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Work and Energy Test - 44

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Work and Energy Test - 44
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  • Question 1
    1 / -0
    $$4$$ bulbs rated $$100$$W each, operate for $$6$$ hours per day. What is the cost of the energy consumed in $$30$$ days at the rate of Rs. $$5$$/kWh?
    Solution
    $$\displaystyle E= Power(in kW)\times time=\frac{4\times 100\times 6}{1000}$$kWh$$=2.4$$kWh
    Consumed in $$30$$ days$$=30\times 2.4=72$$
    Total cost $$=72\times 5=360$$Rs.
  • Question 2
    1 / -0
    Energy can neither be created nor be destroyed, but it can be changed from one form to another. This law is known as:
    Solution
    The law of conservation of energy states that the total energy of an isolated system remains constant, it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed from one form to another.
    For instance, chemical energy is converted to kinetic energy when a stick of dynamite explodes. If one adds up all the forms of energy that were released in the explosion, such as the kinetic energy of the pieces, as well as heat and sound, one will get the exact decrease of chemical energy in the combustion of the dynamite.
     Classically, conservation of energy was distinct from conservation of mass; however, special relativity showed that mass could be converted to energy and vice versa by E = mc2, and science now takes the view that mass–energy is conserved.
  • Question 3
    1 / -0
    Which of the following is not an example of potential energy?
    Solution
    Whenever a body is at some height from the mean (base) position or at a state of extension (or contraction), the body will have some potential energy. But when the pendulum is passing the mean position, the height from the base is 0. Hence a vibrating pendulum at the mean position has no potential energy.
  • Question 4
    1 / -0
    In which of the following cases work is said to be done?
    Solution
    In the first case, the man is pushing a roller and it is moving so work is said to be done
    and in the other two both are sitting they are not moving so no work is said to be done. 
    $$\text{ work } = \text{ force} \times \text{displacement} $$
  • Question 5
    1 / -0
    The diagram below shows the path taken by a ball when Sundram kicks it. The potential energy of the ball is highest at ______________

    Solution

  • Question 6
    1 / -0
    What is the formula for electric power?
    Solution
    Formula for Electric Power is $$P = E/t$$
    where $$E$$= Energy in $$J$$ and $$t$$ is the time in second.
  • Question 7
    1 / -0
    When the speed of a body is doubled, its kinetic energy becomes 
    Solution
    The kinetic energy of a body is given by:
    $$KE=\dfrac{1}{2}mv^2$$

    Hence, Kinetic energy depends on the velocity as:
    $$KE\propto v^2$$

    So, if we double the velocity, then $$KE$$  becomes four times.

    Hence, option C is correct.
  • Question 8
    1 / -0
    A ball bounces to $$80$$% of its original height. What fraction of its potential energy is lost in each bounce?
    Solution
    Given,
    $$h'=\dfrac{80h}{100}$$
    where, $$h=$$ original height
    The kinetic energy of a ball striking the ground is equal to the potential energy of a ball at height $$h$$,
    $$K=mgh$$
    The kinetic energy due to rebounce is given by
    $$K_r=mgh'$$
    Loss of kinetic energy, $$\Delta K=K-K_r$$
    $$\Delta K=mgh-mgh'$$
    $$\Delta K=mg(h-\dfrac{80h}{100})$$
    $$\Delta K=\dfrac{20}{100}mgh=0.20mgh$$
    Fractional loss, $$\dfrac{\Delta K}{K}=\dfrac{0.20mgh}{mgh}=0.20$$
    The correct option is A.
  • Question 9
    1 / -0
    Which of the following statements is incorrect?
    Solution
    The kinetic energy of a body of mass mm which is moving with velocity vv is $$ K = \dfrac{1}{2}mv^2$$
    Mass is always positive and if the velocity is either positive or negative, the kinetic energy is always positive due to the square of velocity.
  • Question 10
    1 / -0
    The correct relation between joule and erg is:
    Solution

    Joule and erg both are units of work done. An erg is the amount of work done by applying a force of one dyne for a distance of one centimeter. In the CGS base units, it will be one gram centimeter-squared per second-squared. Whereas joule is the amount of work done by applying a force of one newton for a distance of one meter.

    Thus,

    $$1 joule = 1 newton \times 1 m\\$$

    $$1 joule = \dfrac{1 kg \times 1 m}{1 s ^{2}} \times 1 m\\$$

    $$1 J = \dfrac{1000 g \times 100 cm}{1 s ^{2}} \times 100 cm \\$$

    $$1 J = 10^{7} \ erg$$

    Thus option D is correct.

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