Title 2. Units and Measurements .pdf ✅

2 2.2 The International System of Units 1. The unit of thermal conductivity is (a) W m–1 K–1 (b) J m K–1 (c) J m–1 K–1 (d) W m K–1 (NEET 2019) 2. The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are (a) kg m s–1 (b) kg m s–2 (c) kg s–1 (d) kg s (2012) 3. The unit of permittivity of free space, e0, is (a) coulomb/newton-metre (b) newton-metre2 /coulomb2 (c) coulomb2 /newton-metre2 (d) coulomb2 /(newton-metre)2 (2004) 2.6 Accuracy, Precision of Instruments and Errors in Measurement 4. A screw gauge has least count of 0.01 mm and there are 50 divisions in its circular scale. The pitch of the screw gauge is (a) 0.01 mm (b) 0.25 mm (c) 0.5 mm (d) 1.0 mm (NEET 2020) 5. In an experiment, the percentage of error occurred in the measurement of physical quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Then the maximum percentage of error in the measurement X, where X A B C D = 2 1 2 1 3 3 / / , will be (a) 10% (b) (3/13)% (c) 16% (d) –10% (NEET 2019) 6. The main scale of a vernier callipers has n divisions/cm. n divisions of the vernier scale coincide with (n–1) divisions of main scale. The least count of the vernier callipers is (a) 1 (n + 1) (n − 1) cm (b) 1 n cm (c) 1 2 n cm (d) 1 n n( + 1) cm (Odisha NEET 2019) 7. A student measured the diameter of a small steel ball using a screw gauge of least count 0.001 cm. The main scale reading is 5 mm and zero of circular scale division coincides with 25 divisions above the reference level. If screw gauge has a zero error of –0.004 cm, the correct diameter of the ball is (a) 0.521 cm (b) 0.525 cm (c) 0.053 cm (d) 0.529 cm (NEET 2018) 8. In an experiment, four quantities a, b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity P is calculated as follows P a b cd = 3 2 . % error in P is (a) 7% (b) 4% (c) 14% (d) 10% (NEET 2013) 9. A student measures the distance traversed in free fall of a body, initially at rest, in a given time. He uses this data to estimate g, the acceleration due to gravity. If the maximum percentage errors in measurement of the distance and the time are e1 and e2 respectively, the percentage error in the estimation of g is (a) e2 – e1 (b) e1 + 2e2 (c) e1 + e2 (d) e1 – 2e2 (Mains 2010) 10. If the error in the measurement of radius of a sphere is 2%, then the error in the determination of volume of the sphere will be (a) 8% (b) 2% (c) 4% (d) 6% (2008) 11. The density of a cube is measured by measuring its mass and length of its sides. If the maximum error in the measurement of mass and lengths are 3% and 2% respectively, the maximum error in the measurement of density would be (a) 12% (b) 14% (c) 7% (d) 9%. (1996) 12. Percentage errors in the measurement of mass and speed are 2% and 3% respectively. The error in the estimate of kinetic energy obtained by measuring mass and speed will be (a) 8% (b) 2% (c) 12% (d) 10%. (1995) Units and Measurements 2 CHAPTER EduHulk.COM
Units and Measurements 3 13. A certain body weighs 22.42 g and has a measured volume of 4.7cc. The possible error in the measurement of mass and volume are 0.01 g and 0.1 cc. Then maximum error in the density will be (a) 22% (b) 2% (c) 0.2% (d) 0.02%. (1991) 2.7 Significant Figures 14. Taking into account of the significant figures, what is the value of 9.99 m – 0.0099 m? (a) 9.9801 m (b) 9.98 m (c) 9.980 m (d) 9.9 m (NEET 2020) 2.8 Dimensions of Physical Quantities 15. Dimensions of stress are (a) [MLT–2] (b) [ML2 T–2] (c) [ML0 T–2] (d) [ML–1 T–2] (NEET 2020) 16. The pair of quantities having same dimensions is (a) Impulse and Surface Tension (b) Angular momentum and Work (c) Work and Torque (d) Young’s modulus and Energy (Karnataka NEET 2013) 17. The dimensions of (m0e0)–1/2 are (a) [L1/2T–1/2] (b) [L–1T] (c) [LT–1] (d) [L1/2T1/2] (Mains 2012, 2011) 18. The dimension of 1 2 0 2 ε E ,where e0 is permittivity of free space and E is electric field, is (a) ML2 T–2 (b) ML–1T–2 (c) ML2 T–1 (d) MLT–1 (2010) 19. If the dimensions of a physical quantity are given by Ma Lb Tc , then the physical quantity will be (a) velocity if a = 1, b = 0, c = – 1 (b) acceleration if a = 1, b = 1, c = – 2 (c) force if a = 0, b = – 1, c = – 2 (d) pressure if a = 1, b = –1, c = –2 (2009) 20. Which two of the following five physical parameters have the same dimensions ? 