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CBSE Questions for Class 9 Physics Motion Quiz 2 - MCQExams.com
CBSE
Class 9 Physics
Motion
Quiz 2
A skier starting from rest accelerates down a slope at $$1.6\ ms^{-2}$$. How far has he gone at the end of $$5\ s$$?
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$$20\ m$$
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$$40\ m$$
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$$15\ m$$
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$$60\ m$$
Explanation
Given:
Initial velocity: $$u=0\ m/s$$
Acceleration: $$a = 1.6\ m/s^2$$
Time: $$t = 5\ s$$
Using second equation of motion,
$$s = ut + \cfrac{1}{2} at^{2}$$
$$s = 0 + \cfrac{1}{2} \times 1. 6 \times 25$$
$$s = 20\ m$$
Slope of distance-time graph of a moving body is equal to
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velocity of the body
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speed of the body
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acceleration of the body
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none of these
Explanation
Speed of a body$$=$$slope of the distance-time graph of the body
$$tan\theta=\frac{y}{x}=speed of a body$$
Here x,y are co-ordinates distance of distance- time graph.
Hence,option B is correct.
In Figure as shown above BC represents a body moving:
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backward with uniform velocity
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forward with uniform velocity
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backward with non-uniform velocity
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forward with non-uniform velocity
Explanation
Velocity of an object from displacement-time graph is given by its slope.
ie $$\dfrac{dy}{dt}=V$$ so slope from B to C is constant and negative so body is moving backward and with a constant velocity.
so best possible option is option A.
A stone is dropped in the river from a bridge. The stone takes 1 second to touch the water surface in the river. What is the height of the bridge from the water level?
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9.8 m
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19.6 m
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4.9 m
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1.96 m
Explanation
When particle is dropped from bridge the initial velocity $$u$$ is zero.
Let the height of bridge from water level is $$h$$
Then the displacement traveled by particle is $$h$$.
Acceleration is $$g=9.8 \ m/s^2$$ downwards.
Then applying second equation of motion:
$$s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }$$ where $$a=g$$ and $$u=0$$
then,
$$h=\dfrac { 1 }{ 2 } g{ (1) }^{ 2 }=\dfrac { 1 }{ 2 } \times 9.8=4.9m$$
Thus, option C is correct.
The velocity of an object can be changed by
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changing the speed
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changing the direction of motion
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changing both the speed and direction of motion
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all (a), (b) and (c) are true
Explanation
The velocity of an object can be changed by
A.changing the speed.
B.changing the direction of motion.
C.changing both the speed and direction of motion.
hence,the option D is correct.
A body will have uniform acceleration if its
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speed changes at uniform rate
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velocity changes at uniform rate
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speed changes at non-uniform rate
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velocity remains constant
Explanation
If the velocity of an object changes at a uniform rate, then the acceleration that causes the change in velocity is called uniform acceleration or constant acceleration.
For example, the force of gravity imparts an acceleration uniformly which is called acceleration due to gravity.
Hence,option B is correct.
The body will speed up if ___________ .
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Velocity and acceleration are in same direction
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Velocity and acceleration are in opposite direction
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Velocity and acceleration are in perpendicular direction
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None of these
Explanation
A body will speed up if both velocity and acceleration are in same direction. If they are in opposite directions it results in slowing down the motion. And if they are perpendicular then there will be no effect on magnitude of velocity.
Which one of the following is most probably not a case of uniform circular motion?
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Motion of a racing car on a circular track
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Motion of the moon around the earth
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Motion of a toy train on a circular track
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Motion of seconds hand on the circular dial of a watch
Explanation
Answer is A.
An object moving in uniform circular motion is moving in a circle with a uniform or constant speed. The velocity vector is constant in magnitude but changing in direction.
Hence, the motion of a racing car on a circular track changes it speed, that is, it increases or decreases speed during the drive not a case of uniform circular motion.
(a) Find the acceleration of the car
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$$\, 2\, m/s^2$$
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$$4 m/s^2$$
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$$6 m/s^2$$
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$$8 m/s^2$$
Explanation
Given:
Initial velocity $$u=5\ m/s$$
Final velocity $$v=25\ m/s$$
Time $$t=10\ s$$
From the equation of motion $$v=u+at$$ we get,
acceleration $$a=\dfrac{v-u}{t}=\dfrac{25-5}{10}=2\ m/s^2$$
The displacement versus time graph shows that the body is moving with constant speed.
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True
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False
Explanation
As we can see in the figure, Displacement of the object is not changing with time. Its mean that body is not moving from its initial location.
So, object is in rest ie Speed $$S=0\ m/s$$
Given statement is false.
