Self Studies

Units and Measurements Test - 29

Result Self Studies

Units and Measurements Test - 29
  • Score

    -

    out of -
  • Rank

    -

    out of -
TIME Taken - -
Self Studies

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Self Studies Self Studies
Weekly Quiz Competition
  • Question 1
    1 / -0
    The dimensions of universal gas constant are
    Solution

  • Question 2
    1 / -0
    Dimensions of impulse are
    Solution
    Impulse == Force ×\times time =MLT2T=MLT1 =MLT^{-2}T=MLT^{-1}
  • Question 3
    1 / -0
    Which of the following set have different dimensions?
    Solution
    Electric dipole moment=charge ×\times distance between charges=[AT]×[L]=[ALT]=[AT]\times [L]=[ALT]
    Electric flux == Electric field strength x area=[MLT3A1][L2]=[ML3T3A1]=[MLT^{-3}A^{-1}][L^2]=[ML^3T^{-3}A^{-1}]
    Electric field == Force per unit charge=[MLT3A1]=[MLT^{-3}A^{-1}]
  • Question 4
    1 / -0
    Which of these is dimensionless?
    Solution
    i) Forceacceleration=kg.m.s2m.s2=Kg\displaystyle \frac {Force}{acceleration}=\frac {kg.m.s^{-2}}{m.s^{-2}}=Kg

    ii) Velocityacceleration=m.s1m.s2=s\displaystyle \frac {Velocity}{acceleration}=\frac {m.s^{-1}}{m.s^{-2}}=s

    iii) VolumeArea=m3m2=m\displaystyle \frac {Volume}{Area}=\frac {m^3}{m^2}=m

    iv) EnergyWork=JJ=1\displaystyle \frac {Energy}{Work}=\frac {J}{J}=1
  • Question 5
    1 / -0
    The dimensions of Rydberg's constant are
    Solution
    Using the relation, 1λ=R(1n121n22)\dfrac {1}{\lambda}=R\left (\dfrac {1}{n_1^2}-\dfrac {1}{n_2^2}\right ) where n1& n2 { n }_{ 1 }\&  { n }_{ 2 }denote orbit number  and so are dimensionless
    Therefore dimensions of R=1L=L1=[M0L1T0]R=\frac {1}{L}=L^{-1}=[M^0L^{-1}T^0]
  • Question 6
    1 / -0
    Which one of the following represents the correct dimensions of the coefficient of viscosity?
    Solution
    From Stokes' law: Fd=6πηRvF_d=6\pi \eta Rv
    η=Fd6πRv=MLT2(L)(LT1)=ML1T1\displaystyle \eta =\frac {F_d}{6\pi Rv}=\frac {MLT^{-2}}{(L)(LT^{-1})}=ML^{-1}T^{-1}
  • Question 7
    1 / -0
    Identify the pair whose dimensions are equal.
    Solution

  • Question 8
    1 / -0
    If LL denotes the inductance of an inductor through which a current ii is flowing, the dimensions of Li2Li^2 are
    Solution
    Energy stored in an inductor =12Li2=[Energy]=[ML2T2]=\frac {1}{2}Li^2=[Energy]=[ML^2T^{-2}]
  • Question 9
    1 / -0
    The dimensions of Hubble's constant are
    Solution
    Hubble's constant, H=velocitydistance=[LT1][L]H=\dfrac {velocity}{distance}=\dfrac {[LT^{-1}]}{[L]}
    The Hubble Constant is the unit of measurement used to describe the expansion of the universe. 

  • Question 10
    1 / -0
    [ML2T2][ML^2T^{-2}] are dimensions of
    Solution
    Force =[ML1T2] =[ML^1T^{-2}]
    Moment of force =r×F=[L][MLT2]=[ML2T2]=r\times F=[L][MLT^{-2}]=[ML^2T^{-2}].....correct
    Momentum  =[ML1T1] =[ML^1T^{-1}]
    Power  =[ML2T3] =[ML^2T^{-3}]
Self Studies
User
Question Analysis
  • Correct -

  • Wrong -

  • Skipped -

My Perfomance
  • Score

    -

    out of -
  • Rank

    -

    out of -
Re-Attempt Weekly Quiz Competition
Self Studies Get latest Exam Updates
& Study Material Alerts!
No, Thanks
Self Studies
Click on Allow to receive notifications
Allow Notification
Self Studies
Self Studies Self Studies
To enable notifications follow this 2 steps:
  • First Click on Secure Icon Self Studies
  • Second click on the toggle icon
Allow Notification
Get latest Exam Updates & FREE Study Material Alerts!
Self Studies ×
Open Now