| ⇦ |
| ⇨ |
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: - Time taken by light to cross a glass slab of thickness 4 mm and refractive index 3 is
- In electromagnetic wave, according to Maxwell, changing electric field gives
- In an a.c circuit the e.m.f. (e) and the current (i) at any instant are given
- Which of the following pairs of quantities does not have same dimensional
- A bomb of mass 30 kg at rest exploded into two pieces of masses 18 kg
Topics: Motion in Plane
(93)
Subject: Physics
(2479)
Important MCQs Based on Medical Entrance Examinations To Improve Your NEET Score
- Time taken by light to cross a glass slab of thickness 4 mm and refractive index 3 is
- In electromagnetic wave, according to Maxwell, changing electric field gives
- In an a.c circuit the e.m.f. (e) and the current (i) at any instant are given
- Which of the following pairs of quantities does not have same dimensional
- A bomb of mass 30 kg at rest exploded into two pieces of masses 18 kg
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