State whether given statement is True or False.
Speed is the other name of velocity.
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True
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False
Explanation
Velocity is defined as the displacement of a body in unit time whereas speed of the body is defined as the distance moved by the body in unit time. Velocity is a vector quantity while speed is a scalar quantity.
Therefore, speed and velocity are two separate entities.
Velocity-time graph of a body with uniform velocity is a straight line
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Parallel to x-axis
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Parallel to y-axis
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Inclined to x-axis
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Inclined to y-axis
Explanation
Velocity-time graph of an object moving with uniform velocity. The slope of a Velocity–time graph of an object moving in rectilinear motion with uniform velocity is straight line and parallel to x-axis when velocity is taken along y-axis and time is taken along x-axis.
A boy is running along the circumference of a stadium with constant speed. Which of the following is changing in this case?
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Centripetal force acting on the boy
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Distance covered per unit time
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Direction in which the boy is running
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Magnitude of acceleration
Explanation
Answer is C.
As the boy is running along the circumference of a stadium with constant speed, the centripetal force remains the same as it is a circular path, the distance covered and the magnitude of acceleration is same as he runs with constant speed.
Therefore, the direction in which the boy is running changes direction as the path is a circular path.
In the equation of motion, $$s = ut + \dfrac 12 at^2$$, $$s$$ stands for
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displacement in $$t$$ seconds
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maximum height reached
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displacement in the $$t^{th}$$ second
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None of these
Explanation
In the second equation of motion $$ s=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }$$
$$s =$$ displacement in $$t$$ seconds
$$u =$$ Initial velocity of the motion
$$a =$$ acceleration or retardation of the motion
Hence, option A is correct
Distance is a
vector quantity. State whether true or false.
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True
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False
Explanation
Distance is a scalar quantity whereas displacement is a vector quantity. So the given statement is false.
Area under a speed-time graph gives :
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The time taken by a moving object
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The distance travelled by a moving object
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Tthe acceleration of a moving object
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Tthe retardation of a moving object
Explanation
The area under the velocity-time graph gives the distance traveled by a moving object during that time interval.
While the area under the velocity-time graph gives displacement of the particle.
The ratio of SI units to CGS units of retardation is
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$$\displaystyle { 10 }^{ -2 }$$
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$$\displaystyle { 10 }^{ 2 }$$
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$$\displaystyle 10$$
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$$\displaystyle { 10 }^{ -1 }$$
Explanation
SI unit of retardation is $$ms^{-2}$$ and CGS unit is $$cms^{-2}$$.
Also, we know that $$1 \ m = 100 \ cm$$
Thus ratio,
$$ratio = \dfrac{ms^{-2}}{cms^{-2}}$$
$$\Rightarrow ratio = \dfrac{100\ cm\ s^{-2}}{cm\ s^{-2}}$$
$$\Rightarrow ratio = 10^2$$
A scooter increases its speed from 36 km/h to 72 km/h in 10 s. What is its acceleration?
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$${7.2 m/s^2}$$
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$${1 m /s^2}$$
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$${3.6 m/s^2}$$
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$${10 m/s^2}$$
Explanation
in the given question we have
u=36 km/hr =$$36 \times \frac{5}{18}m/s=10 m/s$$
v= 72 km/hr=$$ 72 \times \frac{5}{18}m/s=20 m/s$$
t=10 s
a=?
we know the formula
v= u +at
20 = 10+ a 10
20-10 =10 a
10 = 10 a
a = 1 $$m/s^{2}$$
What is the velocity of vertically projected body at its maximum height (h)?
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$$\displaystyle \sqrt { 2\ gh } $$
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zero
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$$\displaystyle \frac { { h }^{ 2 } }{ g } $$
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$$\displaystyle \sqrt { \frac { 2h }{ g } } $$
Explanation
When particle is at maximum height it's Potential energy = $$mgh$$
also maximum .
But when it was at ground it's Potential energy =$$0$$
It means all of it's Kinetic energy at ground was converted into Potential energy at top most point so Kinetic energy at top point=$$0$$
$$\frac{mv^2}{2}=0$$
so$$v=0$$
Hence velocity at maximum height=
$$0$$
When the distance an object travels is directly proportional to the time, it is said to travel with
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Constant speed
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Zero velocity
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Constant acceleration
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Uniform velocity
Explanation
When the distance an object travels is directly proportional to time, it is said to travel with Constant Speed.
$$ Speed = \dfrac{distance \ travelled}{time}$$
Retardation means
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negative acceleration
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positive acceleration
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zero acceleration
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none of these
Explanation
If the velocity of a moving object is decreasing with respect to time then you can say that body is in retardation or deacceleration or negative acceleration.
