Physics Part I Chapter 7 System Of Particles And Rotational Motion

1 mole of any gas occupies 22.4 dm3 at stp (standard temperature and pressure, taken as 0°C and 1 atmosphere pressure). You may also have used a value of 24.0 dm3 at room temperature and pressure (taken as about 20°C and 1 atmosphere).

These figures are actually only true for an ideal gas, and we'll have a look at where they come from.

We can use the ideal gas equation to calculate the volume of 1 mole of an ideal gas at 0°C and 1 atmosphere pressure.

First, we have to get the units right.

0°C is 273 K. T = 273 K

1 atmosphere = 101325 Pa. p = 101325 Pa

We know that n = 1, because we are trying to calculate the volume of 1 mole of gas.

And, finally, R = 8.31441 J K-1 mol-1.

Slotting all of this into the ideal gas equation and then rearranging it gives:

And finally, because we are interested in the volume in cubic decimetres, you have to remember to multiply this by 1000 to convert from cubic metres into cubic decimetres.

The molar volume of an ideal gas is therefore 22.4 dm3 at stp.

And, of course, you could redo this calculation to find the volume of 1 mole of an ideal gas at room temperature and pressure - or any other temperature and pressure.

1. The Carnot cycle consists of the following four processes: A reversible isothermal gas expansion process. In this process, the ideal gas in the system absorbs qin amount heat from a heat source at a high temperature Th, expands and does work on surroundings.

2. As you know , absolute zero means the temperature in Kelvin scale is 0K or in Celcius scale is -273.15°C . actually, in this temperature, Gaseous molecule be rest , there is no motion of molecule in their position. due to this reason we say that energy of Gaseous molecule at this is zero. means absolute zero correspond to zero energy .

   well, it's hard to achieve Absolute zero or 0K or -273.15°C temperature. it is just theoretical. molecular can't achieve Absolute zero.

D.

C.

A.

D.

(2/3)ma²

A.

1/3

In this question distance of centre of mass of new disc is αR not α/R.

A.

C.

Zero Torque

C.

A.

B.

4.58 × 10⁻⁶ J/m³

D.

3m

I = 2m (

B.

C.

4

The point P must be the centre of mass of the T-shaped object since the force F does not produce any rotational motion of the object. So, we have to find the distance of the centre of mass from the point C.

The horizontal part of the T-shaped object has length L. If the mass of the horizontal portion is ‘m’, the mass of the vertical portion of the T- shaped object is 2m since its length is 2L. For finding the centre of mass of the T shaped object, it is enough to consider two point masses m and 2m located respectively at the midpoints of the horizontal and vertical portions of the T.

Therefore, the T-shaped object reduces to two point masses m and 2m at distances 2L and L respectively from the point C. The distance ‘r’ of the centre of mass of the system from the point C is given by

r = (m₁r₁ + m₂r₂)/(m₁ + m₂) = (m×2L + 2m×L)/(m + 2m) = 4L/3

[ Note that we have used the equation, r = (m₁r₁ + m₂r₂)/(m₁ + m₂) for the position vector r of the centre of mass in terms of the position vectors r₁ and r₂ of the point masses m1 and m₂. We could use the simple equation involving the distances from C since the points are collinear].

D.

$T\propto R^{\frac{n+1}{2}}$

$$\begin{gathered} m{\omega }^{2}R = Force \propto \frac{1}{R^n} \\ (Force = \frac{m{v}^2}{R}) \\ \Rightarrow {\omega }^2 \propto \frac{1}{R^{n+1}} \\ \Rightarrow \omega \propto \frac{1}{R^{{\displaystyle \frac{n+1}{2}}}} \\ Time Period T = \frac{2\pi }{\omega } \\ Time Period T\hspace{0.17em}\propto R^{\frac{n+1}{2}} \end{gathered}$$

C.

7.1 N/m

Given frequency f  = 10¹²/sec

Angular frequency ω = 2πf = 2π x 10¹² /sec
Force constant k = mω²

= $$\begin{gathered} \frac{108 x {10}^{-3}}{6.02 x {10}^{23}} x 4{\pi }^2 x {10}^{24} \\ k = 7.1 N/m \end{gathered}$$

A.

T

Centripetal force mv²/l is provided by tension so the net force will be equal to tension i.e., T.

B.

