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Units and Measurements Test - 53

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Units and Measurements Test - 53
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  • Question 1
    1 / -0
    Instrument used to measure the temperature is
    Solution
    A thermometer is a device that measures the temperature of a body.
  • Question 2
    1 / -0
    What is the dimensional formula of gravitational constant?
    Solution

    $$F = \dfrac {G.M.M}{L^{2}} = MLT^{-2}$$

    $$G = \dfrac {MLT^{-2} L^{2}}{M^{2}} = M^{-1} L^{3} T^{-2}$$

  • Question 3
    1 / -0
    The dimensional formula of magnetic flux is
    Solution

    $$F = Bqv \rightarrow F = \dfrac {\phi}{A} qv$$

    $$\phi = \dfrac {F . A}{qv} = \dfrac {MLT^{-2} L^{2}}{ATLT^{-1}} = ML^{+2} A^{-1} T^{-2}$$

  • Question 4
    1 / -0
    Write the dimensional formula of Force.
    Solution
    Force $$=$$ mass $$\times $$ acceleration
    $$[F]=[M]\times [LT^{-2}]=MLT^{-2}$$
  • Question 5
    1 / -0
    Dimensions of electrical resistance are
    Solution

  • Question 6
    1 / -0
    Using mass ($$M$$), length ($$L$$), time ($$T$$) and current ($$A$$) as fundamental quantities, the dimensional formula of permittivity is
    Solution

    $$F = \dfrac {1}{4\pi \epsilon_{0}} \cdot \dfrac {q_{1} q_{2}}{r^{2}} \Rightarrow \epsilon_{0} = \dfrac {q_{1}q_{2}}{4\pi Fr^{2}}$$

    $$\dfrac {AT \cdot AT}{MLT^{-2} L^{2}} = \dfrac {A^{2}T^{2}}{ML^{3} T^{-2}} = M^{-1} L^{-3} A^{2} T^{4}$$

  • Question 7
    1 / -0
    The dimensions of torque are
    Solution

  • Question 8
    1 / -0
    Using mass $$(M)$$, length $$(L)$$, time $$(T)$$ and current $$(A) $$as fundamental quantities, the dimensions of permeability are
    Solution

    $$B = \mu ni; F = Bqv$$

    $$F = \mu ni qv \Rightarrow \mu = \dfrac {F}{niqv}$$

    $$\mu = \dfrac {MLT^{-2}}{\dfrac {1}{L} A.AT.LT^{-1}}$$ [n is no. of turns per unit length]

       $$= MLA^{-2} T^{-2}$$

  • Question 9
    1 / -0
    The relation between force $$F$$ and density $$d$$ is $$F = \dfrac{x}{\sqrt{d}}$$. The dimensions of $$x$$ are
    Solution
    Given, $$F = \dfrac{x}{\sqrt{d}}$$
    Substituting dimensions
    $$ML{T}^{-2} = \dfrac{x}{\sqrt{M{L}^{-3}}}$$
    $$\Rightarrow          x = \left[{M}^{{3}/{2}}{L}^{{-1}/{2}}{T}^{-2}\right]$$
  • Question 10
    1 / -0
    The dimensional formula for torque is
    Solution

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