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The CORRECT choice is (b) -3/2, -3/2, -3/2

Easy explanation: Sum of MASSES = 10 + 20 + 30 + 40 = 100

Since initial centre of mass is at (1,1,1), we can assume an OBJECT of 60 kg is at (1,1,1).

x-coordinate;

(60*1 + 40*x)/100 = 0

x = -3/2

y-coordinate;

(60*1 + 40*y)/100 = 0

y = -3/2

z-coordinate;

(60*1 + 40*z)/100 = 0

z = -3/2.

The CORRECT answer is (b) 2R

To ELABORATE: Distance between first and last SPHERE = R + 2R + R = 4R

Since the SPHERES are identical and lie in a straight LINE, the centre of mass will lie exactly in the middle.

Hence the centre of mass lies at a distance of 2R from the centre of the first sphere.

Correct option is (c) Solid SPHERE

Explanation: The equation of motions will be: Mgsinθ – f = Ma, where f is the friction & a is acceleration. Also, fR = Ia/R. From this equation ‘f’ will be substituted in the FIRST equation: Mgsinθ – Ia/R^2 = Ma.

∴ a = Mgsinθ / (I/R^2+ M). Thus, more the value of I lesser will be the acceleration.

Moment of inertia of RING = MR^2

Moment of inertia of DISC= MR^2/2

Moment of inertia of solid sphere = 2MR^2/5

The solid sphere has the least moment of inertia, so it will have the maximum acceleration & HENCE it will take the least time. Note that if there was no friction their accelerations would be the same and they would reach together.

The correct option is (d) √ (gh) m/s

Explanation: Friction will be acting on the ring, which is the reason it is able to roll purely. But this friction will not do any work on the ring. The potential ENERGY at the top will be CONVERTED into kinetic energy at the bottom. Kinetic energy is GIVEN by:1/2MV^2 + 1/2Iw^2. I for a ring = MR^2 & w = V/R.

∴ kinetic energy = 1/2MV^2 + 1/2MV^2 = MV^2.

Potential energy at the top = MGH.

Therefore, Mgh = MV^2 OR V = √(gh) m/s.

Correct CHOICE is (a) 19/3 RAD

For explanation: Let W be the angular velocity & a be the angular acceleration.Given: dw/dt = a = 2t.On INTEGRATING both sides, we get:

w = t^2(no INTEGRATION constant because w=0 at t=0).

Now, dθ/dt = w = t^2

∴ θ = t^3/3.

To find angular distance covered from t=2 to t=3, we can subtract distance covered in 2s from distance covered in 3s.

∴ θ covered = 3^3/3 – 2^3/3= 19/3 rad.

Correct OPTION is (B) 16kgm^2

Easy explanation: The moment of inertia about any diametrical AXIS will be the same. So, we can consider two perpendicular diameters and use the perpendicular axis THEOREM to get the moment of inertia about the axis which is perpendicular to the plane.

Thus, moment of inertia = 8+8

= 16kgm^2.

Correct answer is (a) 0.00083kgm^2

Best EXPLANATION: Let the moment of inertia be I.

And M & I be the mass and LENGTH of rod RESPECTIVELY.

For the given CONDITION,

I = MI^2/12

= 1*0.01/12 = 0.00083kgm^2.

The correct answer is (a) dP/dt = 0

The BEST explanation: The LAW of conservation of linear momentum is derived from Newton’s SECOND law. It states that total linear momentum is constant when external forces ADD up to zero. Newton’s second law: dP/dt = Fext. Here, Fext must be zero for conservation of linear momentum, thus dP/dt = 0.

Right choice is (b) False

The best I can explain: For pure rolling, V = Rw at every instant of time, ( where V is the speed of the centre of the sphere & w is its angular speed).On an INCLINED surface a component of GRAVITY is CONVERTING its potential energy to kinetic energy & thus ‘V’ is increasing, so ‘w’ should also INCREASE. This requires presence of torque, which in this case is to be provided by friction. Hence, pure rolling on an incline requires the surface to be rough.

Correct answer is (C) 500/7π N

Best explanation: LET the force be ‘F’. Work done will be = Tθ,

where T is the TORQUE &is the angular displacement = 2π rad.

Work = Tθ = 10

∴ T(2π) = 10

∴ T = 5/π Nm.

Also, T = r*F

∴ 5/π = 0.07*F

∴ F = 500/7π N.

Correct option is (a) 16.67kgm^2

Easiest explanation: Moment of INERTIA of SOLID sphere ‘I1’ = 2/5 MR^2 = 10kgm^2.

Moment of inertia of THIN spherical shell ‘I2’= 2/3 MR^2.

∴I2 = 5/3I1

= 5/3 * 10 = 16.67kgm^2.

The CORRECT OPTION is (C) The centre of mass is stationary

To explain I would SAY: The velocity of the centre of mass = (M1*v1 + m2*v2)/(m1 + m2)

m1 = 5kg

m2 =3 kg

v1 = 3 m/s

v2 = -5 m/s [opposite direction)

Velocity of centre of mass = (5*3 – 3*5)/(5+3)

 = 0.

The correct option is (a) (Mgsinθ)/3

To elaborate: Let the angular ACCELERATION of the disc be ‘α’. And the linear acceleration along the incline be ‘a’. For pure rolling, a = Rα.Mgsinθ – F = Ma———-(1), where f is the friction.

fR = Iα————–(2), where I is the MOMENT of inertia = MR^2/2.

fR = Ia/R, we substitute the value of a into the first equation,

Mgsinθ – f = (MfR^2)/( MR^2/2) = 2f

∴ Mgsinθ – f = 2f Or f = (Mgsinθ)/3.

