13.COMBINED TEST - 1-SOLUTIONS

SOLUTIONS TO COMBINED TEST - 1 (PAPER - 1) 1. 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 2R GM m  dx R GMm 3. F =    = 2 2 2R R 4. Pressure in the air inside the column of mercury is equal to the weight of mercury over the air divided by the internal cross sectional area of the tube. When the temperature increases, the 1weight of the upper part of the mercury column does not change. That is why the pressure in the air is also constant. For the isobaric process, the change in volume is proportional to the change in temperature. The same is true for the lengths of the air column. 𝑙  T  𝑙  𝑙 0 T = 11 cm. 𝑙 0 T0 T0  2  420 2 5. V = = k  T = =  k2   21  × 0.2 = 80 N ( = mass per unit length) 6. Immediately after string connected to end B is cut, the rod has tendency to rotate about point A. Torque on rod AB about axis passing through A and normal to plane of paper is - m𝑙2 3  = mg 𝑙 2   = 3g 2𝑙 Alternative Applying Newton’s law on center of mass mg – T = ma (i) Writing  = I about center of mass T 𝑙 2 m𝑙2 = 12 ....(ii) Also a = 𝑙  (iii) 2 From (i) , (ii) and (iii) 3g  = 2𝑙 7. AB = mr2 + mr 2 + 2 (2m)r 2 2 = 5 mr2 2 8. Displacement of aircraft in time t = horizontal displacement of projectile  972 × 5 t = V cos.t 18 0 54  5 1  cos = 540 = 2   = 60º . 5 9. 108 km/h = 108 × 18 = 30 m/s 30m/s Train Cart M m 10m/s v1 v2 Train Cart M m From momentum conservation M × 30 + m × 10 = Mv1 + mv2 ...(i) e × velocity of approach = velocity of separation  1 × (30 – 10) = v 2 2 – v1  10 = v2 – v1 ...(ii) On solving , v2 = M 40  m  10 M  m m/s = 40  m . 10 M 1 m M  40 m/s 10. Applying energy conservation : mgh = mLf Lf  h = = 80cal / gm = 80  4.2  1000 J/kg  336  103  N  s2 = = 33.6 km. Ans g 10m/ s2 10 m/s2  10  kg 11. Initial state is same for all three processes (say initial internal energy = E ) In the final state, V = V = V and P > P > P  P V > P V > P V  E > E > E if T > T then E > E for all three processes and hence (E – E ) < (E – E ) < (E – E )  E  < E  < E  If T < T , then E < E for all three processes and hence (E – E ) > (E – E ) > (E – E )  E  > E  > E  12. 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 13. As the object moves from A to C via B the angle between acceleration vector and velocity vector decreases from 90° and then increases back to 90°. Since the angle between velocity and acceleration is acute, the object speeds up. x As the object moves from C to A via D the angle between acceleration vector and velocity vector increases from 90° and then decreases back to 90°. Since the angle between velocity and acceleration is obtuse, the object slows down. 14. mgsin  + mg cos  = ma a = g sin  + g cos   3 4  = 10  5  5  = 14 m/sec2. If fr = mg sin 3 3mg = mg × 5 = 5 fr < fr max < fr max 3mg 5 4mg < 5 hence insect can move with constant velocity. Insect has to generate internal friction upward force or acceleration 14 m/sec2 to climb up with constant velocity. During its downwards journey, force mg sin is less than the maximum friction force mg cos, he can easily walk down with constant velocity without generating its internal force. 15. (A,C) V3kg (Juet before collision) = = 30 During willsion : 3×30 = (3+b) Vf  Vf = 10 ms–1 Amplitude after willsion : (10)2 = wN2 (AN2 – x2)  100 = 900 (An2 –1)  AN = m 9 t 4 16. Displacement =  vdt = (6  2t)dt = 2m 0 0 4 3 4 Distance = | v | dt = (6  2t)dt  (2t  6)dt = 10 m 0 0 3  Distance – displacements = 10 – 8 = 2 17. 