Laws of Motion Test

Result

Laws of Motion Test - 42

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 person is moving on a rough horizontal surface towards east. Then the direction of friction force is towards

A
west

B
east

C
north

D
cant say depends on the situation.

Solution

If man is moving in any direction then there must be some force acting on the man due to which he is able to move, that force is frictional force.
man applies frictional force on ground in backward direction and ground applies force in forward direction.
so option B is correct.
Question 2
1 / -0

A plane travelling at $$300 {m}/{s}$$ banks its wings to enter into a horizontal circular turn. The circular path has a radius of $$2.7 km$$.
Which of these values best represents the angle of the wings relative to the horizontal if the plane experiences no change in its altitude during the turn?

A
$${8}^{o}$$

B
$${16}^{o}$$

C
$${20}^{o}$$

D
$${25}^{o}$$

E
$${32}^{o}$$

Solution

In the flight of a plane the angle of banking with vertical is given by ,
$$\tan \theta=v^{2}/rg$$
given $$v=300m/s , r=2.7km=2700m$$
therefore $$\tan\theta=\frac{300^{2}}{2700\times9.8}=3.4$$
or $$\tan\theta=\tan74$$
or $$\theta=74^{o}$$
so the angle with horizontal will be , $$\theta'=90-74=16^{o}$$
Question 3
1 / -0

A ball of mass 100 g is moving with a velocity of 10 m $$s^{-1}$$. The force of the blow by the bat acts for 0.01 seconds. What is the force exerted on the ball by the bat if the ball retraces its path with same speed?

A
100 N

B
200 N

C
300 N

D
400 N

Solution

Initial velocity of ball   $$u = -10 \ m/s$$
Initial momentum   $$P_i = mu = 0.1\times (-10) = -1 \ kg \ m/s$$
Final velocity of ball   $$v = 10 \ m/s$$
Final momentum $$P_f = mv = 0.1\times 10 = 1 \ kg \ m/s$$
Time of action $$t = 0.01 \ s$$
Using   $$F\times t = P_f - P_i$$
Or   $$F\times 0.01 = 1-(-1) = 2$$
$$\implies \ F = 200 \ N$$
Question 4
1 / -0

Match the entries in Column I with those in Column II.
Column - I Column - II
a Force 1 N $$m^{-1}$$
b Weight 2 kg
c Pressure 3 N
d Impulse 4 N s
e Mass 5 kg wt

A
a-3,b-2,c-4, d-1, e-5

B
a-5,b-4,c-3, d-1, e-2

C
a-3,b-5,c-1, d-4, e-2

D
a-4,b-2,c-1, d-3, e-5

Solution

Si unit of force is Newton denoted by $$N$$, that of weight is $$kg \ wt$$, pressure is $$N m^{-2}$$, impulse is $$N \ s$$ and mass is $$kg$$.
Thus option C is correct.
Question 5
1 / -0

A mass of 2 kg at rest travels for 4 seconds with an acceleration of 1.5 m $$s^{-2}$$. Find the gain of the momentum of the body.

A
5 kg m $$s^{-1}$$

B
10 kg m $$s^{-1}$$

C
12 kg m $$s^{-1}$$

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

Solution

$$v=u+at=1.5\times 4=6ms^{-1}$$
Gain in momentum=$$mv=2\times 6=12kgms^{-1}$$
Question 6
1 / -0

To avoid slipping while walking on ice, one should take smaller steps because:

A
frictional force of ice is large

B
of larger normal reaction

C
frictional force of ice is small

D
of smaller normal reaction

Solution

Hint :- To avoid slipping friction should be sufficient.
Explanation:-
$$\bullet$$To avoid slipping one should take smaller steps because the friction coefficient of ice is small. So we didn’t get enough friction by the ice.
$$\bullet$$Smaller steps will give larger normal force and more the normal force will give more friction and hence we get sufficient friction to avoid slipping.
$$\textbf{Hence option C correct}$$
Question 7
1 / -0

A cricket ball of mass 500 g is moving with a speed of 36 km $$h^{-1}$$. It is reflected with the same speed. What is the impulse applied on it?

