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MHT CET 2025 (25 Apr, Shift 2) · Power

Power question from MHT CET 2025 (25 Apr, Shift 2)

Q6MHT CET 2025 · 25 APR · SHIFT 2Easy

Power

A constant force F⃗=3i^−2j^−k^\vec{F} = 3\hat{i} - 2\hat{j} - \hat{k} newton has a displacement r⃗=2i^−3j^−3k^\vec{r} = 2\hat{i} - 3\hat{j} - 3\hat{k} metre in 2 second. The work done and the power are respectively

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Answer: (b)

  1. W=F⃗⋅r⃗=(3)(2)+(−2)(−3)+(−1)(−3)=6+6+3=15 JW = \vec{F} \cdot \vec{r} = (3)(2) + (-2)(-3) + (-1)(-3) = 6 + 6 + 3 = 15\text{ J}
  2. P=Wt=152=7.5 WP = \frac{W}{t} = \frac{15}{2} = 7.5\text{ W}
  3. So the answer is (b) 15 joule, 7.5 watt.
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