Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Moment of inertia of a body about a given axis is $1.5\:kg\:m^{2}$ . Initially the body is ate rest. In order to produce a rotational kinetic energy of $1200\:J$ , the angular acceleration of $20\:rad/s^{2}$ must be applied about the axis for a duration of :
Option 1)
$2.5\:s$
Option 2)
$2\:s$
Option 3)
$5\:s$
Option 4)
$3\:s$

Answers (1)

best_answer

Given

$K.E.= 1200 J$

$\Rightarrow \frac{1}{2}I\:w^{2}=K.E.$

$\Rightarrow w=\sqrt{\frac{2K.E.}{I}}$

$\tau\:t=I\omega$

$\Rightarrow \tau\: t=I \sqrt{ \frac{2K.E.}{I}}$

$\Rightarrow \: t=\frac{I \sqrt{ \frac{2K.E.}{I}}}{I \:\alpha }$

$\Rightarrow \: t= \sqrt{\frac{2K.E.}{I}}\times \frac{1}{\alpha }$

$\Rightarrow \: t= \sqrt{\frac{2\times 1200}{1.5}}\times \frac{1}{20 }=2\:\:seconds$

Option 1)

$2.5\:s$

Option 2)
$2\:s$
Option 3)

$5\:s$

Option 4)

$3\:s$

Posted by

Aadil

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’

anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year

anam-khan

What after JEE Main results 2024? All you need to know about JEE Adv, CSAB, JoSAA counselling

Latest Question