IIT JEE Main 2017 Physics Question 6 SetA:

A slender uniform rod of mass M and length *l* is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle θ with the vertical is:

(1) (3g/2l) sin θ |

(2) (2g/3l) sin θ |

(3) (3g/2l) cos θ |

(4) (2g/3l) cos θ |

#### Solution

Torque at angle θ

τ = Mg sin θ (*l*/2)

τ = *Iα*

*Iα* = Mg sin θ (*l*/2)

as I = (M*l*^{2})/3

[(M*l*^{2})/3]*α* = Mg sin θ (*l*/2)

*lα*/3 = g sin (θ/2)

=> *α* = (3g sinθ)/2*l*

#### Answer to IIT JEE Main 2017 Physics Question 6 SetA

Correct answer to given problem No 6 of IIT JEE Main Physics paper of 2017 SetA is option number (1) that is angular acceleration of the rod when it makes an angle θ with the vertical is (3g sinθ)/2*l*.

how u take l/2?