Title Units and measurements_Exercise - 1.pdf ✅

38 UNITS AND MEASUREMENTS & BASIC MATHEMATICS EXERCISE – 1: Basic Objective Questions Fundamental and derived Units and Measurements 1. A suitable unit for gravitational constant is (a) kg metre sec−1 (b) Newton metre−1 sec (c) Newton metre2 kg−2 (d) kg metre sec−1 Ans. (c) Sol. As 2 1 2 2 1 2 Gm m Fr F G r m m =  = Substituting the unit of above quantities unit of G = Newton metre2kg−2 . 2. A force F is given by 2 F at bt , = + where t is time. What are the dimensions of a and b (a) MLT−3 and ML2T −4 (b) MLT−3 and MLT−4 (c) MLT−1 and MLT0 (d) MLT−4 and MLT1 Ans. (b) Sol. From the principle of dimensional homogeneity       2 F MLT 3 F at a MLT t T − −     =  = = =             Similarly     2 2 4 2 2 F MLT F bt b MLT . t T − −     =  = = =                 3. Number of particles is given by 2 1 2 1 n n n D x x − = − − crossing a unit area perpendicular to X-axis in unit time, where n1 and n2 are number of particles per unit volume for the value of x meant to be x2 and x1 respectively. Find dimensions of D called as diffusion constant (a) M0LT2 (b) M0L 2T −4 (c) M0LT−3 (d) M0L 2T −1 Ans. (d) Sol. (n) = Number of particle passing from unit area in unit time   0 0 0 2 1 2 No. of particle M L T L T A t L T − −     = = =          n n 1 2  = =   No. of particle in unit volume 3 L − =     Now from the given formula          2 1 2 1 2 1 3 2 1 n x x L T L D L T . n n L − − − − −     = = =     −     4. A physical quantity is measured and its value is found to be nu where n = numerical value and u = unit. Then which of the following relations is true (a) 2 n u  (b) n u  (c) n u  (d) 1 n u  Ans. (d) Sol. We know P = nu = constant  = n u n u 1 1 2 2 or 1 n . u  5. To determine the Young’s modulus of a wire, the formula is F L Y . ; A l =  where L = length, A = area of cross – section of the wire, ΔL = Change in length of the wire when stretched with a force F. The conversion factor to change it from CGS to MKS system is (a) 1 (b) 10 (c) 0.1 (d) 0.01 Ans. (c) Sol. We know that the dimension of young’s modulus is [ML−1T −2 ] C.G.S. unit: gm cm 1 sec 2 and M.K.S. unit: kg. m 1 sec 2 . By using the conversion formula: 1 1 2 1 1 2 1 1 1 2 1 2 2 2 M L T gm cm sec n n M L T kg meter sec − − − −             = =                          conversion factor 1 1 2 2 3 2 1 n gm cm sec 1 0.1 n sec 10 10 gm 10 cm − −       = = =             6. If the present units of length, time and mass (m, s, kg) are changed to 100m, 100s and 1 kg 10 then (a) The new unit of velocity is increased 10 times (b) The new unit of force is decreased 1 1000 times (c) The new unit of energy is increased 10 times (d) The new unit of pressure is increased 1000 times Ans. (b) Sol. Unit of velocity = m/sec; in new system ( ) 100 m m same 100 sec sec = = Unit of force 2 kg m ; sec  = in new system 2 1 100 m 1 kg m kg 10 100 sec 100sec 1000 sec  =  = 
UNITS AND MEASUREMENTS & BASIC MATHEMATICS 39 Unit of energy 2 2 kg m ; sec  = in new system 2 2 1 100 m 100 m 1 kg m kg 10 100 sec 100 sec 10 sec   =  =  Unit of pressure 2 kg ; m sec =  in new system 7 2 1 1 1 kg kg 10 10 100 m 100 sec 100 sec m sec − =   =   7. If 1gm cms−1 = x Ns, then number x is equivalent to (a) 1 1 10−  (b) 2 3 10−  (c) 4 6 10−  (d) 5 1 10−  Ans. (d) Sol. 1 3 2 1 gm cm s 10 kg 10 m s − − − − − =   5 1 5 10 kg m s 10 Ns − − − =   = 8. If velocity v, acceleration A and force F are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v, A and F would be (a) FA−1v (b) Fv3A−2 (c) Fv2A−1 (d) F2v 2A−1 Ans. (b) Sol. Given, v = velocity = [LT−1 ], A = Acceleration = [LT−2 ], F = force = [MLT−2 ] By substituting, the dimension of each quantity we can check the accuracy of the formula [Angular momentum] 3 2 Fv A− = [ML2T 1 ] = [MLT 2 ] [LT 1 ] 3 [LT 2 ] 2 = [ML2T 1 ] L.H.S. = R.H.S. i.e., the above formula is Correct. 