TS EAMCET · Physics · Laws of Motion
An iron rod of length \(1.5 \mathrm{~m}\) lying on a horizontal table is pulled up form one end along a vertical line so as to move it with a constant velocity \(3 \mathrm{~m} / \mathrm{s}\), while the other end of the rod slides along the floor. After how much time the speed of the end sliding on the floor equals to the speed of the end being pulled up.
- A \(\frac{1}{2 \sqrt{2}} \mathrm{~s}\)
- B \(\frac{1}{{2}} \mathrm{~s}\)
- C \(3 \sqrt{2} \mathrm{~s}\)
- D \(\frac{1}{4} \mathrm{~s}\)
Answer & Solution
Correct Answer
(B) \(\frac{1}{{2}} \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
According to question. Given, \(A B\) is an iron rod whose length is given by \( l=1.5 \mathrm{~m} \) velocity of end \(\mathrm{A}\), \( v=3 \mathrm{~m} / \mathrm{s} \) Let \(t\) be the time when both the ends have same speed. Since, end \(A\) has always a constant velocity,…
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