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Units and Measurements Test - 82

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Units and Measurements Test - 82
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  • Question 1
    1 / -0
    In the following list, the only pair which have different dimensions,is
    Solution
    Linear momentum is given as:
    p=$$Mas \times Velocity =[MLT^{-1}]$$

    Moment of a force is given as:
    Moment =$$ Force\times Distance=[ML^2T^{-2}]$$
  • Question 2
    1 / -0
    The dimensions of inter atomic force constant are
    Solution
    We know that inter atomic force is given as:
    $$K=Y \times r_0=[ML^{-1}T^{-2}]\times [L]=[MT^{-2}]$$
    Y = Young's modulus and $$r_0$$ = Interatomic distance
  • Question 3
    1 / -0
    If R and L represent respectively resistance and self inductance,which of the following combinations has the dimensions of frequency
    Solution
    We know that ohm law is represented as:
    $$V=IR\Rightarrow R=\dfrac VI$$

    Whereas Inductance $$L$$ is given as:
    $$L=V\times\dfrac TI$$

    So, the ratio is given as:
    $$\dfrac{R}{L}=\dfrac{V/I}{V \times T /I}=\dfrac{1}{T}$$=Frequency.
  • Question 4
    1 / -0
    The dimensions of surface tension are
    Solution
    The surface tension is given as:
    Surface tension =$$\dfrac{Force}{Length}$$

    $$=\dfrac{[MLT^{-2}]}{L}=[MT^{-2}]$$
  • Question 5
    1 / -0
    The dimensions of $$CV^2$$ matches with the dimensions of
    Solution
    The energy of the capacitor is given by $$E=\dfrac 12 CV^2$$

    So,  $$CV^2$$ has the dimension of energy. Similarly, $$\dfrac12 Li^2$$ also has the dimension of energy.

    So, Both are the formula of energy.$$(E=\dfrac{1}{2}CV^2=\dfrac{1}{2}LI^2)$$

    Thus, $$Li^2$$ and $$CV^2$$ have the same dimensions.
  • Question 6
    1 / -0
    The pair having the same dimensions is
    Solution
    The work done is defined as the product of work done and displacement.
    $$Work = Force\times displacement$$

    The torque is also defined as the product of force and perpendicular distance.
    $$\tau=Force\times perpendicular\ distance$$

    So, Dimension of work and torque =$$[ML^2T^{-2}]$$
  • Question 7
    1 / -0
    Which physical quantities have the same dimension  
    Solution
    Couple of force =$$|\underset{r}{\rightarrow} \times \underset{F}{\rightarrow}|=[ML^2T^{-2}]$$

    Work=$$[\underset{r}{\rightarrow}. \underset{F}{\rightarrow}]=[ML^2T^{-2}]$$

    So, option $$A$$ is correct
  • Question 8
    1 / -0
    The dimensional formula for young's modulus is
    Solution

  • Question 9
    1 / -0
    Which of the following quantities is dimensionless
    Solution

  • Question 10
    1 / -0
    The dimensions of pressure is equal to
    Solution

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