Units and Measurements - Test Papers

 CBSE Test Paper 01

Chapter 2 Units and Measurements

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1. The number of significant digits in 6.320 J is 1

   1. 6
   2. 4
   3. 3
   4. 5

2. Newton is the SI unit of 1

   1. acceleration
   2. work
   3. power
   4. force

3. The dimensions of Kinetic energy is same as that of 1

   1. Pressure
   2. Work
   3. Momentum
   4. Force

4. A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance? 1

   1. 500
   2. 450
   3. 600
   4. 550

5. The number of significant digits in 0.0006032 m2${\mathrm{m}}^2$ is 1

   1. 4
   2. 6
   3. 5
   4. 3

6. Define S.I. unit of solid angle? 1

7. How many light years make 1 parsec? 1

8. Is it Possible to have length and velocity both as fundamental quantities? Why? 1

9. The principle of 'parallax' in section 2.3.1 is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth's two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth's orbit = 3×1011m$3\times {10}^{11}\mathrm{m}$. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1" (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1" (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of meters? 2

10. The resistance R is the ratio of potential difference V and current I. What is the percentage error in R if V is (100 ± 5) V and I is (10 ± 2) A? 2

11. The Sun's angular diameter is measured to be 1920". The distance r of the sun from the earth is 1.496 ×$\times$ 1011 m. What is the diameter of the Sun? 2

12. A great physicist of this century (P.A.M. Dirac) loved playing with numerical values of Fundamental constants of nature. This led him to an interesting observation. Dirac found that from the basic constants of atomic physics (c, e, mass of electron, mass of proton) and the gravitational constant G, he could arrive at a number with the dimension of time. Further, it was a very large number, its magnitude being close to the present estimate on the age of the universe (15 billion years). From the table of fundamental constants in this book, try to see if you too can construct this number (or any other interesting number you can think of). If its coincidence with the age of the universe were significant, what would this imply for the constancy of fundamental constants? 3

13. The farthest objects in our Universe discovered by modern astronomers are so distant that light emitted by them takes billions of years to reach the Earth. These objects (known as quasars) have many puzzling features, which have not yet been satisfactorily explained. What is the distance in km of a quasar from which light takes 3.0 billion years to reach us? 3

14. A book with many printing errors contains four different formulas for the displacement y of a particle undergoing a certain periodic motion:
    (a = maximum displacement of the particle, v = speed of the particle. T = time-period of motion). Rule out the wrong formulas on dimensional grounds. 3

    1. y=αsin(2πtT)$y=\alpha \sin \left(\frac{2\pi t}{T}\right)$
    2. y = a sin vt
    3. y=(aT)sinta$y=\left(\frac{a}{T}\right)\sin \frac{t}{a}$
    4. y=(a2–√)(sin2πtT+cos2πtT)$y=(a\sqrt{2})\left(\sin \frac{2\pi t}{T}+\cos \frac{2\pi t}{T}\right)$

15. The diameter of a wire as measured by a screw found to be 1.328,1.330,1.325, 1.326,1.334 and 1.336 cm. Calculate 5

    1. mean value of diameter
    2. absolute error in each measurement
    3. mean absolute error
    4. fractional error
    5. percentage error
    6. diameter of wire.

CBSE Test Paper 01
Chapter 2 Units and Measurements

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Answer

1. 2. 4
      Explanation: There are three rules on determining how many significant figures are in a number:

      • Non-zero digits are always significant.
      • Any zeros between two significant digits are significant.
      • A final zero or trailing zeros in the decimal portion ONLY are significant.

      So keeping these rules in mind, there are 4 significant digit.

2. 4. force
      Explanation: The newton is the SI unit for force; it is equal to the amount of net force required to accelerate a mass of one kilogram at a rate of one meter per second squared.

3. 2. Work
      Explanation: Work

4. 1. 500
      Explanation: Distance between the Sun and the Earth = Speed of light ×$\times$ Time taken by light to cover the distance
      Given that in the new unit,
      speed of light = 1 unit
      Time taken, t = 8 min 20 s = 500 s Distance between the Sun and the Earth = 1 x 500 = 500 units

5. 1. 4
      Explanation: There are three rules on determining how many significant figures are in a number:

      • Non-zero digits are always significant.
      • Any zeros between two significant digits are significant.
      • A final zero or trailing zeros in the decimal portion ONLY are significant.

      So keeping these rules in mind, there are 4 significant digit.

6. One steradian is defined as the angle made by a spherical plane of area 1 square meter at the centre of a sphere of radius 1m.

