Title 01. Units and Dimensions Med Ans.pdf ✅

1. (b) Solid angle, 2 2 2 (5cm) 1cm r dA d = = = 0.04 steradian = 2 4 10−  steradian 2. (c) The SI system of units was developed and recommended by General Conference on Weights and Measures in 1971. 3. (d) Amongs the given physical quantities work is not a fundamental quantity whereas all other three physical quantities are fundamental quantities. 4. (d) Pressure gradient Distan ce Pressure =  The SI units of pressure is 2 Nm− and distances is m.  The SI units of pressure gradient is 3 Nm− . 5. (c) Relative density Relative density Densityof water Densityof a substan ce =  Density of lead = 3 3 11.3 1 10 kgm−   = 3 3 11.3 10 kgm−  = 4 3 1.12 10 kgm−  6. (c) Unit of time (ie., second) is same in all the three ystem of units. 7. (d) Calorie, kilowatt hour, joule all are the units of energy whereas watt is the units of power. 8. (d) Amongs the given units pascal is the derived units whereases others are the fundamental or base units. 9. (d) SI is not based on uits of mass, length and times alone. 10. (a) Physical quanity Names of the units Conductance Siemens Magnetics inductions Tesla Absorbed does Gray Luminous flux lumen 11. (a) The SI units of mass is kilogram in which ‘kilo’ is the prefix. 12. (c) The speed of sound in water is 1 1450ms− . 13. (d) The SI units is based on seven units. These are metre, kilogram, seconds, ampere, Kelvin, mole and candela. 14. (b) Form figure, =     = = =  2 180 30 1 rad 30 1 60R 2R E E  Diameter of the earth as seen from the moon is about 2 . 15. (b) Distances between earth and sun – Light year interatomic distance in a solid – Angstrom Size of a nucleus – Fermi Wavelength of infrared laser – Micron 16. (c) 1 MSD = 0.5 mm 50 VSD = 49 MSD 1 VSD = MSD 50 49 Least count = 1 MSD 1 VSD MSD 50 1 MSD 50 49 =1MSD − = 0.5mm 0.01mm 50 1 =  = 17. (c) Light year is the distances travelled by light in free space in one year. 18. (a) Here, Distance of the sun from the earth, d = 11 1.49610 m Sun’s angular diameter,  =1920''  1'' 4.85 10 rad −6 =   1920'' 4.85 10 rad −6  =   9.31 10 rad −3 =  Sun’s diameter, D = d D (9.31 10 )(1.496 10 )m 3 11 =   − 1.39 10 m 9 =  19. (a) 1 micron ( 1m ) = 10 m −6 1 nanometre (1nm) = 10 m −9 Their ratio is 3 9 6 10 10 m 10 m 1nanometre 1micron = = − − 20. (b) 1 barn = 100 2 fm = 28 2 10 m − ( 1fm 10 m −15  = ) 21. (d) Alpha centauri is 4.29 light years away from the earth. 22. (a) Vernier calipers, screw gauge and sphereometer are used for the measurement of length whereas atomic clock is used for the measurement of time. 23. (c) Here, Speed of light in vacuum,
C = 1 new unit of length 1 s − Time taken by light of sun to reach the earth t = 8 min and 20 seconds = (860+ 20)s = 500s  Distance between the sun and the earth S = ct = 1 new unit of length s 500s 1  − = 500 new unit of length 24. (b) Radius of the atom , R 1A 10 m 10 a − =  = Volume of the atoms 3 10 3 3 a a (10 ) m 3 4 R 3 4 V − =  =  Redius of the nucleus, Rn =1 Fermi = 10 m −15 Their corresponding ratio is 15 45 30 15 3 10 3 n a 10 10 10 (10 ) (10 ) V V = = = − − − − 25. (c) One mole of an ideal gas at standard temperature and pressure (STP) occupies 22.4 liter. 26. (a) 1 light year 9.46 10 m 15 =  (Distance travelled by light in 1 year) 27. (d) Here, ,r 30cm 6 =   = As, l r r l  =  =  30cm 15.7cm 6 3.14 30cm 6  =  =  = 28. (c) 1 mm = 10 m −3 1A 10 m −10 =  1fm 10 m −15 = 29. (b) Among the given quantities displacement gradient is a nutiless quantity. 