Test: Rotational Motion - 1 - NEET MCQ

# Test: Rotational Motion - 1 - NEET MCQ

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## 30 Questions MCQ Test Physics Class 11 - Test: Rotational Motion - 1

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Test: Rotational Motion - 1 - Question 1

### The moment of inertia of a body depends upon -

Detailed Solution for Test: Rotational Motion - 1 - Question 1

I = mr2, therefore it depends on body mass and its mass distribution.

Test: Rotational Motion - 1 - Question 2

### Two spheres of same mass and radius are in contact with each other. If the moment of inertia of a sphere about its diameter is I, then the moment of inertia of both the spheres about the tangent at their common point would be -

Detailed Solution for Test: Rotational Motion - 1 - Question 2

The moment of inertia of a sphere about its diameter is given as,
I=2​/5 MR2
The moment of inertia of the sphere about the tangent is given as,
I′=2/5​MR2+MR2
I′=7/5​MR2
The total moment of inertia of both spheres about the common tangent is given as,
​It=2I′
It​=2×7/5​MR2
It​=7I

Test: Rotational Motion - 1 - Question 3

### A disc of metal is melted to recast in the form of a solid sphere. The moment of inertias about a vertical axis passing through the centre would -

Detailed Solution for Test: Rotational Motion - 1 - Question 3

Moment of inertia will decrease, because
Id=1/2mr2 and Is=2/5mr2,
the radius of sphere formed on recasting the disc will also decrease.

Test: Rotational Motion - 1 - Question 4

The M.I. of a disc about its diameter is 2 units. Its M.I. about axis through a point on its rim and in the plane of the disc is

Detailed Solution for Test: Rotational Motion - 1 - Question 4

We know that for a disc of mass m and radius r
MI of a disc about its diameter = mr2/4 = 2
And also MI about a point on its rim = mr2/4 + mr2
= 5mr2/4
= 5 x 2 = 10

Test: Rotational Motion - 1 - Question 5

A solid sphere and a hollow sphere of the same mass have the same moments of inertia about their respective diameters, the ratio of their radii is

Detailed Solution for Test: Rotational Motion - 1 - Question 5

We know moment of inertia of solid sphere Is​=2​/5ms​Rs2​ and
moment of inertia of hollow sphere IH​=2/3​mH​RH2 ​As per question Is​=IH​
Now,
2/5​ms​Rs2​=2/3​mH​RH2​
as the masses are equal the ratio of their radii will be
​Rs2 /RH2 ​​=2/3​/​2/5​=√5/3​​=(5)1/2: (3)1/2

Test: Rotational Motion - 1 - Question 6

A stone of mass 4kg is whirled in a horizontal circle of radius 1m and makes 2 rev/sec. The moment of inertia of the stone about the axis of rotation is

Detailed Solution for Test: Rotational Motion - 1 - Question 6

Moment of Inertia = MR²
= 4 × 1
= 4 kg-m²

Test: Rotational Motion - 1 - Question 7

Three rings, each of mass P and radius Q are arranged as shown in the figure. The moment of inertia of the arrangement about YY' axis will be

Detailed Solution for Test: Rotational Motion - 1 - Question 7

For ring 1
(MOI)1 = MOI about diameter + PQ2
MOI about diameter = ½ PQ2
(MOI)1 = 3/2 PQ2
Similarly, (MOI)2 = 3/2 PQ2
(MOI)3 = ½ PQ2
Total MOI = 7/2 PQ2

Test: Rotational Motion - 1 - Question 8

A circular disc A of radius r is made from an iron plate of thickness t and another circular disc B of radius 4r is made from an iron plate of thickness t/4. The relation between the moments of inertia IA and IB is

Detailed Solution for Test: Rotational Motion - 1 - Question 8

Test: Rotational Motion - 1 - Question 9

The moment of inertia of a uniform semicircular wire of mass M and radius r about a line perpendicular to the plane of the wire through the centre is

Detailed Solution for Test: Rotational Motion - 1 - Question 9

The formula for the MOI of a uniform semicircular ring about a perpendicular line is ½ Mr2

Test: Rotational Motion - 1 - Question 10

Let IA and IB be moments of inertia of a body about two axes A and B respectively. The axis A passes through the centre of mass of the body but B does not.