C.

C.

Angular momentum

Angular momentum remains conserved until the torque acting on sphere remains zero. ${\tau }_{ex} = 0$

$So, \frac{dL}{dt} = 0$

i.e. L = constant

So angular momentum remains constant.

A.

W_C > W_B> W_A

B.

W_A > W_B > W_C

Work done required to bring them rest

$$\begin{gathered} ∆W = ∆KE \\ ∆W =\frac{1}{2}I{\omega }^2 \\ ∆W \propto I for same \omega \\ {W}_A:\hspace{0.17em}{W}_B: W_C = \frac{2}{5}M{R}^2:\hspace{0.17em}\frac{1}{2}M{R}^2 :M{R}^2 \\ = \frac{2}{5}:\frac{1}{2} :1 \\ = 4:5:10 \\ \Rightarrow W_C >\hspace{0.17em}{W}_B> W_A \end{gathered}$$

A.

$\sqrt{\left(\frac{6\mathrm{g}}{\mathrm{I}}\right)} \sin \frac{\mathrm{\theta }}{2}$

When the rod rotates through an angle θ, the centre of gravity falls through a distance h. From  △BCG'

$$\begin{gathered} \cos \mathrm{\theta } = \frac{(\mathrm{I}/2)-\mathrm{h}}{{\mathrm{\mu }}_2} \\ \mathrm{h} =\frac{1}{2} (1-\cos \mathrm{\theta }) \\ \mathrm{Decrease} \mathrm{in} \mathrm{PE} \\ = \mathrm{mg} \frac{\mathrm{I}}{2} (1- \cos \mathrm{\theta }) ... \left(\mathrm{i}\right) \\ \mathrm{The} \mathrm{decrease} \mathrm{in} \mathrm{P}.\mathrm{E}. \mathrm{is} \mathrm{equal} \mathrm{to} \mathrm{the} \mathrm{kinetic} \mathrm{enrgy} \mathrm{of} \\ \mathrm{rotation} \left(\frac{1}{2}{\mathrm{I\omega }}^2\right) \\ (\mathrm{KE}{)}_{\mathrm{rotational} } = \frac{1}{2} (\frac{{\mathrm{ml}}^2}{3}){\mathrm{\omega }}^2 .. (\mathrm{ii}) \\ \mathrm{From} \mathrm{equ}. (\mathrm{i}) \mathrm{and} (\mathrm{ii}) \mathrm{we} \mathrm{get} \\ \frac{1}{2} (\frac{{\mathrm{ml}}^2}{3}){\mathrm{\omega }}^2 = \mathrm{mg} \frac{\mathrm{I}}{2} (1- \cos \mathrm{\theta }) \\ \mathrm{\omega } = \sqrt{\frac{6\mathrm{g}}{\mathrm{l}}}\sin \frac{\mathrm{\theta }}{2} \end{gathered}$$

B.

I₂ > I₁> I₃

The moment of inertia of a body about an axis depends not only on the mass of the body but also on the distribution of mass about the axis. For a given body mass is the same, so it will depend only on the distribution of mass about the axis. The mass is farthest from axis BC, so I₂ is maximum mass is nearest to axis AC, so I₃ is minimum.

Hence, the correct sequence will be I₂>I₁>I₃

C.

10%

Let the volume of ice-berg is V and its density is ρ. If this ice-berg floats in water with volume Vin inside it, then Vin σg = Vρg

$$\begin{gathered} {\mathrm{V}}_{\mathrm{in}} = \left(\frac{\mathrm{\rho }}{\mathrm{\sigma }}\right)\mathrm{V} \\ \Rightarrow {\mathrm{V}}_{\mathrm{out}} = \mathrm{V} - {\mathrm{V}}_{\mathrm{in}} = \left(\frac{\mathrm{\sigma } - \mathrm{\rho }}{\mathrm{\sigma }}\right) \mathrm{V} \\ = \left(\frac{1000-900}{1000}\right) \mathrm{V} = \frac{\mathrm{V}}{10} \\ \Rightarrow \frac{{\mathrm{V}}_{\mathrm{out}}}{\mathrm{V}} = 0.1 = 10\% \end{gathered}$$

C.

$2\mathrm{\pi }\sqrt{\frac{3\mathrm{M}}{8\mathrm{K}}}$

Let x be the extension in the spring

Then, x = 2Rθ

Therefore, Torque acting on the cylinder about point contact.

