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(d) The instantaneous acceleration of the earth points towards the sun.

Explanation: 

Since during the mentioned period in (a) and (b) the earth covers a semi circle and there is a displacement between original point to final point, so average velocity can not be zero.

 Acceleration is the rate of change of velocity, here the change in velocity may be 30-(-30)=60 km/s but the time taken is six months i.e. 181*24*60*60 s. Dividing the change in velocity by this time will give the quotient much much less than 60 km/s². 

Since the earth moves with a constant speed of 30 km/s, so average speed will also be 30 km/s not zero. 

Since it is a case of uniform circular motion, the tangential component of the instantaneous acceleration will be zero and only radial component will be there which is directed towards the center (Sun in this case). Only option (d) is correct.

(b) speed remains constant

(d) tangential acceleration remains constant.

Explanation: 

Velocity and acceleration will not be constant because direction changes continuously. Hence option (a) and (c) are not correct. With a constant speed the position vector will sweep equal area in equal time. Thus option (b) is correct. Since speed is constant, it is a case of uniform circular motion in which tangential acceleration is always zero i.e, always constant. Thus option (d) is also correct.

(b) speed remains constant

(d) tangential acceleration remains constant.

Explanation: 

Velocity and acceleration will not be constant because direction changes continuously. Hence option (a) and (c) are not correct. With a constant speed the position vector will sweep equal area in equal time. Thus option (b) is correct. Since speed is constant, it is a case of uniform circular motion in which tangential acceleration is always zero i.e, always constant. Thus option (d) is also correct.

The safety speed of a vehicle on a curve horizontal road is \(\sqrt{μrg}\)

Radius of the circular road is 57.87 m

i. F_CF = \(\cfrac{m\text v^2}r\)

ii. F_CP = mrω²

iii. F_CP = 4π²mrn²

iv. F_CP = \(\cfrac{4\pi^2mr}{T^2}\) 

v. F_CP = μmg = mω²r

θ = tan^-1\(\left(\cfrac{\text v^2}{rg}\right)\)

Height of inclined road

h = l sin θ

Answer is (B) zero

Answer is (D) ω_E = \(\frac{1}{2}\)ω_E

n₁ = 600 r.p.m., n₂ = 1200 r.p.m.,

Using,

Increment in angular velocity, ω = 2π(n₂ - n₁)

ω = 2π(1200 - 600)rad/min

= (2π x 600)/60 rad/s

ω = 20πrad/s

Answer is (A) 1.5 m

n = 2000, distance = 9500 m Distance covered in ‘n’ revolutions = n(2πr) = nπD

2000πD = 9500

D = \(\frac{9500}{2000\times \pi}\) = 1.5 m

Answer is (D) towards south.

The particle performing circular motion flies-off tangentially.

Answer is (C) 100π rad 

f = 300 r.p.m = \(\frac{3000}{60}\) r.p.s

θ = ω.t = 2π x  \(\frac{3000}{60}\) x 1 = 100 π rad

Answer is (C) 10 π rad 

Frequency of wheel, n = \(\frac{300}{60}\) = 5 r.p.s.

Angle described by wheel in one rotation = 2π rad. 

Therefore, angle described by wheel in 1 sec θ = 2π x 5 radians = 10 π rad

Answer is (D) 628.4 cm/s

n = 600 r.p.m. = 

v = rω = r x 2πn = 10 x 2 x 3.142 x 10 

= 628.4 cm/s.

Correct option is (B) 1.4 s

Answer is (A) Disc rotates in clockwise direction in segments OA and AB.

Answer is (B) 4π rad/s 

Correct option is (A) 1 : 12

Correct option is (B) π m/s

Answer is (A) R/4

F = mω²R

∴ R ∝ \(\frac{1}{ω^2}\) (m and F are constant)

If ω is doubled, then radius will become 1/4 times i.e., R/4

The right answer is (a) 1 cm

Explanation:

The friction force to keep it rotating =mω²r. At 4 cm away from the center when it just slips, the static frictional force is at its limiting value. When the angular speed of the turntable is doubled the magnitude of the required centripetal force increases four times. So to keep it to limiting value of friction force, 'r' should be made one fourth. It will be clear as below m(2ω)².r/4 =mω²r. 

