑙2sin2]2 = m𝑙22sin2 . 2 Alternate Solution: K.E. = 2 × 1 × m ( 𝑙sin)2 2 14. Torque about C is :  = 𝑙Fsin + 𝑙F sin = 2𝑙Fsin (90º – ) = 2𝑙(m2r)sin(90º – ) = 2𝑙(m2 𝑙sin) sin(90º – ). = 2m2𝑙2sin cos 15. B Sol. Let V = volume of block ma = FB – mg 5  Va   4 – 3h vg – 5  Vg 2 0 0  h  2 0  a = – 6g 5h0 0  h + 3 g 5 16. A Sol. This is SHM equation 6g 2 = 5h0 T = 2 Matrix - Match Type 17. (B) No work is done during process 4 for because volume is not changing. Work done is maximum in process 1 one because area of PV graph is maximum. For process 1 and 2 temperature will increase because PV is increasing. 18. (C) 19. (D) Sol. Let f1 and f2 be the fundamental frequencies of P1 and P2. 3f1 = 4f2 = 12Hz  f1 = 4Hz & f2 = 3Hz. (A) |5f1 – 4f2| = |20 – 12| = 8 (r) (B) (C) (D) 3f1  3f2 9f1  8f2 5f1  5f2  12  9  3  36  24  12  20 15  5 (q) (s) (t) 20. (A) Sol. ƒ = 2v0 ƒ0 v  vs = 2 10 300 × 6000 = 400 Hz ƒ = 0 ƒ = 2VVs V2  VS2 = 2  350  50  6000 400  300 = 1750 Hz ƒ = ƒ = 4ƒ  ƒ = 3 ƒ = 2000 Hz. 3