In other words; Retardation is a negative rate of change of velocity.
A body executing non-uniform motion:
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Undergoes acceleration
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Never undergoes acceleration
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May or may not undergo acceleration
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Never undergoes constant acceleration
Explanation
If a body undergoes non-uniform motion, its speed changes a
nd a body with changing speed undergoes acceleration.
This acceleration may or may not be constant.
Acceleration of the particle is given as-
$$\text{acceleration} = \dfrac{ \text{ change in velocity } }{ \text{ time } }$$
The SI unit of acceleration is $$ms^{-1}$$.
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True
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False
Explanation
The given statement is false because SI unit of acceleration is $$ms^{-2}$$
$$ms^{-1}$$ is the unit of velocity.
The SI unit of retardation is _____.
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m/s$$\displaystyle ^{2}$$
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m/s$$\displaystyle$$
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m$$\displaystyle ^{2}$$
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s$$\displaystyle ^{2}$$
Explanation
Si unit of retardation is same as that of acceleration as $$m/s^2$$.
State whether given statement is True or False.
The rate of change of displacement with time is called speed
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True
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False
Explanation
The rate of change of displacement with time is called velocity. Hence, the given statement is false.
SI unit of distance is
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$$m$$
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$$cm$$
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$$km$$
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none of these
Explanation
In SI, distance is measured in $$m$$.
$$cm$$ and $$km$$ are respectively cgs and practical units of distance.
A body suffers retardation of $$1 m{s}^{-2}$$ means
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its velocity increases by $$1 m{s}^{-1}$$ per second
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its velocity remains constant
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its velocity decreases by $$1 m{s}^{-1}$$ per second
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none of these
Explanation
Acceleration is the rate of change of the velocity of an object with respect to time. Positive acceleration means increase in velocity with time and negative acceleration means decrease in velocity per unit time . Negative acceleration is also called retardation.
So retardation means rate of decrease of velocity per second.
So, retardation of $$1 m{s}^{-2}$$ = decrease of $$1 m{s}^{-1}$$ velocity per second.
Which of the following is not an example of a motion with a constant speed but variable velocity?
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A car moving at $$80 kmph$$ on a straight road
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A car moving at $$80 kmph$$ on a square track
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A car moving at $$80 kmph$$ on a circular track
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A car moving at $$80 kmph$$ on a zig-zag path
Explanation
Apart from motion on straight road, for all other cases, direction of motion changes although the speed remains uniform.
So, speed remaining constant, velocity is variable for all other cases.
A body, initially at rest, starts moving with a constant acceleration $$2 m{s}^{-2}$$. Calculate the velocity acquired in 5 s.
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$$2 m{s}^{-1}$$
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$$5 m{s}^{-1}$$
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$$7.5 m{s}^{-1}$$
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$$10 m{s}^{-1}$$
Explanation
Given,
Initial velocity, $$u=0$$
Acceleration, $$a=2ms^{-2}$$
Time, $$t=5s$$
$$v=u+at$$
$$v=0+(2\times 5)=10ms^{-1}$$
Choose the correct statement:
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Velocity determines direction of motion of a body
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Acceleration determines direction of motion of a body
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Both velocity and acceleration determine direction of motion of a body
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Neither velocity nor acceleration determine direction of motion of a body
Explanation
Velocity, being a vector, gives the direction of the change of position of a body. Hence it determines the direction of motion of a body. In contrast, acceleration doesn't need to relate to the direction of motion.
Identify the wrong statement about acceleration.
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Acceleration is rate of change of velocity per second
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Acceleration is rate of change of speed per second in a certain direction
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Acceleration is a vector quantity
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Unit of acceleration is $$m{s}^{-1}$$
Explanation
Acceleration= rate of change of velocity = change in velocity velocity / time =$$\dfrac{ms^{-1}}{s}= ms^{-2}$$
Unit of acceleration is $$m{s}^{-2}$$.
Hence, option D is correct
Identify the correct statement:
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Acceleration increases velocity, retardation decreases it
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Acceleration decreases velocity, retardation increases it
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Acceleration and retardation both increase velocity
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Acceleration and retardation both decrease velocity
Explanation
Acceleration is the rate of increase of velocity.
Retardation is the rate of decrease of velocity. It is also called $$\text{Negative Acceleration}$$.
Which of the following terms does not go well with the motion of a bus on a crowded road.
