Question

Two forces $\overrightarrow{A}$ and $\overrightarrow{B}$ have a resultant ${\overrightarrow{R}}_1$. If the magnitude of $\overrightarrow{B}$ is doubled without changing the angle, the new resultant ${\overrightarrow{R}}_2$ is perpendicular to $\overrightarrow{A}$, then

• A. $|{\overrightarrow{R}}_1|=|\overrightarrow{A}|+|\overrightarrow{B}|$
• B. $|{\overrightarrow{R}}_1|=|\overrightarrow{A}||\overrightarrow{B}|$
• C. $|{\overrightarrow{R}}_1|=|\overrightarrow{A}|$
• D. $|{\overrightarrow{R}}_1|=|\overrightarrow{B}|$

Answer 1

The correct option is D $|{\overrightarrow{R}}_1|=|\overrightarrow{B}|$

Let us consider the direction of force $A$ is acting along $x-$axis as shown in the diagram.
Let us consider $|\overrightarrow{A}|=A$ and $|\overrightarrow{B}|=B$


From the diagram,
$$\begin{array}{}{R}_1=\sqrt{A^2+B^2+2AB\cos \theta }& \text{(1)}\end{array}$$
Now, when $B$ is doubled,


$x$ component of $A=A$
$x$ component of $2B=2B\cos \theta$

Since, the resultant of $A$ and $2B$ is perpendicular to $A$, hence $x-$ component of resultant $R_2$ will be $0$, i.e.

$$A+2B\cos \theta =0$$$$\cos \theta =\frac{-A}{2B}$$
After putting the value of $\cos \theta$ in eqn. $\left(1\right)$, we get
$$R_1=\sqrt{A^2+B^2+2AB(\frac{-A}{2B})}$$
$$R_1=\sqrt{A^2+B^2-A^2}=\sqrt{B^2}$$
$$\therefore R_1=B$$
Therefore, option $\left(\text{B}\right)$ is correct.