T = Fx2R = - Kx(2R) = - 4R²Kθ

Thereore, $$\begin{gathered} \mathrm{T} = 2\mathrm{\pi } \sqrt{\frac{\mathrm{I}}{\mathrm{Ke}}} = 2\mathrm{\pi }\sqrt{\frac{{\displaystyle \frac{3}{2}}{\mathrm{MR}}^2}{4{\mathrm{R}}^2\mathrm{K}}} \\ = 2\mathrm{\pi }\sqrt{\frac{3\mathrm{M}}{8\mathrm{K}}} \end{gathered}$$

If either of acceleration or time , is zero
then the relation
s=ut ( where s=displacement, u=initial velocity,t=time,a=acceleration) holds true
as the original formula is s=ut+1/2at^2
and as either a and t are 0 then it becomes s=ut

Newton second law say that , force is directly proportional to rate of change of momentum(say p).
F=(p2-p1)÷t
On solving this we get fundamental definition of force.i.e
F=ma

B.

0.1 kg m²

The moment of inertia of given system that contains 5 particles each of mass = 2 kg on the rim of circular disc of radius 0.1 m and of negligible mass is given by :

MI of disc + MI of particle

Since the mass of the disc is negligible

∴ MI of the system = MI of particle

= 5 × 2 × (0.1)²

= 0.1 Kg m²

C.

35 m

The condition of slipping by the particle from the top is given by

From the figure,

Suppose that the particle will leave the surface at height h from the centre of the sphere then it will create an angle θ with the center.

Hence, the height from the bottom of the sphere

$\mathrm{h} = \mathrm{r} \cos \mathrm{\theta } = 21 \times \frac{2}{3}$ = 14 m

Therefore, the required height at which the particle will leave the sphere is :

21 + h = 21+ 14 = 35 m

A.

3/4 A_o

Torque is the time rate of total angular momentum of a system of particles.

$$\begin{gathered} \mathrm{Torque} \mathrm{T} =\frac{\mathrm{dL}}{\mathrm{dt}} =\frac{\left(4{\mathrm{A}}_{\mathrm{o}} -{\mathrm{A}}_{\mathrm{o}}\right)}{4} \\ =\frac{3{\mathrm{A}}_{\mathrm{o}}}{4} \end{gathered}$$

B.

2/7

The rotational energy of sphere

$$\begin{gathered} {\mathrm{E}}_{\mathrm{R}} = \frac{1}{2}\mathrm{I} {\mathrm{\omega }}^2 \\ \mathrm{For} \mathrm{sphere} \mathrm{moment} \mathrm{of} \mathrm{inertia} \\ \mathrm{I} = \frac{2}{5}{\mathrm{mR}}^2 \\ {\mathrm{E}}_{\mathrm{R}} =\frac{1}{2}\left(\frac{2}{5}\mathrm{m} {\mathrm{R}}^2\right){\left(\frac{\mathrm{v}}{\mathrm{R}}\right)}^2 \\ = \frac{1}{5} {\mathrm{mv}}^2 \\ \mathrm{TRanslational} \mathrm{kinetic} \mathrm{energy} \\ {\mathrm{E}}_{\mathrm{r}} = \frac{1}{2 }{\mathrm{mv}}^2 \\ \mathrm{Total} \mathrm{energy} = \frac{1}{5}{\mathrm{mv}}^2 + \frac{1}{2}{\mathrm{mv}}^2 \\ = \frac{7}{10}{\mathrm{mv}}^2 \end{gathered}$$

$$\begin{gathered} \mathrm{Required} \mathrm{fraction} =\frac{{\displaystyle \frac{1}{5}{\mathrm{mv}}^2}}{{\displaystyle \frac{7}{10}{\mathrm{mv}}^2}} \\ = \frac{2}{7} \end{gathered}$$

B.

