Physics MCQs for NEET — Practice Questions with Answers

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

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

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)

Elastic potential energy of a spring $U=\frac{1}{2}k{x}^2 \therefore U\propto x^2$

So $\frac{U_2}{U_1}={\left(\frac{x_2}{x_1}\right)}^2 \Rightarrow \frac{U_2}{U}={\left(\frac{10 cm}{2 cm}\right)}^2 \Rightarrow U_2 = 25 U$

Work done to stretch the spring from $x_1 to x_2$

$W= \frac{1}{2}k\left(x_2^2-x_1^2\right) =\frac{1}{2}5\times {10}^3[{\left(10\times {10}^{-2}\right)}^2 -{\left(5\times {10}^{-2}\right)}^2] = \frac{1}{2}\times 5\times {10}^3\times 75\times {10}^{-4}= 18.75 N.m.$

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

$F=\frac{mdv}{dt} \therefore F dt= mdv \Rightarrow v= \frac{F}{m}t$

Now $P= F\times v= F\times \frac{F}{m}t =\frac{F^{2}t}{m}$

If force and mass are constants then $P \propto t$

We know, 

$dV=-\overrightarrow{I}.\overrightarrow{dr} \Rightarrow dV=-\frac{k}{x^2}dx \Rightarrow \int dV=-k \int \frac{dx}{x^2} \Rightarrow V=\frac{k}{x}+C$

When $x = \infty , V=0 \Rightarrow C=0$

$\Rightarrow V=\frac{k}{x} \therefore At x = a, V= \frac{k}{a}$

The torque acting on the gate is given by the product of the force acting on it and the perpendicular distance between the line of action of the force and the hinge. The force acting on the gate is the weight of the gate, which is equal to its mass times the acceleration due to gravity.