9. The largest mass (m) that can be moved by a flowing river depends on velocity (v), density (ρ) of river water and acceleration due to gravity (g). The correct relation is (a) 2 4 2 v m g   (b) 6 2 v m g   (c) 4 3 v m g   (d) 6 3 v m g   Ans. (d) Sol. Given, m = mass = [M], v = velocity = [LT−1 ], ρ = density = [ML−3 ], g = acceleration due to gravity = [LT−2 ] By substituting, the dimension of each quantity we can check the accuracy of the formula 6 3 v m K g  =   6 3 1 3 2 ML LT M LT − − −      =         = M L.H.S. = R.H.S. i.e., the above formula is Correct. 10. If the velocity of light (c), gravitational constant (G) and Plank’s constant (h) are chosen as fundamental units, then the dimensions of mass in new system is (a) 1 1 1 2 2 2 c G h (b) 1 1 1 2 2 2 c G h − (c) 1 1 1 2 2 2 c G h − (d) 1 1 1 2 2 2 c G h − Ans. (c) Sol. Let x y z m c G h  or x y z m Kc G h = By substituting the dimension of each quantity in both sides x y z M L T K LT M L T ML T 1 0 0 1 1 3 2 2 1 − − − −         =         y z x 3y 2z x 2y z M L T − + + + − − − =     By equating the power of M, L and T in both sides: − + = + + = − − − = y z 1, x 3y 2z 0, x 2y z 0 By solving above three equations 1 1 x , y 2 2 − = = and 1 z . 2 = 1 1 1 m c G h 2 2 2 −   11. If the time period (T) of vibration of a liquid drop depends on surface tension (S), radius (r) of the drop and density (ρ) of the liquid, then the expression of T is (a) 3 r T K S  = (b) 1 2 3 r T K S  = (c) 3 1 2 r T K S  = (d) None of these Ans. (a) Sol. Let x y z T S r   or x y z T KS r =  By substituting the dimension of each quantity in both sides   x z y M L T K MT L ML 0 0 1 2 3 − −       =       x z y 3z 2x M L T + − − =     By equating the power of M, L and T in both sides x z 0, y 3z 0, 2x 1 + = − = − = By solving above three equations
40 UNITS AND MEASUREMENTS & BASIC MATHEMATICS 1 3 1 x , y , z 2 2 2 −  = = = So the time period can be given as, 1 3 1 3 2 2 2 r T KS r K . S −  =  = 12. The dimensional formula of torque is (a) [ML2T −2 ] (b) [MLT−2 ] (c) [ML−1T −2 ] (d) [ML−2T −2 ] Ans. (a) Sol.  = Force × distance 2 2 2 [MLT ][L] ML T − − = = 13. Of the following quantities, which one has dimension 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. Ans. (d) Sol. For angular momentum, the dimensional formula is ML2T −1 . For the other three, it is ML−1T −2 . 14. The dimensional formula of pressure is (a) [MLT 2 ] (b) [ML 1T 2 ] (c) [ML 1T 2 ] (d) [MLT2 ] Ans. (c) Sol. [Pressure] = [Force] / [Area] 2 1 2 2 MLT ML T L − − − = = 15. The centripetal force (F) acting on an object that is rotating in a circular path depends upon the mass of object (m), radius of the circular path (r), and velocity (v) of the object. Derive the formula for the centripetal force. (a) kmv2 r −1 (b) kmv3 (c) kmv2 r 1 (d) kmv2 Ans. (a) Sol. Let a b c F km r v = Then,     a b c F k m r v =             2 0 0 0 a b c c MLT M L T M L L T − −           =           2 a b c c MLT M L T − + −         =         On comparing, we get a 1, c 2, b c 1 b 1 = = + = → = − So, 2 1 F kmv r− = 16. Expression for time in terms of G (universal gravitational constant), h (Planck constant) and c (speed of light) is proportional to (a) 5 hc G (b) 3 c Gh (c) 5 Gh c (d) 3 Gh c Ans. (c) Sol. Dimension of time, t = [T] Now, t ∝ Gkh q c r t = kGkh q c r ....(1) Dimension of Gravitational constant, G = [M−1L 3T −2 ] Dimension of Planck’s constant h = [ML2T −1 ] Dimension of speed of light c = [LT−1 ] Put the Dimensions in Eq.(1), we get [T] = [M−1L 3T −2 ] p [ML2T −1 ] q [LT−1 ] r [T] = [M − p + q] [L3p+2q+r] [T−2p-q-r ] On comparing the power of both sides: − + =  = p q 0 p q ... 