MHT CET · Physics · Work Power Energy
A rod of length 'L' is hung from its one end and a mass 'm' is attached to its free end. What tangential velocity must be imparted to 'm', so that it reaches the top of the vertical circle? (g = acceleration due to gravity)
- A \(4 \sqrt{\mathrm{gL}}\)
- B \(2 \sqrt{\mathrm{gL}}\)
- C \(5 \sqrt{\mathrm{gL}}\)
- D \(3 \sqrt{\mathrm{gL}}\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{\mathrm{gL}}\)
Step-by-step Solution
Detailed explanation
\(2 \sqrt{g L}\)
\(h=2 L\)
\(\frac{1}{2} \mathrm{mv}^{2}=\mathrm{mg} 2 \mathrm{L}\)
\(v_{2}=4 g L\)
\(v=2 \sqrt{g L}\)
\(h=2 L\)
\(\frac{1}{2} \mathrm{mv}^{2}=\mathrm{mg} 2 \mathrm{L}\)
\(v_{2}=4 g L\)
\(v=2 \sqrt{g L}\)
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