Title Physics KCET Solution.pdf ✅

Physics - CET 3 6. (c) Angular frequency 2 [ ] −1 =  f = T a = 0, b = 0, c = −1. 7. (d) Velocityof lightin vaccum 3 10 / . 8 =  m s 8. (c) dx dv F = −A [ ] [ ] −1 −1   = ML T Aliter : F = 6rv [ ] [ ]. −1 −1  = ML T 9. (a) = − = → − T f ML ML T 1 2 2 1 frequency. 10. (c) Angular momentum [ ] 2 −1 = ML T frequency [ ] −1 = T Both have different dimensions. 11. (a) [ ] [ ][ ] 2 3 1 A ML T A T L dt di e L − − =  = [ ] 2 −2 −2 = ML T A Aliter : 2 2 1 Energy= Li  [ ] [ ]. 2 −2 −2 L = ML T A 12. (a) Mass Heatenergy =  = = m Q Q mL L [ ]. [ ] [ ] [ ] 2 2 2 2 − − = = L T M ML T L 13. (c) (a) qI W I V Resistance(R) = =   [ ] [ ][ ] [ ] 2 3 2 2 2 − − − = = ML T A AT A ML T R (b) Resistivity()→m  = Rl [ ] [ ] 3 −3 −2  = MLT A (c) Electrical conductivity   1 ( ) =     1 3 3 2 M L T A − −  = (d) Electromotive force q W (emf)= Volt = [ ] [ ]. 2 −3 −1 emf = ML T A 14. (a) bx a t P 2 − = [ ] [ ] 2 a = T [ ] [ ] 1 2 2 ML T L b a P bx a bx T − − =       =       =         [ ]. −2 = MT 15. (a) As [ ] [ ] 2 −2 −1  = MLT A and [ ] [ ] −2 −2  = MLT A  [ ] [ ] [ ] [ ] [ ] 0 1 0 1 2 2 2 2 1 M LT A MLT A MLT A = = − − − −   16. (c) Here, V = (100  5)V and I = (10  0.2)A As I V R =  I I V V R R  +  =  The percentage error in R is 100 100       +   =  I I V V R R 100 10 0.2 100 5         = + A A V V = (0.05+0.02)100 = 7%. 17. (b) Physical quantities of different dimensions can not be subtracted. 18. (b) Given : mass m= (0.30.003)g , radius r = (0.50.005)mm and length l = (6 + 0.06)cm Density, r l m V m 2   = =  100 100 2 100 100   +   +   =  l l r r m m   100 6 0.06 100 0.5 0.005 100 2 0.3 0.003 =  +   +  = 1+ 2+1 = 4% . 19. (d) Unit of magnetic flux, () is Weber and of electric current, (I) is ampere. So, quantity corresponding to Weber/ampere will be proportional to I  . It is equal to the mutual inductance and self inductance both. i.e., M = L =  / I . 20. (a) 1 MSD = 0.5mm , 50VSD = 49 MSD 1 VSD 50 49 = MSD Least count = 1 MSD – 1 VSD = 1 MSD 50 49 − MSD 50 1 = MSD 0.5mm 50 1 =  = 0.01mm
4 Physics - CET 21. (b) True value = 3.678cm Accuracy of a measurement is a measure of how close the measured value is to the true value of the length. Here, measured value i.e., 3.5cm is more closer to true value (3.678cm) as compared to 3.38cm . Hence 3.5cm measured by instrument A is more accurate. Now, precision tells us to what resolution the value is calculated. The more number of digits after decimal, the more is the precision of the measurements. Here, 3.38cm has 2 digits after the decimal as compared to 3.5cm which has only 1 digit after the decimal. Hence, the 3.38cm measured by the instrument B is more precise measurement. 22. (a) 4 (Area) time (Temperature) Energy Stefan's constant    = [ ]. [ ][ ][ ] [ ] [ ] 3 4 2 4 2 2 − − − = = MT K L T K ML T  23. (b) Percentage error in x is given by c c b b a a x x  +  +  =  2 2 100 = 22%+ 23%+ 4% =14%  x x . 24. (a) Moment of force is known as torque. Thus, dimensions of force and torque are same. Dimension of torque = Dimension of moment of force [ ] 2 −2 = ML T Dimension of moment of force [ ] 2 −2 = ML T . 25. (b) Gyromagnetic ratio of particle or a system is defined as the ratio of its magnetic moment to its angular momentum. Angular momentum Magnetic moment   = or   [ ] [ ] [ ] 0 1 1 1 2 1 2 L M T I L MT L I − −  = = . 26. (a) Given, force, d y F = where F = force and d = density   2 3 1/2 [ ] [ ][ ] − −  y = F d = MLT ML [ ] 3 / 2 −1/ 2 −2 = M L T 27. (c) We know that, di dt M dt di M dt d   = − = −  = (−)  Dimension of mutual inductance is [ ] [ ] [ ] [ ] 1 2 2 2 2 3 1 − − − − =  = M L T I I ML T I T M 28. (c) Ratio of the dimensions of Planck's constant to that of moment of inertia is, = = = − − [ ] [ ] [ ] [ ] [ ] 1 2 2 1 T ML ML T I h Dimension of frequency. 29. (a) 2 2 1 (1 )(1 ) (1 sec)− J = kg m 2 2 1 ( )( ) ( sec)−  = xkg ym z 2 2 2 2 ( ) 1 1 1 1 xy z z x y J =                =   2 2 1 xy z J = or 2 2 1 4.18 xy z cal = 30. (c) In C.G.S Force = 100 dyne 2 2 1 1 2 1 1 2 1 2 1 −                   = T T L L M M n n 1 1 2 60sec sec 1000 100 100 −                   = cm cm g g 3.6 100000 100 3600 =  = . 31. (c) [ ] 2 −3 −2 ohm = ML T A [ ]. [ ] [ ] 2 3 2 2 2 2 1 2 − − − = =       ML T A A T ML T e h 32. (c) = time constant R L . / / t V I dI dt EMF R L = = 33. (c) Impulse = Change in momentum = F t. 34. (a) [ ] [ ] Time Work done Power 2 −3 =  P = ML T 35. (b) ( ) [ ] [ ]. 2 −1 L = mr v  L = ML T   36. (b) Impulse changein momentum [ ]. −1 = =mv = MLT 37. (d) [ ] [angular momentum] ]. −1 h = = [ML T 2 38. (d) [Stress] =[Pressure]=[Young's Modulus] Pressure [ ]. −1 −2 = = ML T A F 39. (c) Torque( ) [ ] 2 −2  = ML T Angular momentum( ) [ ] 2 −1 L = ML T . dt dL  = 40. (d) = = =  − − [ ] [ ][ ] [ ] Mass Length Energy 2 2 2 LT M L ML T Acceleration. 41. (a) l NI Magnetic fieldintensity(H) = [ ] [ ]. −1 H = AL Level 2