Units and Measurements Test - 14

In a number less than one (i.e; a decimal). The zeroes on the left of non zero number are not significant figures, and zeroes to the right side of a nonzero number are significant figures. In the number 0.06900, two zeroes before six are not significant figure and two zero on right side of 9 are significant figures. Significant figures are underlined, so verifies option (b).

The sum of the numbers can be calculated as 663.821 arithmetically. The number with least decimal places is 227.2 is correct to only one decimal place.
The final result should, therefore be rounded off to one decimal place, i.e., 663.8.

In multiplication or division, the final result should retain as many significant figures as are there in the original number with the least significant figures. the significant figures in given numbers 4.237 g and 2.5 \(cm^3\) are four and two respectively so result must have only two significant figures.

Density = \({mass\over volume}={4.237\over2.5}\), Density = 1.6948=1.7 g \(cm^{-3}\) rounding off upto 2 significant figures.

(i) If the preceding digit of dropped out digit 5 is even there will be no change in rounding off. (ii) if the preceding digit of dropped out digit is 5 is odd then preceding digit is increased by one. 

(i) In given number 2.745 it is round off upto 3 significant figure, \(IV^{th}\) digit is 5 and its preceding digit is even; so no change in 4 and hence the answer is 2.74. (ii) Given figure 2.735 is round off upto 3 significant figure here \(IV^{th}\) i.e., next digit is 5 and its preceding digit is 3 (odd). so 3 is increased by 1 and answer becomes 2.74. Hence, verifies the option(d).

(i) Significant figures in the result multiplication (or division) is the minimum number of significant figures in given number.

(ii) If \(\Delta\)x is error is quantity x then relative error or error is \(\Delta x\over x\).

l = 16.2 cm \(\Delta\)l = 0.1

b = 10.1 cm \(\Delta\)b = 0.1

l = 16.2 \(\pm\) 0.1

b = 10.1 \(\pm\) 0.1

A = Area = l x b = 16.2 x 10.1 =163.62 \(cm^2\)

= 164 \(cm^2\) (in significant figures)

\({\Delta A\over A}={\Delta l\over l}+{\Delta b\over b}={0.1\over16.2}+{0.1\over10.1}\)

\({\Delta A\over 164}={10.1\times 0.1+16.2\times 0.1\over 16.2\times10.1}\)

\(\Delta A=164({1.01+1.62\over 163.62})\)

\(\Delta A=2.63\ cm^2\)

Now rounding off upto significant figures in \(\Delta\)l and \(\Delta\)b i.e., one

\(\Delta\)A = 3 \(cm^2\)

A = (164\(\pm\)3) \(cm^2\). Hence, verifies the option (a).

Tension = force = \([MLT ^{-2} ] \)

Surface tension = \({Force\over l}={[MLT^{-2}]\over[L]}=[ML^0T^{-2}]\)

Dimensions of both are not same

Rules of significant figure in multiplication and addition.

\(A = (2.5 \pm 0.5)ms^{-1}\)

\(B = (0.10\pm 0.01) s\)

X = AB = 2.5 x 0.10 = 0.25 m

\({\Delta X\over X}={\Delta A\over A}+{\Delta B\over B}\)

\({\Delta X\over X}={0.5\over 2.5}+{0.01\over 0.10}\)

\({\Delta X\over X}={0.075\over 0.25},\ \ \Delta X=0.007\cong0.08\)

(Rounding off upto 2 significant figures)

\(\therefore\) AB = \( (0.25 \pm 0.08)\ m\)

Hence, verifies the option (A).

In significant figures of measured quantities zeroes are included in significant figures.

Quantities A and B are measured quantities so number of significant figures in 1.0 m and 2.0 m are two.

\(\sqrt{AB}=\sqrt{1.0\times2.0}=\sqrt{2}=1.414\ m\)

rounding off upto minimum numbers of significant figure in 1.0 and 2.0 result must be in 2 significant figures

x = \(\sqrt{AB}\) = 1.4

\({\Delta X\over X}={1\over2}[{\Delta A\over A}+{\Delta B\over B}]={1\over2}[{0.2\over1.0}+{0.2\over 2.0}]\)

= \({1\over2}\times 0.2[{1\over1.0}+{1\over 2.0}]=0.1[{2.0+1.0\over1.0\times 2.0}]\)

\({\Delta X\over1.4}=0.1({3.0\over2.0})=2.1\ m\)

\(\Delta\)x = 0.2 m rounding off upto 1 place of decimal.