3 cm

Given equation 

$y = 5 \sin \frac{\mathrm{\pi x}}{3}\cos 40 \mathrm{\pi t}$

Compairing with standard equation

y = A sin kx cosωt

$$\begin{gathered} \therefore \mathrm{k} = \mathrm{\pi }/3 \\ \Rightarrow \frac{2\mathrm{\pi }}{3}= \frac{\mathrm{\pi }}{3} \\ \mathrm{\lambda } = 6 \mathrm{cm} \\ \mathrm{Separation} \mathrm{between} \mathrm{two} \mathrm{cosecutive} \mathrm{nodes} \\ \frac{\mathrm{\lambda }}{2} = \frac{6}{2} = 3\mathrm{cm} \end{gathered}$$

D.

zero

$$\begin{gathered} \left(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}\right).\overrightarrow{ \mathrm{A}} \\ \Rightarrow \left(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}\right).\overrightarrow{ \mathrm{A}} = \mathrm{B}\hspace{0.17em}\mathrm{A} \mathrm{cos\theta } \hat{\mathrm{n}} .\overrightarrow{\mathrm{A}} \\ = 0 \end{gathered}$$

Here $\hat{\mathrm{n}}$is perpendicular to both $\overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{B}}$.

Alternative method

$\left(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}\right) .\overrightarrow{\mathrm{A}}$

Interchange the cross and dot we have,

$$\begin{gathered} \left(\overrightarrow{\mathrm{B}} \times \overrightarrow{\mathrm{A}}\right). \overrightarrow{\mathrm{A}} = \overrightarrow{\mathrm{B}} .\left( \overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{A}}\right) \\ = 0 ..... \left(\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{A}}=0 \right) \end{gathered}$$

The volume of a parallelopiped bounded by vectors $\overrightarrow{\mathrm{A}} , \overrightarrow{\mathrm{A}} \mathrm{and} \overrightarrow{\mathrm{A}}$can be obtained by giving formula $\left(\overrightarrow{\mathrm{A}} \times \overrightarrow{ \mathrm{A}}\right). \overrightarrow{\mathrm{A}}$

D.

$\frac{3}{2} {\mathrm{MR}}^2$

We should use parallel axis theorem.
Moment of inertia of disc passing through its centre of gravity and perpendicular to its plane

${\mathrm{I}}_{\mathrm{AB} }= \frac{1}{2}{\mathrm{MR}}^2$

Using theorem of parallel axes, we have

$$\begin{gathered} {\mathrm{I}}_{\mathrm{CD}} = {\mathrm{I}}_{\mathrm{AB}} + {\mathrm{MR}}^2 \\ = \frac{1}{2}\mathrm{M} {\mathrm{R}}^2 + \mathrm{M} {\mathrm{R}}^2 \\ {\mathrm{I}}_{\mathrm{CD}} = \frac{3}{2} {\mathrm{MR}}^2 \end{gathered}$$

The role of moment of inertia in the study of rotational motion is analogous to that of mass in study of linear motion.

C.

$\mathrm{F} \left(\hat{\mathrm{i}} + \hat{\mathrm{j}} \right)$

$\overrightarrow{\mathrm{\tau }} = \overrightarrow{\mathrm{r}} \times \overrightarrow{\mathrm{F}}$

$$\begin{gathered} \overrightarrow{\mathrm{\tau }} =\left( \hat{\mathrm{i}} - \hat{\mathrm{j}}\right) \times \left(-\mathrm{F} \hat{\mathrm{k}} \right) \\ =\mathrm{F} \left[\left(-\hat{\mathrm{i}} \times \hat{\mathrm{k}} \right) + \left(\hat{\mathrm{j}} \times \hat{\mathrm{k}}\right)\right] \\ =\mathrm{F} \left[\hat{\mathrm{j}} + \hat{\mathrm{i}}\right] \\ \overrightarrow{\mathrm{\tau }} = \mathrm{F} \left[\hat{\mathrm{i}} + \hat{\mathrm{j}}\right] \end{gathered}$$

C.

17

Earth rotates about its own axis in 24 h.

