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NVAMediumJEE 2023Uniform Circular Motion

JEE Physics 2023 Question with Solution

A car is moving on a circular path of radius 600 m600 \, \text{m} such that the magnitudes of the tangential acceleration and centripetal acceleration are equal. The time taken by the car to complete the first quarter of the revolution, if it is moving with an initial speed of 54 km/hr54 \, \text{km/hr}, is t(1−e−π/2) st(1-e^{-\pi/2}) \, \text{s}. The value of tt is:

Answer

Correct answer:40

Step-by-step solution

Standard Method

Given: Initial speed is v0=54 km/hr=15 m/sv_0 = 54 \, \text{km/hr} = 15 \, \text{m/s} and radius is R=600 mR = 600 \, \text{m}.

Find: The value of tt in the expression t(1−e−π/2) st(1-e^{-\pi/2}) \, \text{s} for the first quarter revolution.

Since the magnitudes of tangential and centripetal accelerations are equal,

at=ac a_t = a_c

so,

dvdt=v2R\frac{dv}{dt} = \frac{v^2}{R}

Rearranging,

dvv2=dtR\frac{dv}{v^2} = \frac{dt}{R}

Integrating from v0v_0 to vv and from 00 to tt,

∫v0vdvv2=∫0tdtR\int_{v_0}^{v} \frac{dv}{v^2} = \int_0^t \frac{dt}{R}

Therefore,

−1v∣v0v=tR-\frac{1}{v}\Big|_{v_0}^{v} = \frac{t}{R}

which gives

1v−1v0=tR\frac{1}{v} - \frac{1}{v_0} = \frac{t}{R}

Hence,

1v=1v0+tR\frac{1}{v} = \frac{1}{v_0} + \frac{t}{R}

and therefore

v(t)=11v0+tRv(t) = \frac{1}{\frac{1}{v_0} + \frac{t}{R}}

For circular motion,

dθ=vR dtd\theta = \frac{v}{R} \, dt

Substituting v(t)v(t),

dθ=1R(1v0+tR) dtd\theta = \frac{1}{R\left(\frac{1}{v_0} + \frac{t}{R}\right)} \, dt

Integrating for the first quarter revolution,

∫0π/2dθ=∫0t1R(1v0+tR) dt\int_0^{\pi/2} d\theta = \int_0^t \frac{1}{R\left(\frac{1}{v_0} + \frac{t}{R}\right)} \, dt

Using the result from the extracted solution,

θ=−ln⁡(1−tRv0)\theta = -\ln\left(1 - \frac{tR}{v_0}\right)

For θ=π/2\theta = \pi/2,

t=40(1−e−π/2) st = 40\left(1-e^{-\pi/2}\right) \, \text{s}

Therefore, the required value is 4040.

Using the quarter-revolution condition

Given: at=aca_t = a_c, R=600 mR = 600 \, \text{m}, and v0=15 m/sv_0 = 15 \, \text{m/s}.

Find: The constant tt appearing in the given time expression.

The equality of accelerations gives a differential relation between speed and time. After integration, the speed becomes a function of time. Then angular displacement is found from

dθ=vR dtd\theta = \frac{v}{R} \, dt

Applying the limit for a quarter revolution, θ=π/2\theta = \pi/2, the extracted working concludes

t=40(1−e−π/2) st = 40\left(1-e^{-\pi/2}\right) \, \text{s}

so the coefficient asked in the question is 4040.

Common mistakes

  • Using ac=vRa_c = \frac{v}{R} instead of ac=v2Ra_c = \frac{v^2}{R} is incorrect because centripetal acceleration depends on the square of speed. Always write centripetal acceleration as v2R\frac{v^2}{R}.

  • Not converting 54 km/hr54 \, \text{km/hr} into 15 m/s15 \, \text{m/s} leads to inconsistent units. Convert all quantities to SI units before integrating.

  • Confusing the asked quantity with the total time expression is a common error. The question asks for the value of tt in t(1−e−π/2)t(1-e^{-\pi/2}), not the entire time itself.

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