Laws of Motion Test

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Laws of Motion Test - 28

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Weekly Quiz Competition

Question 1
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A body is moving in a vertical circle such that the velocities of body at different points are critical. The ratio of velocities of body at angular displacements $$60^{0}$$ and $$120^{0}$$ from the lowest point is:

A
$$\sqrt{5}:\sqrt{2}$$

B
$$\sqrt{3}:\sqrt{2}$$

C
$$\sqrt{3}:1$$

D
$$\sqrt{2}:1$$

Solution

Given that velocities at different points are critical
so velocity at top point ($$T$$)
$$\Rightarrow$$ $${ V }_{ T }$$$$=\sqrt { gR } ...(1) $$
From the diagram
in right triangle $$\triangle OMA$$
$$\Rightarrow$$$$MA=OP=OP\cos { \theta } $$
At point $$A$$, $$\theta =60°$$
$${ \Rightarrow H }_{ A }=OG-OP\\ { \Rightarrow H }_{ A }=R-R\cos { 60° } =\frac { R }{ 2 } ...(2) $$
in right triangle $$\triangle OMB$$
$$\Rightarrow$$$$MB=QP=QP\cos { (180-\theta) } $$
At point $$B$$, $$\theta =120°$$
$${ \Rightarrow H }_{ B }=OG+OQ\\ { \Rightarrow H }_{ B }=R+R\cos { 60° } =\frac { 3R }{ 2 } ...(4) $$
Using energy conservation equation at point $$T$$ and $$A$$
$$\Rightarrow \frac { 1 }{ 2 } m{ { V }_{ T } }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ A } }^{ 2 }+mg{ H }_{ 1 }...(2)$$
putting the value of eqn (1), eqn (2) in eqn (4)
$$\Rightarrow \frac { 1 }{ 2 } m{ (\sqrt { gR } ) }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ A } }^{ 2 }+mg\frac { R }{ 2 } $$
$$\Rightarrow V_{ A }=2\sqrt { gR } $$
Using energy conservation equation at point $$T$$ and $$B$$
$$\Rightarrow \frac { 1 }{ 2 } m{ { V }_{ T } }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ B } }^{ 2 }+mg{ H }_{2}...(2)$$
putting the value of eqn (1), eqn (3) in eqn (4)
$$\Rightarrow \frac { 1 }{ 2 } m{ (\sqrt { gR } ) }^{ 2 }+mg(2R)=\frac { 1 }{ 2 } m{ { V }_{ B} }^{ 2 }+mg\frac { 3R }{ 2 } $$
$$\Rightarrow V_{ B }=\sqrt { 2gR } $$
Ratio of $${ V }_{ A }$$ and $${ V }_{ B }$$$$=\sqrt { 2 } :1$$
So the correct option is ($$D$$)
Question 2
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The bob of a simple pendulum at rest position is given a velocity $$V$$ in horizontal direction so that the bob describes vertical circle of radius equal to length of pendulum $$l$$ . If the tension in string is $$4$$ times weight of bob when the string is horizontal, the velocity of bob when it is crossing highest point of vertical circle is:

A
$$\sqrt{\dfrac{gl}{2}}$$

B
$$\sqrt{gl}$$

C
$$\sqrt{\dfrac{3gl}{2}}$$

D
$$\sqrt{2gl}$$

Solution

At pt B.

$$T = \dfrac{mv^{2}} {L} and T = 4mg.$$

$$4\dfrac{mg}{m}L = v^{2}$$

$$ V= \sqrt{4gL}$$

Now WET between B and C.

$$mgL = \dfrac{1}{2}mv^{2} - \dfrac{m}{2}v_{0}^{2}$$

$$mgL = 2mgL - \dfrac{m}{2}v_{0}^{2}$$

$$v_{0}= \sqrt{2gL}$$
Question 3
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A ball of mass 0.6kg attached to a light inextensible string rotates in a vertical circle of radius 0.75m such that it has speed of 5 m/s when the string is horizontal. Tension in string when it is horizontal on other side is:
$$(g=10ms^{-2})$$

