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QJEEAdv16Ph04a1: JEE Advanced 2016 Physics - Class 11 JEEAdv16Ph04a1 Solution: A uniform wooden stick of mass 1.6 kg and length rests in an…

A uniform wooden stick of mass 1.6 kg and length ll rests in an inclined manner on a smooth, vertical wall of height h ( < ll ) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30∘30^\circ with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio hl{h \over l} and the frictional force f at the bottom of the stick are ( g =10 ms-2 )

Multiple Choice Options

  • A hl=316,f=1633N{h \over l} = {{\sqrt 3 } \over {16}},f = {{16\sqrt 3 } \over 3}N
  • B hl=316,f=1633N{h \over l} = {3 \over {16}},f = {{16\sqrt 3 } \over 3}N
  • C hl=3316,f=833N{h \over l} = {{3\sqrt 3 } \over {16}},f = {{8\sqrt 3 } \over 3}N
  • D hl=3316,f=1633N{h \over l} = {{3\sqrt 3 } \over {16}},f = {{16\sqrt 3 } \over 3}N

Step-by-Step Verified Solution

JEE Advanced 2016 Paper 1 Offline Physics - Rotational Motion Question 56 English Explanation

∑Fx=0\sum {{F_x} = 0} , N1cos⁡30∘−f=0{N_1}\cos 30^\circ - f = 0 .... (i)

∑Fy=0\sum {{F_y} = 0} , N1sin⁡30∘+N2−mg=0{N_1}\sin 30^\circ + {N_2} - mg = 0 .... (ii)

∑τ0=0\sum {{\tau _0} = 0}

mg12cos⁡60∘−N1hcos⁡30∘=0mg{1 \over 2}\cos 60^\circ - {N_1}{h \over {\cos 30^\circ }} = 0 .... (iii)

Also, given N1=N2{N_1} = {N_2} .... (iv)

Solving Eqs. (i), (ii), (iii) and (iv) we have

hl=3316{h \over l} = {{3\sqrt 3 } \over {16}} and f=1633f = {{16\sqrt 3 } \over 3}

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