JEE Advanced · Physics · 5. Laws of Motion

A student skates up a ramp that makes an angle $30 °$ with the horizontal. He/she starts (as shown in the figure) at the bottom of the ramp with speed $v_{0}$ and wants to turn around over a semicircular path $x y z$ of radius $R$ during which he/she reaches a maximum height $h$ (at point $y$ ) from the ground as shown in the figure. Assume that the energy loss is negligible and the force required for this turn at the highest point is provided by his/her weight only. Then ( $g$ is the acceleration due to gravity)

A $v_{0}^{2} - 2 g h = \frac{1}{2} g R$
B $v_{0}^{2} - 2 g h = \frac{\sqrt{3}}{2} g R$
C the centripetal force required at points $x$ and $z$ is zero
D the centripetal force required is maximum at points $x$ and $z$

Verified Solution

Answer & Solution

Correct Answer

(D) the centripetal force required is maximum at points $x$ and $z$

Step-by-step Solution

Detailed explanation

Speed at

x

and

z

is equal and also maximum in circular track. From energy conversation,

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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From JEE Advanced

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

Column 1	Column 2	Column 3
(I) $x^{2} + y^{2} = a^{2}$	(i) $m y = m^{2} x + a$	(P) $(\frac{a}{m^{2}}, \frac{2 a}{m})$
(II) $x^{2} + a^{2} y^{2} = a^{2}$	(ii) $y = m x + a \sqrt{m^{2} + 1}$	(Q) $(\frac{- m a}{\sqrt{m^{2} + 1}}, \frac{a}{\sqrt{m^{2} + 1}})$
(III) $y^{2} = 4 a x$	(iii) $y = m x + \sqrt{a^{2} m^{2} - 1}$	(R) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} + 1}}, \frac{1}{\sqrt{a^{2} m^{2} + 1}})$
(IV) $x^{2} - a^{2} y^{2} = a^{2}$	(iv) $y = m x + \sqrt{a^{2} m^{2} + 1}$	(S) $(\frac{- a^{2} m}{\sqrt{a^{2} m^{2} - 1}}, \frac{- 1}{\sqrt{a^{2} m^{2} - 1}})$

The tangent to a suitable conic (Column 1) at

(\sqrt{3}, \frac{1}{2})

is found to be

\sqrt{3} x + 2 y = 4

, then which of the following options is the only Correct combination?

JEE Advanced 2017 Hard

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