
A car of mass $1000\,kg$ negotiates a banked curve of radius $90\,m$ on a frictionless road. If the banking angle is ${45^ \circ }$ , the speed of the car is:
A. $5\,m{s^{ - 1}}$
B. $10\,m{s^{ - 1}}$
C. $20\,m{s^{ - 1}}$
D. $30\,m{s^{ - 1}}$
Answer
477.3k+ views
Hint: Here we need to find the centripetal velocity required to overcome the banked curve. The concept of uniform circular motion is applied here.
Complete step by step answer:Given,
Radius of banked curve, $r = 3\,m$
Mass of car, $m = 1000\,kg$
Acceleration due to gravity, $g = 10m{s^{ - 2}}$
Banking angle of the frictionless road,
$\theta = {45^ \circ }$
Speed of the car =?
We know that,
The velocity due to centripetal action is given by
$
v = \sqrt {rg} \tan \theta \\$
$ v = \sqrt {90 \times 10} \times 1 = 30m{s^{ - 1}} \\
$
Therefore, the speed of the car when the banking angle is ${45^ \circ }$
is $30\,m{s^{ - 1}}$
.
Hence, option D is correct.
Additional information:
Coefficient of friction: Coefficient of friction is defined as the force ratio needed to transfer sliding surfaces over each other and it is also the force holding them together.
Uniform circular motion: Uniform circular motion describes the motion of a body at constant speed, traversing a circular path. Since the body describes circular motion, at all times the distance from the axis stays unchanged.
Since, the direction of the velocity varies continuously in uniform circular motion, so acceleration is still present but the speed may not change.
Centripetal force: A particular type of force is required to go around a curved path. This force is known as centripetal force. As the centripetal force acts at constant speed on an object travelling in a circle, the force often acts inward as the object’s velocity is tangent to the circle.
Note:Here we have to see that the road is frictionless. So, we need not apply the coefficient of friction. Also since the banking angle is given. So, we have to use the banking angle in this question.
Complete step by step answer:Given,
Radius of banked curve, $r = 3\,m$
Mass of car, $m = 1000\,kg$
Acceleration due to gravity, $g = 10m{s^{ - 2}}$
Banking angle of the frictionless road,
$\theta = {45^ \circ }$
Speed of the car =?
We know that,
The velocity due to centripetal action is given by
$
v = \sqrt {rg} \tan \theta \\$
$ v = \sqrt {90 \times 10} \times 1 = 30m{s^{ - 1}} \\
$
Therefore, the speed of the car when the banking angle is ${45^ \circ }$
is $30\,m{s^{ - 1}}$
.
Hence, option D is correct.
Additional information:
Coefficient of friction: Coefficient of friction is defined as the force ratio needed to transfer sliding surfaces over each other and it is also the force holding them together.
Uniform circular motion: Uniform circular motion describes the motion of a body at constant speed, traversing a circular path. Since the body describes circular motion, at all times the distance from the axis stays unchanged.
Since, the direction of the velocity varies continuously in uniform circular motion, so acceleration is still present but the speed may not change.
Centripetal force: A particular type of force is required to go around a curved path. This force is known as centripetal force. As the centripetal force acts at constant speed on an object travelling in a circle, the force often acts inward as the object’s velocity is tangent to the circle.
Note:Here we have to see that the road is frictionless. So, we need not apply the coefficient of friction. Also since the banking angle is given. So, we have to use the banking angle in this question.
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