So,            T =24 x 60 x 60 s

So, angular speed of earth about its own axis

              $$\begin{gathered} \mathrm{\omega } = \frac{2\mathrm{\pi }}{\mathrm{T}} \\ \mathrm{\omega } = \frac{2\mathrm{\pi }}{24 \times 60 \times 60}\mathrm{rad}/\mathrm{s} \end{gathered}$$   

$\mathrm{At} \mathrm{equator} \mathrm{g}\text{'} = \mathrm{g} - {\mathrm{R}}_{\mathrm{e}} \mathrm{\omega }{\text{'}}^2$

               $$\begin{gathered} 0 = \mathrm{g} - {\mathrm{R}}_{\mathrm{e}}\mathrm{\omega }{\text{'}}^2 \\ \Rightarrow \mathrm{\omega }\text{'} = \sqrt{\frac{\mathrm{g}}{{\mathrm{R}}_{\mathrm{e}}}} \\ = \frac{2\mathrm{\pi }}{84.6 \times 60} \\ \therefore \frac{\mathrm{\omega }\text{'}}{\mathrm{\omega }} = \frac{24 \times 60 \times 60}{84.6 \times 60} \end{gathered}$$

                          ≈ 17

                     ω' = 17ω

So, value of x is 17.

C.

$\frac{3}{2} {\mathrm{MR}}^2$

The moment of inertia of circular disc about an axis through its centre and normal to its plane

             I_CG $= \frac{1}{2} {\mathrm{MR}}^2$

Where M is the mass of the disc and R its radius.

According to the theorem of parallel axis, the moment of inertia of the uniform circular disc about an axis passing from the edge of the disc and normal to the disc,

         I = I_CG + MR² 

          = $\frac{1}{2} {\mathrm{MR}}^2 + {\mathrm{MR}}^2$

         I = $\frac{3}{2} {\mathrm{MR}}^2$

A.

43 kg-m²

Moment of inertia of the whole system about the axis of rotation will be equal to the sum of the moments of inertia of all the particles.

           I = I₁ + I₂ + I₃ + I₄

           I = ${\mathrm{m}}_1 {\mathrm{r}}_1^2 + {\mathrm{m}}_2 {\mathrm{r}}_2^2 + {\mathrm{m}}_3 {\mathrm{r}}_3^2 + {\mathrm{m}}_4 {\mathrm{r}}_4^2$

           I = ( 1 × 0 ) + ( 2 × 0 ) + ( 3 × 3² ) + 4 (-2 )²

           I = 0 + 0 + 27 + 16

           I = 43 kg-m³  

B.

8 cm

The moment of inertia of a body about any axis is equal to the sum of the moment of inertia of the body about a parallel axis passing through its centre of mass and the product of its mass and the square of the distance between the two parallel axes.

From the theorem of parallel axis, the moment of inertia I is equal to

   $\mathrm{I} = {\mathrm{I}}_{\mathrm{CM}} + \mathrm{M} {\mathrm{a}}^2$

where ${\mathrm{I}}_{\mathrm{CM}}$ is moment of inertia about the centre of mass and  'a'  the distance of axis from the centre.

    I = MK² + M × (6)²

    $\mathrm{M} {\mathrm{K}}_1^2 = \mathrm{M} {\mathrm{K}}^2 + 36 \mathrm{M}$

⇒     ${\mathrm{K}}_1^2 = {\mathrm{K}}^2 + 36$

⇒       (10)² = K² + 36

⇒        K²  = 100 - 36

⇒         K² = 64

⇒          K = 8 cm

A.

2 kg-m²

Torque is defined as the moment of force or turning effect of force or turning effect of force about the given axis or point. It is measured as the cross product of position and force vector while as angular momentum is the rotational analogue of linear momentum.

Given:-

 r = 0.4 m

 Angular acceleration $\mathrm{\alpha } = 8 \mathrm{rad} {\mathrm{s}}^{-2}$

 m = 4 kg

 l = ?

 Torque $\mathrm{\tau } = \mathrm{I} \mathrm{\alpha }$

But we know Τ =  Fr 

and  The equation for the force of gravity is ( F = mg ) where g is the acceleration due to gravity.

          mgr = $\mathrm{I} \mathrm{\alpha }$

         4 × 10 × 0.4 =  I × 8

⇒         $\mathrm{I} = \frac{16}{8}$

           I  = 2 kg.m²

A.

both will reach with same speed

Both start with the same potential energy component ( gh )

When they both reach the ground, both have converted that identical energy component kinetic energy.

Thus, the object sliding down a frictionless incline will reach the same velocity as the one dropped straight down.

The times to reach the speed to reach the speed will not be the same, the dropped one will reach it faster.

C.

