What should be the angle between force and displacement for maximum work?
A. $0^\circ $
B. $30^\circ $
C. $60^\circ $
D. $90^\circ $

Hint: In order to solve this question we need to understand the force and the work. So force is defined as the push or pull on a body which changes state of motion, for example if it is at rest then force applied could bind it to move and thereby changing state and vice versa. Newton's second law force is defined mathematically as the rate of change of linear momentum. Work is defined as the energy that is stored in the body and it is equal to applied force causing unit displacement of the body. So we need to minimize work done and this is done if the body has a circular shape or it is placed on something which can be easily moved.

Complete step by step answer:
Consider a body on which Force acts on it. Force is inclined to the body, making an angle $\theta $ with horizontal. Now suppose the body displaced by some displacement say,
$\vec s = \vec d$
Then the work is defined as,
$W = \vec F.\vec d$
$\Rightarrow W = Fd\cos \theta $

So to maximize this we need to differentiate work with respect to angle and equate it to zero we have,
$\dfrac{{dW}}{{d\theta }} = \dfrac{d}{{d\theta }}(Fd\cos \theta )$
since $\dfrac{d}{{d\theta }}(\cos \theta ) = - \sin \theta $
$\dfrac{{dW}}{{d\theta }} = - Fd\sin \theta $
Equating it to zero we have, $\dfrac{{dW}}{{d\theta }} = 0$
$ - Fd\sin \theta = 0$
Since $F \ne 0{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} ,d \ne 0{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \Rightarrow \sin \theta = 0$
$as{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} \sin {0^0} = 0$
$\therefore \theta = 0^\circ $

So the correct answer is A.

Note: It should be remembered that to move a certain body upward or to lift something inclined plane prove to be better as it requires less work to perform. Inclined plane is defined as a plane which is inclined to some angle with the ground hence making a simple sloping surface. It is used to lift heavy objects.