A
$$30N$$

B
$$26N$$

C
$$20N$$

D
$$6N$$

Solution

Given: $$m=0.6 kg; R=0.75 m$$

Force equilibrium
$$T=\dfrac{mv^{2}}{R}$$

$$=\dfrac{(0.6)(5)^{2}}{(0.75)}$$

$$=20 N$$
Question 4
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A simple pendulum is oscillating with an angular amplitude $$60^{0}$$ . If mass of bob is $$50\ g$$, the tension in the string at mean position is:
Consider: $$g=10ms^{-2}$$, length of the string, $$L = 1\ m$$.

A
$$0.5\ N$$

B
$$1\ N$$

C
$$1.5\ N$$

D
$$2\ N$$

Solution

Apply work energy theorem between A and B
$$mg L(1-cos\theta)=\dfrac{1}{2}mv^{2}$$

$$10\times 1\times (1-\dfrac{1}{2})=\dfrac{1}{2}mv^{2}$$

$$v^{2}=10$$

Then, the tension in, $$T=mg+\dfrac{mv^{2}}{R}$$

$$=\dfrac{50\times 10}{1000}+\dfrac{50}{1000}\times \dfrac{10}{1}$$

$$=1 N$$
Question 5
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A water bucket of mass '$$m$$' is revolved in a vertical circle with the help of a rope of length '$$r$$'. If the velocity of the bucket at the lowest point is $$\sqrt{7gr}$$ . Then the velocity and tension in the rope at the highest point are:

A
$$\sqrt{3gr},2mg$$

B
$$\sqrt{2gr},mg$$

C
$$\sqrt{gr},mg$$

D
$$Zero$$ , $$Zero$$

Solution

$$V_{b} = \sqrt{7gr}$$
Work Energy Theorem applied between top and bottom.
$$mg(2R) = \dfrac{1}{2}m (V_{b} )^{2} - \dfrac{1}{2}m (V_{T} )^{2}$$
$$2mgr = \dfrac{7}{2}mgr - \dfrac{1}{2}m (V_{T} )^{2}$$
$$V_{T} = \sqrt{3gr}$$

At top point:
$$T = \dfrac{mV_{T}^{2}}{r} - mg$$
$$=m \dfrac{3gr}{r} -mg$$
$$= 2mg$$
Question 6
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A person carries a hammer on his shoulder and holds the other end of its light handle in his hand. Let $$y$$ be the distance between his hand and the point of support. If the person changes $$y$$, the pressure on his hand will be proportional to:

A
$$y$$

B
$$y^{2}$$

C
$$\dfrac{1}{y}$$

D
$$\dfrac{1}{y^{2}}$$

Solution

As it is clear from the Free body diagram, for the hammer to be in equilibrium condition following should be followed :
$$ F \times y = w \times (l-y) $$
which gives;
$$ F \times y = w \times l - w \times y $$
$$ y = \dfrac {w \times l}{F+w} $$
$$ \dfrac {F+w}{w \times l} = \dfrac {1}{y} $$
Let $$ w \times l =d $$ and $$ \dfrac {1}{l} = c $$
Hence the new equation becomes;
$$ \dfrac {F}{d} + c = \dfrac {1}{y} $$
which gives $$ F = \dfrac {d}{y} - c\times d $$
Let $$ c \times d =a $$ , which gives
$$ F+ a = \dfrac {d}{y} $$
As "$$a$$" being a constant does not effect proportionality we have $$F$$ inversely proportional to $$y$$.
Question 7
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An impulse is supplied to a moving object with the force at an angle of $$120^{0}$$ with the velocity vector. The angle between the impulse vector and the change in momentum vector is

A
$$120^{0}$$

B
$$0^{0}$$

C
$$60^{0}$$

D
$$240^{0}$$

Solution

Impulse of a force which is defined as force times its duration of action is equal to the change in the momentum of the body. Hence the direction of impulse will be along the direction of change in momentum.
Question 8
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A pilot of mass $$m$$ can bear a maximum apparent weight $$7$$ times of $$mg$$. The aeroplane is moving in a vertical circle. If the velocity of aeroplane is $$210\ m/s$$ while driving up from the lowest point of vertical circle, the minimum radius of vertical circle should be:

