Laws of Motion Test

Result

Laws of Motion Test - 70

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Question 1
1 / -0

A uniform rod AB mass m= 1.12 kg and length l= 100 cm is placed on a sharp support O such that AO = a = 40 cm and OB = b = 60 cm. To keep the rod horizontal, its end A is tied with a thread. Calculate reaction of support O on the rod when the thread is burnt. ( g 10 $$m{ s }^{ -1 }$$)

A
40 N

B
10 N

C
15.5 N

D
20 N

Solution
Question 2
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A uniform rod of length $$4L$$ and mass $$M$$ is suspended form a horizontal roof by two light strings of length $$L$$ and $$2L$$ as shown. Then the tension in the left string of length $$L$$ is:

A
$$\dfrac{{Mg}}{2}$$

B
$$\dfrac{{Mg}}{3}$$

C
$$\dfrac{3}{5}Mg$$

D
$$\dfrac{{Mg}}{4}$$

Solution

$$AC = 4L$$
$$BC = L$$
$$AB = \sqrt {A{C^2} - B{C^2}} $$
$$ = \sqrt {{{\left( {4L} \right)}^2} - {L^2}} $$
$$ = \sqrt {16{L^2} - {L^2}} $$
$$ = \sqrt {15{L^2}} $$
$$AB = \sqrt {15} L$$
Also, $$\cos \theta = \dfrac{{AB}}{{AC}} = \dfrac{{AE}}{{AD}}$$
$$ = \dfrac{{\sqrt {15} L}}{{4L}} = \dfrac{{AE}}{{2L}}$$
$$AE = \dfrac{{\sqrt {15} }}{4} \times 2L$$
$$AE = \dfrac{{\sqrt {15} }}{2}L$$
taking moment at $$A$$
$${T_1} \times AB = Mg \times AE$$
$${T_1} \times \sqrt {15} L = Mg \times \dfrac{{\sqrt {15} }}{2}L$$
$${T_1} = \dfrac{{Mg}}{2}$$
Question 3
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If a street light of mass $$M$$ is suspended from the end of uniform rod of length $$L$$ in the different possible patterns as shown in figure, then

A
Pattern $$A$$ is more sturdy

B
Pattern $$B$$ is more sturdy

C
Pattern $$C$$ is more sturdy

D
none of these

Solution

$$\text { Torque croated due to weight of streat light remains } \\$$
$$\text { same in all the three cases. } 9 t \text { is balanced by } \\$$
$$\text { torque created by tension in the string. So if } \gamma \\$$
$$\text { be the torque created by weight of lamp and } T \text { ' } \\$$
$$\text { be tension in the string and ' } d \text { ' be perpendicular } \\$$
$$\text { distance of cable from the axis thou, }$$
$$\tau=T \cdot d$$
$$\text { Tension will be least for laxgest'd'. This is } \\$$
$$\text { in pattern } A \text { is more sturdy. } \\$$
Question 4
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A particle of mass $$m$$ strikes a wall elastically with speed $$v$$ at an angle $${30}^{o}$$ with the wall as shown in the figure. The magnitude of impulse imparted to the ball by the wall is:

A
$$mv$$

B
$$\cfrac{mv}{2}$$

C
$$2mv$$

D
$$\sqrt {3} mv$$

Solution

Before the particle strike,

Linear momentum perpendicular to the wall$$= mvsin(30^o)= \dfrac{mv}{2}$$

Linear momentum along to the wall$$= mvcos(30^o)$$

After the particle strike,

Linear momentum perpendicular to the wall$$= -mvsin(30^o)= -\dfrac{mv}{2}$$

Linear momentum along to the wall$$= mvcos(30^o)$$

Impulse=change in linear momentum,

Linear Momentum changed only perpendicular to the wall.

$$\Rightarrow Impulse= -\dfrac{mv}{2}-\dfrac{mv}{2}=-mv$$

Magnitude of Impulse is $$mv$$
Question 5
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Directions For Questions

The following problems illustrate the effect of a time-dependent force of a large average magnitude which acts on a moving object only for a short duration. Such forces are called 'impulsive' forces. According to the impulse-momentum theorem, impulse delivered to a body is equal to the change of linear momentum of the body.
A ball of mass $$250 g$$ is thrown with a speed of 30 m/s. The ball strikes a bat and is hit straight back along the same line at a speed of 50 m/s. Variation of the interaction force, as long as the ball remains in contact with the bat, is as shown in Fig.

...view full instructions

Maximum force exerted by the bat on the ball is:

A
$$2500 N$$

B
$$5000 N$$

C
$$7500 N$$

D
$$1250 N$$

Solution

Mass of balls
$$\Rightarrow { m }_{ b }=250g$$
$${ u }_{ { b }_{ 1 } }=30m/s$$
abler strike $${ u }_{ { b }_{ 2} }=50m/s$$
$$\Rightarrow \triangle u=20m/s$$
$$\Rightarrow \dfrac{ \triangle u}{t}=20\times 10^3m/s$$
$$\Rightarrow F=m\times \dfrac{ \triangle u}{t}=20\times 10^3\times .250$$
$$=5000N$$
Hence, the answer is $$5000N.$$
Question 6
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A uniform hollow hemisphere of mass m and radius r is released from rest on a smooth horizontal surface with its open face vertical initially. Find out the Maximum normal reaction between hemisphere and ground during motion is

A
$$\dfrac { 10mg }{ 5 }$$

B
$$\dfrac { 9mg }{ 5 }$$

C
$$\dfrac { 11mg }{ 5 }$$

D
$$\dfrac { 12mg }{ 5 }$$

Solution
Question 7
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For stable equilibrium, which of the following is/are true?

A
$$\cfrac { dU }{ dx } =0$$ and $$\cfrac { { d }^{ 2 }U }{ d{ x }^{ 2 } } >0$$

B
$$\cfrac { dU }{ dx } =0\quad $$ only

C
$$\cfrac { dU }{ dx } =0$$ and $$\cfrac { { d }^{ 2 }U }{ d{ x }^{ 2 } } < 0$$

D
None of these

Solution
Question 8
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What angle a bicycle and its rider must make with the vertical when travelling at $$54\ km/hr$$ around a horizontal curve of radius $$40\ m$$? $$(g=10\ m/s^{2})$$:

A
$$\sin^{-1}\ (0.56)$$

B
$$\cot^{-1}\ (1.78)$$

C
$$\cos^{-1}\ (0.56)$$

D
$$\tan^{-1}\ (0.56)$$

Solution
Question 9
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A string wraps around a fat pipe as a bob attached to the string is made to move in a circular path in the horizontal. Assuming the speed is somehow held constant as the radius diminishes due to the wrapping, how will the centripetal force change?

A
It will stay the same.

B
It will diminish.

C
It will increase

D
None of the above (this is a jokeits got to be one of the three above)

Solution

As the string wraps, its radius decreases. Thus the centripetal force increases.

The correct option is (c)
Question 10
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A small block of mass m is pushed on a smooth track from position A with a velocity $$2\sqrt5$$ times the minimum velocity required to reach point D. The block will leave the contact with track at the point where normal force between them becomes zero.
At what angle $$\theta$$ with horizontal does the block gets separated from the track?

A
$$sin^{-1}(\frac{1}{3})$$

B
$$sin^{-1}(\frac{3}{4})$$

C
$$sin^{-1}(\frac{2}{3})$$

D
never leaves contact with the track

Solution