Statement I : A cyclist is moving on an un-banked road with a speed of $7 \mathrm{kmh}^{-1}$ and takes a sharp circular turn along a path of radius of 2 m without reducing the speed. The static friction coefficient is 0.2 The cyclist will not slip and pass the curve. ( $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$ ) Statement II: If the road is at an angle of $45^{0}$, cyclist can cross of the curve of 2 m radius with the speed of $18.5 \mathrm{kmh}^{-1}$ without slipping In the light of the above statements, choose the correct the answer from the options given below.

(B) Both statement I and statement II are true

(A) Statement I is incorrect and statement II is correct

(C) Statement I is correct and statement II is incorrect

(D) Both statement I and statement II are false

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Laws of Motion

Statement I : A cyclist is moving on an un-banked road with a speed of

7 kmh^{- 1}

and takes a sharp circular turn along a path of radius of 2 m without reducing the speed. The static friction coefficient is 0.2 The cyclist will not slip and pass the curve. (

g = 9.8 m / s^{2}

)

Statement II: If the road is at an angle of

4 5^{0}

, cyclist can cross of the curve of 2 m radius with the speed of

18.5 kmh^{- 1}

without slipping

In the light of the above statements, choose the correct the answer from the options given below.

(A)Statement I is incorrect and statement II is correct
(B)Both statement I and statement II are true✓ Correct
(C)Statement I is correct and statement II is incorrect
(D)Both statement I and statement II are false

Answer: (B)

Solution

On an unbanked road only friction supplies the centripetal force, so the turn is negotiated without slipping while

\frac{v ^{2}}{r} \leq μg

With

v = 7 km h^{- 1} = 1.94 m s^{- 1}

and

r = 2 m

the left side is

\frac{( 1.94 ) ^{2}}{2} = 1.89 m s^{- 2}

, while the right side is

0.2 \times 9.8 = 1.96 m s^{- 2}

. The condition holds, so the cyclist crosses the curve without slipping.
On a road banked at

θ

with the same friction, the greatest safe speed is

v_{m a x} = r g \frac{tan θ + μ}{1 - μ tan θ} = 2 \times 9.8 \times \frac{1 + 0.2}{1 - 0.2} = 29.4 = 5.42 m s^{- 1}

which is

19.5 km h^{- 1}

, above the stated

18.5 km h^{- 1}

, so that turn can also be taken. Statement I is true and Statement II is true.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Laws of Motion

Solution

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