A
$$375\ m$$

B
$$420\ m$$

C
$$750\ m$$

D
$$840\ m$$

Solution

Given maximum apparent weight is $$7mg$$
Apparent weight at the lowest point is ($$mg + m \dfrac{v^2}{r} $$)
So, $$7mg = mg + m \dfrac{v^2}{r} $$
or, $$6mg = m \dfrac{v^2}{r} $$
or, $$r = \dfrac{v^2}{6g} $$
or, $$r = \dfrac{210^2}{6g} $$
or, $$ r = 750\ m$$
So , minimum radius of circle is $$750\ m $$
Question 9
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A body of mass $$2\ kg$$ attached at one end of light string is rotated along a vertical circle of radius $$2\ m$$. If the string can withstand a maximum tension of $$140.6\ N$$, the maximum speed with which the stone can be rotated is:

A
$$22\ m/s$$

B
$$44\ m/s$$

C
$$33\ m/s$$

D
$$11\ m/s$$

Solution

$$ m= 2 Kg \quad R = 2m $$
$$T_{max} = 1406\ N $$
The tension is max at bottom point.
At bottom:
$$ T - mg = \dfrac{mv^{2}}{R}$$

$$ \dfrac{2(140.6- 2X9.81)}{2} = V^{2}$$

$${V = 10.99\ m/s}$$
Question 10
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A body is revolving in a vertical circle of radius r such that the sum of its $$K.E.$$ and $$P.E.$$ is constant. If the speed of the body at the highest point is $$\sqrt{2gr}$$ then the speed of the body at the lowest point will be:

A
$$\sqrt{7gr}$$

B
$$\sqrt{6gr}$$

C
$$\sqrt{8gr}$$

D
$$\sqrt{9gr}$$

Solution

Applying WET between A and B.

$$\dfrac{m(\sqrt{2gR})^{2}}{2} + mg(2R) = \dfrac{m}{2}v_{0}^{2}$$

$$mgr + 2 mgR = \dfrac{m}{2}v_{0}^{2}$$

$$v_{0} = \sqrt{6gr}$$

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Re-Attempt Weekly Quiz Competition

Laws of Motion Test - 28

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Laws of Motion Test - 28

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SHARING IS CARING

Question 1

$$\sqrt{5}:\sqrt{2}$$

$$\sqrt{3}:\sqrt{2}$$

$$\sqrt{3}:1$$

$$\sqrt{2}:1$$

Solution

Question 2

$$\sqrt{\dfrac{gl}{2}}$$

$$\sqrt{gl}$$

$$\sqrt{\dfrac{3gl}{2}}$$

$$\sqrt{2gl}$$

Solution

Question 3

$$30N$$

$$26N$$

$$20N$$

$$6N$$

Solution

Question 4

$$0.5\ N$$

$$1\ N$$

$$1.5\ N$$

$$2\ N$$

Solution

Question 5

$$\sqrt{3gr},2mg$$

$$\sqrt{2gr},mg$$

$$\sqrt{gr},mg$$

$$Zero$$ , $$Zero$$

Solution

Question 6

$$y$$

$$y^{2}$$

$$\dfrac{1}{y}$$

$$\dfrac{1}{y^{2}}$$

Solution

Question 7

$$120^{0}$$

$$0^{0}$$

$$60^{0}$$

$$240^{0}$$

Solution

Question 8

$$375\ m$$

$$420\ m$$

$$750\ m$$

$$840\ m$$

Solution

Question 9

$$22\ m/s$$

$$44\ m/s$$

$$33\ m/s$$

$$11\ m/s$$

Solution

Question 10

$$\sqrt{7gr}$$

$$\sqrt{6gr}$$

$$\sqrt{8gr}$$

$$\sqrt{9gr}$$

Solution

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