Laws of Motion Test

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Laws of Motion Test - 45

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Question 1
1 / -0

A $$500 g$$ particle tied to one end of a string is whirled in a vertical circle of circumference $$14 m$$.If the tension at the highest point of its path is $$2 N$$, what is its speed? (in m/s)

A
$$4.365$$

B
$$5.369$$

C
$$5.546$$

D
$$4.331$$

Solution

as the diagram shows
balancing the force at point $$P$$
given that $$T=2N$$, $$m=0.5kg$$ and $$R=\dfrac { 14 }{ 2\pi }=2.23m $$
$$\Rightarrow T+mg=m\dfrac { { v }^{ 2 } }{ R } $$
$$\Rightarrow v=\sqrt { R\dfrac { (T+mg) }{ m } } $$
$$\Rightarrow v=\sqrt { 2.23\times \dfrac { (2+0.5\times 10) }{ 0.5 } } =5.546m/s$$
Question 2
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Three charged particles are collinear and are in equilibrium, then

A
all the charged particle have the same poarity

B
the equilibrium is unstable

C
all the charged particles cannot have the same polarity

D
both (b) and (c) are correct.
Solution
- If a charged particle is in equilibrium under electrostatic force, then that particle never be in stable equilibrium.
  Hence option B is correct.
- Since all charges are in equilibrium, net force an any one of the charges are in equilibrium,so net force an any one of charges should be zero-for this two equal and opposite forces will act on any charge.
  Hence option C is correct.
Question 3
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A car moves at a speed of $$20\ ms^{-1}$$ on a banked track and describes an arc of a circle of radius $$40\sqrt {3}$$. The angle of banking is (take, $$g = 10ms^{-2})$$.

A
$$25^{\circ}$$

B
$$60^{\circ}$$

C
$$45^{\circ}$$

D
$$30^{\circ}$$

E
$$40^{\circ}$$

Solution

Given, $$v = 20\ ms^{-1}$$
$$r = 40\sqrt {2} m$$
$$\Rightarrow g = 10\ ms^{-2}$$
We know that, $$\tan \theta = \dfrac {v^{2}}{rg}$$
$$\Rightarrow \tan \theta = \dfrac {(20)^{2}}{40\sqrt {3}\times 10}$$
$$\Rightarrow \tan \theta = \dfrac {1}{\sqrt {3}}$$ or $$\theta = \tan^{-1}\left (\dfrac {1}{\sqrt {3}}\right )$$
$$\Rightarrow \theta = 30^{\circ}$$.
Question 4
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A $$200 g$$ mass is whirled in a vertical circle making $$60$$ revolutions per minute. What is the tension in the string at the top of the circle if the radius of the circle is $$0.8 m$$? (in $$N$$)

A
$$6.3$$

B
$$2.36$$

C
$$4.35$$

D
$$4.23$$

Solution

$$ w = 60 \space \text {revolutions per minute }= 60 * 2\pi / 60 \space\space\text {rad/ sec} = 2\pi \space\space rad/ sec $$
$$ 200 gm = 0.2 kg $$
$$ T = mrw^2 - mg $$
$$ T = 0.2* 0.8 * (2\pi)^2 - 0.2* 9.8 $$
$$ T = 4.35 N$$
Question 5
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Mark correct option or options.

A
The body of greater mass needs more forces to move due to more inertia

B
Force versus time graph gives impulse

C
Microscopic area of contact is about $$10^{-4}$$ times actual area of the contact

D
All of the above

Solution

When the mass of the body is higher, then it needs more force to move because of inertia. Impulse is defined as force per unit time.
Microscopic area of contact is about $$10^{-4}$$ times actual area of the contact, It is known facts.
Therefore all the given statements are correct.
So the correct option is $$D$$
Question 6
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Keeping the banking angle same to increase the maximum speed with which a vehicle can travel on the curved road by $$10\%$$, the radius of curvature of the road has to be changed from $$20\ m$$ to-

A
$$16\ m$$

B
$$18\ m$$

C
$$24.2\ m$$

D
$$30.5\ m$$

Solution

For speed v, we have
$$\displaystyle \tan { \theta } =\frac { { v }^{ 2 } }{ rg } =\frac { { v }^{ 2 } }{ 20g } $$......(i)
If $$\displaystyle { v }^{ \prime }=v+0.1 v=1.1v$$, then
$$\displaystyle \tan { \theta } =\frac { { v }^{ \prime 2 } }{ { r }^{ \prime }g } =\frac { { \left( 1.1v \right) }^{ 2 } }{ { r }^{ \prime }g } $$....(ii)
Equating (1) and (ii), we get $$\displaystyle \quad \frac { 1 }{ 20 } =\frac { 1.21 }{ { r }^{ \prime } } $$
$$\displaystyle \Rightarrow { r }^{ \prime }=20\times 1.21=24.2m$$
Question 7
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A stone of mass m is tied to a string and is moved in a vertical circle of radius r making n revolution per minute. The total tension in the string when the stone is at its lowest point is