30. (b) The number of wavelength of 86 Kr in 1m is 1650763.73. 31. (d) The most precise device is one whose least count is the least. (a) Least count of vernier calipers MSD 20 19 =1MSD −1VSD =1MSD − cm 0.005cm 200 1 mm 20 1 MSD 20 1 = = = = ( 1MSD =1mm ) (b) Least count of screw gauge No.of division oncircularscale Pitch = cm 0.001cm 1000 1 mm 100 1 = = = (c) Least count of spherometer No.of divisions oncircularscale Pitch = mm 0.0001cm 1000 1 100 0.1mm = = = (d) Wavelength of light , 5 10−   cm= 0.00001cm Clearly the optical instrument is the most precise. 32. (a) Volume occupied by 1 mole of an ideal gas at STP is known as molar volume.  Molar volume = 22.4 liter 3 3 22.4 10 m − =  Radius of hydrogen atom is r 0.5A 0.5 10 m −10 =   =  Atomic volume A 3 r N 3 4 =  10 3 23 (0.5 10 ) 6.023 10 7 22 3 4 =      − 7 3 3.15 10 m − =  Their corresponding ratio is 4 7 3 3 3 7.1 10 3.15 10 m 22.4 10 m Atomic volume Molar volume =    = − − 33. (b) Fathom is a unit of length equal to six feet. It is used to measure depth of water. 34. (a) By definition of parsec 1 parsec 1arc second IAU = 1 = 3600arc second rad 180 1   =  1 arc second rad 3600180  =  1 parsec AU 3600 180   = 206265AU 2 10 AU 5 =   35. (c) A SONAR uses ultrasonic waves to detect and locate objects under water. 36. (d) 1 micron = 10 m −6  10 m −6 space is occupied by 1 bacteria  1 m space will be occupied by 10 m −6 bacteria 6 10 bacteria =1 million bacteria
37. (d) A laser is a source of very intense, monochromatic and unidirectional beam of light. These properties of a laser light can be exploited to measure long distances. 38. (b) 39. (c) The order of magnitude of the diameter of the earth is 7. 40. (a) 1 AU = 1.496 10 m 1.496 10 m 11 8  =  41. (c) Angstrom, Fermi and parsec are the units of length whereas barn is the units of nuclear cross-sections. 42. (c) Here, F = 100 N,L = 10 m, T = 100s Dimensional formula of force is [MLT ]. −2 10 kg (10m)(100s) 100N LT F M 5 2 2  = = = − − 43. (c) The device used for measuring the mass of atoms and molecules is mass spectrograph. 44. (b) 1 milligram (mg) = 3 10− gram (g) 45. (d) Mass does not depend on the temperature, pressure or location of the object in space. 46. (b) 1u 1.66 10 kg −27 =  47. (a) Life time of an excited state of an atom : 10 s −8 Average human life – span : 10 s 9 Age of Egyptian pyramides : 10 s 11 Age of the universe : 10 s 17 48. (c) Times taken by light to reach the earth from the sun is T = 8 min 20s=8×60s + 20s = 480s+ 20s = 500s 49. (d) The second is the duration of 9192631770 periods of the radiations corresponding to the transitions between the two hyperfine levels of the ground state of the cesium – 133 atoms. 50. (d) The magnitude of the difference between the true value of the quantity and the individual measurement value is called the absolute error of the measurement. Hence, options (d) is an incorrect statement while all other statement are correct. 51. (b) An atomic clock is the most precise time measuring device because atomic oscillation are repeated with a precision of 1 s in 10 s. 13 52. (c) Screw gauge has minimum least count of 0.001cm. Hence, it is most precise instrument. 53. (b) As per rule of significant figures, 4 4.800010 has 5 significant figures and 48000.50has 7 significant figures. 54. (a) Here Length of the cube , L 1.2 10 m −2 =  Volume of the cube, 2 3 V (1.2 10 m) − =  6 3 1.728 10 m − =  As the result can have only two significant figures therefore, on rounding off, we get 6 3 V 1.7 10 m − =  55. (d) The power of 10 is irrelevant to the determinations of significant figures. 