Verified Solution • 2025–26 Session moderate

QJEEM21Ph535b: JEE Main 2021 Physics - Class 11 JEEM21Ph535b Solution: A modern grand - prix racing car of mass m is travelling on a…

A modern grand - prix racing car of mass m is travelling on a flat track in a circular arc of radius R with a speed v. If the coefficient of static friction between the tyres and the track is μ\mu s, then the magnitude of negative lift FL acting downwards on the car is : (Assume forces on the four tyres are identical and g = acceleration due to gravity)

JEE Main 2021 (Online) 17th March Morning Shift Physics - Circular Motion Question 49 English

Multiple Choice Options

  • A m(g−v2μsR)m\left( {g - {{{v^2}} \over {{\mu _s}R}}} \right)
  • B −m(g+v2μsR)- m\left( {g + {{{v^2}} \over {{\mu _s}R}}} \right)
  • C m(v2μsR−g)m\left( {{{{v^2}} \over {{\mu _s}R}} - g} \right)
  • D m(v2μsR+g)m\left( {{{{v^2}} \over {{\mu _s}R}} + g} \right)

Step-by-Step Verified Solution

JEE Main 2021 (Online) 17th March Morning Shift Physics - Circular Motion Question 49 English Explanation
f=mv2Rf = {{m{v^2}} \over R}

⇒\Rightarrow μN=mv2R\mu N = {{m{v^2}} \over R}

μ(mg+FL)=mv2R\mu (mg + {F_L}) = {{m{v^2}} \over R}

⇒\Rightarrow mg+FL=mv2Rμmg + {F_L} = {{m{v^2}} \over {R\mu }}

⇒\Rightarrow FL=mv2μR−mg{F_L} = {{m{v^2}} \over {\mu R}} - mg

=m(v2μR−g)= m\left( {{{{v^2}} \over {\mu R}} - g} \right)

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