Tag: physics

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Questions Related to physics

Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

A small bucket containing water is rotated in a vertical circle of radius $R$ by means of a rope. $v$ is velocity of bucket at highest point. Then water does not fall down if:

  1. $v \leq \sqrt{gR}$

  2. $v \leq \sqrt{\frac{gR}{2}}$

  3. $v \geq \sqrt{gR}$

  4. $v \geq \sqrt{\frac{gR}{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

At highest point the centripetal acceleration should be greater than g.
$\therefore \dfrac{v^{2}}{R}\geq g$


or, $v\geq \sqrt{Rg}$

Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

A bucket filled with water is tied to a rope of length $0.5\ m$ and is rotated in a circular path in vertical plane. the least velocity it should have at the lowest point of circle so that water does not spill is $(g=10ms^{-2})$:

  1. $\sqrt{5}\ m/s$

  2. $\sqrt{10}\ m/s$

  3. $5\ m/ s$

  4. $2\sqrt{5}\ m/s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Minimum velocity at bottom $V=\sqrt{5g\ell}$ $=\sqrt{5\times 10\times 0.5}$$=5:m/s$


Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

A very small particle rests on the top of a hemisphere of radius $20\ cm$. The smallest horizontal velocity to be given to it, if it is to leave the hemisphere without sliding down its surface taking $g = 9.8 m/s^{2}$ is:

  1. $\sqrt{9.8}\ m/s$

  2. $\sqrt{4.9}\ m/s$

  3. $\sqrt{1.96}\ m/s$

  4. $\sqrt{3.92}\ m/s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Critical velocity at the top most point is $\sqrt{g\ell}$
$\therefore$ smallest velocity, such that the particle just leaves the surface =$\sqrt{9.8\times 0.2}=\sqrt{1.96}=1.4\ ms^{-1}$

Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

The minimum centripetal force required to rotate a body mass $m$ in a vertical circle of radius $r$ is

  1. $mg$

  2. $2mg$

  3. $\dfrac{mg}{2}$

  4. Zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The tension at the top-most point would be zero in the critical case that the particle just completes the verticle circle without being slake. The minimum centripetal force would then be equal to the force due to gravity $=mg$


Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

A particle of mass m is rotating by means of a string in a vertical circle. The difference in the tensions at the bottom and the top towards completion of a full revolution would be

  1. 6 mg

  2. 4 mg

  3. 5 mg

  4. 2 mg
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The tension at any point on the circular path is given by $T=(m/R)(u^2+gr-3gh)$; h is the height of the point from the horizontal

For a full circle, $h =2R$ Substituting, we get, $T _H=(m/R)(u^2-5gR)$ and 
at the bottom point of the circle , h=0 and thus, $T _L=(m/R)(u^2+gR)$

The difference in tension will be  $(T _L-T _H)=(m/R)(6gR) = 6mg$

Thus, the correct option is (a)

Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

Four objects of masses m, 2m , 3m and 4m are attached to a string of length L=2 metres and rotated vertically with a speed of $\sqrt(5Lg)$, pick out the most appropriate one:

  1. m

  2. 2m

  3. 4m

  4. All the objects will complete the circle
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For an object to complete one full circle, the initial velocity should be atleast $\sqrt{5Lg}$. Thus, it is independent of the mass used

The correct option is (d)

Multiple choice examples of circular motion uniform circular motion circular motion and gravitation physics

A block of mass $m$ at the end of a string is whirled round in a vertical circle of radius $R$. The critical speed of the block at the top of its swing below which the string would slacken before the block reaches the top is:

  1. $Rg$

  2. ${(Rg)}^{2}$

  3. $R/g$

  4. $\sqrt {Rg}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

At the top if the block is moving with critical speed the tension in the string would be zero and therefore force of gravity would provide the necessary centripetal force:

$mg=\cfrac { m{ v }^{ 2 } }{ R } \ \Rightarrow v=\sqrt { Rg } $