56. (a) The mean period of oscillations of the determinations of significant figures. n T T n i 1 i mean = = s 5 (2.63 2.56 2.42 2.741) Tmean + + + = s 2.624s 2.62s 5 13.12 = = = (Rounded off to two decimal places The absolute errors in the measurement are T 2.62s 2.63s 0.01s  1 = − − T 2.62s 2.56s 0.06s  2 = − = T 2.62s 2.42s 0.20s  3 = − = T 2.62s 2.71s 0.09s  4 = − = − T 2.62s 2.80 0.18s  5 = − = − Means absolute error is n | T | T n i 1 i mean =   = 5 (0.01 0.06 0.20 0.09 0.18) Tmena + + + +  = s 0.11s 5 0.54 = = 57. (d) Among the given physical quantities angle has a unit but no dimensions. Angle = [M L T ] 0 0 0 The SI units of angle is radian. 58. (a) Physical quantities having different dimensions cannot be added or subtracted. As P, Q and R are physical quantities having different dimensions, therefore they can neither be added nor be subtracted. Thus, (a) can never a meaningful quantity. 59. (d) [E] [ML T ],[M] [M] 2 2 = = −
[l] [ML T ],[G] [M L T ] 2 −1 −1 3 −2 = = [M ][M L T ] [ML T ][M L T ] m G Et 5 2 6 4 2 2 2 4 2 5 2 2 − − − −  =       [M L T ] 0 0 0 = As angle has no dimensions, therefore 5 2 2 m G El has the same dimensions as that of angle. 60. (b) Relative density is the ratio of two like quantities therefore, it ahs neither unit nor dimensions. 61. (c) 62. (a) Pressure [ML T ] [L ] [MLT ] Area Force 1 2 2 2 − − − = = = [ML T ] [L ] [ML T ] Volume Energy 1 2 3 2 2 − − − = = [ML T ] [L ] [ML T ] Volume Energy 0 2 2 2 2 − − = = [ML T ] [L ] [MLT ] Area Force 2 2 3 2 − − − = = [ML T ] [L ] [MLT ] Volume Momentum 2 1 3 1 − − − = = Out of the four given ratios only energy/volume express the pressure. 63. (d) Energy of a photon, E = h Where h is the Planck’s constant and  is the frequency. [ML T ] [T ] [ML T ] [ ] [E] [h] 2 1 1 2 2 − − − − = =   = Angular momentum = moment of inertia × Angualr velocity [Angular momentum ] = [ML ][T ] [ML T ] 2 −1 2 −1 = 64. (c) (A) Pa s = [ML T ][T] [ML T ] −1 −2 −1 −1 = A – r (B) [ML T K ] [K] [MLT ][L] NmK 2 2 1 2 1 − − − − = = B – s (C) [M L T K ] [M][K] [ML T ] J kg K 0 2 2 1 2 2 1 1 − − − − − = = C – p (D) [MLT K ] [L][K] [ML T ] Wm K 3 1 2 3 1 1 − − − − − = = D – q 65. (a) kg [AT][A] [ML T ] [A T] [ML T ] m s A [ML T A ] 2 2 2 2 2 2 3 2 2 3 2 − − − − − = = = Resistan ce Current Voltage Ch arge current Work = =  = 66. (a) Power Time Work = [ML T ] [T] [ML T ] [P] 2 3 2 2 − − = = Surface tension Length Force = [ML T ] [L] [MLT ] [S] 0 2 2 − − = = Planck’s constant Frequency Energy = [MT T ] [T ] [ML T ] [h] 2 1 1 2 2 − − − = = The ascending order of dimensions of time in their dimensional formulae is P,S,h. 67. (d) Dipole moment = charge × distance Electric flux = electric field × area  Dipole moment, electric flux and electric field have different dimensions. 68. (c) Permittivity of free space 2 4 electricalforce distan ce Ch arg e ch arg e    = A – s Radiant flux [ML T ] [T] [ML T ] Time Energy emitted 2 3 2 2 − − = = = B – r Resistivity Length Re sistan ce Area = [ML T A ] [L] [ML T A ][L ] [ ] 3 3 2 2 3 2 2 − − −  = = − C – q Hubble constant Distan ce Re cessionspeed = [M L T ] [L] [LT ] 0 0 1 1 − − = = D – p 69. (c) A + B = 2.331 + 2.1 = 4.431 = 4.4 cm Since B has 2 significant figures  A + B must have only 2 significant figures 70. (b) Joule is a units of energy SI New system n 5 1 = n ? 2 = M1 = 1lg M kg 2 =  L1 = 1m L2 = m T 1s 1 = T s 2 =  Dimensional formula of energy is [ML T ] 2 −2 . Comparing with [M L T ], we get a b c