That is why it should be placed at 1 cm.

Correct option is (A) 1 cm

The right answer is (a) 1 cm

Explanation:

The friction force to keep it rotating =mω²r. At 4 cm away from the center when it just slips, the static frictional force is at its limiting value. When the angular speed of the turntable is doubled the magnitude of the required centripetal force increases four times. So to keep it to limiting value of friction force, 'r' should be made one fourth. It will be clear as below m(2ω)².r/4 =mω²r. 

That is why it should be placed at 1 cm.

Answer is (C) changes in direction only.

Correct option is (C) zero

Correct option is (C) rω²

Correct option is (C) 45 m/s

v = rω = r (αt) = 3 x 3 x 5 = 45 m/s

i. Angular velocity of a particle performing circular motion is defined as the time rate of change of limiting angular displacement.

OR

The ratio of angular displacement to time is called angular velocity.

ii. Instantaneous angular velocity is given by,

\(\omega=\cfrac{\theta}t\)

iii. It is a vector quantity.

iv. Direction: The direction of angular velocity is given by right hand rule and is in the direction of angular displacement.

v. Unit: rad s^-1

vi. Dimensions: [M⁰ L⁰ T^-1]

Note: Magnitude of angular velocity is called angular speed.

Correct option is (C) Force experienced by a person standing on a merry-go- round.

Correct option is (B) 0.628 rad/s²

Correct option is (C) 2R, πR

In half the period, particle is diametrically opposite to its initial position. Hence, its displacement is 2R. It has covered a semicircle, hence distance covered by particle is πR.

Correct option is (C) F = mg - \(\cfrac{m\text v^2}r\) 

Answer is (B) at the bottom of the circle

Using, 

\(\frac{mv^2}{r}\) = \(\frac{2\times(4)^2}{1}\) = 32N

It is clear that tension will be 52 N at the bottom of the circle because we know that,

T_Bottom = mg + \(\frac{mv^2}{r}\)

Answer is (C) zero

α = \(\frac{dω}{dt}\) = 0 ...(∵ω = constant)

Answer is (D) 1.6π² m/s²

Using,

ω = 2πn = 2π x 1 = 2π rad/s

a = rω² = 0.4 x (2π)² = 0.4 x 4 π² 

a = 1.6 π² m/s²

Answer is (B) a_r = 0 and a_t ≠ 0 

If a_r = 0, there is no radial acceleration and circular motion is not possible 

So a_r ≠ 0

If a_t ≠ 0 the motion is not uniform as angular velocity will change

So a_r ≠ 0 and a_t = 0 for uniform circular motion

Correct Answer is: (a) in all cases

There are no external horizontal forces acting on the ‘man plus boat’ system. (The forces exerted by the man and the boat on each other are internal forces for the system.) Therefore, the centre of mass of the system, which is initially at rest, will always be at rest.

Correct Answer is: (a) its centre of mass

Correct option is (C) 50 m/s

Answer is (B) 0.24

For banking of road,θ = tan^-1\(\Big(\frac{v^2}{rg}\Big)\)

θ = tan^–1 (0.24)

∴ tan θ = 0.24 

Also, tan θ = \(\frac{v^2}{rg}\) ⁼ μ ⇒ μ = 0.24

Correct option is (B) 120 m

Maximum number of revolutions is 3.150 rev/s

Answer is (C) centripetal acceleration

Correct option is (D) m [g + (π² n² r)/900]

Answer is (A) 4 m/s

Critical velocity at highest point = \(\sqrt{gR}\) 

= \(\sqrt{10\times1.6}\) 

= 4 m/s

When a bird takes left or right turn it does not remain horizontal but tilts like a plane. Thus pressing the air with greater area of the underside. As per Newton's Third Law of Motion the air underneath also exerts equal and opposite force on the bird, like normal force on a banked circular road. The horizontal component of this force is the required centripetal force for the bird.

Correct option is (C) 10π² cm/s²

Answer is (B) 3 g

v = \(\sqrt{3gr}\) and  a = \(\frac{v^2}{r}\) = \(\frac{3gr}{r}\) = 3g

Correct option is (C) frictional force between the surface of the road and the tyres of the car.