1. energy density 2. refractive index 3. dielectric constant 4. Young’s modulus 5. magnetic field (a) 1 and 4 (b) 1 and 5 (c) 2 and 4 (d) 3 and 5 (2008) 21. Dimensions of resistance in an electrical circuit, in terms of dimension of mass M, of length L, of time T and of current I, would be (a) [ML2 T–2] (b) [ML2 T–1I–1] (c) [ML2 T–3I–2] (d) [ML2 T–3I–1] (2007) 22. The ratio of the dimensions of Planck’s constant and that of moment of inertia is the dimensions of (a) time (b) frequency (c) angular momentum (d) velocity. (2005) 23. The dimensions of universal gravitational constant are (a) [M–1L3 T–2] (b) [ML2 T–1] (c) [M–2L3 T–2] (d) [M–2L2 T–1] (2004,1992) 24. The dimensions of Planck’s constant equals to that of (a) energy (b) momentum (c) angular momentum (d) power. (2001) 25. Which pair do not have equal dimensions ? (a) Energy and torque (b) Force and impulse (c) Angular momentum and Planck’s constant (d) Elastic modulus and pressure. (2000) 26. The dimensions of impulse are equal to that of (a) pressure (b) linear momentum (c) force (d) angular momentum (1996) 27. Which of the following dimensions will be the same as that of time? (a) L R (b) C L (c) LC (d) R L (1996) 28. The dimensions of RC is (a) square of time (b) square of inverse time (c) time (d) inverse time. (1995) 29. Which of the following has the dimensions of pressure? (a) [MLT–2] (b) [ML–1T–2] (c) [ML–2T–2] (d) [M–1L–1] (1994, 1990) 30. Of the following quantities, which one has dimensions different from the remaining three ? (a) Energy per unit volume (b) Force per unit area (c) Product of voltage and charge per unit volume (d) Angular momentum. (1989) 2.9 Dimensional Formulae and Dimensional Equations 31. The dimensional formula of magnetic flux is (a) [M0 L–2T–2A–2] (b) ML0 T–2A–2] (c) [ML2 T–2A–1] (d) [ML2 T–1A3 ] (1999) 32. The dimensional formula of permeability of free space m0 is EduHulk.COM
4 (a) [MLT–2A–2] (b) [M0 L1 T] (c) [M0 L2 T–1A2 ] (d) none of these. (1991) 33. According to Newton, the viscous force acting between liquid layers of area A and velocity gradient Dv/DZ is given by F A v Z = − η ∆ ∆ , where h is constant called coefficient of viscosity. The dimensional formula of h is (a) [ML–2T–2] (b) [M0 L0 T0 ] (c) [ML2 T–2] (d) [ML–1T–1]. (1990) 34. Dimensional formula of self inductance is (a) [MLT–2A–2] (b) [ML2 T–1A–2] (c) [ML2 T–2A–2] (d) [ML2 T–2A–1] (1989) 35. The dimensional formula of torque is (a) [ML2 T–2] (b) [MLT–2] (c) [ML–1T–2] (d) [ML–2T–2]. (1989) 36. If C and R denote capacitance and resistance, the dimensional formula of CR is (a) [M0 L0 T1 ] (b) [M0 L0 T0 ] (c) [M0 L0 T–1] (d) not expressible in terms of MLT. (1988) 37. The dimensional formula of angular momentum is (a) [ML2 T–2] (b) [ML–2T–1] (c) [MLT–1] (d) [ML2 T–1]. (1988) 2.10 Dimensional Analysis and its Applications 38. A physical quantity of the dimensions of length that can be formed out of c, G and e 2 0 4πε is [c is velocity of light, G is the universal constant of gravitation and e is charge] (a) c G 2 e 2 0 1 2 4πε         / (b) 1 4 2 2 0 1 2 c e G πε         / (c) 1 4 2 0 c G e πε (d) 1 4 2 2 0 1 2 c G e πε         / (NEET 2017) 39. Planck’s constant (h), speed of light in vacuum (c) and Newton’s gravitational constant (G) are three fundamental constants. Which of the following combinations of these has the dimension of length ? (a) hG c 3 2/ (b) hG c 5 2/ (c) hc G (d) Gc h3 2/ (NEET-II 2016) 40. If dimensions of critical velocity vc of a liquid flowing through a tube are expressed as [hx ry rz ] where h, r and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values of x, y and z are given by (a) –1, –1, –1 (b) 1, 1, 1 (c) 1, –1, –1 (d) –1, –1, 1 (2015) 41. If force (F), velocity (V) and time (T) are taken as fundamental units, then the dimensions of mass are (a) [FVT–1] (b) [FVT–2] (c) [FV–1T–1] (d) [FV–1T] (2014) 42. The density of a material in CGS system of units is 4 g cm–3. In a system of units in which unit of length is 10 cm and unit