Laws of Motion Test

Result

Laws of Motion Test - 44

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A force acts on a body of mass 3 kg such that its velocity changes from $$4 \,m\, s^{-1}$$ to $$10 \,m\, s^{-1}$$. The change in momentum of the body is:

A
$$42 \,kg\, m\, s^{-1}$$

B
$$2 \,kg\, m\, s^{-1}$$

C
$$18 \,kg\, m\, s^{-1}$$

D
$$14 \,kg\, m\, s^{-1}$$

Solution

Given, $$m=3 kg$$, $$v_1=4 m/s$$ and $$v_2=10 m/s$$
Change in momentum $$=p_2-p_1=mv_2-mv_2=m(v_2-v_1)=3(10-4)=18 $$ $$kg $$ $$ms^{-1}$$
Question 2
1 / -0

From Newton's third law, Action and reaction____.
Which of the following will not fit in the above sentence.

A
acts on the same body

B
are equal

C
are opposite in direction

D
acts on different bodies

Solution

According to newton's third law of motion, every action has an equal and opposite reaction. The action-reaction pair always acts on different bodies.
So option A is wrong and rest three options are correct.
Question 3
1 / -0

On which one of the following conservation laws, does a rocket work?

A
Mass

B
Energy

C
Linear momentum

D
Angular momentum

Solution

Rocket works on the principle of conservation of linear momentum.
Question 4
1 / -0

An aeroplane executes a horizontal loop at a speed of 720 kmph with its wings banked at $$45^o$$. What is the radius of the loop?
Take g = 10 $$ms^{-2}$$.

A
4 km

B
4.5 km

C
7.2 km

D
2 km

Solution

Given : $$v = 720$$ km/h $$ = 720\times \dfrac{5}{18} = 200$$ $$m/s$$
Angle of banking $$\theta = 45^o$$
Radius of loop $$r = \dfrac{v^2}{g\tan\theta}$$
$$\therefore$$ $$r = \dfrac{200^2}{10\times 1} = 4000$$ m $$(\because \tan45^o =1)$$
$$\implies$$ Radius of loop $$r = 4$$ km
Question 5
1 / -0

A stone of mass $$10 kg$$ tied with a string of length $$0.5 m$$ is rotated in a vertical circle. Find the total energy of the stone at the highest position. (in $$J$$)

A
$$123.36$$

B
$$122.5$$

C
$$154.3$$

D
$$185.6$$

Solution

assume that potential energy at lowest point is zero and stone is rotating with critical velocity
so the magnitude of velocity at highest point is $$v=\sqrt { gl } $$ where $$l=$$length of string
total energy $$E=K.E.+P.E.$$
$$\Rightarrow K.E.=\dfrac { 1 }{ 2 } m{ v }^{ 2 }=\dfrac { 1 }{ 2 } 10\times { (\sqrt { 9.8\times 0.5 } ) }^{ 2 }=24.5J$$
$$P.E.=mgh=mg(2l)=10\times 9.8\times \left( 2\times 0.5 \right) =98J$$
so the total energy $$E=98+24.5=122.5J$$
Question 6
1 / -0

A small stone tied to an inextensible string of negligible mass is rotated in a circle of radius $$2 m$$ in a vertical plane. Find the speed at a horizontal point on the circle. (in $$m/s$$)

A
$$7.67$$

B
$$2.36$$

C
$$6.32$$

D
$$8.32$$

Solution

$$ v= \sqrt {3gr}$$
$$ v= \sqrt {3 * 9.8* 2} = 7.67 m/s $$
Question 7
1 / -0

A car executes a turn of radius $$22 m$$ on a banked road while travelling at a speed of $$45 km/h$$. If the height of the outer edge above the inner edge of the road is $$1.1 m$$, what is the breadth of the road? (in $$m$$)

