Home Maths JEE - Advanced Previous Year Paper JEE-Advanced-2023-Paper-1 SECTION-2 : (Maximum Marks : 12) • This sect…
Maths JEE - Advanced Previous Year Paper JEE-Advanced-2023-Paper-1 Single Correct MCQ
Published on: August 13, 2026

SECTION-2 : (Maximum Marks : 12)

• This section contains FOUR (04) questions.

• Each question has FOUR options

A
, = 6.98 rad s -1 and v = 4.30 m s -1
B
, = 3.75 rad s -1 and v = 4.30 m s -1
C
and =3.75 rad s -1 and v = 10.0 m s -1
D
. ONLY ONE of these four options is the correct answer. • For each question, choose the option corresponding to the correct answer. • Answer to each question will be evaluated according to the following marking scheme: Full Marks :+3 If ONLY the correct option is chosen; Zero Marks : 0 If none of the options is chosen (i.e. the question is unanswered); Negative Marks :-1 In all other cases. A bar of mas M= 1.00 kg and length L = 0.20 m is lying on a horizontal frictionless surface. One end of the bar is pivoted at a point about which it is free to rotate. A small mass m = 0.10 kg is moving on the same horizontal surface with 5.00 m s -1 speed on a path perpendicular to the bar. It hits the bar at a distance L/2 from the pivoted end and returns back on the same path with speed v. After this elastic collision, the bar rotates with an angular velocity . Which of the following statement is correct? = 6.80 rad s -1 and v = 4.10 m s -1

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Text Solution

Verified by Experts
The correct answer is:
A

Applying angular momentum conservation about hinge

....(i)

Also from eq. of restitution

....(ii)

Solving (i) & (ii)

6.98 rad/sec & v = 4.30 m/s

Hence option

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