of mass is 100 g, the value of density of material will be (a) 0.04 (b) 0.4 (c) 40 (d) 400 (Mains 2011) 43. The velocity v of a particle at time t is given by v at b t c = + + , where a, b and c are constants. The dimensions of a, b and c are (a) [L], [LT] and [LT–2] (b) [LT–2], [L] and [T] (c) [L2 ], [T] and [LT–2] (d) [LT–2], [LT] and [L]. (2006) 44. An equation is given here P a V b V +       = 2 θ where P = Pressure, V = Volume and q = Absolute temperature. If a and b are constants, the dimensions of a will be (a) [ML–5T–1] (b) [ML5 T1 ] (c) [ML5 T–2] (d) [M–1L5 T2 ]. (1996) 45. Which of the following is a dimensional constant? (a) Relative density (b) Gravitational constant (c) Refractive index (d) Poisson’s ratio. (1995) 46. Turpentine oil is flowing through a tube of length l and radius r. The pressure difference between the two ends of the tube is P. The viscosity of oil is given by η = P r − x vl ( ) 2 2 4 where v is the velocity of oil at a distance x from the axis of the tube. The dimensions of h are (a) [M0 L0 T0 ] (b) [MLT–1] (c) [ML2 T–2] (d) [ML–1T–1] (1993) 47. The time dependence of a physical quantity p is given by p = p0 exp (–at2 ), where a is a constant and t is the time. The constant a (a) is dimensionless (b) has dimensions [T–2] (c) has dimensions [T2 ] (d) has dimensions of p (1993) 48. P represents radiation pressure, c represents speed of light and S represents radiation energy striking per EduHulk.COM
Units and Measurements 5 unit area per sec. The non zero integers x, y, z such that Px Sy cz is dimensionless are (a) x = 1, y = 1, z = 1 (b) x = –1, y = 1, z = 1 (c) x = 1, y = –1, z = 1 (d) x = 1, y = 1, z = –1 (1992) 49. The frequency of vibration f of a mass m suspended from a spring of spring constant k is given by a relation f = amx ky , where a is a dimensionless constant. The values of x and y are (a) x = = y 1 2 1 2 , (b) x = − y = − 1 2 1 2 , (c) x = = y − 1 2 1 2 , (d) x = − y = 1 2 1 2 , (1990) 50. If x = at + bt2 , where x is the distance travelled by the body in kilometers while t is the time in seconds, then the units of b is (a) km/s (b) km s (c) km/s2 (d) km s2 (1989) ANSWER KEY 1. (a) 2. (c) 3. (c) 4. (c) 5. (c) 6. (c) 7. (d) 8. (c) 9. (b) 10. (d) 11. (d) 12. (a) 13. (b) 14. (b) 15. (d) 16. (c) 17. (c) 18. (b) 19. (d) 20. (a) 21. (c) 22. (b) 23. (a) 24. (c) 25. (b) 26. (b) 27. (a) 28. (c) 29. (b) 30. (d) 31. (c) 32. (a) 33. (d) 34. (c) 35. (a) 36. (a) 37. (d) 38. (d) 39. (a) 40. (c) 41. (d) 42. (c) 43. (b) 44. (c) 45. (b) 46. (d) 47. (b) 48. (c) 49. (d) 50. (c) 1. (a) : K Qx A T T t = ( − ) , 1 2 where Q is the amount of heat flow, x is the thickness of the slab, A is the area of cross- section, and t is the time taken. K = J m m K s 2 = = W W 1 1 − − 1 1 m K m K 2. (c) : Damping force, F ∝ v or F = kv where k is the constant of proportionality ∴ = = = = − − − − k F v N m s kg m s m s kg s 1 2 1 1 3. (c) : Force between two charges F q r q Fr = ⇒ = = 1 4 1 4 0 2 2 0 2 πε 2 ε π C /N-m 2 2 4. (c) : Given : least count = 0.01 and number of circular scale divisions = 50. \ Pitch= L.C × No. of circular scale division = 0.01 × 50 = 0.5 mm. 5. (c) : X = A B C D 2 1 2 1 3 3 / / Maximum percentage error in X dX X dA A dB B dC C dD D       × = + + +      100 2  × 1 2 1 3 3 100 = 2 1 1 2 2 1 3 × + × + × +3 3 × 4 = 16% 6. (c) : If n divisions of vernier scale coincides with (n – 1) divisions of main scale. Therefore, n VSD = (n – 1) MSD ⇒ 1 VSD = (n ) n − 1 MSD \ Least count = 1 MSD – 1 VSD = − − 1 1 MSD MSD (n ) n = 1 − 1 + 1 MSD MSD MSD n = 1 n MSD = × = =       1 1 1 1 1 2 n n n n cm  MSD cm 7. (d) : Diameter of the ball = MSR + CSR × (Least count) – Zero error = 5 mm + 25 × 0.001 cm – (–0.004) cm = 0.5 cm + 25 × 0.001 cm – (–0.004) cm = 0.529 cm. 8. (c) : As P a b cd = 3 2 % error in P is ∆P ∆ ∆ ∆ ∆ P a a b b c c d d × =       +       + +       100 3 2 ×100 = [3 × 1% + 2 × 2% + 3% + 4%] = 14% 9. (b) : From the relation, h u = +t gt 1 2 2 h gt g h t = ⇒ = 1 2 2 2 2 (Q body initially at rest) Taking natural logarithm on both sides, we get ln g = ln h – 2 ln t Differentiating, ∆g ∆ ∆ g h h t t = − 2 Hints & Explanations EduHulk.COM