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QJEEM19Phb64a: JEE Main 2019 Physics - Class 11 JEEM19Phb64a Solution: A rod of length 50 cm is pivoted at one end. It is raised such…

A rod of length 50 cm is pivoted at one end. It is raised such that if makes an angle of 30o from the horizontal as shown and released from rest. Its angular speed when it passes through the horizontal (in rad s −- 1) will be (g = 10 ms −- 2)

JEE Main 2019 (Online) 9th January Evening Slot Physics - Rotational Motion Question 175 English

Multiple Choice Options

  • A 307\sqrt {{{30} \over 7}}
  • B 30\sqrt {30}
  • C 203{{\sqrt {20} } \over 3}
  • D 302{{\sqrt {30} } \over 2}

Step-by-Step Verified Solution

JEE Main 2019 (Online) 9th January Evening Slot Physics - Rotational Motion Question 175 English Explanation
When this rod move from initial position to final position then,

Gain in kinetic energy = loss in potential energy

∴\therefore    12I{1 \over 2}I ω\omega 2 = mgh

12(ml33){1 \over 2}\left( {{{m{l^3}} \over 3}} \right) ω\omega 2 = mg (l2sin⁡30∘)\left( {{l \over 2}\sin {{30}^ \circ }} \right)

⇒\Rightarrow    ω\omega 2 = 3g2l{{3g} \over {2l}}

⇒\Rightarrow    ω\omega = 3g2l\sqrt {{{3g} \over {2l}}}

⇒\Rightarrow    ω\omega = 3×102×0.5\sqrt {{{3 \times 10} \over {2 \times 0.5}}}

⇒\Rightarrow    ω\omega = 30\sqrt {30} rad/sec

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