\(\sqrt{AB}\) = \(1.4\ m\pm0.2\ m\)

Verifies the option (d).

In these problems units must be least and in digits number of digits including zeroes after decimal must be zero.

All the measurements are upto two places of decimal, least unit is mm. so 5.00 mm measurement is most precise. Hence, verifies answer (a).

Absolute error (\(\bar a -a_i \)) must be minimum for more accuracy.

Error or absolute error

\(|\Delta a_1|=|5-4.9|=0.1\ cm,\ \\|\Delta a_2|=|5-4.805|=0.195\ cm\)

\(|\Delta a_3|=|5-25|=0.25\ cm,\ \\|\Delta a_4|=|5-5.4|=0.4\ cm\)

\(|\Delta a_1|\) is minimum. 

Y = \(1.9\times 10^{11} N/m^2 \)

Y = \({1.9\times 10^{11}N\over 1M^2}={1.9\times 10^{11}\times 10^5\ dynes\over 10^4 \ cm^2}\)

Y = \(1.9\times10^{11+5-4}\)

Y = \(1.9\times10^{12}\ dyne/cm^2\)

Verifies the option (c).

Let the dimensional formula for energy in fundamental quantities P, A and T is \([P ^a A ^b T ^c ]\).

Dimensional formula of E = \([P ^a A ^b T ^c ]\) dimensional formula of momentum P = mv = \([MLT ^{-1} ]\).

Area A = \([L ^2 ]\)

Time T = \([T ^1 ]\)

\(\therefore\) Energy = f.s = \([MLT ^{-2}L]\) = \([ML ^2T ^{-2} ]\)

\(\therefore\) \([ML ^2T ^{-2} ]\) = \([MLT ^{-1} ]^a\) \([L ^2 ] ^b\) \([T] ^c\)

\([M^1L ^2T ^{-2} ]\) = \([M^aL ^{a+2b}T ^{-a+c}]\)

Comparing the powers

a = 1           a + 2b = 2          -a + c = -2

                   1 + 2b = 2          -1 + c = -2

                   2b = 2 - 1            c = -2 + 1

                     b = \(1\over2\)                 c = -1

\(\therefore\) Dimensional formula of energy is \([P^1A^{1/2}T^{-1}]\)

Verifies the option (d).

Displacement y and amplitude a has same dimensions, angle of sin and cos is dimensionless. So dimensions in both side must be equal by the principle of homogeneity.

In option (a) and (d) the dimensions of y and a in L.H.S and R.H.S. are equal to L and while in option (b) angle is v.t (where v is velocity)

\(\therefore\) dimension of v.t is \([LT^{-1}][T] = [L]\)

So sin vt is not dimensionless so option (b) is wrong. In option (c) in R.H.S. dimension of amplitude \({a\over T}={[L]\over [T]}=[LT^{-1}]\) which not equal to the dimension of y i.e., L and angle \({t\over a}={[T]\over [L]}=[LT^{-1}]\) is not dimensionless. Hence, verifies the option (b) and (c).

Addition or subtraction may be possible of the same physical quantity. But multiplication can be possible for different physical quantities or different dimensions. After multiplication or division of two quantities their dimension may be equal to IIIrd and can be added or subtracted.

In option (a) and (e) there is term (P – Q) and (R +Q) as different physical quantities can never be added or subtracted so option (a) and (e) can never be meaningful. In option (b), the dimension of PQ may be equal to dimension of R so option (b) can be possible . similarly dimensions of PR and \(Q ^2\) may be equal and gives the possibility of option (d). In option (c), there is no addition subtraction gives the possibilities of option (c). 

\(\therefore\) E = hv

\(h={E\over v}={[ML^2T^{-2}]\over[T^{-1}]}=[ML^2T^{-1}]\)

Linear impulse = F.t = \(dp\over dt\). dt = dp

= mv = \([MLT^{-1}] \)

Angular impulse = \(\tau\).dt = \(dL\over dt\).dt = dL = mvr

= \( [M][LT^{-1}][L] = [ML^2T^{-1}]\)

Linear momentum = mv = \([MLT^{-1}] \)

Angular momentum L = mvr = \([ML^2T^{-1}]\)

So the dimensional formulae of h, Angular impulse and Angular momentum are same. 

