Physics MCQs for NEET — Practice Questions with Answers

$\text{Stopping distance}=\frac{\text{kinetic energy}}{\text{retarding force}}$ ⇒ $s=\frac{1}{2}\frac{m{u}^2}{F}$

If lorry and car both possess same kinetic energy and retarding force is also equal then both come to rest in the same distance. 

Potential energy of the particle $U=k(1-e^{-x^2})$

Force on particle $F=\frac{-dU}{dx}=-k[-e^{-x^2}\times (-2x\left)\right]$

F $=\text{\hspace{0.17em}}-2kx{e}^{-x^2}=-2kx\left[1-x^2+\frac{{\displaystyle x^4}}{{\displaystyle 2!}}-.....\right]$

For small displacement $F=-2kx$

⇒ $F\left(x\right)\propto -x$ i.e. motion is simple harmonic motion. 

Kinetic energy acquired by the body

= Force applied on it × Distance covered by the body

K.E. = F × d

If F and d both are same then K.E. acquired by the body will be same

$P=Fv=mav=m\left(\frac{{\displaystyle dv}}{{\displaystyle dt}}\right)v$ ⇒ $\frac{P}{m}dt=vdv$

⇒ $\frac{P}{m}\times t=\frac{v^2}{2}$ ⇒ $v={\left(\frac{{\displaystyle 2P}}{{\displaystyle m}}\right)}^{1\text{}/\text{}2}{\left(t\right)}^{1/2}$

Now $s={\int }_{}^{}vdt={\int }_{}^{}{\left(\frac{2P}{m}\right)}^{1/2}{t}^{1/2}dt$

∴ $s={\left(\frac{{\displaystyle 2P}}{{\displaystyle m}}\right)}^{1/2}\left[\frac{{\displaystyle 2t^{3/2}}}{{\displaystyle 3}}\right]$ ⇒ $s\propto t^{3/2}$ 

Displacement of the particle,

      $$\begin{gathered} ∆\overrightarrow{s}={\overrightarrow{r}}_2-{\overrightarrow{r}}_1 \\ =4\hat{j}+3\hat{k}-\left(-2\hat{i}+5\hat{j}\right) \\ =2\hat{i}-\hat{j}+3\hat{k} \end{gathered}$$

Force on the particle,

$$\begin{gathered} \overrightarrow{\mathrm{F}}=4\hat{\mathrm{i}}+3\hat{\mathrm{j}} \mathrm{N} \\ \therefore \mathrm{Work} \mathrm{done}, \mathrm{W}=\overrightarrow{\mathrm{F}}.∆\overrightarrow{\mathrm{s}} \\ =\left(4\hat{\mathrm{i}}+3\hat{\mathrm{j}}\right).\left(2\hat{\mathrm{i}}-\hat{\mathrm{j}}+3\hat{\mathrm{k}}\right) \\ =8-3=5\mathrm{J} \end{gathered}$$

(c) According to question, a body of mass 1kg begins to move under the action of time dependent force,

    $F=\left(2t \hat{i}+3t^2 \hat{j}\right)$

where  $\hat{i}$ and $\hat{j}$ are unit vectors along X and Y axis

$$\begin{gathered} \because F=ma \\ \Rightarrow a=\frac{F}{m} \\ \Rightarrow a=\frac{\left(2t\hat{i}+3t^2\hat{j}\right)}{1} \left\{\because m=1 kg\right\} \\ \Rightarrow a=\left(2t\hat{i}+3t^2\hat{j}\right)m/s \end{gathered}$$

$\because acceleration, a =\frac{dv}{dt}$

$\Rightarrow dv=adt$                ...(i)

integrating both sides, we get

              $$\begin{gathered} \int dv=\int adt \\ =\int \left(2t\hat{i}+3t^2\hat{j}\right)dt \\ v=t^2\hat{i}+t^3\hat{j} \end{gathered}$$

$\because$Power developed by the force at the time t will be given by as

                            $$\begin{gathered} P=F.v=\left(2t\hat{i}+3t^2\hat{j}\right).\left(t^2\hat{i}+t^3\hat{j}\right) \\ =\left(2t.t^2+3t^2.t^3\right) \\ P=\left(2t^3+3t^5\right)W \end{gathered}$$

$$\begin{gathered} Given that; K_P>K_Q. \\ Case \left(a\right): \\ The elongation is same i.e. x_1=x_2=x \\ So, W_P=\frac{1}{2}{K}_P{x}^2 \& W_Q=\frac{1}{2}{K}_Q{x}^2 \\ So, W_P>W_Q\Rightarrow \frac{W_P}{W_Q}>1. \\ Case \left(b\right): \\ The spring force is same i.e. F_1=F_2=F. \\ So, x_1=\frac{F}{K_P} \& x_2=\frac{F}{K_Q} \\ \Rightarrow W_P=\frac{1}{2}{K}_P{x_1}^2=\frac{F^2}{2K_P} \& W_Q=\frac{1}{2}{K}_Q{x_2}^2=\frac{F^2}{2K_Q} \\ \Rightarrow W_P<W_Q\Rightarrow \frac{W_P}{W_Q}<1. \end{gathered}$$

_{(1)
=>Kf=W+Ki=$$\begin{gathered} \mathrm{From} \mathrm{work}-\mathrm{energy} \mathrm{theorem} : \\ \mathrm{Work} \mathrm{done}=\mathrm{Change} \mathrm{in} \mathrm{K}.\mathrm{E}. \\ \Rightarrow \mathrm{W}={\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}} \\ \Rightarrow {\mathrm{K}}_{\mathrm{f}}=\mathrm{W}+{\mathrm{K}}_{\mathrm{i}}={\int }_{x_1}^{x_2}{\mathrm{F}}_{\mathrm{x}}\mathrm{dx}+\frac{1}{2}{\mathrm{mv}}^2 \\ \Rightarrow {\mathrm{K}}_{\mathrm{f}}=-{\int }_{20}^{30}0.1\mathrm{xdx}+\frac{1}{2}\times 10\times {10}^2=500-0.1{\left[\frac{{\mathrm{x}}^2}{2}\right]}_{20}^{30} \\ \Rightarrow {\mathrm{K}}_{\mathrm{f}}=475 \mathrm{J} \end{gathered}$$}

As the machine delivers a constant power, so F.v=constant=k(watts)
=> m$\frac{\mathrm{dv}}{\mathrm{dt}}$.v=k

=>$\int$vdv=$\frac{k}{m}$$\int$dt

=> v²/2=$\frac{kt}{m}$

=>v=$\sqrt{\frac{2kt}{m}}$

Now, the force on the particle is given by

F=m$\frac{\mathrm{dv}}{\mathrm{dt}}$

=$\sqrt{\left(\frac{mk}{2}\right)}{t}^{-1/2}$

Total initial energy=$\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂²

Since, after collision one particle absorb energy $\epsilon$.

∴ Total final energy=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₁²+$\epsilon$

From conservation of energy,

$\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u_2²=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²+$\epsilon$

=> $\frac{1}{2}$m₁u₁²+$\frac{1}{2}$m₂u₂²-$\epsilon$

=$\frac{1}{2}$m₁v₁²+$\frac{1}{2}$m₂v₂²