
A bucket full of water is revolved in a vertical circle of radius 2 m. What should be the maximum time-period of revolution, so that the water doesn't fall out of the bucket?
A.1 sec
B.2 sec
C.3 sec
D.4 sec
Answer
475.5k+ views
Hint: As the bucket rotates, centrifugal acts on the bucket. So, use the formula for centrifugal force. But, gravitational force acts on the water in the bucket. Thus, use the formula for gravitational force. Equate the equations of both the forces and obtain the velocity with which the bucket is revolving. And using the obtained value of velocity calculate the maximum time period of revolution.
Complete answer:
Given: radius r= 2m
Let v be the velocity of a vertical circle.
As the bucket is rotating, centrifugal force acts on it.
Centrifugal Force should be greater than gravitational force so that the water doesn’t fall out of the bucket.
$\Rightarrow {F}_{c}= F$ ...(1)
Centrifugal force is given by,
${F}_{c}= \dfrac{m{v}^{2}}{r}$
Where, m is the mass of the object
Gravitational Force is given by,
F=mg
Where, m is the mass of the object
g is the acceleration due to gravity
Substituting these values in the equation. (1) we get,
$\dfrac{m{v}^{2}}{r}= mg$
$\Rightarrow {v}^{2}= rg$
$\Rightarrow v= \sqrt{rg}$
Circular velocity is given by,
$v= \dfrac{2\pi r}{T}$
Rearranging the above equation we get,
$T= \dfrac{2\pi r}{v}$
Substituting value of r and g in above equation we get,
$T= \dfrac{2\pi r}{\sqrt{rg}}$
Now, substituting value of v in above equation we get,
$T= \dfrac{2\pi × 2}{ \sqrt{2 × 9.8}}$
$\Rightarrow T=\dfrac {4\pi}{ \sqrt{19.6}}$
$\Rightarrow T=\dfrac {12.57}{ 4.43}$
$\Rightarrow T= 2.84 sec$
Therefore, the maximum time period of revolution should be 2.84 sec $\simeq 3 sec$.
Hence, the correct answer is option C i.e. 3sec.
Note:
When a bucket of water is raised and inverted, the water is strongly pulled by the force of gravity of earth's surface and therefore it falls.
When a bucket of water is raised in the vertical circle, the water is pushed away from the hand, towards the base of the bucket, by a force which is directed away from the hand.
This force when balances gravity, the water would just be in the bucket. But when you spin the bucket this outward force named as centrifugal force overcomes earth's gravitational pull and pushes the water to the end of the bucket. So, it does not drop from the open end of the bucket.
Complete answer:
Given: radius r= 2m
Let v be the velocity of a vertical circle.
As the bucket is rotating, centrifugal force acts on it.
Centrifugal Force should be greater than gravitational force so that the water doesn’t fall out of the bucket.
$\Rightarrow {F}_{c}= F$ ...(1)
Centrifugal force is given by,
${F}_{c}= \dfrac{m{v}^{2}}{r}$
Where, m is the mass of the object
Gravitational Force is given by,
F=mg
Where, m is the mass of the object
g is the acceleration due to gravity
Substituting these values in the equation. (1) we get,
$\dfrac{m{v}^{2}}{r}= mg$
$\Rightarrow {v}^{2}= rg$
$\Rightarrow v= \sqrt{rg}$
Circular velocity is given by,
$v= \dfrac{2\pi r}{T}$
Rearranging the above equation we get,
$T= \dfrac{2\pi r}{v}$
Substituting value of r and g in above equation we get,
$T= \dfrac{2\pi r}{\sqrt{rg}}$
Now, substituting value of v in above equation we get,
$T= \dfrac{2\pi × 2}{ \sqrt{2 × 9.8}}$
$\Rightarrow T=\dfrac {4\pi}{ \sqrt{19.6}}$
$\Rightarrow T=\dfrac {12.57}{ 4.43}$
$\Rightarrow T= 2.84 sec$
Therefore, the maximum time period of revolution should be 2.84 sec $\simeq 3 sec$.
Hence, the correct answer is option C i.e. 3sec.
Note:
When a bucket of water is raised and inverted, the water is strongly pulled by the force of gravity of earth's surface and therefore it falls.
When a bucket of water is raised in the vertical circle, the water is pushed away from the hand, towards the base of the bucket, by a force which is directed away from the hand.
This force when balances gravity, the water would just be in the bucket. But when you spin the bucket this outward force named as centrifugal force overcomes earth's gravitational pull and pushes the water to the end of the bucket. So, it does not drop from the open end of the bucket.
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