9th Class Science Work and energy Question Bank Work, Energy and Power

  • question_answer
    An elevator is designed to lift a load of 1000 kg, through 6 floors of a building averaging \[\text{3}\text{.5 m}\]per floor in 6 sec. Power of the elevator, neglecting other losses, will be

    A) \[\text{3}\text{.43 }\times \text{ 1}{{0}^{\text{4}}}\text{ watt}\]  

    B)         \[\text{4}\text{.33 }\times \text{ 1}{{0}^{\text{4}}}\text{ watt}\]

    C) \[\text{2}\text{.21 }\times \text{ 1}{{0}^{\text{4}}}\text{ watt}\]

    D)         \[\text{5}\text{.65 }\times \text{ 1}{{0}^{\text{4}}}\text{ watt}\]

    Correct Answer: A

    Solution :

     \[P=\frac{mgh}{t}=\frac{1000\times 9.8\times (6\times 3.5)}{6}\]         \[=\mathbf{3}\mathbf{.45\times 1}{{\mathbf{0}}^{\mathbf{4}}}\mathbf{watt}\]


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