A
$$mg$$

B
$$m(g+\pi n r^2)$$

C
$$m(g+ n r)$$

D
$$m(g+ \frac{\pi^2 n^2 r} {900})$$

Solution

Tension in string when it reach lower-most point
$$\displaystyle T=mg+m \omega^2 r$$
$$\displaystyle =m(g+4 \pi^2 n^2 r)$$ [as $$\omega=2 \pi n$$]
$$\displaystyle = m \left( g + 4 \pi^2 \left( \frac{n}{60} \right)^2 r \right)$$
$$\displaystyle = m \left( g + \left( \frac{\pi^2 n^2 r}{900} \right) \right)$$
Question 8
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A particle is moving in a vertical circle. The tension in the string when passing through two platform at angles $$\displaystyle 30^o$$ and $$\displaystyle 60^o$$ vertical (lowest position) are $$T_1$$ and $$T_2$$ respectively

A
$$\displaystyle T_1 = T_2$$

B
$$\displaystyle T_2 > T_1$$

C
$$\displaystyle T_1 > T_2$$

D
Tension in the string always remain the same

Solution

Tension, $$\displaystyle T= \frac{mv^2}{r} + mg cos \theta$$
For $$\displaystyle \theta =30^o$$
$$\displaystyle T_1= \frac{mv^2}{r} + mg cos 30^o$$
$$\displaystyle = \frac{mv^2}{r} + \frac{\sqrt{3}}{2} mg$$
For $$\displaystyle \theta =60^o$$
$$\displaystyle T_2= \frac{mv^2}{r} + mg cos 60^o$$
$$\therefore$$ $$\displaystyle \frac{mv^2}{r} + \frac{1}{2} mg$$
$$T_1> T_2$$
Question 9
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A weightless thread can bear tension up to $$3.7$$kg wt. A stone of mass $$500$$g is tied to it and revolved in a circular path of radius $$4$$m in a vertical plane. If $$g=10m/s^2$$, then the maximum angular velocity of the stone will be :

A
$$4rad/s$$

B
$$2rad/s$$

C
$$6rad/s$$

D
none of these

Solution

Maximum tension $$=\displaystyle\frac{mv^2}{r}+mg$$
$$3.7\times 10=\displaystyle\frac{0.5v^2}{4}+0.5\times 10$$
$$\Rightarrow v=16m/s$$
$$\therefore \omega =\displaystyle\frac{v}{r}=\frac{16}{4}=4rad/s$$.
Question 10
1 / -0

Railway tracks are banked on curves

A
necessary centrifugal force may be obtained from the horizontal component weight of the train

B
to avoid frictional force between the tracks and wheels

C
necessary centripetal force may be obtained from the horizontal component of the weight of the train

D
the train may not fly off in the opposite direction

Solution

The banking is provided to counteract the centrifugal force which acts (away from the center of the curve) on the train. Here component of weight is towards center providing necessary centripetal force as shown in diagram.

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Re-Attempt Weekly Quiz Competition

Laws of Motion Test - 45

Result

Laws of Motion Test - 45

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SHARING IS CARING

Question 1

$$4.365$$

$$5.369$$

$$5.546$$

$$4.331$$

Solution

Question 2

all the charged particle have the same poarity

the equilibrium is unstable

all the charged particles cannot have the same polarity

both (b) and (c) are correct.

Solution

Question 3

$$25^{\circ}$$

$$60^{\circ}$$

$$45^{\circ}$$

$$30^{\circ}$$

$$40^{\circ}$$

Solution

Question 4

$$6.3$$

$$2.36$$

$$4.35$$

$$4.23$$

Solution

Question 5

The body of greater mass needs more forces to move due to more inertia

Force versus time graph gives impulse

Microscopic area of contact is about $$10^{-4}$$ times actual area of the contact

All of the above

Solution

Question 6

$$16\ m$$

$$18\ m$$

$$24.2\ m$$

$$30.5\ m$$

Solution

Question 7

$$mg$$

$$m(g+\pi n r^2)$$

$$m(g+ n r)$$

$$m(g+ \frac{\pi^2 n^2 r} {900})$$

Solution

Question 8

$$\displaystyle T_1 = T_2$$

$$\displaystyle T_2 > T_1$$

$$\displaystyle T_1 > T_2$$

Tension in the string always remain the same

Solution

Question 9

$$4rad/s$$

$$2rad/s$$

$$6rad/s$$

none of these

Solution

Question 10

necessary centrifugal force may be obtained from the horizontal component weight of the train

to avoid frictional force between the tracks and wheels

necessary centripetal force may be obtained from the horizontal component of the weight of the train

the train may not fly off in the opposite direction

Solution

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