NEET Practice Questions (MCQs) with Answers & Solutions

(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

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

The tension in the supporting cable is the force required to provide the necessary acceleration to the combined mass of the lift and the person. Using Newton's second law, F = ma = (60 + 940) kg × (10 + 1) m/s^2 = 11000 N.

The tension in the supporting cable is the force required to accelerate the lift upwards. Using Newton's second law, F = ma, where F is the tension (28000 N), m is the mass of the lift (2000 kg), and a is the acceleration. Therefore, a = 28000 N / 2000 kg = 14 m/s^2 upwards.

According to Newton's second law of motion, force = mass $\times$ acceleration.

Here,  $\overrightarrow{F}=6\hat{i}-8\hat{j}+10\hat{k}$

         $\left|F\right|=\sqrt{36+64+100}$

             = $10\sqrt{2}N$

        $a=1 m{s}^{-2}$

$\therefore m=\frac{10\sqrt{2}}{1}=10\sqrt{2}kg$

Balancing the forces, we get

        $Mg-N=M\frac{v^2}{R}$

For weightlessness, N = 0

$\therefore \frac{M{v}^2}{R}=Mg$

where R is the radius of curvature and v is the speed of car.

Therefore,                $v=\sqrt{Rg}$

Putting the values, R=20m, g=10.0m/$s^2$

So,  $v=\sqrt{20\times 10.0}=14.14m/s^2$

Thus, the speed of the car at the top of the hill is between 14m/s and 15m/s.

Note: The roller coaster is a popular amusement ride developed for amusement parks and modern theme parks.

To keep the belt moving at the same speed, the force required will be equal to change in its momentum.

Force required, F=$\frac{d\left(mv\right)}{dt}$

                        =v$\left(\frac{dm}{dt}\right)=Mv$

As velocity v is constant, hence,

                       F = Mv newton