Answer
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Hint: When we go around a curved path we need some force to act as centripetal force to prevent us from sliding. When the roads are banked, a component of normal from the road can act as a centripetal force. Thus ensuring safe ride even at higher speed.
Complete step by step answer:
Let us consider a scenario in which a car is going around a curved path and the road is flat. Here, only frictional force will provide centripetal acceleration. Centripetal acceleration is proportional to square of velocity of the object (here, a car). So friction will force will have to increase for greater velocities, but it has some limitations and cannot exceed beyond a value. So there will exist a velocity beyond which the car will not be able to go through the curved path and eventually skid. This value of velocity is equal to
$\sqrt{\mu gr}$, where
$\mu $ is coefficient of friction, g is acceleration due to gravity and r is radius of curvature
Now, if the road is banked in addition to friction there will be a component of normal from the road which will aid in providing this centripetal force. So now the limiting value of velocity beyond which the car will not be able to go through the curved path, will increase. This value of velocity is equal to
$v=\sqrt{\dfrac{rg\left( \sin \theta +\mu \cos \theta \right)}{\cos \theta -\mu \sin \theta }}$ , where
$\theta $ is banking angle, g is acceleration due to gravity, and r is radius of curvature.
Hence the correct option is B.
Note: Some students might find that after banking only a component of friction is acting as centripetal force, however before banking complete friction was acting as centripetal force. But even after taking that into consideration, calculations have shown that limiting velocity after banking is much greater than limiting velocity before banking. Also after banking a minimum velocity should also be maintained else car will slip down; however there is no such limit for unbanked roads
Complete step by step answer:
Let us consider a scenario in which a car is going around a curved path and the road is flat. Here, only frictional force will provide centripetal acceleration. Centripetal acceleration is proportional to square of velocity of the object (here, a car). So friction will force will have to increase for greater velocities, but it has some limitations and cannot exceed beyond a value. So there will exist a velocity beyond which the car will not be able to go through the curved path and eventually skid. This value of velocity is equal to
$\sqrt{\mu gr}$, where
$\mu $ is coefficient of friction, g is acceleration due to gravity and r is radius of curvature
Now, if the road is banked in addition to friction there will be a component of normal from the road which will aid in providing this centripetal force. So now the limiting value of velocity beyond which the car will not be able to go through the curved path, will increase. This value of velocity is equal to
$v=\sqrt{\dfrac{rg\left( \sin \theta +\mu \cos \theta \right)}{\cos \theta -\mu \sin \theta }}$ , where
$\theta $ is banking angle, g is acceleration due to gravity, and r is radius of curvature.
Hence the correct option is B.
Note: Some students might find that after banking only a component of friction is acting as centripetal force, however before banking complete friction was acting as centripetal force. But even after taking that into consideration, calculations have shown that limiting velocity after banking is much greater than limiting velocity before banking. Also after banking a minimum velocity should also be maintained else car will slip down; however there is no such limit for unbanked roads
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