The CORRECT CHOICE is (a) 5m/s

To ELABORATE: The speed of any POINT w.r.t the centre is equal to R⍵,

where ‘ω’ is the angular speed & ‘R’ is the distance of that point from the centre.

In PURE rolling motion, we know that V = Rω,

where V is the velocity of centre of mass.

∴ The velocity of centre of mass = 5m/s, since Rω = 5.

The correct CHOICE is (b) 0.9 RAD

To explain: Let w be the angular velocity & a be the angular acceleration.

Using, w^2 = w0^2 + 2aθ, we get:

0 = 9 – 2*5*θ

Or 9 = 10θ

∴θ = 0.9 rad.

Correct answer is (c) 5.25kgm^2

For EXPLANATION I WOULD say: Moment of inertia about an axis perpendicular to length and passing through COM is equal to MI^2/12.To find I about given axis, we use parallel axis theorem,

I = ICOM + ma^2, where a is the distance between the axis and m is MASS of the body.

I = MI^2/12 + Ma^2

= 1*36/12 + 1*2.25

= 3 + 2.25 = 5.25kgm^2.

Correct choice is (B) Torque

The EXPLANATION: Let L, r & p be the angular momentum, radius & momentum vectors respectively.Angular momentum of a body is ∑(r X p). On DIFFERENTIATING w.r.t time,dL/dt = ∑{dr/dt X p + r X dp/dt} = ∑{v X mv + r X F} = ∑{r X F} = total torque ‘τ’.

Here, F is the total force vector. And v X mv is ZERO because the cross product of two vectors in the same DIRECTION is zero.

Right choice is (a) 1:1

For EXPLANATION I would SAY: Moment of INERTIA of ring = MR^2& moment of inertia of disc = MR^2/2.

Mass ratio = 2:1, Radii ratio = 1:2.

Thus ratio of moment of inertia = 2MrRr^2/MdRd^2

= (2*2*1)/(1*4)

= 1:1.

Correct option is (b) (b . C)(E . f) [ d + a ]

Easiest EXPLANATION: [ a(b . c) X (e . f)d ] X [ a – d ]

= (b . c)(e . f) { [ a X d ] X [ a – d ] }

= (b . c)(e . f) { [ (a X d) X a ] – [ (a X d) X d ] }

Now, (a X d) is perpendicular to the plane of a & d. Refer to the DIAGRAM below.

We see that (a X d) X a = d&(a X d) X d = −a.

Therefore, our expression = (b . c)(e . f) [ d + a ].

Correct CHOICE is (b) v = w X r

To explain: For a particle rotating about a fixed axis, its ANGULAR velocity vector points along the axis. Velocity of a particle is a vector so it will be a cross PRODUCT and not a dot product. Now if we keep our fingers along the direction of the angular velocity vector and curl them in the direction of the RADIUS we get the direction of velocity for that radius vector. Therefore, v = w X r. Refer to the diagram.

Correct option is (a) a*DT

The explanation: The EQUATION of rotational motion is GIVEN by: W = W0 + a*dt, where dt is the time interval. Therefore, the change in angular speed = w – w0 = a*dt.

Correct choice is (a) 0s

Explanation: When the balls are released in their respective directions, the ball going downwards is HEAVIER and is covering more distance per UNIT time. This is because both the balls have the same initial SPEED and the heavier ball is gaining speed downwards while the other ball is losing speed as it moves upwards. Their mid-point will be below the point of release and centre of mass, will THEREFORE, be even below that point, SINCE heavier mass is moving downwards.

Correct option is (c) 1.25 rad/s, spheres will move to opposite ends of rod

To explain I would say: When the rod gets an IMPULSE, it gets an initial angular velocity. Now, the two spheres will start going outwards TILL they reach the end of the rod on either side as the walls of rod’s ends will provide the REQUIRED centripetal force. Also, to find final angular velocity we can conserve angular momentum as there is no external torque.

I1w1 = I2w2,

where I1, I2, w1, w2 are the initial & final moment of INERTIAS & angular velocities resp.

I1 = MI^2/12 + 0 = 2*0.01/12 = 0.01/6 kgm^2.

I1 = MI^2/12 + 2*mI^2/4

= (0.01/6) + (0.5*1*0.01)

= (0.01/6) + (0.01/2)

= 0.04/6 = 0.02/3 kgm^2.

∴ I1w1 = I2w20.01/6 * 5 = 0.02/3 *w2

∴ w2 = 1.25 rad/s.

The CORRECT OPTION is (b) 0.707m

To explain I WOULD say: Moment of inertia of disc = MR^2/2 = 10*1/2 = 5kgm^2.

Radius of gyration ‘k’ is given by the expression: I = Mk^2

∴ k = √(I/M)

= √(5/10) = 0.707m.

The correct choice is (c) The CENTRE of mass REMAINS in the same trajectory as before breaking APART of the asteroid

To explain I would say: Since internal forces do not change the position of the centre of mass of a system as their net force is zero, the centre of mass remains in the same trajectory as before breaking apart of the asteroid. However, it is an INFLUENTIAL factor for the trajectories of the TWO pieces after breaking apart.

Correct ANSWER is (b) False

The explanation is: No, the centre of mass cannot VARY upon the application of internal force. This is because internal force do not produce any net force. HENCE the STATE of centre of mass in this scenario will remain unchanged.

Right ANSWER is (B) 2.4 m

Explanation: Sum of MASSES = 3 + 2 = 5 kg

Centre of mass from 3 kg ball = (6 X 2) / 5

 = 2.4 m.