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Uniform velociity
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Variable velocity
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Variable acceleration
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Variable speed
Explanation
On a crowded road, depending on traffic conditions, the bus driver has to change its speed and direction of motion many times. The driver will have to frequently apply brakes as well. He might also have to accelerate the bus with different rates.
So, the velocity changes both by magnitude and direction.
Thus, uniform velocity is not attained in this case.
Velocity of a moving body reduces when it undergoes
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acceleration
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retardation
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either acceleration or retardation
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neither acceleration nor retardation
Explanation
Retardation means rate of decrease of velocity with time. So, velocity of a body reduces when it undergoes retardation.
Direction of motion of a body is given by
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its speed
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its velocity
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both its speed and velocity
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neither of its speed and velocity
Explanation
Velocity is a vector. So, it gives the magnitude of speed and the direction of motion of the body as well.
Speed being a scalar, does not give the direction of motion.
Pick out the wrong statement about retardation.
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Retardation is negative acceleration
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Retardation is rate of decrease of velocity with time
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SI unit of retardation is $$m{s}^{-2}$$
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Retardation is a scalar quantity
Explanation
Retardation, or negative acceleration, is nothing but an acceleration which decreases the speed. Its SI unit is same as that of acceleration. Just like acceleration, it has both magnitude and direction, i.e. it is a vector quantity. So option D is the wrong statement about retardation.
The units of speed and average speed are
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same across all systems of units
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same only for SI
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different across all systems of units
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different only for SI
Explanation
Units for speed and average speed are the same for all systems of units. For example, SI unit is $$m{s}^{-1}$$, practical unit is $$kmph$$ etc.
Formula:
$$speed = \dfrac{distance}{time}$$
$$velocity = \dfrac{displacement}{time}$$
When particles move in a circle at a constant speed then the motion is said to be
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Uniform circular motion
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Non uniform motion
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Projectile motion
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All
Explanation
When an object moves in a circular path with uniform speed, its motion is called uniform circular motion.
Identify in which of the following graphs velocity of the moving object remains constant?
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0%
0%
0%
0%
Explanation
Except graphs A and B, others are Velocity time graphs, As we can see that velocity of particles is changing with time.
Now,
Slope of the curve under the position-times graph gives the instantaneous velocity of the object.
As slope of the graph shown in option A is constant, thus velocity of the object is constant in graph of option A.
A jeep is accelerating uniformly in a straight line with $$5 m/s^2$$. Calculate the time required to cover a distance of 200 m if jeep starts from rest :
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9.0 s
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10.5 s
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12.0 s
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15.5 s
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20.0 s
Explanation
Given : $$S = 200m$$, $$a = 5 m/s^2$$, $$u = 0$$
Using $$2^{nd}$$ equation of motion, $$s = ut + \dfrac{1}{2}at^2$$
$$\therefore$$ $$200 = 0 + \dfrac{1}{2}\times 5t^2$$ $$\Rightarrow t^2 = \cfrac{200\times 2}{5}$$
$$\implies t \approx 9 $$ s
When a body moves in a circular motion :
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its direction constantly changes
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its direction remains constant
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its velocity vector is always perpendicular to the direction of motion
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all of the above
Explanation
As an object moves in a circle, it is constantly changing its direction. In all instances, the object is moving tangent to the circle, and the direction of the velocity vector is the same as the direction of the object's motion.
Which of the following best define the acceleration of a particle:
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the rate of change of velocity.
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only experienced during a change of direction.
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only experienced during a change of speed.
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calculated by multiplying speed by velocity.
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always constant.
Explanation
Acceleration is defined as the rate of change of velocity.
acceleration $$a = \dfrac{change \ in \ velocity}{time \ interval}$$
Hence, any change in the speed or direction would cause a change in velocity, and hence will produce acceleration.
So, the correct option is A.
A car is moving with speed $$2 \ m/s$$ crosses a electric post and starts accelerating at a constant rate of $$2 \ m/s^2$$. How far past the electric post will the car be after $$3 \ s$$?
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$$8 m$$
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$$9 m$$
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$$12 m$$
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$$15 m$$
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$$18 m$$
Explanation
Given :
Initial velocity $$u =2 $$ m/s
Acceleration $$a = 2$$ $$m/s^2$$
Time $$t = 3$$ s
Using second equation of motion:
$$S = ut + \dfrac{1}{2}at^2$$
Thus the distance of electric post from the car after $$3 \ s$$,
$$S= 2 \times 3 + \dfrac{1}{2}\times 2 \times 3^2 =15m$$
A body is moving vertically upwards. Its velocity changes at a constant rate from $$50 m{s}^{-1}$$ to $$20 m{s}^{-1}$$ in $$3 s$$. What is its acceleration?
