AP EAMCET · PHYSICS · Laws of Motion
Two smooth inclined planes \(A\) and \(B\) each of height 20 m have angles of inclination \(30^{\circ}\) and \(60^{\circ}\) respectively. If \(t_1\) and \(t_2\) are respectively the times taken by two blocks to reach the bottom of the planes \(A\) and \(B\) from the top, then \(t_1-t_2=\) (Acceleration due to gravity \(=10 \mathrm{~m} \mathrm{~s}^{-2}\) )
- A \(\frac{\sqrt{3}-1}{\sqrt{3}} \mathrm{~s}\)
- B \(3(\sqrt{3}-1) \mathrm{s}\)
- C \(4\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) \mathrm{s}\)
- D \((3 \sqrt{3}-2) \mathrm{s}\)
Answer & Solution
Correct Answer
(C) \(4\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right) \mathrm{s}\)
Step-by-step Solution
Detailed explanation
\(t = \frac{1}{\sin\theta}\sqrt{\frac{2h}{g}}\) \(t_1 = \frac{1}{\sin 30^{\circ}}\sqrt{\frac{2 \times 20}{10}} = \frac{1}{1/2}\sqrt{4} = 2 \times 2 = 4 \mathrm{~s}\)…
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