NEET Practice Questions (MCQs) with Answers & Solutions

$$\begin{gathered} v=\frac{dx}{dt}=3t^2-6t+12 \\ a=\frac{dv}{dt}=6t-6 \\ When a=0, 6t-6=0\Rightarrow t=1 sec \\ v at 1 sec.=9m/s \end{gathered}$$

Resultant force = $\overrightarrow{F_1}+\overrightarrow{F_2}=2\hat{i}+5\hat{j}+4\hat{k}$

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-

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

The area of a disk of radius 'a' can be calculated by integrating the area of a thin ring of radius 'r' from r=0 to r=a. The area of a thin ring is 2πrdr. Integrating this from 0 to a gives the total area as πa^2.

Given that, $a=\frac{dv}{dt}=\alpha t+\beta$

${}_0{}^v\int dv={}_0{}^t\int (\alpha t+\beta )dt={\left[\frac{\alpha t^2}{2}+\beta t\right]}_0^t \Rightarrow v=\frac{\alpha t^2}{2}+\beta t$

$v=k\sqrt{x}$  $\Rightarrow \frac{dx}{dt}=k\sqrt{x}$

$\Rightarrow \int \frac{dx}{\sqrt{x}}=\int kdt \Rightarrow \frac{x^{+1/2}}{1/2}=kt+c$

Given that, at t=0, x=0      ${.}^{.}.$ c=0

Now, $2x^{1/2}=kt \Rightarrow \sqrt{x}=(1/2)kt,$

$\because x=\frac{k^2{t}^2}{4}$

Now, $v=k(1/2 kt)=k^{2}t/2$

Thus velocity varies with time. Hence correct answer is (1)

$a=3x^2 \Rightarrow v \frac{dv}{dx}=3x^2$

$\Rightarrow vdv=3x^2 dx \Rightarrow \frac{v^2}{2}=3\frac{x^3}{3}+c$

At t=0, v=0, x=0;

$\therefore c=0 Now, \frac{v^2}{2}=x^3$

    $v^2=2x^3\Rightarrow v=\sqrt{2} x^{3/2} ...\left(1\right)$

$\Rightarrow \frac{dx}{dt}=\sqrt{2} x^{3/2} \Rightarrow dx=\sqrt{2} x^{3/2} dt \Rightarrow \frac{dx}{x^{3/2}}=\sqrt{2} dt$

Integrating both sides, we get $\frac{-2}{\sqrt{x}}=\sqrt{2}t+c\text{'}$

At t=0, x=0, v=0       ${.}^{.}.$ c'=0

Now $\frac{-2}{\sqrt{x}}=\sqrt{2}t$   $\Rightarrow 4=2x{t}^2 \Rightarrow x=\frac{2}{t^2} ...\left(2\right)$

From (1) and (2) $v=\sqrt{2}{\left(\frac{2}{t^2}\right)}^{3/2}$

At t=2 s, v=1/2 m/sec.

Hence correct answer is (2).

$$\begin{gathered} a=3t \Rightarrow \frac{dv}{dt}=3t \Rightarrow \int dv=\int 3tdt \\ \Rightarrow v=\frac{3t^2}{2}+c \end{gathered}$$

Substituting the initial conditions, at t=0, v=0 and x=0

$\therefore c=0 Hence, v=\frac{3t^2}{2}$

Velocity at t= 2 sec is $\frac{3\times 2^2}{2}=6 m/s$

Also, $$\begin{gathered} \frac{dx}{dt}=\frac{3t^2}{3} \Rightarrow \int dx=\frac{3}{2}\int t^{2}dt \\ \Rightarrow x=\frac{3}{2}\frac{t^3}{3}+c\text{'} \end{gathered}$$

at t=0, x=0 ${.}^{.}.$ c'=0, ${.}^{.}.$$x=\frac{t^3}{2},$

Now displacement at t= 2 sec is $\frac{2^3}{2}$=4 m

Hence correct answer is (1)