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NEET · Physics · STD 11 - 5. work,energy,power and collision
A force \(\mathrm{F}=20+10 \mathrm{y}\) acts on a particle in \(y\) direction where \(\mathrm{F}\) is in newton and \(\mathrm{y}\) in meter. Work done by this force to move the particle from \(y=0\) to \(y=1 \;\mathrm{m}\) is......\(J\)
- A \(30\)
- B \(5\)
- C \(25\)
- D \(20\)
Answer & Solution
Correct Answer
(C) \(25\)
Step-by-step Solution
Detailed explanation
\(\mathrm{W}=\int_{\mathrm{y}_{\mathrm{1}}}^{y_{\mathrm{2}}} \mathrm{F} \mathrm{dy}\) \(\Rightarrow \mathrm{W}=\int_{0}^{1}(20+10 \mathrm{y}) \mathrm{dy}\)
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