𝑙 = xB + 3xA d2 x d2x  0 = B +3 A dt2 dt2  0 = – a + 3a B A  a = 3aA (1) For B T = maB ....................... (2) For A 3mg – 3T = 3maA (3) mg – T = maA By (1) , (2) & (3) 3mg  a = 15 2 and aA = 5 ms2 2  2a = 3 5 18. Time to go from P to A and back to A = 0.55 And time to go from P to B and back to B = 1.5 s T = Time period = 0.5 + 1.5 = 2.05 Time from O to P 0.5 = 0.5 – 2 0.5 = 2 s Let equation of SHM be x = A sin 2 .t T 2 v = A. T 2 . cos T .t  A. 2 = v =  = 3 2 = 6 m/s. T max cos 2 .t cos 2  0.5 T 2 2 19. 2(2T𝑙) cos53º = Kx 4T𝑙3 5K = x. 20. For floating condition buoyant force = weight 1  h 2 3 A .  H  1 h . dw . g = 3 (AH) dw. d . g A d = Relative d density    27 1/ 3 d h H h = H (d)1/3 = 4   h = 3 m Ans. 3  64  SOLUTIONS TO COMBINED TEST - 1 (PAPER - 2) x 3 4 → → 1. y = 4  y = 3 x and a = – (100 )2 r 2. As, → acm  0 and → = 0 →  Fext.  0  ext about any point will be same. Which is equal to zero. 3. Initially they moves in opposite direction v1 M so mv1 = mv2 As ; Ans. BCD v = m Initially means at t = 0 both have zero velocity. Ans. ABC Each block will oscillate about its mean position not about COM of system. Ans. BC 4. P3V1 < P1V3  T < TD So A & C are correct. 5. Zero error (excess reading) = 0.3 mm. observed thickness of block = 13.8 mm. Actual thickness = 13.8 – 0.3 = 13.5 mm. 1 6. Let m be mass of first particle hence 2m will be mass of stationary particle. K.E. = 2 mv2 = 3J For maximum deformation state (from conservation of linear momentum) v mv + 0 = 3m v'  v' = 3 1 Minimum kinetic energy of the system = 2 (3m)v'2 1  v 2 1  1 2  1 = (3m).    =  mv  = ×3 = 1 J 2  3  3  2  3 Maximum elastic potential energy of the system = Total K.E. – Min. K.E. = 3 – 1 = 2 J As the external force on the system is zero, hence its momentum is conserved at every instant. As the collision is elastic, hence total energy will also be conserv ed at every instant. Ratio of K.E. to P.E. of the system first decreases, as the K.E. decreases & P.E. increases upto maximum deformation state after that K.E. increases & P.E. decreases hence, the ratio of K.E. to P.E. then increases. 7. 20  3 / 5 t1 = 10 (12)2 = 1.2 sec. H = 2 10 = 7.2 m = r 1.6  16  16 F = 7.2 512 = 9 N For circle t = 9. 9 (A), 10 (A) 3.14  7.2 2 16 < 1.2 sec. 0 x a  v.dv = (g sin –  x g cos)   vdv  (gsin  – 0xgcos) dx dx 0 x0 0 0 Mgcostan2  H = 2 0xMgcos.dx = 0 20 11. Heat given : Q  n1CV1T  For gas A [As V = constant  dW = 0] & for Gas B & Q = n2CV2 T ( For same heat given, temperature rises by same value for both the gases.)  n1CV  n2CV ................(1) Also, (PB)V = n2RT and (PA)V = n1RT n1  n2 = PA PB = 2.5 5 1.5 = 3  n1 = 5 3 n2 Substituting in (1) 5 Cv 5 ( 5 R) 3 n2 CV = n2Cv  2  C 3 1  2 ( 3 R) Hence, Gas B is diatomic and Gas A is monoatomic. 5 125 5  60  Since n1 = n2 Therefore =   3 MA 3  MB  (From experiment 1 : WA = 125 gm & wB = 60 gm)  5MB = 4MA The above relation holds for the pair–Gas A : Ar and Gas B : O2 . 125 12. No. of molecules in 'A' = nNA = 40 125 NA = 3.125 NA (Since n = 40 for Ar) 14. 13 (A),14 (C) y (x,t) = (0.1)sin (4t + 8x +  ) At x = 2, t = 1sec, y = 0 dy = 0.4 m/s   = 0 dt 16. A from tB0 ls x T = y a d2x   y x dt 2 am t B0 simply tB0 other ext. = blh çdkj tB0 vU; çlkj ds fy, = Timeperiod vkorZdky = s m (  Energy conservation ÅtkZlaj{k.kls yb2  yc2 2a 2b c = b Distancenwjh  b (ext. to ext.) (çlkjlsçlkj) a  b + c = b 17. A C A 18. Ans : B 19. s (A) F = 0 N = 1200 N (B) F = 100 N N = 1100 N (C) F = 500 N N = 700 N (D) F = 700 N N = 500 N 20. (A) At centre of thin spherical shell V  0, E = 0. (B) At centre of solid sphere V  0 , E = 0. (C) At centre of spherical cavity inside solid sphere V  0, E  0. (D) At centre of two point masses V  0, E=0.