Title Units and measurements_Exercise - 2.pdf ✅

64 UNITS AND MEASUREMENTS & BASIC MATHEMATICS (c) 2 2 2 ML T A− −     (d) ML T A 2 3 1 − −     Ans. (d) Sol. Electromotive force = Potentialdifference   2 2 W ML T 2 3 1 V ML T A q AT − − −      = = =     6. Velocity v is given by 2 v at bt c, = + + where t is time. What are the dimensions of a, b and c respectively? [UP CPMT 2011] (a) 3 2 LT , LT − −         and 1 LT−     (b) 1 2 LT , LT − −         and 3 LT−     (c) 2 3 LT , LT − −         and 1 LT−     (d) 1 3 LT , LT − −         and 2 LT−     Ans. (a) Sol. Dimensions of velocity are     1 v L T− =     So, dimensions of 2 1 at LT−     =         2 1 3 a T LT a LT − −  =          =     Dimensions of   1 bt LT− =          1 2 b T LT b LT − −  =      =     Dimensions of   1 c LT− =     7. If RT V/RT p e , V b − = − then dimensional formula of  is [UP CPMT 2011] (a) p (b) R (c) T (d) V Ans. (a) Sol. Given, RT V/RT p e V b − = − So, V RT  is dimensionless Hence,       2 2 1 RT ML T 1 2 ML T V L − − − −          = = =         This is dimensional formula ofpressure (p). 8. The relation Z k p e −   =  , where p is pressure, Z is distance, k is Boltzmann constant and  is temperature. The dimensional formula of  will be [AFMC 2011] (a) 0 2 0   M L T   (b) 2   ML T   (c) ML T0 1−     (d) M L T 0 2 1−     Ans. (a) Sol. In the given equation, Z k   should be dimensionless. k Z    =      2 2 1 ML T K K 2 MLT L − − −       = =     and   2 1 2 MLT p p ML T − − −        =   = =          M L T 0 2 0 =     9. If C be the capacitance and V be the electric potential, then the dimensional formula ofCV 2 is [KCET 2011] (a) ML T A 2 2 0 −     (b) MLT A− − 2 1     (c) M LT T 0 2 0 −     (d) 3 ML TA −     Ans. (a) Sol. We know that, energy, 1 2 E CV 2 = Dimensions of CV2 = Dimensions ofenergy, E = [ML2T -2A0 ] 10. The dimensions of impulse are [NEET 2011] (a) 1 MLT−     (b) ML T2 1−     (c) 1 1 ML T− −     (d) 1 MT−     Ans. (a) Sol. The dimensions of impulse are Ft = [MLT-2 ][T] = [MLT-1 ] 11. What is the dimensions of surface tension? [NEET 2011] (a) 0   MLT   (b) 4 MLT−     (c) ML T0 2−     (d) ML T0 1−     Ans. (c) Sol. We know that, F T L = Dimensions of
UNITS AND MEASUREMENTS & BASIC MATHEMATICS 65   2 Dimensions of F MLT 0 2 T ML T Dimensions of L L − −     = = =     12. Surface tension has the same dimensions as that of [KERALA CEE 2011] (a) coefficient of viscosity (b) impulse (c) momentum (d) spring constant Ans. (d) Sol. Surface tension, S 2 Force MLT 2 MT Length L − − = = =     Spring constant, R   2 2 Force MLT Elongation L MT − −     = = =     13. The density of a material in the 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: [PMT 2011] (a) 0.04 (b) 0.4 (c) 40 (d) 400 Ans. (c) Sol. Density = 4gcm-3 In new system, 3 g 4 100 cm 10 4 1000 100 1       =       =  = 40 units 14. A physical quantity x is given by 3 2 2k l x m n = . The percentage error in the measurements of k, l, m and n are 1%, 2%, 3% and 4% respectively. The value of x is uncertain by [AMU 2012] (a) 8% (b) 10% (c) 12% (d) None of these Ans. (c) Sol. Given, 3 2 2k l x m n = Percentage error in x 1 