System of Particles and Rotational Motion Online Objective Test | System of Particles and Rotational Motion Online Objective Test

# System of Particles and Rotational Motion # CBSE 11TH PHYSICS Prepare / Learn

Q1. If the radius of earth contracts to half of its present value, the mass remaining unchanged, the duration of the day will be

A. 24 Hrs

B. 12 Hrs

C. 6 hrs

D. 48 hrs

Answer : Option C
Explaination / Solution:

As the Moment of inertia of earth considered as sphere is I = 2/5 MR2 , thus according to law of conservation of angular momentum as the radius contracts to half , thus new moment of inertia of earth will be I/4 , thus the angular velocity will increase 4 times and making the length of the day to 6 hrs.

# System of Particles and Rotational Motion # CBSE 11TH PHYSICS Prepare / Learn

Q2. A boy comes running and sits on a merry-go-round. What is conserved?

A. Angular momentum

B. Kinetic energy of rotation

C. None of these

D. Linear momentum

Answer : Option A
Explaination / Solution:

According to law of conservation of angular momentum if no external torque is applied on a body in rotation than its angular momentum remains conserved.

# System of Particles and Rotational Motion # CBSE 11TH PHYSICS Prepare / Learn

Q3. A ring of radius r and mass m rotates about its central axis. The kinetic energy is

m r^{2}

ω^{2}

B. mr

ω^{2}

m r^{2}

ω^{2}

D. mr

ω^{2}

Answer : Option C
Explaination / Solution:

The kinetic energy of body in rotational motion is $K E = \frac{1}{2} I ω^{2} = \frac{1}{2} m r^{2} ω^{2}$ as moment of inertia of ring about its central axis is I = mr2

# System of Particles and Rotational Motion # CBSE 11TH PHYSICS Prepare / Learn

Q4. A thin circular ring of mass M and radius R is rotating about its central axis with angular velocity. Four point objects each of mass m are attached gently to the opposite ends of two perpendicular diameters, the angular velocity of the ring is given by

A. MM+4m⋅ω

B. M+4mM⋅ω

C. M−4mM+4m⋅ω

D. MM+m⋅ω

Answer : Option A
Explaination / Solution:

Let $ω$ be the angular velocity of the Ring of Mass M , thus the moment of inetia about given axis is I1 = $M R^{2}$ and the four point objects are genlty placed at perpendicular diameters at opposite end, so thus the distance of each object from axis of rotation is R, so total moment of inetia of ring and four objects is I2 = $M R^{2} + 4 m R^{2}$

According to law of conservation of angular momentum I1 $ω$ = I2 $ω$ 2 , So on solving $ω_{2} = (\frac{M R^{2}}{M R^{2} + 4 m R^{2}}) ω$ = $\frac{M}{M + 4 m} \cdot ω$

# System of Particles and Rotational Motion # CBSE 11TH PHYSICS Prepare / Learn

Q5. What is the moment of inertia of a thin rod of length L and mass M about an axis passing through one end and perpendicular to its length?

A. 1/2 ML²

B. ML2

1/12 ML2

D. 1/3 ML2

Answer : Option D
Explaination / Solution:

Using theorem of parallel axis, the axis is shifted by L/2 distance from centre of Mass where the Moment of Inertia is ML2/12, thus the moment of inertia of a thin rod about an axis passing through one end and perpendicular to its length is ML2/12 + ML2/4 =

\frac{1}{3}

M L^{2}

# System of Particles and Rotational Motion # CBSE 11TH PHYSICS Prepare / Learn

Q6. The radius of gyration of a solid disc about one of its diameter is given by

A. 2R

B. √2 R

\sqrt{}

C. R/√2

D. R/2

Answer : Option D
Explaination / Solution:

Moment of inetia of axis passing through its center and perpendicular to its plane:

I = ½ MR2

Using perpendicular axis theorem Ix +Iy = Iz ,so 2Id = ½ MR2

Moment of inetia of along its diameter Id = ¼MR2

And the radius of gyration MK2 = ¼MR2

$R 2$

# System of Particles and Rotational Motion # CBSE 11TH PHYSICS Prepare / Learn

Q7. The moment of inertia of a solid sphere of density ρ and radius R is given by

A. 176105ρR3

B. 176105ρR5

C. 105176ρR2

D. 176105ρR2

Answer : Option B
Explaination / Solution:

\begin{aligned} I = \frac{2}{5} (M R^{2}) \\ = \frac{2}{5} [(\frac{4}{3} π R^{3}) . ρ . R^{2}] \\ = \frac{2}{5} [(\frac{4}{3} \frac{22}{7} R^{3}) . ρ . R^{2}] \end{aligned}

As Mass = Density x Volume of Sphere

$= \frac{176}{105} ρ R^{5}$

# System of Particles and Rotational Motion # CBSE 11TH PHYSICS Prepare / Learn

Q8. A thin uniform rod of length 2l and mass M is acted upon a constant torque. The angular velocity changes from zero to ωω in time t. The value of torque is

A. 2Ml2ω3t

B. Ml2ω3t

C. Ml2ω12t

D. Ml2ωt

Answer : Option B
Explaination / Solution:

As Torque(

τ

) is equal to product of Moment of Inertia (I) and Angular acceleration

(α)

$τ = I α τ = I Δ ω Δ t τ = [M ( 2 l ) 2 12] [ω t] τ = M l 2 ω 3 t$

# System of Particles and Rotational Motion # CBSE 11TH PHYSICS Prepare / Learn

Q9. A particle is orbiting in a vertical plane. Its linear momentum will be directed

A. horizontally

B. at 45

^{\circ}

to the vertical

C. vertically

D. tangential to the orbit.

Answer : Option D
Explaination / Solution:

As the direction of velocity at any point moving in circular orbit is tangent at that point, thus momentum would be tangential to the orbit.

# System of Particles and Rotational Motion # CBSE 11TH PHYSICS Prepare / Learn

Q10.

Three thin uniform rods each of mass M and length L are placed along the three axis of a Cartesian coordinate system with one end of each rod at the origin. The M.I. of the system about z- axis is

M L^{2}

M L^{2}

(2/3)ML2

M L^{2}

M L^{2}

M L^{2}

M L^{2}

Answer : Option B
Explaination / Solution:

Moment of inertia about an axis passing through edge

$I = \frac{M L^{2}}{3}$

Moment of inertia of system about z-axis

$\begin{aligned} I = I_{x} + I_{y} + I_{z} \\ I = \frac{M L^{2}}{3} + \frac{M L^{2}}{3} + 0 \\ I = \frac{2}{3} M L^{2} \end{aligned}$

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