Laws of Motion Test

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Laws of Motion Test - 41

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Question 1
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Directions For Questions

As shown above, a ball is attached to the string and it is swinging in vertical circle. The three position of ball are given as I, II and III on the circle.

...view full instructions

If the string were to break at point II, what would be the path of the ball?

A

B

C

D

E
None of these

Solution

When the bob is swinging in vertical circle the centripetal force and the pseudo force varies. Centripetal force causes the tension in the string.
Let centripetal force be $$C$$, pseudo force be $$P$$ and weight of bob be $$W$$.
At position 2, $$C = W$$
If the string were to break at this position, the only external force acting on the bob is $$W$$ (downward). Hence the bob would move vertically down.

Therefore option D is correct.
Question 2
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A $$50 kg$$ acrobat is swinging on a rope with a length of $$15 m$$ from one horizontal platform to another. Both platforms are at equal height.
If the maximum tension the rope can support is $$1200 N$$, which of the following answer best represents the maximum velocity the acrobat can reach without breaking the rope?

A
$$25 {m}/{s}$$

B
$$19 {m}/{s}$$

C
$$20 {m}/{s}$$

D
$$15 {m}/{s}$$

E
$$5 {m}/{s}$$

Solution

Given : $$T =1200$$ N $$m = 50$$ kg $$L =15$$ m
Let the maximum velocity of the acrobat be $$v$$.
From figure, $$\dfrac{mv^2}{L} =T - mg$$

$$\therefore$$ $$\dfrac{50 \times v^2}{15} =1200 - 50 \times 9.8$$ $$\implies v =\sqrt{213} \approx 15 $$ $$m/s$$
Question 3
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A plane banks its wings $${30}^{o}$$ relative to the horizontal to enter into a circular turn. The circular path has a radius of $$1500 m$$.
Which of these values best represents the velocity of the plane if the lift force exerted on the wings is equal to twice the weight of the plane?

A
$$120 {m}/{s}$$

B
$$140 {m}/{s}$$

C
$$160 {m}/{s}$$

D
$$240 {m}/{s}$$

E
$$320 {m}/{s}$$

Solution

In the flight of a plane the angle of banking with vertical is given by ,
$$\tan \theta=v^{2}/rg$$
given $$\theta=90-30=60^{o} , r=1500m$$
therefore $$v^{2}=rg\tan60$$
or $$v^{2}=1500\times9.8\times\sqrt3=25460.4$$
or $$v=160m/s$$
Question 4
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Directions For Questions

As shown above, a ball is attached to the string and it is swinging in vertical circle. The three position of ball are given as I, II and III on the circle.

...view full instructions

Which of the following statements is true?

A
The tension in the string is greater at points III than at point I

B
The tension in the string is greater at points I than at point III

C
The tension at point I is equal to the tension at point III

D
The tension in the string is greatest at point II

E
The tension in the string is same at points I, II and III

Solution

When the bob is swinging in vertical circle the centripetal force and the pseudo force varies. Centripetal force causes the tension in the string.
Let centripetal force be $$C$$, pseudo force be $$P$$ and weight of bob be $$W$$.
At position 1, $$C = P - W$$
At position 2, $$C = W$$
At position 3, $$C = P + W$$

As C at position 3 is greatest, the tension in the string is greatest at this position.
Therefore option A is correct.
Question 5
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A ball of mass $$m$$ is attached with the string of length $$R$$, rotating in circular motion, with instantaneous velocity $$v$$ and centripetal acceleration $$a$$.
Calculate the centripetal acceleration of the ball if the length of the string is doubled?

A
$${a}/{4}$$

B
$${a}/{2}$$

C
$$a$$

D
$$2a$$

E
$$4a$$

Solution

Centripetal acceleration $$a = \dfrac{v^2}{R}$$
Now the length of the string is doubled keeping the instantaneous velocity constant i.e $$R' = 2R$$
$$\therefore$$ New centripetal acceleration $$a' = \dfrac{v^2}{R'} = \dfrac{v^2}{2R} =\dfrac{a}{2}$$
Question 6
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A uniform rod of mass 1 kg is hanging from a thread attached at the midpoint of the rod. A block of mass $$m$$=3kg hangs at the left end of the rod and block of mass $$M$$ hangs at the right side at 80 cm from block $$m$$. If system is in equilibrium. Calculate $$M$$.

