
A bend in a level road has a radius of 10m. Calculate the maximum speed which a car turning this bend may have without skidding. ($\mu = 0.81$)
A. 6m/s
B. 3m/s
C. 4m/s
D. 9m/s
Answer
559.2k+ views
Hint: Given question refers to the phenomena of banking of roads. In banking of roads, edges are raised for the curved roads above the inner edge to provide the required centripetal force to the vehicle so that they do not skid and take a safe turn. Here centripetal force is equal to the frictional force. So equate these forces to find the value of maximum speed.
Complete step by step answer:
We are given that a bend in a level road has a radius of 10m.
We have to calculate the maximum speed which a car turning this bend may have without skidding.
So for the car to have maximum speed and not skid, centripetal force acting on it must be equal to frictional force it is exerting.
Centripetal force can be calculated using $\dfrac{{m{v^2}}}{r}$, where ms is the mass of the car, v is the speed and r is the radius of the bend.
Frictional force can be calculated using $\mu mg$, where g is the acceleration due to gravity and $\mu $ is the coefficient of friction.
Therefore,
$
\dfrac{{mv_{\max }^2}}{r} = \mu mg \\
\Rightarrow v_{\max }^2 = \mu gr \\
\therefore {v_{\max }} = \sqrt {\mu gr} \\
r = 10m,g = 10m/{s^2},\mu = 0.81 \\
\Rightarrow{v_{\max }} = \sqrt {10 \times 0.81 \times 10} = \sqrt {81} \\
\therefore {v_{\max }} = 9m/s \\
$
Therefore, the correct option is Option D, 9m/s.
Note: Do not confuse centripetal force with centrifugal force. Centripetal force is the force that puts an object in a curved or circular path whereas centrifugal force is the apparent force felt by an object moving in a curved path. Centripetal acts inwards and centrifugal act outwards.
Complete step by step answer:
We are given that a bend in a level road has a radius of 10m.
We have to calculate the maximum speed which a car turning this bend may have without skidding.
So for the car to have maximum speed and not skid, centripetal force acting on it must be equal to frictional force it is exerting.
Centripetal force can be calculated using $\dfrac{{m{v^2}}}{r}$, where ms is the mass of the car, v is the speed and r is the radius of the bend.
Frictional force can be calculated using $\mu mg$, where g is the acceleration due to gravity and $\mu $ is the coefficient of friction.
Therefore,
$
\dfrac{{mv_{\max }^2}}{r} = \mu mg \\
\Rightarrow v_{\max }^2 = \mu gr \\
\therefore {v_{\max }} = \sqrt {\mu gr} \\
r = 10m,g = 10m/{s^2},\mu = 0.81 \\
\Rightarrow{v_{\max }} = \sqrt {10 \times 0.81 \times 10} = \sqrt {81} \\
\therefore {v_{\max }} = 9m/s \\
$
Therefore, the correct option is Option D, 9m/s.
Note: Do not confuse centripetal force with centrifugal force. Centripetal force is the force that puts an object in a curved or circular path whereas centrifugal force is the apparent force felt by an object moving in a curved path. Centripetal acts inwards and centrifugal act outwards.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