2( ) 3p 2q r 0 5p r 0 ... 3 + + =  + = ( ) − − − =  − − = 2p q r 1 3p r 1 ... 4( ) On solving the above Eq. (2), (3) and (4), we get 1 p q 2 = = and 5 r 2 − = Put these values in Eq. (1), we get 1 1 5 2 2 2 t G h c − = 5 Gh t k c = where, k = constant 17. Let l, r, c and v represent inductance, resistance, capacitance and voltage, respectively. The dimension of 1/rcv in SI units will be (a) [LA−2 ] (b) [A−1 ] (c) [LTA] (d) [LT2 ] Ans. (b) Sol. Dimension of inductance = [M1L 2T −2A−2 ] = [l] Dimension of capacitance = [M−1L −2T 4A2 ] = [c] Dimension of resistance = [M1L 2T −3A−2 ] = [r] Dimension of voltage = [M1L 2T −3A−1 ] = [v] Dimension of l/rcv = M1L 2T −2A−2 ] / [M−1L 2T 4A2 ][M1L 2T −3A−2 ] [M1L 2T −3A−1 ] = [ML2T −2A−2 ]/[ML2T −2A−1 ] = [A−1 ] 18. If force [F], acceleration [A] and time [T] are chosen as the fundamental physical quantities. Find the dimensions of energy. (a) [F][A][T−1 ] (b) [F][A−1 ][T]
UNITS AND MEASUREMENTS & BASIC MATHEMATICS 41 (c) [F][A][T] (d) [F][A][T2 ] Ans. (d) Sol. a b c E F A T    a b c M L T M LT LT T 1 2 2 1 1 2 2 − − −              a 1 = a b 2 + =  = b 1 − − + = − 2a 2b c 2  =c 2 a 1 = b 1 = c 2 =    2 E F A T      19. Time (T), velocity (C) and angular momentum (h) are chosen as fundamental quantities instead of mass, length and time. In terms of these, the dimensions of mass would be: (a) [M] = [T–1 C –2h] (b) [M] = [T C–2h] (c) [M] = [T–1C –2h –1 ] (d) [M] = [T–1 C 2h] Ans. (a) Sol. M ∝ T x v y h z M1 L 0T 0 = (T)x (L1T −1 ) y (M1L 2T −1 ) z M1L 0T 0 = Mz L y+2z + Tx−y−z z = 1 y + 2z = 0 x – y – z = 0 y = –2 x + 2 – 1 = 0 x = –1 M ⇒ T –1 C –2 h 1 20. The dimensional formula for permeability μ is given by (a) MLT−2A−2 (b) M0L 1T (c) M0L 2T −1A2 (d) None of the above. Ans. (a) Sol. B 2 1 ; [B] MT A , nI − −  = = 1 [n] L , [I] A − = = B MLT A2 2 nI   − − =         21. P represents radiation pressure, c represents speed of light and S represents radiation energy striking unit area per sec. The non zero integers x, y, z such that x y z P S c 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 Ans. (c) Sol. Try out the given alternatives. When x = 1, y = 1, z = 1 x y z 1 1 1 Pc P S c P S c S − = = 1 2 1 0 0 0 2 2 2 ML T LT M L T ML T L T − − − −         = =           22. The time dependence of a physical quantity p is given by o p p = exp ( ) 2 −at , 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 Ans. (b) Sol. In o p p = exp ( ) 2 2 −at , at dimension less 2 2 2 1 1 T t T −   = = =     23. In a particular system, the unit of length, mass and time are chosen to be 10 cm, 10 g and 0.1 s respectively. The unit of force in this system will be equivalent to (a) 0.1 N (b) 1 N (c) 10 N (d) 100 N Ans. (a) Sol.   ( )( )( ) 2 2 F MLT 10g 10 cm 0.1s − − = = ( )( )( ) 2 2 1 1 1 10 kg 10 m 10 s 10 N. − − − − − = = 24. Which of the following is a dimensional constant? (a) Refractive index (b) Poissons ratio (c) Relative density (d) Gravitational constant Ans. (d) Sol. A quantity which has dimensions and a constant value is called dimensional constant. Therefore, gravitational constant (G) is a dimensional constant. 25. The dimensions of Planck’s constant are same as (a) energy (b) power (c) momentum (d) angular momentum Ans. (d) Sol. We know that E hv = 2 2 2 1 1 E ML T h ML T v T − − −     = = =         Angular momentum = I 2 1 2 1 ML T ML T − − = =             26. Which of the following pairs of physical quantities does not have same dimensional formula? (a) Work and torque. (b) Angular momentum and Planck’s constant. (c) Tension and surface tension. (d) Impulse and linear momentum. Ans. (c) Sol. (a) Work = Force × displacement = [MLT−2 ][L]=[ML2T −2 ] Torque = force × distance = [MLT−2 ][L]=[ML2T −2 ]