Detailed Solution for Test: Rotational Motion - 1 - Question 10

If the axes are parallel, we use the formula:
Io = Icm + md2
For the first body the distance between the axis and the axis passing through C.O.M is 0
Therefore, Io = Icm + m(0)2 = lcm since it is mentioned in question
Whereas for the second body it is not passing through the C.O.M therefore, there is some distance between the axes, say 'd'
So, Io = Icm + md2

Test: Rotational Motion - 1 - Question 11

For the same total mass which of the following will have the largest moment of inertia about an axis passing through its centre of mass and perpendicular to the plane of the body

Detailed Solution for Test: Rotational Motion - 1 - Question 11

Moment of inertia depends on distribution of mass around axis. The more the mass near the axis lesser is the moment of inertia. Four rods forming a square have more moment of inertia because of less mass near axis.

Test: Rotational Motion - 1 - Question 12

Moment of inertia of a thin semicircular disc (mass = M & radius = R) about an axis through point O and perpendicular to plane of disc, is given by :

Detailed Solution for Test: Rotational Motion - 1 - Question 12

Mass of semicircular disc = M
Suppose there is a circular disc of mass 2M, then
Moment of intertia of circular disc = ½ (2M)R2
Moment of intertia of circular disc = ½ (2M)R2 = MR2
=> So, Moment of intertia of semi-circular disc = ½ MR2

Test: Rotational Motion - 1 - Question 13

A rigid body can be hinged about any point on the x-axis. When it is hinged such that the hinge is at x, the moment of inertia is given by

I = 2x2 - 12x + 27 The x-coordinate of centre of mass is

Detailed Solution for Test: Rotational Motion - 1 - Question 13

Moment of inertia is minimum at centre of mass of the body.
I = minimum for centre of mass
Therefore x co-ordinate of centre of mass is 3 unit.

Test: Rotational Motion - 1 - Question 14

Consider the following statements

Assertion (A) : The moment of inertia of a rigid body reduces to its minimum value as compared to any other parallel axis when the axis of rotation passes through its centre of mass.

Reason (R) : The weight of a rigid body always acts through its centre of mass in uniform gravitational field. Of these statements :

Detailed Solution for Test: Rotational Motion - 1 - Question 14

Moment of inertia about its centroidal axis has a minimum value as the centroidal axis has mass evenly distributed around it thereby providing minimum resistance to rotation as compared to any other axis. Also the statement-2 is correct but is not the correct explanation for statement-1.

Test: Rotational Motion - 1 - Question 15

A body is rotating uniformly about a vertical axis fixed in an inertial frame. The resultant force on a particle of the body not on the axis is

Detailed Solution for Test: Rotational Motion - 1 - Question 15

Since the axis is vertical, the force will be horizontal and since it is centripetal force it will intersect the axis.

Test: Rotational Motion - 1 - Question 16

One end of a uniform rod of mass m and length I is clamped. The rod lies on a smooth horizontal surface and rotates on it about the clamped end at a uniform angular velocity w. The force exerted by the clamp on the rod has a horizontal component

Detailed Solution for Test: Rotational Motion - 1 - Question 16

Consider a rod of length l and mass m rotating with an angular velocity of w
dF = dm l w2
dF = (m/l) dμ l w2
Integrating
F = ½mlw2

Test: Rotational Motion - 1 - Question 17

A rod of length 'L' is hinged from one end. It is brought to a horizontal position and released. The angular velocity of the rod when it is in vertical position is

Detailed Solution for Test: Rotational Motion - 1 - Question 17

Test: Rotational Motion - 1 - Question 18

A disc of radius 2m and mass 200kg is acted upon by a torque 100N-m. Its angular acceleration would be

Detailed Solution for Test: Rotational Motion - 1 - Question 18

F = ma
a = Wr
So, F = mwr
a = wr = acceleration
m = 200kg
F = 100 n-m, r = 2m, w = to be calculated
100 = 200 x w2

*Multiple options can be correct
Test: Rotational Motion - 1 - Question 19

On applying a constant torque on a body

Detailed Solution for Test: Rotational Motion - 1 - Question 19

If a constant torque is applied it is possible that a positive angular acceleration gets generated which can generate a positive acceleration and hence increasing both velocity and angular velocity.