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$$10 m{s}^{-2}$$
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$$- 10 m{s}^{-2}$$
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$$1 m{s}^{-2}$$
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$$- 1 m{s}^{-2}$$
Explanation
Initial velocity $$= 50 m{s}^{-1}$$
Final velocity $$= 20 m{s}^{-1}$$
Time $$= 3 s$$
Acceleration = $$\dfrac{20-50}{3}$$ m$${s}^{-2}$$ $$= - 10 m{s}^{-2}$$
A car accelerates at a rate of $$5 m{s}^{-2}$$. Find the increase in its velocity in $$2 s$$.
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$$10 m{s}^{-1}$$
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$$2.5 m{s}^{-1}$$
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$$5 m{s}^{-1}$$
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$$2 m{s}^{-1}$$
Explanation
Acceleration $$= 5 m{s}^{-2}$$
Time $$= 2 s$$
Increase in velocity $$=$$ acceleration $$\times$$ time $$=(5 \times 2)m{s}^{-1}=10m{s}^{-1}$$
A car moving with speed $$2 \ m/s$$ crosses an electric post and starts accelerating at a constant rate of $$2 \ m/s^2$$. What will be the speed of the car after $$3 \ s$$?
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$$8 \ m/s$$
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$$9 \ m/s$$
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$$12 \ m/s$$
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$$15 \ m/s$$
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$$18 \ m/s$$
Explanation
Given : initial speed $$u =2 \ m/s $$
acceleration $$a = 2\ m/s^2$$
Calculate speed after time $$t = 3 \ s$$
Using the first equation of motion $$v = u+at$$
$$\therefore$$ $$v =2 + 2\times 3 =8 \ m/s$$
A car moving is with a particular velocity, which has been measured each second and the data is recorded as per the table below.
Velocity Time
$$0$$ $$0$$
$$2 \ {m}/{s}$$ $$1 \ s$$
$$4 \ {m}/{s}$$ $$2 \ s$$
$$6 \ {m}/{s}$$ $$3 \ s$$
$$8 \ {m}/{s}$$ $$4 \ s$$
Calculate the acceleration of the car ?
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$$1\ {m}/{{s}^{2}}$$
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$$2 \ {m}/{{s}^{2}}$$
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$$4 \ {m}/{{s}^{2}}$$
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$$6 \ {m}/{{s}^{2}}$$
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$$8 \ {m}/{{s}^{2}}$$
Explanation
Acceleration is defined as the ratio of change in velocity to the change in time interval.
Acceleration $$a=\dfrac{\Delta v}{\Delta t}$$
The given data shows that the velocity of car changes by $$2 \ m/s$$ in $$1$$ $$s$$ each.
Thus acceleration of the car $$a = \dfrac{\Delta v}{\Delta t} = \dfrac{2}{1} =2 \ m/s^2$$
In which of the following option could represent the ball increasing its speed?
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0%
0%
0%
0%
Explanation
Option B; Since the distance between the constant time intervals is increasing, the ball is covering more distance per second each second, which is an increase in speed (acceleration).
Option A: Represents constant speed as same distance covered in each second.
Option C: Represents retardation.
Option D: First accelerate then retardation
Option E: First accelerate then retardation
A car starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 m. Determine the acceleration of the car.
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6.10 m $$s^{-2}$$
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8.10 m $$s^{-2}$$
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10.10 m $$s^{-2}$$
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12.10 m $$s^{-2}$$
Explanation
using second eq of motion $$s=ut+1/2at^2$$
as u=0, $$a=\dfrac{2s}{t^2}=\dfrac{2*110}{5.21^2}=8.10m/s^2$$
The position-time graphs above represent the motions of cars 1 toHow do they rank according to their speeds (greatest first)?
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$$1,\ 2,\ 3, 4,\ 5$$
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$$1,\ 2,\ 3\ and\ 5\ tie,\ 4$$
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$$1,\ 2,\ 4,\ 3\ and\ 5\ tie$$
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$$3\ and\ 5\ tie,\ 4,\ 2,\ 1$$
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$$5,\ 4,\ 3,\ 2,\ 1$$
Explanation
We know,
$$speed=\dfrac{distance}{time}$$
If time is constant, then speed will depend on distance travelled.
Consider a time interval from $$t_1$$ to $$t_2$$.
The distance travelled by $$car\ 1$$ is greatest, followed by $$car\ 2$$ and $$car\ 4$$.
The distance travelled by $$car\ 3$$ and $$car\ 4$$ is $$0$$.
$$\therefore$$ The descending order of their speed is $$1 \gt 2\gt 4\gt 3=5$$
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