A force of 40N acts on a point B at the end of an L -shaped object, as shown in the figure. The angle \theta that will produce maximum moment of the force about point A is given by :

Rotational Motion PYQ — JEE Main 2018 · 15 Apr · Shift 1 · Q51 with Solution

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Correct answer: B: $\TAN\THETA = 2$

From the standard $L$ -shaped figure, point $B$ is horizontally $2$ units and vertically $1$ unit from point $A$ .

So the position vector of $B$ relative to $A$ is

\vec r = 2\hat i + 1\hat j

with magnitude

|\vec r| = \sqrt{2^2+1^2}=\sqrt{5}.

The moment of a force about $A$ is

\tau = |\vec r \times \vec F| = rF\sin\phi,

where $\phi$ is the angle between $\vec r$ and $\vec F$ .

For fixed $r$ and fixed force magnitude $F=40\,\text{N}$ , the torque is maximum when

\sin\phi = 1 \quad \Rightarrow \quad \phi = 90^\circ.

So, the force must be perpendicular to $\vec r = 2\hat i+\hat j$ .

The line $AB$ has slope

m_{AB} = \frac{1}{2}.

If the force is perpendicular to $AB$ , then its slope must be

m_F = -\frac{1}{m_{AB}} = -2.

Hence, the acute angle $\theta$ that the force makes with the horizontal satisfies

\tan\theta = 2.

Therefore, the correct option is

\boxed{\text{B: } \tan\theta=2}.

The stored correct answer is A, but from torque maximization, the force must be perpendicular to $AB$ , giving

\boxed{\tan\theta=2}.

So I do not agree with the stored answer.