Test: Rotational Motion - 1 - Question 20

A wheel starting with angular velocity of 10 radian/sec acquires angular velocity of 100 radian/sec in 15 seconds. If moment of inertia is 10kg-m2, then applied torque (in newton-metre) is

Detailed Solution for Test: Rotational Motion - 1 - Question 20

I(ωf - ωi)/t = τ
Hence, τ = 60

Test: Rotational Motion - 1 - Question 21

An automobile engine develops 100H.P. when rotating at a speed of 1800 rad/min. The torque it delivers is

Detailed Solution for Test: Rotational Motion - 1 - Question 21

100 HP = 74570 W or 74.57 KW Now, P = 2*π*N*T/60 where, P is the power (in W), N is the operating speed of the engine (in r.p.m.) and T is the Torque (in N.m). Therefore, 74570 = 2*π*1800*T/60 i.e. T = 395.606 N.m

Test: Rotational Motion - 1 - Question 22

The moment of inertia and rotational kinetic energy of a fly wheel are 20kg-mand 1000 joule respectively. Its angular frequency per minute would be -

Detailed Solution for Test: Rotational Motion - 1 - Question 22

Rotational K.E. = ½ × moment of inertia × square of angular velocity = ½Iw2

Test: Rotational Motion - 1 - Question 23

Calculate the M.I. of a thin uniform ring about an axis tangent to the ring and in a plane of the ring, if its M.I. about an axis passing through the centre and perpendicular to plane is 4 kg m2.

Detailed Solution for Test: Rotational Motion - 1 - Question 23

At tangent I =MR2/2 + MR2 --- (1)
MR2/2 = 4, MR2 = 8 then substitute in (1) you will get I = 12, this is for tangent perpendicular to plane then divide by 2 you will get tangent along the plane

Test: Rotational Motion - 1 - Question 24

A torque of 2 newton-m produces an angular acceleration of 2 rad/sec2 a body. If its radius of gyration is 2m, its mass will be:

Detailed Solution for Test: Rotational Motion - 1 - Question 24

Test: Rotational Motion - 1 - Question 25

A particle is at a distance r from the axis of rotation. A given torque t produces some angular acceleration in it. If the mass of the particle is doubled and its distance from the axis is halved, the value of torque to produce the same angular acceleration is

Detailed Solution for Test: Rotational Motion - 1 - Question 25

We know that from some torque t, angular acceleration a produced can be find by,
t = I a, where I is moment of inertia = mr2
Thus we get a = t / mr2
Now we have 2m and r/2
Thus a =  t / mr2 = T / 2m(r/2)2
Thus we get T = t/2

Test: Rotational Motion - 1 - Question 26

A weightless rod is acted on by upward parallel forces of 2N and 4N ends A and B respectively. The total length of the rod AB = 3m. To keep the rod in equilibrium a force of 6N should act in the following manner :

Detailed Solution for Test: Rotational Motion - 1 - Question 26

Test: Rotational Motion - 1 - Question 27

A right triangular plate ABC of mass m is free to rotate in the vertical plane about a fixed horizontal axis through A. It is supported by a string such that the side AB is horizontal. The reaction at the support A is :

Detailed Solution for Test: Rotational Motion - 1 - Question 27

The distance of Centre Of Mass of the given right angled triangle is 2L/3​ along BA and L/3​ along AC from the point B.
Force of magnitude mg is acting downwards at its COM.
Moment balance around B gives:
mg(2L/3​)−FA​(L)=0
(Moment=  × =rFsin(θ)=F(rsin(θ))=Fr⊥​)
∴FA​=2​mg/3

Test: Rotational Motion - 1 - Question 28

In an experiment with a beam balance on unknown mass m is balanced by two known mass m is balanced by two known masses of 16 kg and 4 kg as shown in figure.

The value of the unknown mass m is

Detailed Solution for Test: Rotational Motion - 1 - Question 28

Test: Rotational Motion - 1 - Question 29

A homogeneous cubical brick lies motionless on a rough inclined surface. The half of the brick which applies greater pressure on the plane is :

Detailed Solution for Test: Rotational Motion - 1 - Question 29

As weight of the body (mg) acts along the left side i.e. mg sinx acts along flow of object.
mgcosx acts perpendicular to flow of the object.
The pressure of right half comes on left half hence left half has maximum pressure.

Test: Rotational Motion - 1 - Question 30

Consider the following statements

Assertion (A) : A cyclist always bends inwards while negotiating a curve

Reason (R) : By bending he lowers his centre of gravity Of these statements,

Detailed Solution for Test: Rotational Motion - 1 - Question 30

B is correct. The reason is a part of the question.

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