5

From the formula

          ${\mathrm{E}}_{\mathrm{rotational} }= \frac{1}{2}\mathrm{I} {\mathrm{\omega }}^2$ 

                      = $\frac{{\mathrm{L}}^2}{2\mathrm{I}}$                   $\left(\because \mathrm{L} = \mathrm{I\omega }\right)$

Therefore,  L² = 2 EI

Hence       ${\mathrm{L}}_{\mathrm{A}}^2 = 2{\mathrm{E}}_{\mathrm{A}} {\mathrm{I}}_{\mathrm{A}}$                  .....(i)

and           ${\mathrm{L}}_{\mathrm{B}}^2 = 2{\mathrm{E}}_{\mathrm{B}} {\mathrm{I}}_{\mathrm{B}}$                 ......(ii)

From equatin (i) and (ii)

               $\frac{{\mathrm{L}}_{\mathrm{A}}}{{\mathrm{L}}_{\mathrm{B}}} = \sqrt{\frac{2{\mathrm{E}}_{\mathrm{A}} {\mathrm{I}}_{\mathrm{A}}}{2{\mathrm{E}}_{\mathrm{B}} {\mathrm{I}}_{\mathrm{B}}}}$

                      = $\sqrt{\frac{100}{4}}$

               $\frac{{\mathrm{L}}_{\mathrm{A}}}{{\mathrm{L}}_{\mathrm{B}} } = 5$

        $\left(\because \frac{{\mathrm{E}}_{\mathrm{B}}}{{\mathrm{E}}_{\mathrm{A}}} = 100 \mathrm{and} \frac{{\mathrm{I}}_{\mathrm{A}}}{{\mathrm{I}}_{\mathrm{B}}} = 4\right)$

B.

surface tension

The principle of working of ball pen corresponds to surface tension. As the pen moves along the paper, the ball rotates picking up ink from the ink cartridge and leaving it on the paper.

C.

90^o

As it is clear that the phase difference between cosθ and sinθ representing two simple harmonic motion is 90^o.

B.

α = 0

τ   the torque on a given axis is product of moment of inertia and angular acceleration. 

              $\mathrm{\tau } = \mathrm{I} \mathrm{\alpha }$

    I = moment of inertia

    α = angular acceleration

     If     $\mathrm{\tau } = 0 \Rightarrow \mathrm{I\alpha } = 0$

     But       I ≠ 0

                 α = 0

When the torque acting upon a system is zero, the angular momentum is constant and hence conserved. This is analogous to the linear counterpart i.e when the force on the system is zero, the momentum is conserved.

B.

A - 4, B - 3

Dimensions of Torque = [M² L² T^-2]

Dimension of angular momentum = [ M¹ L² T^-1]

∴ Relation A-4 ,B -3 is right. 

A.

$\frac{10 \mathrm{F}}{7 \mathrm{m}}$

Let a be the acceleration of centre of sphere. The angular acceleration about the centre of the sphere is $\mathrm{\alpha } = \frac{\mathrm{\theta }}{\mathrm{r}}$, as there is no slipping.

For the linear motion of the centre

   f + F = ma                  .....(i)

and for the rotational motion about the centre

  Fr - fr = I α

            = $\left(\frac{2}{5} {\mathrm{mr}}^2\right) \left(\frac{\mathrm{a}}{\mathrm{r}}\right)$

 ⇒  F - f = $\frac{2}{5} \mathrm{ma}$           ....(ii)

From Eqs. (i) and (ii)

   2F = $\frac{7}{5} \mathrm{ma}$

 ⇒  a  = $\frac{10 \mathrm{F}}{7 \mathrm{m}}$

A.

$\frac{3\mathrm{v}}{5}$

If we take moment at A, then external torque will be zero.

Therefore,

the initial angular momentum= the angular momentum after rotation stop

(i.e. only linear velocity exist )

(i) Initial angular momentum about point A

      L = m ( v × R ) - Iω

⇒      Mv × R - $\left(\frac{2}{5}\right)$Iω = M v₀ × R

⇒    MvR - $\frac{2}{5}$ MR² × $\frac{\mathrm{v}}{\mathrm{R}}$ = Mv₀ R

⇒                    v₀ = $\frac{3 \mathrm{v}}{5}$

A.

${\left[\frac{{\mathrm{m}}_1 {\mathrm{m}}_2}{{\mathrm{m}}_1 + {\mathrm{m}}_2}\right]}^{\frac{1}{2}}$ v₀

The centre of mass is the location of particles within a system where the total mass of the system can be considered concentrated. When the system of particles is moving, the center of mass moves along with it. 