Dimension of

\(h={E\over v}={[ML^2T^{-1}]\over [T^{-1}]}=[ML^2T^{-1}]\)

\(c={s\over t}=[LT^{-1}]\)

\(G={Fr^2\over M_1M_2}={[ML^3T^{-2}]\over [M][M]}=[M^{-1}L^3T^{-2}]\)

hc = \([ML^2T^{-1}]\times [LT^{-1}]=[ML^3T^{-2}]\)

\({hc\over G}={[ML^3T^{-2}]\over [M^{-1}L^3T^{-2}]}=[M^2]\)

M = \(\sqrt{hc\over G}=[h^{1/2}c^{1/2}G^{-1/2}]\)

\({h\over c}={[ML^2T^{-1}]\over [LT^{-1}]}=[ML]=\sqrt{hc\over G}\times L\)

\(L={h\over c}\times \sqrt{G\over hc}={\sqrt{Gh}\over c^{3/2}}=[G^{1/2}h^{1/2}c^{-3/2}]\)

\(c=[LT^{-1}]=[G^{1/2}h^{1/2}c^{-3/2}T^{-1}]\)

\(T=[G^{1/2}h^{1/2}c^{-{3\over2}-1}]=[G^{1/2}h^{1/2}c^{-5/2}]\)

Hence, physical quantities (a, b and d) can be used to represent L, M,T in terms of the choosen fundamental quantities

Ratio can be express as pressure (P) if the dimension of ratio is same as P.

Dimension of pressure \(P={F\over A}={[MLT^{-2}]\over [L^2]}\)

Dimension of pressure \(P=[ML^{-1}T^{-2}]\)

(a) Dimension of \({F\over A}={[MLT^{-2}]\over [L^2]}=[ML^{-1}T^{-2}]\) = dimension of P

(b) Dimension of \({E\over V}={[ML^2T^{-2}]\over [L^3]}=[ML^{-1}T^{-2}]\) = dimension of P

(c) Dimension of \({E\over A}={[ML^2T^{-2}]\over [L^2]}=[ML^{0}T^{-2}]\ne\) dimension of P

(d) Dimension of \({F\over V}={[MLT^{-2}]\over [L^3]}=[ML^{-2}T^{-2}]\ne\) dimension of P

Hence, required option are (a) and (b).

Dimension of time is [T].

the second, and year measures the time so their dimension is \([M^0L^ 0T^ 1 ]\) or unit of time. But Parsec and Light year measures the distance and dimension of distance is [L] which is not equal to the dimension of Time [T]. Hence, (b) and (d) are not unit of time.

(a) Volume of cube is

\(V = (a)^3 = (1 cm)^3 = (1 \times 10^{–2} m)^3 = 10^{–6} m^3\)

(b) Surface Area of solid cylinder

\(S=2\pi r(r+h)\)

= \(2\times3.14\times2(2+10)\ cm^2\)

= \(12.56\times12\ cm^2\)

S = 150.72 \(cm^2\)

= 15072 \(mm^2\)            [as 1 \(cm^2\) = 100 \(mm^2\)]

(c) v = 18 \(kmh^{-1}\) = \(18\times {5\over 18}\, m \, sec^{-1}\)

\(=5\ m\ sec^{-1}=5\ m\ in\ 1\ sec\)

(d) Relative density = 11.3

density = \(11.3\times 1\, g\, cm^{-3}=11.3\, g\, cm^{-3}\)

\(=11.3\times {1\over1000}kg\times({1\over100}m)^{-3}\)

\(=11.3\times{1\over10^3}\times 10^{-6}=11300\ kg\, m^{-3}\)

(a) \(1\, kg\, m^2s^{-2}=1000g\times(100\ cm)^2\times 1\, s^{-2}\)

\(= 10^7\, g\, cm^2\, s^{-2}\)

(b) \(1\, m={1\over9.46\times10^{15}}=1.057\times10^{-16}\, light\, year\)

(c) \(3.0\ ms^{-2}=3.0\times({1\over1000}km)\times[(3600)^2h^{-2}]\)

\(= {3\times3600\times3600\over 1000}\, km\, h^{-2}\)

\(= 3.888\times10^4\, km\, h^{-2}\)

(d) \(G=6.67\times10^{-11}N\, m^2\, kg^{-2}\)

\(=6.67\times10^{-11}\, m^3s^{-2}\, kg^{-1}\)

\(=6.67\times10^{-11}\times\ (100\ cm)^3\times s^{-2}\times\ (1000\ g)^{-1}\)

\(={6.67\times10^{11}\times10^6\over10^3}cm^3\, s^{-2}\, g^{-1}\)

\(=6.67\times10^{-8}cm^3\, s^{-2}\, g^{-1}\)