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Answer (Detailed Solution Below)

Option 3 : 18.0 J / 18.0 जूल

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Detailed Solution

सही उत्तर 18.0 J / 18.0 जूल है।

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Question:

A hoop of mass $2$ kg rolls without slipping at $3$ m/s. Its total kinetic energy is / $2$ kg का एक छल्ला बिना फिसले $3$ m/s से लुढ़कता है। इसकी कुल गतिज ऊर्जा है

1.
9.0 J / 9.0 जूल
2.
12.6 J / 12.6 जूल
3.
18.0 J / 18.0 जूल
4.
27.0 J / 27.0 जूल

Correct Answer: 18.0 J / 18.0 जूल

$K = \dfrac{1}{2}mv^{2} + \dfrac{1}{2}I\omega^{2} = \dfrac{1}{2}mv^{2} + \dfrac{1}{2}mv^{2} = mv^{2} = 2 \times 9 = 18.0$ J. / $K = \dfrac{1}{2}mv^{2} + \dfrac{1}{2}I\omega^{2} = \dfrac{1}{2}mv^{2} + \dfrac{1}{2}mv^{2} = mv^{2} = 2 \times 9 = 18.0$ जूल।

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