Tag: physics

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Questions Related to physics

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

An inclined plane ends into vertical loop of radius $R$. A particle is released from height $3R$. Can it loop the loop?

  1. yes

  2. no

  3. cannot say

  4. yes if fraction is present
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The minimum speed required at the bottom of the circle(r) to complete the circular motion is $\sqrt {5gR}$
For this speed, the minimum height required is
$0.5mv^2=mgh$
$\Rightarrow 0.5(5gR)=gh$
$\Rightarrow h=\dfrac {5}{2}R$
So, the minimum height from which the body has to released is 2.5R. We have a height of 3R, so the particle completes the vertical loop.
Option A.

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 block reaches the top is

  1. $\sqrt{2rg}$

  2. $\sqrt{3rg}$

  3. $\sqrt{rg}$

  4. $\sqrt{5rg}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

the string will not slack if the centrifugal force on the particle is greater then or equal to the weight of mass i.e.
$\dfrac{mv^2}{r} \ge mg \ \Rightarrow v \ge \sqrt{rg} \ \Rightarrow v _c=\sqrt{rg} $

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

The maximum tension that an inextensible ring of radius 1 m and mass density 0.1 kg ${ m }^{ -1 }$ can bear is 40 N . The maximum angular velocity with which it can be rotated in a  circular path is 

  1. 20 rad/s

  2. 18 rad/s

  3. 16 rad/s

  4. 15 rad/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a rotating ring, the tension T = lambda * v^2 = lambda * (omega * R)^2. Given T = 40 N, lambda = 0.1 kg/m, R = 1 m: 40 = 0.1 * omega^2 * (1)^2. omega^2 = 400, so omega = 20 rad/s.

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

A body is allowed to slide on a frictionless track from rest under-gravity.The track ends in a circular of diameter D. What should be the mini-mum height of the body in terms of D.          So that it may successfully complete the loop?

  1. $\dfrac{9}{4}D $

  2. $ \frac {5}{4} D $

  3. D

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

Velocity at top of the circle is $\sqrt{5gr}$ 

Energy lost to reach height D from height H  $ = mg(H-D)$

Energy gained in the form of $K.E= \dfrac12 mv^2= \dfrac12 m(5gr) = \dfrac{5mgD}{4}$ 

Equating K.E. and P.E. ,

$\dfrac{5mgD}{4} = mg(H-D)$

$H=\dfrac{9D}{4}$

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

A body is a allowed to slide down a frictionless track from rest position at its top under gravity. The track ends in a circular loop of diameter $D$. Then, the minimum height of the inclined track  (in terms of $D$ ) so that it may complete successfully the loop is:

  1. $\dfrac{7D}{4}$

  2. $\dfrac{9D}{4}$

  3. $\dfrac{5D}{4}$

  4. $\dfrac{3D}{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$V _{0}$ required for a complete vertical circle: $V _{0} =\sqrt{\dfrac{5}{2}gD}$
WET between top and bottom:
              $mgH  = \dfrac{1}{2}m V _{0}^{2}$


                 $gH  = \dfrac{5}{4}gD$

                   $H  = \dfrac{5}{4}D$

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

The length of simple pendulum is $1m$ and mass of its bob is $50 gram$. The bob is given  sufficient velocity so that the bob describes vertical circle whose radius equal to length of pendulum, the tension in the string at lowest extreme position is:

  1. $2.5 N$

  2. $1N$

  3. $1.5 N$

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

Using conservation of energy, 

$mg \times 2R = mv^2/2$
$v=2 \sqrt{Rg}$ .........(1)

At lower most point, from free body digram,
$T - mg = m v^2/R$ ...........(2)
From (1) and (2), 
$T = 5mg$
$T = 2.5N$

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

A weightless thread can bear tension up to $3.7kg-wt$. A stone of mass $500g$ is tied to it and revolved in a circular path of radius $4$m in a verticle plane. If $g=10m/s^2$, then the maximum angular velocity of the stone will be:

  1. $2 rad/s$

  2. $4 rad/s$

  3. $16 rad/s$

  4. $\sqrt{21} rad/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Tension is maximum at lowest point of a verticle circle $T _{max}=mr\omega^2+mg$
$3.7g=0.5\times 4\times \omega^2+0.5g$
$2\omega^2=3.2g$
$\omega=\sqrt{1.6g}=\sqrt{1.6\times 10}$
$\omega=4rad/s$

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

A body attached to a string of length describes a vertical circle such that it is just able to cross the highest point. Find the minimum velocity at the bottom of the circle.

  1. <span>$\sqrt { 2gL } $</span>

  2. <span>$\sqrt { gL } $</span>

  3. <span>$\sqrt { 3gL } $</span>

  4. <span>$\sqrt { 5gL } $</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Radius of the circle traced will be equal to the length of the string
In circular motion tension in the string is balanced by the centripetal force
hence we know that,
$T=\dfrac { m{ v }^{ 2 } }{ r } $ 
$T=\dfrac { m{ v }^{ 2 } }{ L } $
Velocity at any point,$V=\sqrt { \dfrac { TL }{ m }  } $
And at lowest point the tension is balanced by the weight of the block
∴∴ tension = weight = mg
$V=\sqrt { \dfrac { mgL }{ m }  } =\sqrt { gL } $

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

A bucket is whirled in a vertical circle with a string attached to it. The water in the bucket does not fall down even when the bucket is inverted at the top of its path. We can say that in this position.

  1. $mg=\displaystyle\frac{mv^2}{r}$

  2. mg is greater than $\displaystyle\frac{mv^2}{r}$

  3. mg is not greater than $\displaystyle\frac{mv^2}{r}$

  4. mg is not less than $\displaystyle\frac{mv^2}{r}$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

When the bucket is at the top of its path, the forces acting on the water are gravity and centripetal force. Since, the water does not fall down. Therefore, centripetal force is greater than the force of gravity.

Therefore, $\cfrac{mV^2}{R}\ge mg$