wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A wooden cube (density of wood 'd') of side '' floats in a liquid 'ρ' with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period 'T'. Then, 'T' is equal to :-

A
2π=ρ(ρd)g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
2π=dρg
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2π=ρρg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2π=d(ρd)g
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 2π=dρg
Let at any instant, the cube be at a depth x from the equilibrium position,
then Net force acting on the cube = Upthrust on the portion of length x .
Therefore F=ρl2xg-------(i)
where F is the force exerted , ρ is the density of the liquid l2×x is the volume and g is the acceleration due to gravity .
It performs simple harmonic motion and it can be seen that ,
Fx
As the sign is negative it is implied that the force is applied in the opposite direction .
Therefore the equation for simple harmonic motion is
F=kx where k is a constant . --------(ii)
Comparing the value of k with (i)
We get , k=ρl2g
Now time period T is equal to
T=2πmk
Or, T=2πl3dρl2g \\ where d is the density of the cube
Or,T=2πldρg (ANSWER)


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Archimedes' Principle
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon