100 rad s^-1 . The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder?

What is the magnitude of angular momentum of the cylinder about its axis?

Solution:

Explanation

Final Answer

• The kinetic energy associated with the rotation of the cylinder is 3125 J.
• The magnitude of angular momentum of the cylinder about its axis is 62.5 Js.

 If the radius of the sphere is increased keeping mass samewhich one of the following will not be affected?

A: Moment of inertia

B: Angular momentum

C: Angular velocity

D: Rotational kinetic energy

Solution:

Explanation

• In free space, a solid sphere rotates.
• Because there is no external tension, angular momentum will be conserved. 
• As r varies, the moment of inertia will vary as.1= 2/5 mr²
• As,1ω =constant the angular velocity will vary.
• As 1/21ω², rotational kinetic energy will also vary.

Final Answer

The correct option is B i.e Angular momentum.

 by brakes after atleast 6 m. If the same car is moving at a speed of 100 km h^-1 the minimum stopping distance is :

A: 6 m

B: 12 m

C: 18 m

D: 24 m

Solution:

Final Answer

The correct answer is D i.e.24m.

A: mr^2

B: mr^2/2

C: 2mr^25

D: mr^25

Solution:

Explanation

• We can use the folding technique to get the moment of inertia of a solid cylinder along its height.

If we compress the cylinder along its height (central axis), it is changed into a disc, but the mass distribution remains the same around the central axis.

• As a result, the moment of inertia of the disc at an axis running through its centre and perpendicular to its plane is equal to the moment of inertia of the disc about its height.

1 _{solid cylinder} =1 _disc

⇒1_{solid cylinder} =1/2_mr ²

Final Answer

The correct option is B i.e 1/2_mr².

A: impulse and momentum

B: specific heat and latent heat

C: moment of inertia and force

D: thrust and surface tension

Solution:

Explanation

• We have impulse and momentum here.
• Now, dimensional formula of impulse will be,[ MLT ^-1]
• And the dimensional formula for the momentum is [ MLT ^-1].
• So, impulse and momentum have the same dimensions.
• We have specific heat and latent heat
• Now, the dimensional formula of specific heat will be,[ M⁰L²T ^-2 K^-1 ].
• And The dimensional formula for latent heat is,[ M⁰L²T ^-2  ].
• As a result, the dimensional formulas for specific heat and latent heat differ.
• We have now moment of inertia and force
• The dimensional formula of moment of inertia is,[ ML^-2 ].
• And the dimensional formula of force is [ MLT^-2 ].
• As a result, inertia and force moments have distinct dimensions.
• Here we have thrust and surface tension. 
• The dimensional formula for thrust is same as force i.e  [ MLT^-2 ].
• And the dimensional formula for surface tension is  [ ML⁰T^-2 ]...
• As a result, the dimensional formulas for thrust and surface tension are different.

Final Answer

The correct answer is A i.e impulse and momentum have the same dimensions.