The centre of mass of velocity equation is the sum of each particle's momentum ( mass times velocity ) divided by the total mass of the system.

The velocity of the centre of mass of two particles

     v_cm = $\frac{{\mathrm{m}}_1 {\mathrm{v}}_1 + {\mathrm{m}}_2 {\mathrm{v}}_2}{{\mathrm{m}}_1 + {\mathrm{m}}_2}$

When v₁ =0  and  v₂ =v₀, then

    ${\mathrm{v}}_{\mathrm{cm}} = \frac{{\mathrm{m}}_2 {\mathrm{v}}_0}{{\mathrm{m}}_1 + {\mathrm{m}}_2}$

Now, let 'x' be the elongation in the spring.

Change in potential energy = potential energy stored in spring

⇒   $\frac{1}{2} {\mathrm{m}}_2 {\mathrm{v}}_0^2 - \frac{1}{2} \left({\mathrm{m}}_1 + {\mathrm{m}}_2\right)$${\left(\frac{{\mathrm{m}}_2 {\mathrm{v}}_0}{{\mathrm{m}}_1 + {\mathrm{m}}_2}\right)}^2 = \frac{1}{2} {\mathrm{kx}}^2$

⇒    m₂ ${\mathrm{v}}_0^2$ $\left\{1 - \frac{{\mathrm{m}}_2}{{\mathrm{m}}_1 + {\mathrm{m}}_2}\right\}$ = kx² 

⇒  ${\mathrm{m}}_2 {\mathrm{v}}_0^2 \left[\frac{{\mathrm{m}}_1 + {\mathrm{m}}_2 - {\mathrm{m}}_2}{{\mathrm{m}}_1 + {\mathrm{m}}_2}\right]$  = kx²

This gives

          x = ${\left(\frac{{\mathrm{m}}_1 {\mathrm{m}}_2}{{\mathrm{m}}_1 + {\mathrm{m}}_2}\right)}^{\frac{1}{2}}$ v₀

D.

$\frac{\left(2\mathrm{n} + 1\right) \mathrm{L}}{3}$

Centre of mass formula

  centre of mass =  $\frac{\mathrm{sum} \mathrm{of} \mathrm{all} \left( \mathrm{position} \times \mathrm{mass} \right)}{\mathrm{sum} \mathrm{of} \mathrm{all} \mathrm{masses}}$

$\because$       X_CM  = $\frac{\mathrm{mL} + 2\mathrm{m} \times 2\mathrm{L} + .... + \mathrm{nm} \times \mathrm{nL}}{\mathrm{m} + 2\mathrm{m} + .... + \mathrm{nm}}$

                  = $\frac{\mathrm{L} \left(1^2 + 2^2 + ..... + {\mathrm{n}}^2\right)}{\left(1 + 2 + ..... + \mathrm{n}\right)}$

                  = L $\frac{\mathrm{\Sigma } {\mathrm{n}}^2}{\mathrm{\Sigma } \mathrm{n}}$

                  = L  $\frac{{\displaystyle \frac{\mathrm{n} \left(\mathrm{n} + 1\right) \left(2\mathrm{n} + 1\right)}{6}}}{{\displaystyle \frac{\mathrm{n} \left(\mathrm{n} + 1\right)}{2}}}$

           X_CM  = $\frac{\left(2 \mathrm{n} + 1\right) \mathrm{L}}{3}$

A.

$\frac{{\mathrm{K}}^2}{{\mathrm{K}}^2 + {\mathrm{R}}^2}$

 Kinetic energy of rotation is

  K_rot = $\frac{1}{2}{\mathrm{I\omega }}^2$                  

          = $\frac{1}{2} {\mathrm{MK}}^2 \frac{{\mathrm{v}}^2}{{\mathrm{R}}^2}$

 where, k is radius of gyration.