A
4 kg

B
5 kg

C
6 kg

D
8 kg

E
9 kg

Solution

Given : $$m = 3 kg$$
As the rod is in equilibrium, thus net torque about the point O is zero i.e $$\tau_o = 0$$
$$\therefore$$ $$mg (0.5 ) -Mg (0.3) = 0$$

$$\implies$$ $$M = \dfrac{m(0.5)}{0.3} = \dfrac{3 \times 0.5}{0.3} = 5 kg$$
Question 7
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The given rod is uniform and has a mass m. Find the tension in the string.

A
$$\dfrac { \left( mg\sin { \theta } \right) }{ 2 } $$

B
$$\dfrac { mg\sin { } }{ 2 } $$

C
$$\dfrac { \left( mg\cos { \theta } \right) }{ 2 } $$

D
$$\dfrac { mg }{ 2 } $$

E
$$mg$$

Solution

Let the tension in the string be $$T$$ and $$L$$ be the length of the rod.
As the rod is in equilibrium, thus the torque about point O is zero i.e $$\tau_o = 0$$
$$\therefore$$ $$T (L sin\theta) - mg \times \dfrac{L}{2} sin\theta = 0$$ $$\implies T =\dfrac{mg}{2}$$
Question 8
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A $$110 kg$$ panda is riding on a $$3.0 m$$ long swing whose mass can be considered negligible. The highest point of its arc occurs when the swing makes a $${20}^{o}$$ angle with the vertical.
What is the magnitude of the total tension in the ropes of the swing at that point?

A
$$0N$$

B
$$103 N$$

C
$$369 N$$

D
$$1013 N$$

E
$$1078 N$$

Solution

Given : $$m =110$$ kg
Tension in the string $$T = mg$$ $$cos20^o = 110 \times 9.8 \times 0.94 = 1013$$ N
Question 9
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A uniform bar is lying on a flat horizontal table. The bar is acted upon by $${F}_{1}$$ and $${F}_{2}$$,besides the gravitational and normal forces (which cancel), $${F}_{1}$$ and $${F}_{2}$$ are parallel to the surface of the table. If the net force on the rod is zero, then which one of the following statement is true?

A
The net torque on the bar must also be zero

B
The bar can accelerate translationally if $${F}_{1}$$ and $${F}_{2}$$ are not applied at the same point

C
The net torque will be zero if $${F}_{1}$$ and $${F}_{2}$$ are applied at the same point

D
The bar cannot accelerate translationally or rotationally

E
None of the above

Solution

The bar cannot accelerate translationally or rotationally because net force on it ,is zero. It is not compulsory that if net force is zero ,net torque will also be zero ,infact torque is produced by two equal and opposite forces on a body when their lines of action are not on the same point if these forces are acting on the same point then net torque would be zero because in this situation forces will cancel each other.
Question 10
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Two objects rest on a seesaw. The first object has a mass of 3 kg and rests 10 m from the pivot. The other rests 1 m from the pivot. What is the mass of the second object if the seesaw is in equilibrium?

A
0.3 kg

B
3 kg

C
10 kg

D
30 kg

E
50 kg

Solution

Torque is given by,
$$Torque,\quad \tau =\vec { r } \times \vec { F } =rF\sin { \theta } \\ Since\ \vec { r } \ perpendicular\quad to\ \vec { F } ,\quad \theta ={ 90 }^{ \circ }\\ So,\quad \tau =rF\\ For\quad the\quad seesaw\quad to\quad be\quad in\quad equilibrium,\quad \\ { r }_{ 1 }{ F }_{ 1 }={ r }_{ 2 }{ F }_{ 2 }\\ { r }_{ 1 }{ m }_{ 1 }g={ r }_{ 2 }{ m }_{ 2 }g\\ 1\times { m }_{ 1 }=10\times 3\\ { \quad \quad m }_{ 1 }=30kg$$