Solids And Fluids of Class 11
Question
A uniform cylindrical block of length l density d1 and area of cross section A floats in a liquid of density d2 contained in a vessel (d2 > d1). The bottom of the cylinder just rests on a spring of constant k. The other end of the spring is fixed to the bottom of the vessel. Determine the weight (in N) that may be placed on top of the cylinder such that the cylinder is just submerged in the liquid. (Take l = 1m, 
, d2 = 2000 kg/m3, K = 1000 N/m, A = 0.1 m2 and g = 10 m/s2)
, d2 = 2000 kg/m3, K = 1000 N/m, A = 0.1 m2 and g = 10 m/s2)
Frequently Asked Questions
A solid cylinder of mass 20 kg rotates about its axis with angular speed
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
.png)
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.
A solid sphere is rotating in free space
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.
A car moving with a speed of 50 km h^-1 can be stopped
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.
What is moment of inertia of a solid cylinder of radius r along its height
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
⇒1solid cylinder =1/2mr 2
Final Answer
The correct option is B i.e 1/2mr2.
Which of the following have the same dimensions
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.
