A solid cylinder of mass 2 Kg and radius 4 cm is rotating about its axis at the rate of 3 rpm . The torque required to stop after 2 \pi revolution is : 

 

  • Option 1)

    2 \times 10 ^{-6} Nm

  • Option 2)

    2 \times 10 ^{-3} Nm

  • Option 3)

    12 \times 10 ^{-4} Nm

  • Option 4)

    2 \times 10 ^{6} Nm

 

Answers (1)
V Vakul

\omega _2 ^2 = \omega _1 ^2 + 2 \alpha \theta \\\\ 0 = ( 2 \pi*\frac{3}{60} ) ^2 - 2 \times 2 \pi \times 2 \pi \alpha \\\\ \alpha = \frac{1}{800} \\\\ T = I\alpha = \frac{mr ^2}{2} d \\\\ = 2 \times 10 ^{-6}


Option 1)

2 \times 10 ^{-6} Nm

Option 2)

2 \times 10 ^{-3} Nm

Option 3)

12 \times 10 ^{-4} Nm

Option 4)

2 \times 10 ^{6} Nm

Preparation Products

Knockout NEET May 2021 (One Month)

An exhaustive E-learning program for the complete preparation of NEET..

₹ 14000/- ₹ 6999/-
Buy Now
Knockout NEET May 2021

An exhaustive E-learning program for the complete preparation of NEET..

₹ 22999/- ₹ 14999/-
Buy Now
Test Series NEET May 2021

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 6999/- ₹ 4999/-
Buy Now
Rank Booster NEET 2021

This course will help student to be better prepared and study in the right direction for NEET..

₹ 13999/- ₹ 9999/-
Buy Now
Knockout NEET May 2022

An exhaustive E-learning program for the complete preparation of NEET..

₹ 34999/- ₹ 24999/-
Buy Now
Boost your Preparation for NEET 2021 with Personlized Coaching
 
Exams
Articles
Questions