Explanation
$$\bar A \cdot \bar B = \left| A \right|\left| B \right|1\cos {90^ \circ }$$
$$ = 0$$
So , $${{1 - \bar A \cdot \bar B} \over {1 + \bar A \cdot \bar B}} = {{1 - 0} \over {1 + 0}}$$
$$ = 1$$
Hint:
Third Equation of Uniform Motion is,
$${\left( {{\omega _f}} \right)^2}\, - \,{\left( {{\omega _i}} \right)^2} = 2\alpha \theta $$
Where,
$${\omega _f}$$ is final angular velocity,
$${\omega _i}$$ is initial angular velocity,
$$\alpha $$ is angular acceleration,
$$\theta $$ is angular displacement.
Note that, $$1rpm=\dfrac{2\pi }{60}rad{{s}^{-1}}$$
Step 1: Note all the given values.
Initial Angular velocity is, $${{\omega }_{i}}=210rpm=\dfrac{210\times 2\pi }{60}rad{{s}^{-1}}$$
Final Angular velocity, $${{\omega }_{f}}=420rpm=\dfrac{420\times 2\pi }{60}rad{{s}^{-1}}$$
$$No. of rotations = 21$$
Therefore, angular displacement $$\theta =21\times 2\pi \,=42\pi \,rad$$
Step 2: Calculate what is required.
Using Third Equation of motion,
$${{\left( {{\omega }_{f}} \right)}^{2}}\,-\,{{\left( {{\omega }_{i}} \right)}^{2}}=2\alpha \theta $$
$$\Rightarrow {{\left( 14\pi \right)}^{2}}\,-\,{{\left( 7\pi \right)}^{2}}=2\alpha 42\pi $$
$$\Rightarrow 196{{\pi }^{2}}\,-\,49{{\pi }^{2}}=84\pi \alpha $$
$$\Rightarrow 147\pi \,=84\alpha $$
$$\Rightarrow \alpha =5.5\,rad{{s}^{-2}}$$
Thus, acceleration of disc is, $$\alpha =5.5\,rad{{s}^{-2}}$$
The value is closest to option A, thus Option A is correct.
Change in velocity will be vector quantity $$ {V}_2 - {V}_1 $$, whereas all other options are scalar quantities.
Out of the following options, the resultant of which pair cannot be 4N?
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