A
20 kg m $$s^{-1}$$

B
5 kg m $$s^{-1}$$

C
25 kg m $$s^{-1}$$

D
10 kg m $$s^{-1}$$

Solution

Initial velocity of the ball   $$u = -36 \ km/h =- 36\times \dfrac{5}{18} = -10 \ m/s$$
Initial momentum   $$P_i = mu = -0.5\times 10 = -5 \ kg \ m/s$$
Final velocity of the ball   $$v = 36 \ km/h = 36\times \dfrac{5}{18} = 10 \ m/s$$
Initial momentum   $$P_f = mv = 0.5\times 10 = 5 \ kg \ m/s$$
Impulse on the ball   $$I = P_f - P_i = 5-(-5) = 10 \ kg \ m/s$$
Question 8
1 / -0

Due to an impulse, the change in the momentum of a body is 1.8 kg m $$s^{-1}$$. If the duration of the impulse is 0.2 s, then what is the force produced in it?

A
9 N

B
8 N

C
7 N

D
6 N

Solution

Impulse=$$\dfrac{F}{\Delta t}=\dfrac{1.8}{0.2}=9N$$
Question 9
1 / -0

For the plank depicted above, a man who weighs $$\dfrac{1}{8}$$ the weight of the plank stands halfway between the pivot and the center of mass of the plank.
If the plank weighs $$500 N$$ and $$\theta={30}^{o}$$, which of the following values best represents the tension in the cable?

A
$$500 N$$

B
$$471 N$$

C
$$531 N$$

D
$$1800 N$$

E
$$1500 N$$

Solution

Given : $$F_w = 500$$ N
As the plank is in rotational equilibrium, thus net torque acting on the plank about the point O is zero i.e $$\tau_o = 0$$
$$\therefore$$ $$F_t sin\theta \times L - F_w \times \dfrac{L}{2} - \dfrac{F_w}{8} \times \dfrac{L}{4} = 0$$
$$\implies $$ $$\dfrac{17}{32}F_w =F_t$$ $$sin\theta$$

OR $$\dfrac{17}{32} \times 500 = F_t \times 0.5$$ $$\implies F_t \approx 531$$ N
Question 10
1 / -0

China wares are wrapped in straw or paper before packing. What is the basis of this application?

A
Impulse

B
Momentum

C
Acceleration

D
Force

Solution

Because the sharp impulse given to objects may cause damage.

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

	Column - I		Column - II
a	Force	1	N $$m^{-1}$$
b	Weight	2	kg
c	Pressure	3	N
d	Impulse	4	N s
e	Mass	5	kg wt

Laws of Motion Test - 42

Result

Laws of Motion Test - 42

Score

-

Rank

-

SHARING IS CARING

Question 1

west

east

north

cant say depends on the situation.

Solution

Question 2

$${8}^{o}$$

$${16}^{o}$$

$${20}^{o}$$

$${25}^{o}$$

$${32}^{o}$$

Solution

Question 3

100 N

200 N

300 N

400 N

Solution

Question 4

a-3,b-2,c-4, d-1, e-5

a-5,b-4,c-3, d-1, e-2

a-3,b-5,c-1, d-4, e-2

a-4,b-2,c-1, d-3, e-5

Solution

Question 5

5 kg m $$s^{-1}$$

10 kg m $$s^{-1}$$

12 kg m $$s^{-1}$$

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

Solution

Question 6

frictional force of ice is large

of larger normal reaction

frictional force of ice is small

of smaller normal reaction

Solution

Question 7

20 kg m $$s^{-1}$$

5 kg m $$s^{-1}$$

25 kg m $$s^{-1}$$

10 kg m $$s^{-1}$$

Solution

Question 8

9 N

8 N

7 N

6 N

Solution

Question 9

$$500 N$$

$$471 N$$

$$531 N$$

$$1800 N$$

$$1500 N$$

Solution

Question 10

Impulse

Momentum

Acceleration

Force

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.