| ⇦ | 
| ⇨ |
A small object of uniform density rolls up a curved surface with an initial velocity v’.It reaches upto a maximum height of 3v² / 4g with respect to the initial position.The object is
Options
(a) ring
(b) solid sphere
(c) hollow sphere
(d) disc
Correct Answer:
disc
Explanation:
No explanation available. Be the first to write the explanation for this question by commenting below.
Related Questions: - A body starts from rest, what is the ratio of the distance travelled by the body
- The velocity v of waves produced in water depends on their wavelength λ, the density
- Which one of the following is a vector?
- Two identical circular coils A and B are kept on a horizontal tube
- Four blocks of same mass connected by cords are pulled by a force F
Topics: Motion in Plane
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- A body starts from rest, what is the ratio of the distance travelled by the body
- The velocity v of waves produced in water depends on their wavelength λ, the density
- Which one of the following is a vector?
- Two identical circular coils A and B are kept on a horizontal tube
- Four blocks of same mass connected by cords are pulled by a force F
Topics: Motion in Plane (93)
Subject: Physics (2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
18000+ students are using NEETLab to improve their score. What about you?
Solve Previous Year MCQs, Mock Tests, Topicwise Practice Tests, Identify Weak Topics, Formula Flash cards and much more is available in NEETLab Android App to improve your NEET score.
Share this page with your friends
Conservation of mechanical energy:
Initial total energy=Final total energy
i.e. (K.E.+P.E.)1=(K.E.+P.E.)2
1/2mv^2{1+K^2/R^2}+0=0+mgh
1/2mv^2{1+K^2/R^2}=mg(3v^2/4g)
1/2{1+K^2/R^2}=3/4
1+K^2/R^2=3/2
Hence,K^2/R^2=1/2
which implies that the small object is disc