A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ, where θ is the angle by which it has rotated, is given as kθ2. If its moment of inertia is I then the angular acceleration of the disc is:

\(\frac{{\rm{k}}}{{4{\rm{I}}}}{\rm{\theta }}\)

\(\frac{{\rm{k}}}{{{\rm{I}}}}{\rm{\theta }}\)

\(\frac{{\rm{k}}}{{2{\rm{I}}}}{\rm{\theta }}\)

\(\frac{{\rm{2k}}}{{{\rm{I}}}}{\rm{\theta }}\)

A stationary horizontal disc is free to rotate about its axis. When a

1.	A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ, where θ is the angle by which it has rotated, is given as kθ2. If its moment of inertia is I then the angular acceleration of the disc is:
A.	\(\frac{{\rm{k}}}{{4{\rm{I}}}}{\rm{\theta }}\)
B.	\(\frac{{\rm{k}}}{{{\rm{I}}}}{\rm{\theta }}\)
C.	\(\frac{{\rm{k}}}{{2{\rm{I}}}}{\rm{\theta }}\)
D.	\(\frac{{\rm{2k}}}{{{\rm{I}}}}{\rm{\theta }}\)
Answer» E.

Discussion

No Comment Found

Related MCQs

A slab is subjected to two forces F1 and F2 of same magnitude as shown in the figure. Force F2 is in xy-plane while force F1 acts along Z-axis at the point \(\left( {2\hat i + 3\hat j} \right).\) The moment of these forces about point O will be
Particle undergoes uniform circular motion. About which point on the plane of the circle, the angular momentum of the particle will remains conserved?
A solid sphere & a hollow sphere of radius R are rolling down in inclined lane of height h. The ratio of velocities of solid sphere to Hollow sphere on reaching the bottom is
If a disc has moment of inertia I about the axis which is tangential and in plane of the disc, then the moment of inertia about the axis which is tangential and perpendicular to its plane will be:
If the radius of earth is 6400 km. then angular velocity for a point on its equator will be -
A uniform rectangular thin sheet ABCD of mass M has length ‘a’ and breadth ‘b’, as shown in the figure. If the shaded portion HBGO is cut-off, the coordinates of the centre of mass of the remaining portion will be:
Moment of inertia of a thin circular ring of mass M and R rotating about an axis, passing through its centre and perpendicular to the plane is
Charge is distributed within a sphere of radius R with a volume charge density \(\rho \left( r \right) = \frac{{\rm{A}}}{{{{\rm{r}}^2}}}{{\rm{e}}^{ - 2{\rm{r}}/{\rm{a}}}}\), where A and a are constants. If Q is the total charge of this charge distribution, the radius R is:
A circular disc of radius b has a hole of radius ‘a’ at its centre (see figure). If the mass per unit area of the disc varies as \(\left( {\frac{{{\sigma _0}}}{{\rm{r}}}} \right)\), then the radius of gyration of the disc about its axis passing through the centre is:
In an experiment, electrons are accelerated, from rest by applying a voltage of 500 V. Calculate the radius of the path, if a magnetic field 100 mT is then applied.(Take, charge of the electron = 1.6 × 10-19 C and mass of the electron = 9.1 × 10-31 kg)

A stationary horizontal disc is free to rotate about its axis. When a torque is applied on it, its kinetic energy as a function of θ, where θ is the angle by which it has rotated, is given as kθ2. If its moment of inertia is I then the angular acceleration of the disc is:

Discussion

No Comment Found

Related MCQs

Reply to Comment