Work Energy And Power MCQs for NEET — Physics Questions with Answers

If A and B lie in a plane, and C lies outside this plane, then A + B + C cannot be zero. This is because the sum of any two vectors lying in a plane must also lie in that plane, and adding a non-zero vector from outside the plane cannot make the resultant zero.

When a set of vectors are added in a closed polygon, their resultant is a null vector. This is because the vectors form a closed loop, and their sum cancels out to zero.

(D)

$$\begin{gathered} R=\sqrt{{\left(P\right)}^2+{\left(Q\right)}^2+2PQ\cos \theta } \\ When \theta =\pi , \cos \theta = -1 \\ then R is minimum \end{gathered}$$

$$\begin{gathered} R^2 = P^2+Q^2 + 2PQ\cos \theta \\ When \theta =0, \cos =1 \\ So R is maximum when \theta =0 \end{gathered}$$

$As \overrightarrow{C}=\overrightarrow{A}+\overrightarrow{B}$

Let angle between A and B is $\theta$

$$\begin{gathered} \Rightarrow C^2=A^2+B^2+2ABCos\theta \\ \Rightarrow {\left(13\right)}^2={\left(12\right)}^2+{\left(5\right)}^2+2\times 12+5+\cos \theta \\ \Rightarrow \cos \theta =0 \\ \Rightarrow \theta =90° \end{gathered}$$

$\overrightarrow{\mathrm{P}}+\overrightarrow{\mathrm{Q}}=\overrightarrow{\mathrm{P}}-\overrightarrow{\mathrm{Q}}$

$2\overrightarrow{\mathrm{Q}} = 0$

angle between ${\overrightarrow{\mathrm{F}}}_1 \& {\overrightarrow{\mathrm{F}}}_2$ is given by-

$\cos \theta =\frac{{\overrightarrow{F}}_1.{\overrightarrow{F}}_2}{F_1.F_2}=\frac{6}{2\sqrt{2}\times 5}=\frac{3}{5\sqrt{2}}\Rightarrow \theta ={\cos }^{-1}\left(\frac{3}{5\sqrt{2}}\right)$

Magnitude of the forcee $\overrightarrow{F_1} along \overrightarrow{F_2}=F_1\cos \theta$ where $\theta$ is the angle between two forces

$$\begin{gathered} \Rightarrow F_1\cos \theta =F_1\times \frac{\overrightarrow{F_1}.\overrightarrow{F_2}}{F_1{F}_2} \\ =\frac{\overrightarrow{F_1}.\overrightarrow{F_2}}{F_2}=\frac{6}{5}N \end{gathered}$$

The angle between two vector will be given by-

$$\begin{gathered} \cos \theta =\frac{\overrightarrow{A}.\overrightarrow{B}}{AB}=\frac{(4\hat{i}+4\hat{j}-4\hat{k}).(3\hat{i}+\hat{j}+4\hat{k})}{\sqrt{{\left(4\right)}^2+{\left(4\right)}^2+{(-4)}^2}\times \sqrt{{\left(3\right)}^2+{\left(1\right)}^2+{\left(4\right)}^2}} \\ =\frac{0}{4\sqrt{3}\times \sqrt{26}}=0 \\ \cos \theta =\cos 90° \\ \theta =90° \end{gathered}$$

Velocity

As Fv= P= constant

i.e.    $m\frac{dv}{dt}v=P$
                                          $\left[As F= \frac{mdv}{dt}\right]$
or     $\int vdv= \int \frac{P}{m}dt$

By integrating both sides we get $\frac{v^2}{2}=\frac{P}{m}t+C_1$

As initially the body is at rest i.e. v= 0 at t= 0, so $C_1= 0 \therefore v={\left(\frac{2Pt}{m}\right)}^{1/2}$

Position

From the above expression $v={\left(\frac{2Pt}{m}\right)}^{1/2}$

or $\frac{ds}{dt}={\left(\frac{2Pt}{m}\right)}^{1/2}$         $\left[As v= \frac{ds}{dt}\right]$

i.e.  $\int ds= \int {\left(\frac{2Pt}{m}\right)}^{1/2}dt$

By integrating both sides we get $s={\left(\frac{2P}{m}\right)}^{1/2}.\frac{2}{3}{t}^{3/2}+C_2$

Now as at t= 0, s= 0, so $C_2=0 \therefore s={\left(\frac{8P}{9m}\right)}^{1/2}{t}^{3/2}$