System of Particles and Rotational Motion - Result Question 78

«««< HEAD:content/english/neet-pyq-chapterwise/physics/system-of-particles-and-rotational-motion/system-of-particles-and-rotational-motion-result-question-78.md

84. A small object of uniform density rolls up a curved surface with an initial velocity ’ $v$ ‘. It reaches upto a maximum height of $\frac{3 v^{2}}{4 g}$ with respect to the initial position. The object is a

======= ####84. A small object of uniform density rolls up a curved surface with an initial velocity ’ $v$ ‘. It reaches upto a maximum height of $\frac{3 v^{2}}{4 g}$ with respect to the initial position. The object is a

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/system-of-particles-and-rotational-motion/system-of-particles-and-rotational-motion—result-question-78.md (a) solid sphere

(b) hollow sphere

(c) disc

(d) ring

[2013]

Show Answer

Answer:

Correct Answer: 84. (c)

Solution:

  1. (c)

From law of conservation of mechanical energy,

$\Rightarrow \frac{1}{2} I \omega^{2}+0+\frac{1}{2} mv^{2}=mg \times \frac{3 v^{2}}{4 g}$

$\Rightarrow \frac{1}{2} I \omega^{2}=\frac{3}{4} mv^{2}-\frac{1}{2} mv^{2}$

$\Rightarrow \frac{1}{2} I \omega^{2}=\frac{mv^{2}}{2}(\frac{3}{2}-1)$

or, $\frac{1}{2} I(\frac{V^{2}}{R^{2}})=\frac{mv^{2}}{4}$ or, $I=\frac{1}{2} mR^{2}$

Hence, object is a disc.