TS EAMCET · Physics · Laws of Motion
At time \(t=0\), a force \(\mathrm{F}=\alpha t\), where \(t\) is time in seconds, is applied to a body of mass \(1 \mathrm{~kg}\), resting on a smooth horizontal plane. If the direction of the force makes an angle of \(45^{\circ}\) with the horizontal, then the velocity of the body at the moment of its breaking off the plane is
- A \(\frac{100}{\alpha} \mathrm{m} / \mathrm{s}\)
- B \(\frac{50 \sqrt{2}}{\alpha} \mathrm{m} / \mathrm{s}\)
- C \(\frac{50 \alpha}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)
- D \(\frac{50}{\alpha} \mathrm{m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(\frac{50 \sqrt{2}}{\alpha} \mathrm{m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
Let at \(\mathrm{t}=\mathrm{t}_0\), body leaves the plate Then, at \(\mathrm{t}=\mathrm{t}_0 \stackrel{\mathrm{N}}{\Rightarrow}=0\) and \(\mathrm{V}=\mathrm{V}_{\text {max }}\) So, \(\Delta_{\mathrm{O}}+\alpha \mathrm{t}_0 \sin 45^{\circ}=\mathrm{mg}\)…
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