Physics MCQs for NEET — Practice Questions with Answers

When ball falls vertically downward from height $h_1$ its velocity ${\overrightarrow{v}}_1=\sqrt{2g{h}_1}$

and its velocity after collision ${\overrightarrow{v}}_2=\sqrt{2g{h}_2}$

Change in momentum

$\Delta \overrightarrow{P}=m({\overrightarrow{v}}_2-{\overrightarrow{v}}_1)=m(\sqrt{2g{h}_1}+\sqrt{2g{h}_2})$

(because ${\overrightarrow{v}}_1$ and ${\overrightarrow{v}}_2$ are opposite in direction)

$$\begin{gathered} When the spring is cut into pieces, they will have the new force cons\tan t .The spring is divided into 1:2:3 ratio. \\ Let the l_1=x, then l_2=2x and l_3=3x\Rightarrow x+2x+3x=l\Rightarrow x=\frac{l}{6} \\ Since spring cons\tan t is inversely proportional to length, new cons\tan ts are: \\ For springs, k_1{l}_1= k_2{l}_2= k_3{l}_3= k_n{l}_n \\ {k}_1 =\frac{kl}{{\displaystyle \raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$6$}\right.} }= 6k \\ {k}_2 =\frac{kl}{{\displaystyle \raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$3$}\right. }}= 3k \\ {k}_3 =\frac{kl}{{\displaystyle \raisebox{1ex}{$l$}\!\left/ \!\raisebox{-1ex}{$2$}\right. }}= 2k \end{gathered}$$

When the pieces are connected in series, the resultant force constant 

   $$\begin{gathered} \frac{1}{v^{\text{'}}}=\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3} \\ \frac{1}{v^{\text{'}}}=\frac{1}{6k}+\frac{1}{2k}+\frac{1}{3k} \\ {v}^{\text{'}}=k \end{gathered}$$

In parallel,the net force constant 

$K^{\text{'}\text{'}}=6k+3k+2k=11k$

The requried ratio$\frac{K}{K^{\text{'}\text{'}}}=\frac{k}{11k}=1:11$

For a vehicle on a banked road, the maximum safe velocity is given by √(gR((μs + tanθ)/(1 - μs tanθ))), where g is the acceleration due to gravity, R is the radius of curvature, μs is the coefficient of static friction, and θ is the angle of banking. This expression takes into account the centripetal force and the frictional force acting on the vehicle.

(d) Given, mass of particle m=0.01 kg.
Radius of circle along which particle is moving, r=6.4cm.

 ∴ Kinetic energy of particle, K.E=8x10^-4 J

=>  $\frac{1}{2}$mv²=8x10^-4 J

=> v²=$\frac{16\times {10}^{-4}}{0.01}$=16x10^-2 …(i)

As it is given that K.E of particle is equal to 8x10^-4 J by the end of second revolution after the beginning of motion of particle. It means, it’s initial velocity (u) is 0 m/s at this moment.

∴ By Newton’s 3^rd equation of motion,

v²= u²+2$a_{t}s$

v²= 2$as$ or v²= 2$a$ (4πr)
(∴ particle covers 2 revolutions)

a= v²/8πr = 16x10^-2/8x3.14x6.4x10^-2
(∴ from equation (i), v²=16x10^-2)

∴ $a_t$=0.1m/s²

Velocity v= $\sqrt{2\mathrm{gh}}$

and momentum p=mv

From Eqs. (i) and (ii), we have 

                 $p\propto \sqrt{h}$

Hence          $$\begin{gathered} \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\sqrt{\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}} \\ \mathrm{so}, \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\sqrt{\frac{2\mathrm{h}}{\mathrm{h}}}=\sqrt{2} \\ {\mathrm{P}}_2=1.414 {\mathrm{p}}_1 \end{gathered}$$

% change =$\frac{{\mathrm{P}}_2-{\mathrm{P}}_1}{{\mathrm{P}}_1}\times 100=41\%$

In this condition, centripetal force is equal to 

static frictional force road and tyres,

so           

                 $$\begin{gathered} {\mathrm{\mu }}_{\mathrm{s}}\mathrm{mg}=\frac{{\mathrm{mv}}^2}{\mathrm{R}} \\ {\mathrm{v}}_{\max }=\sqrt{{\mathrm{\mu }}_{\mathrm{s}}\mathrm{Rg}} \end{gathered}$$

Given, r= 5 cm=$5\times {10}^{-2}$m

and $\mathrm{T}=0.2 \mathrm{\pi s}$

We know that acceleration

              $$\begin{gathered} \mathrm{a}={\mathrm{r\omega }}^2 \\ =\frac{4{\mathrm{\pi }}^2}{{\mathrm{T}}^2}\mathrm{r} \\ =\frac{4\times {\mathrm{\pi }}^2\times 5\times {10}^{-2}}{(0.2\mathrm{\pi }{)}^2}=5 {\mathrm{ms}}^{-2} \end{gathered}$$