12.PRACTICE TEST-3 (Solution-Paper-1)

PAPER - 1 PART-I (Physics) → 1. f U U = – x ˆi – y → ˆ ˆ ˆj = – [6 ˆi ] + [8] ˆj = – 6 ˆi + 8 ˆj →  3ˆi  4ˆj  →   a = – 3 i + 4 j has same direction as that of u    a  2 2   | → | = 5 , | → | = 5/2 a u Since → and → are in same direction, particle will move along a straight line u a  S = 5 × 2 + 2 1 × 5 × 22 = 5 + 10 = 15 m. 15 m. Ans 2 2. The given structure can be broken into 4 parts For AB :  =  + m × d2 = m𝑙2  5m 𝑙2 ; AB = 4 3 ml2 CM m𝑙2 12 4 For BO :  = 3  For composite frame : (by symmetry)  = 2[ +  ] =  4m𝑙2 2 m𝑙2    10 = m𝑙2.] AB OB  3 3  3 3. Substituting x = 0 we have given wave y = A sin wt at x = 0 other should have y = – A sin wt equation so displacement may be zero at all the time Hence (B) is correct option. 4. In an adiabatic expansion, internal energy decreases and hence temperature decreases.  from equation of state of ideal gas PV = nRT  The product of P and V decreases. 5. Rate of radiation per unit area is proportional to (T4)  P  AT4  P  r2. dT Also ms dt dT  AT4  dt = R  1 r (because m = (v)  r3 and A  r2)  P R  r 6. Speed will be maximum when acceleration becomes zero. ie when kx = EQ  X = EQ K 1 1 By work–energy theorum : w Substituting x = EQ/K : Vmax = QE/ mk all =  KE  EQX – 2 K X2 = 2 m v2 compression will be maximum when velocity becomes zero. 1 2EQ all =  KE  EQX – 2 KX2 = 0 ; X max K 7. Voltage across each bulb will be Vi - iR = V . R (V/n) nR V V V 2 so power developed by each bulb = iV1 = nR . n P = n2R so power consumed by one bulb = n 8. (A) Magnitude of velocity is changing hence acceleration is present. (C) Velocity is changing hence it can happen by change in direction also as in a uniform circular motion. Hence acceleration is present. 9. There is no horizontal force on block A, therefore it does not move in x-direction, whereas there is net downward force (mg – N) is acting on it, making its acceleration along negative y-direction. Block B moves downward as well as in negative x-direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B, A moves in positive x-direction. Due to the component of normal exterted by C on B, it moves in negative x-direction. Mg The force acting vertically downward on block B are mg and NA(normal reaction due to block A). Hence the component of net force on block B along the inclined surface of B is greater than mg sin. Therefore the acceleration of 'B' relative to ground directed along the inclined surface of 'C' is greater than g sin  10. V at origin  0 E(r = 2 m) = K(q)r  (R2  r 2 )3 / 2 K.Q.r (R2  r 2 )3 / 2  = K.rq   1 103 / 2   23 / 2.103 / 2  = 0 From origin to r = 2, field is towards origin. 11. V = V  x = x  f = = 125 Hz = v0 = 0 R 200 100 ( X = 0  Z = R) = 2A V1 = V2 = X = .(L) = 2 × 2 × 125 × 2/ = 1000 volt Ans. 12. to 13. In the first case : From the figure it is clear that VRM is 10 m/s downwards and VM is 10 m/s towards right. In the second case : Velocity of rain as observed by man becomes  New velocity of rain VM times in magnitude. 10 3 m/s 10 m/s → → → VR'  VR'M  VM  The angle rain makes with vertical is 10 VRM tan  = 10 3 or  = 30°  Change in angle of rain = 45 – 30 = 15°. M 14. RA = R . RV < R R  RV 15. RB = R + RG > R 16. % error in case A. RA  R  RV  R R × 100 =  R  R  1 × 100 =  R  RV × 100  - 1% % error in case B RB  R × 100 = RG× 100  10% R R Hence percentage error in circuit B is more than that in A. 17. GM (2R)2 GM' = R2 M = M 4 4 4  4 3  3 R3  + 3 (8R3 – R3) = 4   . R . 1  + 7 = 4 1 7 2 = 3 18. Focal length of water–lens :  1  𝑙  m 1 2 𝑙  = + 30 cm The combination of the lens and the mirror will act like a mirror, whose equivelent focus distance will be 1 feq = 1 = 1  2 fm f𝑙 1  2 feq  5  30 feq =  15 cm 4 Image distance is given by, 1  1 = 1 V U 1  1 feq 1 = . V (15)  15 / 4 V = – 3 cm. 19. Displacement till it comes to rest V2 = u2 + 2as O2 = (10)2 + 2 (–(g sin +  gcos)) (s) 100 s = 24.8  4 m time to reach the hishest point V = u + at O = 10 + (12.4) t t = 10  0.8 sec 12.4 and after that the block will remain stationery a becouse tan  <  . So in 2 sec distance travelled is 4 m. 20. We have force per unit length , f  0i1i2 2r 0i1i2  22 = 0.1  i i = 106 We know that for a given product of two quantities their sum is a minimum when they are equal. So sum of currents is min. when i1 = i2 = 103 Now the magnetic field at a point 1m from both the wires is 0i1i2 0i1 + 21 0i2 21 = 0103 21 + 0103 21 = 4 x 10–4 T.  22 = 0.1  i i = 106 21. Speed decreasing to zero uniformly. Let a be a positive constant (acceleration) then v = a (T – t) T s =  v dt 0 2s = aTt   1 T at 2 0 = aT2 – aT2 2 = 1 aT 2 2 a = T 2 2S v = T 2 2S Ans. v = T 2 (T – t) (T – t) = 1 m/s 22. Radius of curvature is minimum at heightst point and that is v2 R = a  (ucos)2 g (10 2  1)2 = 2 g = 5 m Ans. 23. i = f mg L sin + 0 = m g (L  x) sin – mg x + 1 mv2 2 1 2 2 2 L g x2 mg 2 sin = L (L2 + x2 – 2Lx)sin – + v2 g x2 gL sin = L v = (L2 + x2 – 2Lx) sin  – g + v2 L