You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Varying Mass System

A block of ma...

Question

A block of mass $M$ is situated on a smooth horizontal table. A thread tied to the block passes through a hole in the table and carries a mass $m$ at its other end. If the length of the thread above the table is $l$ and $M$ is revolving in a horizontal circle with angular speed $ω$ on the table, then the value of $m$ so that it remains suspended at a constant height $h$ is:

{M g h ω}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{M g l ω}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{M {l ω}^{2}}{g}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

M {l ω}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $\frac{M {l ω}^{2}}{g}$
The mass $m$ will remain in constant height when its weight balance the centripetal force on $M$ .
i.e, $m g = M w^{2} l \Rightarrow m = \frac{M w^{2} l}{g}$

Suggest Corrections

Similar questions

Q. A block of mass

M

is situated on a smooth horizontal friction less table. A thread tied to a block passes through a hole in the table and carries a mass

m

at its other end. If the length of the thread above the table is

l

, what should be the value of

m

so that it may remain suspended at a constant height and the block

M

moves in a circular path with an angular velocity

ω

on the table?

Q. A particle of mass

m

is attached to one end of a weightless and inextensible string of length

L

. The particle is on a smooth horizontal table. The string passes through a hole in the table and to its other end is attached a small particle of equal mass

m

. The system is set in motion with the first particle describing a circle on the table with constant angular velocity

ω_{1}

and the second particle moving in the horizontal circle as a conical pendulum with constant angular velocity

ω_{2}

. Then

ω_{2}^{2} : ω_{1}^{2} =

[Take

r_{1} = 1 m, L = 3 m

]

Q. A puck of mass

100 gm

is tied to a string and allowed to revolve in a horizontal circle of radius

2 cm

on a frictionless horizontal table as shown in the figure. Other end of the string passes through a small hole at the centre of the table and a load of mass

5 kg

is tied to the string. The suspended load remains in equilibrium while the puck revolves on the table. If the tension in the string is

x (N)

and speed of the puck is

y (cm/s)

, then find the values of

x

and

y

respectively. Take

g = 10 {m/s}^{2}

Q. A puck of mass

100 gm

is tied to a string and allowed to revolve in a horizontal circle of radius

2 cm

on a frictionless horizontal table as shown in the figure. Other end of the string passes through a small hole at the centre of the table and a load of mass

5 kg

is tied to the string. The suspended load remains in equilibrium while the puck revolves on the table. If the tension in the string is

x (N)

and speed of the puck is

y (cm/s)

, then find the values of

x

and

y

respectively. Take

g = 10 {m/s}^{2}

Q. A block of mass

m_{1}

rests on a horizontal table. A string tied to the block is passed on a frictionless pulley fixed at the end of the table and to the other end of string is hung another block of mass

m_{2}

. The acceleration of the system is :

Join BYJU'S Learning Program

The correct option is D Mlω2gThe mass m will remain in constant height when its weight balance the centripetal force on M.i.e, mg=Mw2l⇒m=Mw2lg

The correct option is D $\frac{M {l ω}^{2}}{g}$
The mass $m$ will remain in constant height when its weight balance the centripetal force on $M$ .
i.e, $m g = M w^{2} l \Rightarrow m = \frac{M w^{2} l}{g}$