
What torque will increase the angular velocity of a solid disc of mass $16kg$ and diameter $1m$ from $0{\text{ to 2rpm}}$ in$8{\text{ }}\sec ond$?
$(A){\text{ }}\dfrac{\pi }{4}N - m$
$(B){\text{ }}\dfrac{\pi }{2}N - m$
$(C){\text{ }}\dfrac{\pi }{3}N - m$
$(D){\text{ }}\pi N - m$
Answer
214.2k+ views
Hint: Torque is the product of the moment of inertia and the angular and the angular acceleration. We will calculate torque for the solid disk around the center. And for this, we have to first calculate the inertia for the center of the solid disk.
Formula used
Torque,
$ \Rightarrow \tau = I\alpha $; Where $\tau $is the torque, $I$ is the inertia, and $\alpha $is the angular acceleration.
Moment of Inertia for the center of the solid disk,
$ \Rightarrow I = \dfrac{1}{2}M{R^2}$; Where $M$ is mass of the object, $R$ is the radius.
Angular acceleration,
$ \Rightarrow \alpha = \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$ ; Where $n$ is the change in the rpm and $t$ is the time taken.
Complete step by step answer:
Since it is given in the question that there is a circular disk having mass 16kg and radius 1m. So we have to calculate the torque required for this.
As we know
$ \Rightarrow \tau = I\alpha $
And for calculating torque we have to first calculate the Moment of Inertia and angular acceleration.
For this we will use the below two formulas:
$ \Rightarrow I = \dfrac{1}{2}M{R^2}$
And
$ \Rightarrow \alpha = \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$
Combining and putting the equations in one
$ \Rightarrow \tau = I\alpha $
And we already know the values of these. So putting those values and we will get the required torque.
$ \Rightarrow \dfrac{1}{2}M{R^2} \times \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$
After substituting the values,
$ \Rightarrow 16{\left( {\dfrac{1}{2}} \right)^2} \times \dfrac{{\pi \left( {2 - 0} \right)}}{8}$
We get,
$ \Rightarrow \pi N - m$
So the torque required to increase the angular velocity of a solid circular disk will be\[\pi N - m\].
Therefore the Option D will be the right choice for this question.
Note: So when a force is applied to an object it begins to rotate with an acceleration reciprocally proportional to its moment of inertia. This relation is thought of like Newton’s Second Law for rotation. The moment of inertia is the rotational mass, and therefore, the force is rotational. Angular motion obeys Newton’s 1st Law.
Formula used
Torque,
$ \Rightarrow \tau = I\alpha $; Where $\tau $is the torque, $I$ is the inertia, and $\alpha $is the angular acceleration.
Moment of Inertia for the center of the solid disk,
$ \Rightarrow I = \dfrac{1}{2}M{R^2}$; Where $M$ is mass of the object, $R$ is the radius.
Angular acceleration,
$ \Rightarrow \alpha = \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$ ; Where $n$ is the change in the rpm and $t$ is the time taken.
Complete step by step answer:
Since it is given in the question that there is a circular disk having mass 16kg and radius 1m. So we have to calculate the torque required for this.
As we know
$ \Rightarrow \tau = I\alpha $
And for calculating torque we have to first calculate the Moment of Inertia and angular acceleration.
For this we will use the below two formulas:
$ \Rightarrow I = \dfrac{1}{2}M{R^2}$
And
$ \Rightarrow \alpha = \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$
Combining and putting the equations in one
$ \Rightarrow \tau = I\alpha $
And we already know the values of these. So putting those values and we will get the required torque.
$ \Rightarrow \dfrac{1}{2}M{R^2} \times \dfrac{{2\pi \left( {{n_2} - {n_1}} \right)}}{t}$
After substituting the values,
$ \Rightarrow 16{\left( {\dfrac{1}{2}} \right)^2} \times \dfrac{{\pi \left( {2 - 0} \right)}}{8}$
We get,
$ \Rightarrow \pi N - m$
So the torque required to increase the angular velocity of a solid circular disk will be\[\pi N - m\].
Therefore the Option D will be the right choice for this question.
Note: So when a force is applied to an object it begins to rotate with an acceleration reciprocally proportional to its moment of inertia. This relation is thought of like Newton’s Second Law for rotation. The moment of inertia is the rotational mass, and therefore, the force is rotational. Angular motion obeys Newton’s 1st Law.
Recently Updated Pages
Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction, Transfer of Energy Important Concepts and Tips for JEE

JEE Analytical Method of Vector Addition Important Concepts and Tips

Atomic Size - Important Concepts and Tips for JEE

JEE Main 2022 (June 29th Shift 1) Maths Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Equation of Trajectory in Projectile Motion: Derivation & Proof

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Angle of Deviation in a Prism – Formula, Diagram & Applications

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Collision: Meaning, Types & Examples in Physics