3%k 2%l %m %n 2 = + + + k l 3 100 2 100 k l       =  +          m 1 n 100 100 m 2 n       +  +          1 3 1% 2 2% 1 3% 4% 2 =  +  + + +  =12% 1 3%k 2%l %m %n 2 = + + + k l 3 100 2 100 k l       =  +          m 1 n 100 100 m 2 n       +  +          1 3 1% 2 2% 1 3% 4% 2 =  +  + + +  =12% 15. The error in the measurement of radius of sphere is 0.3%, what is the percentage error in the measurement of its volume? [UP CMT 2012] (a) 0.3% (b) 0.6% (c) 0.9% (d) ( ) 4 3 0.3 3  Ans. (c) Sol. Error in radius, r 0.3% r  = Volume of sphere 4 3 r 3 =   Error in volume r 3 r  =  = (3 × 0.3) % = 0.9% 16. 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 3 2 a b P , cd = % error in P is [NEET 2012] (a) 14% (b) 10% (c) 7% (d) 4% Ans. (a) Sol. Given, 3 2 a b P cd = P 100 P          3 a 2 b c d 100 a b c d       = + + +      a b 3 100 2 100 a b       =  +         
66 UNITS AND MEASUREMENTS & BASIC MATHEMATICS c d 100 100 c d       +  +          a b 100% 1%, 100% 2% a b c d 100% 3% and 100% 4% c d      =  =          =  =     = (3 ×1) + (2 × 2) + (1 × 3) + (1 × 4) = 3 + 4 + 3 + 4 = 14% 17. SI unit of permittivity is [AIMS, MANIPAL 2012] (a) 2 2 2 C m N (b) 2 2 1 C m N− − (c) 2 2 1 C m N− (d) 1 2 2 C m N − − Ans. (b) Sol. From Coulomb’s law, 1 2 2 0 1 q q F 4 r =  or 1 2 0 2 q q 4 Fr  =   Unit of 0  (permittivity) 2 2 2 1 2 C C m N Nm − − = = 18. The damping force on an oscillator is directly proportional to the velocity. The unit of the constant of proportionality is [CBSE AIPMT 2012] (a) kgms−1 (b) kgms−2 (c) kgs−1 (d) kgs Ans. (c) Sol. Given, damping force μ velocity F v F kv F k V   =  = unit of 2 1 1 unit of F kgms k kgs unit of v ms − − − = = 19. The damping force on an oscillator is directly proportional to the velocity. The units of the constant of proportionality are: [PMT 2012] (a) kg ms–1 (b) kg ms–2 (c) kg s–1 (d) kg s Ans. (c) Sol. F v   = F kvF k V  = 2 1 1 kgms k kgs ms − −  = = − 20. If C and R stands for capacitance and resistance, then the dimensions of CR is [UP CPMT 2013] (a) 0 0   M L T   (b) 0   ML T   (c) 0 0 2   M L T   (d) not expressible in terms of M, L and T Ans. (a) Sol. The capacitance of a conductor is defined as the ratio of the charge given to the rise in the potential of the conductor. i.e. 2 q q W C V V W q   = = =     2 2 2 2 ampere s C kg m / s − = − Hence, dimensions of C are M L T A − − 1 2 4 2     . From Ohm’s law, V = iR, therefore unit of resistance, V volt R i ampere = = = kg-m2 s −3 ampere−2  Dimensions of 2 3 2 R ML T A− − =      Dimensions of RC M L T A ML T A − − − − 1 2 4 2 2 3 2 =         M L T 0 0 =     Alternative In a CR circuit, the product of CR is the capacitive time constant of the circuit. If C is in farad, R is in ohm, then time constant is in second. Hence, dimensions of CR = [M0L 0T]. 21. The displacement of an oscillating particle is given by y = Asin (Bx +Ct + D). The dimensional formula for (ABCD) is [UP CPMT 2013] (a) M L T 0 1 0 −     (b) M L T 0 0 1−     (c) 0 1 1 M L T− −     (d) 0 0 0   M L T   Ans. (b) Sol. Displacement, y = A sin (Bx + Ct + D) A = y = [L] As each term inside the brackets is dimensionless, so