7. One parsec is equal to about 3.26 light years.

8. No, since length is fundamental quantity and velocity is the derived quantity and is derived from length and time.

9. Diameter of Earth's orbit =3×1011m$=3\times {10}^{11}\mathrm{m}$
   Radius of Earth's orbit, r=1.5×1011m$r=1.5\times {10}^{11}\mathrm{m}$
   Let the distance parallax angle be 1′′=4.847×10−6rad$1^{\mathrm{\prime }\mathrm{\prime }}=4.847\times {10}^{-6}\mathrm{r}\mathrm{a}\mathrm{d}$
   Let the distance of the star be D.
   Parsec is defined as the distance at which the average radius of the Earth's orbit subtends an angle of 1".
   ∴$\therefore$ We have θ=rD$\theta =\frac{r}{D}$
   D=rθ=1.5×10114.847×10−6$D=\frac{r}{\theta }=\frac{1.5\times {10}^{11}}{4.847\times {10}^{-6}}$
   =0.309×1017≈3.09×1016m$=0.309\times {10}^{17}\approx 3.09\times {10}^{16}\mathrm{m}$
   Hence, 1 parsec ≈3.09×1016m$\approx 3.09\times {10}^{16}\mathrm{m}$

10. Percentage error = ΔRR×100=±[ΔVV+ΔII]×100$\frac{\mathrm{\Delta }R}{R}\times 100=\pm \left[\frac{\mathrm{\Delta }V}{V}+\frac{\mathrm{\Delta }I}{I}\right]\times 100$
    =±[5100+210]×100$=\pm \left[\frac{5}{100}+\frac{2}{10}\right]\times 100$
    = ±25%

11. It is given that angular diameter, θ$\theta$ = 1920" and distance of the sun from earth, r =1.496 ×$\times$ 1011 m.
    Now let the diameter of sun = d meters
    θ=1920′′=1920×4.85×10−6$\theta ={1920}^{\mathrm{\prime }\mathrm{\prime }}=1920\times 4.85\times {10}^{-6}$
    = 9.31 ×$\times$ 10-3 rad
    Since, d = θ$\theta$r
    Therefore, d = 9.31×10−3×1.496×1011=1.39×109m$9.31\times {10}^{-3}\times 1.496\times {10}^{11}=1.39\times {10}^9\mathrm{m}$
    Thus, the diameter of the sun is 1.39 ×$\times$ 109 m

12. Paul Dirac was a British theoretical physicist who made fundamental contributions to the development of quantum mechanics, quantum field theory and quantum electrodynamics, and is particularly known for his attempts to unify the theories of quantum mechanics and relativity theory.
    One relation consists of some fundamental constants that give the age of the Universe by:
    =(e24πε0)2×1mpm2ec3G$={\left(\frac{e^2}{4\pi {\epsilon }_0}\right)}^2\times \frac{1}{m_p{m}_e^2{c}^{3}G}$
    Where,
    t = Age of Universe
    e = Charge of electrons = 1.6×10−19C$1.6\times {10}^{-19}C$
    ε0${\epsilon }_0$ = Absolute permittivity
    mp = Mass of protons = 1.67×10−27$1.67\times {10}^{-27}$ kg
    me = Mass of electrons = 9.1×10−31$9.1\times {10}^{-31}$ kg
    c = Speed of light = 3×108$3\times {10}^8$ m/s
    G = Universal gravitational constant =6.67×1011Nm2kg−2$=6.67\times {10}^{11}{\mathrm{N}\mathrm{m}}^2{\mathrm{k}\mathrm{g}}^{-2}$
    Also, 14πε0=9×109$\frac{1}{4\pi {\epsilon }_0}=9\times {10}^9$ Nm2/C2
    Substituting these values in the equation, we get
    t=(1.6×10−19)4×(9×109)2(9.1×10−31)2×1.67×10−27×(3×108)3×6.67×10−11$t=\frac{{\left(1.6\times {10}^{-19}\right)}^4\times {\left(9\times {10}^9\right)}^2}{{\left(9.1\times {10}^{-31}\right)}^2\times 1.67\times {10}^{-27}\times {\left(3\times {10}^8\right)}^3\times 6.67\times {10}^{-11}}$
    =(1.6)4×819.1×1.67×27×6.67×10−76+18+62+27−24+11s$=\frac{(1.6{)}^4\times 81}{9.1\times 1.67\times 27\times 6.67}\times {10}^{-76+18+62+27-24+11}s$
    =(1.6)4×819.1×1.67×27×6.67×365×24×3600$=\frac{(1.6{)}^4\times 81}{9.1\times 1.67\times 27\times 6.67\times 365\times 24\times 3600}$×10−76+18+62+27−24+11years$\times {10}^{-76+18+62+27-24+11}years$
    ≈6×10−9×1018$\approx 6\times {10}^{-9}\times {10}^{18}$ years
    = 6 billion years
    (which is the approximate age of our universe)

13. Time taken by a quasar light to reach Earth = 3 billion years
    =3×109$=3\times {10}^9$ years [since, 1 billion = 109]
    =3×109×365×24×60×60s$=3\times {10}^9\times 365\times 24\times 60\times 60s$[since 1 year = 365 days = 365 ×$\times$ 24 ×$\times$ 60 ×$\times$ 60 sec]
    We know that, speed of light = 3×108$3\times {10}^8$ m/s
    Now, distance between the Earth and a quasar (using the formula, distance = speed of light × time taken by the light to reach the earth)
    =(3×108)×(3×109×365×24×60×60)$=\left(3\times {10}^8\right)\times \left(3\times {10}^9\times 365\times 24\times 60\times 60\right)$
    =283824×1020$=283824\times {10}^{20}$m
    =2.83×1022$=2.83\times {10}^{22}$km
    This is the required distance between earth and a quasar.