Kinetic energy of translation is K_tran = $\frac{1}{2} {\mathrm{Mv}}^2$

Thus, total energy, 

          E = K_rot + K_trans

              = $\frac{1}{2} {\mathrm{MK}}^2 \frac{{\mathrm{v}}^2}{{\mathrm{R}}^2} +\frac{1}{2} {\mathrm{Mv}}^2$

              = $\frac{1}{2} \mathrm{M} {\mathrm{v}}^2 \left(\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2} + 1\right)$

              = $\frac{1}{2} \frac{{\mathrm{Mv}}^2}{{\mathrm{R}}^2} \left( {\mathrm{K}}^2+ {\mathrm{R}}^2 \right)$

       E= $\frac{1}{2} \frac{{\mathrm{Mv}}^2}{{\mathrm{R}}^2} \left( {\mathrm{K}}^2 +{\mathrm{R}}^2 \right)$

Hence 

     $\frac{{\mathrm{K}}_{\mathrm{rot}}}{\mathrm{Total} \mathrm{energy}, \mathrm{E}} = \frac{{\displaystyle \frac{1}{2} {\mathrm{MK}}^2 \frac{{\mathrm{v}}^2}{{\mathrm{R}}^2}}}{{\displaystyle \frac{1}{2}\frac{\mathrm{M} {\mathrm{v}}^2}{{\mathrm{R}}^2} \left({\mathrm{K}}^2 + {\mathrm{R}}^2\right)}}$

      $\frac{{\mathrm{K}}_{\mathrm{rot}}}{\mathrm{Total} \mathrm{energy} \mathrm{E}}=\frac{{\mathrm{K}}^2}{{\mathrm{K}}^2 + {\mathrm{R}}^2}$        

C.

$\sqrt{\frac{10}{7} \mathrm{gh}}$

When solid sphere rolls down on an inclined plane, then it has both rotational and translational kinetic energy

       K = K_rot + K_trans

       K = $\frac{1}{2} {\mathrm{I\omega }}^2 + \frac{1}{2} \mathrm{M} {\mathrm{v}}^2$

where,

I= moment of inertia of solid sphere = $\frac{2}{5}$ MR²

∴     K = $\frac{1}{2} \left(\frac{2}{5} \mathrm{M} {\left( 2\mathrm{R} \right)}^2\right) {\mathrm{\omega }}^2 + \frac{1}{2} {\mathrm{Mv}}^2$                      [$\because$R =2R ]

           = $\frac{4}{5}{\mathrm{MR}}^2 \left(\frac{\mathrm{v}}{2\mathrm{R}}\right) + \frac{1}{2} {\mathrm{Mv}}^2$                          [ $\because$v = Rω and R = 2R ]

        K  = $\frac{1}{5} {\mathrm{Mv}}^2 + \frac{1}{2} {\mathrm{Mv}}^2$

        K = $\frac{7}{10}$ Mv²

Now, gain in KE= loss in PE

          $\frac{7}{10}$ Mv² = Mgh

 ⇒               v = $\sqrt{\frac{10}{7} \mathrm{gh}}$

C.

125 rad/s

Given:-   

            $\mathrm{\alpha } = {\mathrm{\alpha }}_0 \left(1 - \frac{\mathrm{t}}{5}\right)$

At t = 0,  $\mathrm{\alpha } = {\mathrm{\alpha }}_0$

∴          ${\mathrm{\alpha }}_0$= 50 rad/s²

       $\frac{\mathrm{d\omega }}{\mathrm{dt}} = \mathrm{\alpha } \left(1 - \frac{\mathrm{t}}{5}\right)$

∴      ${\int }_0^{\mathrm{\omega }}\mathrm{d\omega } = {\mathrm{\alpha }}_0 {\int }_0^5\left(1 - \frac{\mathrm{t}}{5}\right)$

⇒            ω =  ${\mathrm{\alpha }}_0 {\left[\mathrm{t} - \frac{{\mathrm{t}}^2}{10}\right]}_0^5$ 

⇒                 = 50 $\left(5 - \frac{25}{10}\right)$ rad/s

⇒              ω = 125 rad/s  

D.

If both assertion and reason are false.

During rolling motion, there is radial acceleration towards the centre. Hence, the contact point moves vertically upward.

On smooth surface, sphere will slide without rolling.

C.

3.5 kg m²

M.I. about tangent =  I + MR²   

                             = $\frac{2}{5}$ MR² + MR²

                            = $\frac{7}{5}$ MR²                  

                             = $\frac{7}{5}$ × (10) × (0.5 )²

M.I. about tangent = 3.5 kg m²