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Units and Measurements Test - 57

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Units and Measurements Test - 57
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  • Question 1
    1 / -0
    The ratio of the dimensions of Planck constant and that of moment of inertia has the dimensions of
    Solution
    Step 1: Dimension formula of Planck constant:

    Energy of photon is given by:
    E=hνE=h \nu
    Where hh is Planck's constant and ν \nu is frequency of photon

    Now,  Planck's constant will be    h=Eνh = \dfrac{E}{\nu}

    Dimension of energy [E]=[ML2T2][E] = [ML^2T^{-2}]

    Dimension of frequency [ν]=[T1][ \nu ]=[T^{-1}]        {frequency is reciprocal of time period) 

    Dimensions of Planck's constant [h]=[ML2T2]/[T1]=[ML2T1][h]= [ML^2T^{-2}]/[T^{-1}] = [ML^2T^{-1}]

    Step 1: Dimension formula of moment of inertia:


    The moment of inertia (I)(I) is given by:
    I=MR2I=MR^2
    Where MM is mass of the obect and RR is distance.

    Dimensions of moment of inertia     [I]=[ML2][I] = [ML^2]

    Step 3: Dimension formula of their ratio:

    \therefore Dimensions of  hI\dfrac{h}{I} =[ML2T1][ML2]=[T1] =\dfrac{[ML^2T^{-1}]}{[ML^2]}=[T^{-1}]
    Thus ratio of Planck's constant and moment of inertia has dimensions of frequency.

    Option B is correct.
  • Question 2
    1 / -0
    If the angular distance between the stars turns out to be approximately 1100 arcseconds, or 0.30 degrees. The moon appears to shift 0.3 degrees when we observe it from two vantage points 2360 km apart, then find the distance of the moon from the surface of the earth.Given angular diameter of moon is 0.5 degrees.

    Solution
    The parallax angle in radians is given by:
    θrad=θdeg×π180\theta_{rad}=\theta_{deg}\times \dfrac{\pi}{180}=0.3×π180rad=0.3\times \dfrac{\pi}{180}rad

