Search any question & find its solution
Question:
Answered & Verified by Expert
A body initially at rest and sliding along a frictionless track from a height $\mathrm{h}$ (as shown in the figure) just completes a vertical circle of diameter $\mathrm{AB}=\mathrm{D}$. 'The height $h$ is equal to.
Options:
Solution:
2815 Upvotes
Verified Answer
The correct answer is:
$\frac{5}{4} \mathrm{D}$
The correct option is $\mathbf{3}: \frac{5}{4} \mathrm{D}$
Let total energy is conserved,
Total mechanical energy at start $=$ Total mechanical energy at ' $\mathrm{B}$ ' $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mgD}...(i)$
Minimum speed required to complete rotation is, $\mathrm{v}_{\mathrm{b}}=\sqrt{\mathrm{rg}}=\sqrt{\mathrm{Dg} / 2}$ Hence equation (1) becomes, $\mathrm{mgh}=\frac{1}{2} \mathrm{~m} \frac{\mathrm{D}_2}{2}+\mathrm{mgD}$
hence, $h=\frac{5 \mathrm{D}}{4}$
Let total energy is conserved,
Total mechanical energy at start $=$ Total mechanical energy at ' $\mathrm{B}$ ' $\mathrm{mgh}=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mgD}...(i)$
Minimum speed required to complete rotation is, $\mathrm{v}_{\mathrm{b}}=\sqrt{\mathrm{rg}}=\sqrt{\mathrm{Dg} / 2}$ Hence equation (1) becomes, $\mathrm{mgh}=\frac{1}{2} \mathrm{~m} \frac{\mathrm{D}_2}{2}+\mathrm{mgD}$
hence, $h=\frac{5 \mathrm{D}}{4}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.