JEE Advanced 2016 Paper 1

The free-body diagram of the given situation is shown in the following figure. The balancing force in x and y directions is given by
N+N‌cos‌60∘=mg‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(1)
That is,
N‌sin‌60∘=f‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(2)
From Eq. (1), we get N=‌

16

1+(1∕2)

=‌

32

3

From Eq. (2), we get the frictional force at the bottom of the stick as follows:
(‌

32

3

)(‌

√3

2

)=f‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(3)
⇒f=‌

16√3

3

N
From triangle ABC, we get
‌

h

x

‌‌=sin‌60∘
⇒x‌‌=‌

h(2)

√3

‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(4)
Now, the balancing torque about the bottom of the stick is
Nx−(‌

l

2

)mg‌sin‌30∘=0
⇒(‌

32

3

)[‌

h(2)

√3

]=‌

l

2

(16)(‌

1

2

)
⇒‌

h

l

=‌

3√3

16

Hence, option (D) is the correct choice.