    Thus the distance between moon and surface of earth is: 
    D=dθD=\dfrac{d}{\theta}=2360km0.3×π180=\dfrac{2360km}{0.3\times \dfrac{\pi}{180}}=450642km=450642km
  • Question 3
    1 / -0
    Parallax angles _______ 0.01/arcsec0.01/ arcsec are very difficult to measure from Earth.
    Solution
    Parallax effect depends upon the path of light that travels from object to the observer. For distant objects observed from Earth, the light that reaches Earth has to go through a number of layers in Earth's atmosphere, during which it refracts and disperses and hence decreases the accuracy of the method. Resultantly parallax angles less than 0.01/arcsec0.01/arcsec are very difficult to measure from Earth.
  • Question 4
    1 / -0
    If a star is 5.2×1016 m5.2\times 10^{16}\ m away. What is the parallax angle in degrees?
    Solution
    Given :    11 AU =1.5×1011 = 1.5\times 10^11 m                d=5.2×1015d = 5.2\times 10^{15} m
    Parallax angle:     α=1AUd=1.5×10115.2×1016=0.288×105\alpha = \dfrac{1 AU}{d} =\dfrac{1.5\times 10^{11}}{5.2\times 10^{16}} = 0.288\times 10^{-5}  radians
        \implies   α=180π×0.288×105=1.67×104\alpha = \dfrac{180}{\pi} \times 0.288\times 10^{-5} = 1.67\times 10^{-4}  degrees
  • Question 5
    1 / -0
    A star is 1.45 parsec1.45\ parsec light years away. How much parallax would this star show when viewed from two locations of the earth six months apart in its orbit around the sun?
    Solution
    One light year == speed of light ×\times one year
    or 1ly=3×108×(24×3600)=94608×1011m1 ly=3\times 10^8\times (24\times 3600)=94608\times 10^{11} m
    So, 4.29ly=4.29×(94608×1011)=4.058×1016m4.29 ly=4.29\times (94608\times 10^{11})=4.058\times 10^{16} m
    As 11 parsec =3.08×1016m=3.08\times 10^{16} m
    (Parsec is a unit of length used to measure large distances to objects outside our Solar System)
    Thus, 4.29ly=4.058×10163.08×1016=1.324.29 ly=\dfrac{4.058\times 10^{16}}{3.08\times 10^{16}}=1.32 parsec
    Now angular displacement , θ=dD\theta=\dfrac{d}{D}
    where d=d= diameter of earth's orbit =3×1011m= 3\times 10^{11} m and 
    D=D= distance of star from the earth =4.058×1016m=4.058\times 10^{16} m
    So, θ=3×10114.058×1016=7.39×106\theta=\dfrac{3\times 10^{11}}{4.058\times 10^{16}}=7.39\times 10^{-6} rad
    As 1sec=4.85×106rad1 sec=4.85\times 10^{-6} rad so, 7.39×106rad=7.39×1064.85×106=1.52sec7.39\times 10^{-6} rad=\dfrac{7.39\times 10^{-6}}{4.85\times 10^{-6}}=1.52 sec
  • Question 6
    1 / -0
    Which of the following is a possible dimensionless quantity?
    Solution
    Displacement gradient in x-direction is given by  drdx\dfrac{dr}{dx}
    Since rr and xx have same dimension, this results the displacement gradient to be dimensionless quantity.
  • Question 7
    1 / -0
    5×107μs 5\times 10^7 \mu s is equivalent to ____
    Solution
    We know that  1μs=106 s1\mu s = 10^{-6} \ s
    Thus,  5×107μs=5×107μs×106 sμs=50 s5\times 10^{7}\mu s = 5\times 10^{7}\mu s\times \dfrac{10^{-6} \ s}{\mu s} = 50 \ s
  • Question 8
    1 / -0
    The dimension of hmv\dfrac{h}{mv} is: (hh== planks constant, mm= mass, vv=velocity)
    Solution
    Debroglie wavelength is given by: λ=hmv\lambda=\dfrac{h}{mv}
    Wavelength has dimension of length [L][L].
  • Question 9
    1 / -0
    The dimensional formula of coefficient of viscosity is
    Solution
    Coefficient of viscosity is given by     η=FAdzdv\eta =\dfrac{F}{A}\dfrac{dz}{dv}
    Dimension of force  F=[MLT2]F = [MLT^{-2}]
    Dimension of Area  A=[L2]A = [L^2]
    Dimension of velocity gradient  dvdz=[T1]\dfrac{dv}{dz} = [T^{-1}]
    Thus dimension of coefficient of viscosity   η=[MLT2][L2][T1]=[ML1T1]\eta = \dfrac{[MLT^{-2}]}{[L^2][T^{-1}]} = [ML^{-1}T^{-1}]
  • Question 10
    1 / -0
    A physical quantity QQ is found to depend on observables x,yx, y and zz, obeying relation Q=x2/5z3yQ=\dfrac{x^{2/5}z^{3}}{y}. The percentage error in the measurements of x,yx, y and zz are 1%1\%, 2%2\% and 4%4\% respectively. What is percentage error in the quantity QQ will be :
    Solution
    The given quantity      Q=x3y2zQ = \dfrac{x^3y^2}{z}

    Percentage error in Q   ΔQQ× 100=3Δxx×100\dfrac{\Delta Q}{Q} \times  100 = 3\dfrac{\Delta x}{x}\times 100 +2Δyy× 100+Δzz×100+2\dfrac{\Delta y}{y} \times  100 + \dfrac{\Delta z}{z}\times 100

    Percentage error in x,yx,y and zz are  1%1\%, 2%2\% and 4%4\% respectively.

        \implies   ΔQQ× 100=3(1)+2(2)+1(4)=11\dfrac{\Delta Q}{Q} \times  100 = 3(1) + 2(2)+1(4) = 11
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