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Units and Measurements Test - 62

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Units and Measurements Test - 62
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  • Question 1
    1 / -0
    The dimensional formula of electric potential is
    Solution
    Electric potential,
    $$\displaystyle V = \frac{W}{q} = \frac{[ML^2T^{-2}]}{[AT]} = [ML^2T^{-3}A^{-1}]$$
  • Question 2
    1 / -0
    Which of the following relation is dimensionally incorrect?
    Solution
    As, we know, $$E = mc^2$$
    and energy released can be given as: 
    $$E=931MeV$$
    for, $$m =1u$$
    $$1u = 931.5 MeV/c^{2}$$ - Dimensionally correct [Dimension of mass]
    $$1u=1.67\times 10^{-27}\ kg$$ - Dimensionally correct [Dimension of mass]
    but, $$931.5 \ MeV$$ is having the dimension of energy so it can't be equated to dimension of mass or dimension of $$1u$$.
    Hence, dimensionally correct data is given in $$option(A)$$
    Hence, the correct option will be $$(A)$$
  • Question 3
    1 / -0
    The velocity of a particle (v) at an instant t is given by v =at + bt$$^{2}$$. The dimension of b is the
    Solution
    $$v = at + bt^2$$

    $$[v] = [bt^2]$$ or $$[LT^{-1}] = [bT^2]$$ or $$[b] = [LT^{-3}]$$
  • Question 4
    1 / -0
    Dimensional formula of $$\Delta Q$$, heat supplied to the system is
    Solution
    Heat supplied to a system is in the form of energy.
    $$\therefore$$ Its dimensional formula is $$\displaystyle [ML^2T^{-2}]$$.
  • Question 5
    1 / -0
    The dimensional formula of physical quantity is $$[M^aL^bT^c]$$. Then that physical quantity is
    Solution
    1) Surface Tension $$[S] = \dfrac{Force [F]}{Length [L]}$$

    So, $$S = \dfrac{[M^1 L^1 T^{-2}]}{[L^1]}$$

    $$\therefore$$ Surface Tension $$= S = [M^1 L^0 T^{-2}]$$


    2) Force $$[F] = $$ mass (m) $$\times $$ acceleration (a)

                          $$= [M^1] \times [L^1 T^{-2}]$$

                     $$F = [M^1 L^1 T^{-2}]$$

    3) Angular frequency $$ = w = \dfrac{red [d \phi ]}{\sec [dt]}$$

                                             $$w = [M^0 L^0 T^{-1}]$$

    4) Spring constant $$[K] = \dfrac{Force \ applied [F]}{Extension \ of \ spring [x]}$$

                                           $$= \dfrac{[M^1 L^1 T^{-2}]}{[L^1]}$$

                                      $$K = [M^1 T^{-2}]$$

                                      $$K = [M^1 L^0 T^{-2}]$$

    $$\therefore$$ Option C is correct.
  • Question 6
    1 / -0
    The displacement of a progressive wave is represented by y = A sin($$\omega$$t - kx) where x is distance and t is time. The dimensions of $$\displaystyle \frac{\omega}{k}$$ are same as those of the
    Solution
    $$y = A sin (\omega t - kx)$$
    As $$(\omega t - kx)$$ represents an angle which is dimensionless, therefore
    $$\displaystyle [\omega] = \frac{1}{[t]} = [T^{-1}]$$ and $$\displaystyle [k] = \frac{1}{[x]} = [L^{-1}]$$
    $$\displaystyle \left[ \frac{\omega}{k} \right] = \frac{[T^{-1}]}{[L^{-1}]} = [LT^{-1}]$$ = velocity
  • Question 7
    1 / -0
    The mean length of an object is 5 cm. Which of the following measurements is most accurate. ?
    Solution
    To signify the mean length of an object comprising $$5$$ cm in dimension, it is better to present it with $$4.9$$ cm can be easily rounded up to $$5$$ cm.
    Whereas, other options are either less than or greater than the given option.
  • Question 8
    1 / -0
    The physical quantity to be measured more accurately in the experimental determination of Young's modulus of a wire is
    Solution
    In the experimental determination of Young's modulus, the extension  $$\delta L$$ in the wire is measured and using this Young's modulus is calculated by the relation :
    $$Y = \dfrac{F \ L}{A \ \delta L}$$
  • Question 9
    1 / -0
    The dimensional formula for electric intensity is
    Solution
    Since $$E=\cfrac { F }{ q } ,\left[ E \right] =\cfrac { \left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ -2 } \right]  }{ \left[ AT \right]  } =\left[ { M }^{ 1 }{ L }^{ 1 }{ T }^{ 3 }{ A }^{ -1 } \right] $$
  • Question 10
    1 / -0
    Consider two cylindrical rods of identical dimensions, one of rubber and the other of steel. Both the rods are fixed rigidly at one end to the roof. A mass $$M$$ is attached to each of the free ends at the centre of the rods.
    Solution
    According to the diagram shown below in which a man $$AT$$ is attached at the center of each rod, then both rods will be elongated. But due to different elastic properties of the material the steel rod will elongate without making any perceptible change in shape, but the rubber rod will elongate with the shape of the bottom edge tapered to a tip at the centre. 

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