KCET · Physics · Oscillations
From a fixed support, two small identical spheres are suspended by means of strings of length \(1 \mathrm{~m}\) each. They are pulled aside as shown and then released. \(B\) is the mean position. Then the two spheres collide
- A at \(B\) after \(0.25 \mathrm{~s}\)
- B at \(B\) after \(0.5 \mathrm{~s}\)
- C on the right side of \(B\) after some time
- D on the right side of \(B\) when the strings are inclined at \(15^{\circ}\) with \(B\)
Answer & Solution
Correct Answer
(B) at \(B\) after \(0.5 \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
Irrespective of amplitude, both spheres take same time to reach the mean position, i.e., \(\frac{T}{4} \mathrm{~S}\)
\(\therefore T=2 \pi \sqrt{\frac{l}{g}}=2 \times 3.14 \times \sqrt{\frac{1}{10}}=2 \mathrm{~s}\)
So, they will collide at \(\frac{T}{4}=\frac{2}{4}=0.5 \mathrm{~s}\).
\(\therefore T=2 \pi \sqrt{\frac{l}{g}}=2 \times 3.14 \times \sqrt{\frac{1}{10}}=2 \mathrm{~s}\)
So, they will collide at \(\frac{T}{4}=\frac{2}{4}=0.5 \mathrm{~s}\).
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