14. 1. Correct
       y=αsin2πtT$y=\alpha \sin \frac{2\pi t}{T}$
       Dimension of y = M0L1T0
       Dimension of a = M0L1T0
       Dimension of sin2πtT=M0L0T0$\sin \frac{2\pi t}{T}=M^0{L}^0{T}^0$
       ∵$\because$Dimension of L.H.S = Dimension of R.H.S
       Hence, the given formula is dimensionally correct.
    2. Incorrect
       y = a sin vt
       Dimension of y = M0L1T0
       Dimension of a = M0L-1T0
       Dimension of vt = M0L1T−l×M0L0T1=M0L1T0$M^0{L}^1{T}^{-l}\times M^0{L}^0{T}^1=M^0{L}^1{T}^0$
       But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the given formula is dimensionally incorrect.
    3. Incorrect
       y=(αT)sin(tα)$y=\left(\frac{\alpha }{T}\right)\sin \left(\frac{t}{\alpha }\right)$
       Dimension of y = M0L1T0
       Dimension of αT=M0L1T−1$\frac{\alpha }{T}=M^0{L}^1{T}^{-1}$
       Dimension of ta=M0L−1Tl$\frac{t}{a}=M^0{L}^{-1}{T}^l$
       But the argument of the trigonometric function must be dimensionless, which is not so in the given case. Hence, the formula is dimensionally incorrect.
    4. Correct
       y=(α2–√)(sin2πtT+cos2πtT)$y=(\alpha \sqrt{2})\left(\sin 2\pi \frac{t}{T}+\cos 2\pi \frac{t}{T}\right)$
       Dimension of y = M0L1T0
       Dimension of a = M0L1T0
       Dimension of tT=M0I0T0$\frac{t}{T}=M^0{I}^0{T}^0$
       Since the argument of the trigonometric function must be dimensionless (which is true in the given case), the dimensions of y and a are the same. Hence, the given formula is dimensionally correct.
15. 1. Mean Value of diameter
       Dmean =d1+d2+d3+d4+d5+d66$\frac{d_1+d_2+d_3+d_4+d_5+d_6}{6}$ =1.328+1.330+1.325+1.326+1.334+1.3366$\frac{1.328+1.330+1.325+1.326+1.334+1.336}{6}$
       =7.9796$=\frac{7.979}{6}$= 1.3298 = 1.330 cm
    2. Absolute Error in different observations are
       Δ$\mathrm{\Delta }$D1 = Dmean - d1 =1.330−1.328$1.330-1.328$ =0.002 cm
       Δ$\mathrm{\Delta }$D2 = Dmean - d2 = 1.330−1.330$1.330-1.330$ = 0 cm
       Δ$\mathrm{\Delta }$D3 = Dmean - d3 = 1.330−1.325$1.330-1.325$ = + 0.005 cm
       Δ$\mathrm{\Delta }$D4 = Dmean - d4 =1.330−1.326$1.330-1.326$ = 0.004 cm
       Δ$\mathrm{\Delta }$D5 = Dmean - d5 =1.330−1.334$1.330-1.334$ = - 0.004 cm
       Δ$\mathrm{\Delta }$D6 = Dmean - d6 =1.330−1.336$1.330-1.336$ = - 0.006 cm
    3. Mean absolute error
       ΔD mean $\mathrm{\Delta }{D}_{\text{ mean }}$=|ΔD1|+|ΔD2|+|ΔD3|+|ΔD4|+|ΔD5|+|ΔD6|6$=\frac{\left|\mathrm{\Delta }{D}_1\right|+\left|\mathrm{\Delta }{D}_2\right|+\left|\mathrm{\Delta }{D}_3\right|+\left|\mathrm{\Delta }{D}_4\right|+\left|\mathrm{\Delta }{D}_5\right|+\left|\mathrm{\Delta }{D}_6\right|}{6}$
       =0.002+0+0.005+0.004+0.004+0.0066$=\frac{0.002+0+0.005+0.004+0.004+0.006}{6}$
       =0.0216$=\frac{0.021}{6}$= 0.0035 = 0.004 cm
    4. Fractional Error = δd=ΔD mean D mean =±0.0041.330$\delta d=\frac{\mathrm{\Delta }{D}_{\text{ mean }}}{D_{\text{ mean }}}=\pm \frac{0.004}{1.330}$= ±0.003 cm
    5. Percentage Error = δd×$\delta d\times$100 = ±0.003×$\times$100% = ±0.3%
    6. Diameter of wire = (1.330 ± 0.003) cm
       or D = 1.330 cm ± 0.3 %