JEE Main 20244 AprilEvening Shift
- Exam
- JEE Main
- Year
- 2024
- Date
- 4 April
- Shift
- Evening Shift
- Subject
- Physics
- Chapter
- Work, Power and Energy
JEE Main 2024 · 4 April · Evening Shift · Physics · Work, Power and Energy
A body of m ~kg slides from rest along the curve of vertical circle...
A body of slides from rest along the curve of vertical circle from point to in friction less path. The velocity of the body at is:
Options
- A
- B
- C
- D
Correct answer
D.
Step-by-step solution
Apply W.E.T. from A to B
\begin{aligned}
& \Rightarrow \mathrm{W}_{\mathrm{mg}}=\mathrm{K}_{\mathrm{B}}-\mathrm{K}_{\mathrm{A}} \\
& \Rightarrow \mathrm{mg} \times\left(\frac{\mathrm{R}}{\sqrt{2}}+\mathrm{R}\right)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}^2-0\left\{\mathrm{v}_{\mathrm{A}}=0 \text{ rest }\right\} \\
& \Rightarrow \mathrm{mgR} \frac{(\sqrt{2}+1)}{\sqrt{2}}=\frac{1}{2} \mathrm{mv}_{\mathrm{B}^2 \\
& \Rightarrow \sqrt{\mathrm{gR} \frac{2(\sqrt{2}+1)}{\sqrt{2}}}=\mathrm{v}_{\mathrm{B}} \\
& \Rightarrow \sqrt{\frac{10 \times 14 \times 2(2.4)}{1.4}}=\mathrm{v}_{\mathrm{B}} \\
& \Rightarrow 21.9=\mathrm{v}_{\mathrm{B}}
\end{aligned}