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Work Energy and Power Test - 21

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Work Energy and Power Test - 21
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  • Question 1
    1 / -0
    A body is moving in a vertical circle of radius '$$r$$' by a string. If the ratio of maximum to minimum speeds is $$\sqrt{3}:1$$ , the ratio of maximum to minimum tensions in the string is:
    Solution
    Let $$v_1$$ and $$v_2$$ be the velocities at the lowest and the highest points.
    $$\therefore \dfrac {v_1}{v_2}=\dfrac {\sqrt 3}{1}$$
    Conserving energy between the lowest and highest points gives
    $${v_1}^2={v_2}^2+2gh$$
    $$\Rightarrow 3{v_2}^2={v_2}^2+2gh$$
    $$\Rightarrow {v_2}^2=gh$$
    $$\Rightarrow {v_1}^2=3gh$$
    $$h$$ is distance between the lowest and the highest points which is $$2r$$.
    $$h=2r$$

    Tensions at the lowest and highest point
    $$T_1=\dfrac {m{v_1}^2}{r}+mg$$

    $$T_2=\dfrac {m{v_2}^2}{r}-mg$$

    $$\Rightarrow \dfrac {T_1}{T_2}=\dfrac {{v_1}^2/r+g}{{v_2}^2/r-g}$$

    $$\Rightarrow \dfrac {T_1}{T_2}=\dfrac {6gr/r+g}{2gr/r-g}$$

    $$\Rightarrow \dfrac {T_1}{T_2}=7/1$$
  • Question 2
    1 / -0
    A simple pendulum of length $$50 cm$$ is suspended from a fixed point $$O$$ and a nail is fixed at a point $$P$$ which is vertically below $$O$$ at some distance. The bob is released when string is horizontal. The bob reaches lowest position then it describes vertical circle whose centre coincides with point $$P$$. The minimum distance between $$O$$ and $$P$$ is:
    Solution
    When the ball just reaches the bottom, from conservation of energy, $$v^2=2gL$$
    But when it reaches the bottom, it describes a vertical circle. hence $$v^2=5gr$$
    where $$r$$ is the radius of the circle.
    Hence $$\dfrac{r}{L}=\dfrac{2}{5}$$
    $$\implies r=20cm$$
    Hence minimum distance between O and P=$$50cm-20cm=30cm$$
  • Question 3
    1 / -0
    Mass of the bob of a simple pendulum of length $$L$$ is $$m$$. If the bob is projected horizontally from its mean position with velocity $$\sqrt{4gL}$$ , then the tension in the string becomes zero after a vertical displacement of :
    Solution
    Let the angle $$\theta$$ be when tension becomes zero.
    Under this condition $$mg sin\theta = \dfrac{mv^{2}}{L}$$ ....... eq(1)

    Also, by conservation of energy:
    $$\dfrac { m (\sqrt {4gL})^{2}}{2} = mg(L+L sin \theta) + \dfrac{mv^{2}}{2}$$

    => $$\dfrac{mv^{2}}{L} = 2gmL - 2gmLsin \theta$$ ------- eq(2)

    From eq (1) and eq(2):
    $$mg sin\theta = 2mg - 2mg sin\theta$$
    $$sin \theta = 2/3$$

    So, the tension in the string becomes zero after a vertical displacement of:  $$L+Lsin\theta = 5L/3$$

  • Question 4
    1 / -0
    A mass $$0.1\ kg$$ is rotated in a vertical circle using the cord of length $$1\ m$$, when the cord makes an angle $$60^0$$ with the vertical, the speed of the mass is $$3\  m/s$$ the resultant  radial acceleration of mass in that position is:
    Solution
    Radial acceleration at that position , $$a_r = \dfrac{v^2}{r} - g cos \theta $$
    $$a_r = 9 - 4.9 = 4.1\  m/s^2 $$
  • Question 5
    1 / -0
    A coconut fruit hanging high in a palm tree has ......... owing to its location.
    Solution
    The potential energy possessed by the object is the energy present in it by virtue of its position or configuration. So, the hanging coconut has potential energy due to its location (height).
  • Question 6
    1 / -0
    Water stored in a dam possesses:

    Solution
    The potential energy possessed by the object is the energy present in it by virtue of its position or configuration. Water stored in a dam , when allowed to flow, has kinetic energy which was earlier stored as potential energy.
  • Question 7
    1 / -0
    When velocity of a moving object is doubled its:
    Solution
    $$ K.E = \dfrac{1}{2}mv^2$$

    $$ K.E \propto v^2 $$

    $$ New\ K.E = \dfrac{1}{2}m(2v)^2 = 4 \times K.E$$

    $$ \therefore $$ The K.E. of a body will increase to 4 times if its velocity doubles.
  • Question 8
    1 / -0
     In a hydro power plant:
    Solution
    Hydro power plant uses the potential energy stored in water. When water flows down the dam, potential energy is converted into kinetic energy which is used to rotate the turbine which produce electricity.
  • Question 9
    1 / -0
    Which of the following physical quantities is different from others?
    Solution
    Kinetic energy is the energy possessed by the body by virtue of its motion and potential energy is the energy possessed by the virtue of its position and shape. Energy is required to do work. 
    Hence work, kinetic energy, potential energy all can be measured using same unit, i.e joule. 
    So, force is the physical quantity here which is different from others as given by product of mass and acceleration. S.I unit of force is $$N$$ (newton).
  • Question 10
    1 / -0
    A ball is projected upwards. As it rises, there is increase in its:
    Solution
    When a ball is projected upwards it's height increases.
    As height increases, $$v$$ velocity decreases  (Kinetic Energy Decreases)  so Potential energy increases.
    $$Potential $$ $$ Energy$$= $$mgh$$
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