A
$$1.745$$

B
$$1.569$$

C
$$1.875$$

D
$$1.236$$

Solution

given that $$R=22m$$, $$v=45km/h=12.5m/s$$
as the diagram shows, we have to find the value of $$L$$, we can see there is a right triangle $$\Delta OQP$$ so
$$\Rightarrow { L }=\sqrt { \left( { a }^{ 2 }+1.1^{ 2 } \right) }...(1) $$
and $$\tan { \theta } =\dfrac { 1.1 }{ a } \Rightarrow a=\dfrac { 1.1 }{ \tan { \theta } } ...(2)$$
balancing the force in car we gets two equations
$$\Rightarrow N\cos { \theta } =mg...(3)$$
$$\Rightarrow N\sin { \theta } =\dfrac { m{ v }^{ 2 } }{ R } ...(4)$$
by dividing eqn(3) by eqn(4)
$$\Rightarrow \tan { \theta } =\dfrac { { v }^{ 2 } }{ gR } $$ now putting the value of $$\tan{\theta}$$ in eqn(2)
$$\Rightarrow a =\dfrac { 1.1\times gR }{ { v }^{ 2 } } =\dfrac { 1.1\times 10\times 22 }{ { 12.5 }^{ 2 } } =1.548m$$
now again putting the value of $$a$$ in eqn(1)
$${ L }=\sqrt { \left( { 1.5488 }^{ 2 }+1.1^{ 2 } \right) } =1.875m$$
Question 8
1 / -0

A motorcyclist executes a horizontal loop at a speed of $$65km/h$$ while himself making an angle of $$12^o$$ with the horizontal. What is the radius of the loop? (in $$m$$)

A
$$123.6$$

B
$$156.6$$

C
$$156.3$$

D
$$174.3$$

Solution

as the diagram shows
$$N\cos { \theta } $$ provides it centripetal force and $$N\sin { \theta } $$ balance the weight($$mg$$)
$$\Rightarrow N\cos { \theta } =mg...(1)$$ and $$N\sin { \theta } =\dfrac { m{ v }^{ 2 } }{ R }...(2)$$
dividing eqn(1) by eqn(2)
$$\Rightarrow \tan { \theta } =\dfrac { { v }^{ 2 } }{ gR } $$ or
$$R=\dfrac { { v }^{ 2 } }{ g\tan { \theta } } $$
given $$v=65km/h=18.056m/s$$, $$\tan { \theta } =0.2126$$
$$\Rightarrow R=\dfrac { { 18.056 }^{ 2 } }{ 9.8\times 0.2126 } =156.6m$$
Question 9
1 / -0

The angle of banking of a turn of radius $$75 m$$ on a road is $$30^o$$. What is the speed at which a car can turn along this curve? (in $$m/s$$)

A
$$26.3$$

B
$$63.32$$

C
$$21.3$$

D
$$20.6$$

Solution

$$ v = \sqrt {r gtan\theta }$$
$$ v= \sqrt {75* 9.8* tan 30}$$
$$ v= 20.6 m/s $$
Question 10
1 / -0

The centre of gravity of a car is $$0.62 m$$ above the ground. It can turn along a track which is $$1.24 m$$ wide and has radius $$r$$. If the greatest speed at which the car can take the turn is $$22.02 m/s$$, What is the value of $$r$$? (in $$m$$)

A
$$49.48$$

B
$$58.36$$

C
$$21.36$$

D
$$96.36$$

Solution

assume that critical velocity is $$v$$, if speed of car is more than $$v$$ then car will topple due to torque of centrifugal force $${ F }_{ c }$$ around point $$P$$ , so balancing the torque at point $$P$$
$${ \Rightarrow F }_{ c }\times 0.62=mg\times 0.62\\ \Rightarrow \dfrac { m{ v }^{ 2 } }{ R } \times 0.62=mg\times 0.62\\ \Rightarrow \dfrac { m{ v }^{ 2 } }{ R } =mg\\ \Rightarrow R=\dfrac { { v }^{ 2 } }{ g } =\dfrac { { 22.02 }^{ 2 } }{ 10 } =49.48m$$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Laws of Motion Test - 44

Result

Laws of Motion Test - 44

Score

-

Rank

-

SHARING IS CARING

Question 1

$$42 \,kg\, m\, s^{-1}$$

$$2 \,kg\, m\, s^{-1}$$

$$18 \,kg\, m\, s^{-1}$$

$$14 \,kg\, m\, s^{-1}$$

Solution

Question 2

acts on the same body

are equal

are opposite in direction

acts on different bodies

Solution

Question 3

Mass

Energy

Linear momentum

Angular momentum

Solution

Question 4

4 km

4.5 km

7.2 km

2 km

Solution

Question 5

$$123.36$$

$$122.5$$

$$154.3$$

$$185.6$$

Solution

Question 6

$$7.67$$

$$2.36$$

$$6.32$$

$$8.32$$

Solution

Question 7

$$1.745$$

$$1.569$$

$$1.875$$

$$1.236$$

Solution

Question 8

$$123.6$$

$$156.6$$

$$156.3$$

$$174.3$$

Solution

Question 9

$$26.3$$

$$63.32$$

$$21.3$$

$$20.6$$

Solution

Question 10

$$49.48$$

$$58.36$$

$$21.36$$

$$96.36$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.