MCQs*

Question. Solved by 60% From NCERT NEET - 2021

From a circular ring of mass $M$ and radius $R$ , an arc corresponding to a $90^{0}$ sector is removed The moment of inertia of the remaining part of the ring about an axis passing through the center of the ring and perpendicular to the plane of the ring is $k$ times $M R^{2}$ . Then the value of $k$ is

$\frac{3}{4}$

$\frac{7}{4}$

$\frac{3}{2}$

$\frac{3}{5}$

Answer

Correct option is

$\frac{3}{4}$

Question. Solved by 60% From NCERT Derived Question NEET - 2021

From a complete ring of mass 'm' and radius 'R' a $3 0^{0}$ sector is removed.The M.I. of the remaining ring about an axis through centre of ring and normal to its plane is.

$\frac{3}{4} m R^{2}$

$\frac{1 1}{1 2} m R^{2}$

$\frac{3}{5} m R^{2}$

$m R^{2}$

Answer

Correct option is

$\frac{1 1}{1 2} m R^{2}$

Question. Solved by 60% From NCERT Derived Question NEET - 2021

The moment of inertia of a ring of mass $10 g$ and radius $1 c m$ about a tangent to the ring and normal to its plane is :

$2 \times 1 0^{- 6} k g m^{2}$

$5 \times 1 0^{- 7} k g m^{2}$

$5 \times 1 0^{- 8} k g m^{2}$

$4 \times 1 0^{- 6} k g m^{2}$

Answer

Correct option is

$2 \times 1 0^{- 6} k g m^{2}$

The moment of inertia about a tangent to the ring and normal to its plane is given as,
$I = 2 M R^{2}$
$= 2 \times 10 \times 1 0^{- 3} (1 \times 10^{- 2})^{2}$
$= 2 \times 1 0^{- 6} k g m^{2}$
Thus, the moment of inertia about a tangent to the ring and normal to its plane is $2 \times 1 0^{- 6} k g m^{2}$ .

Question. Solved by 60% From NCERT Derived Question NEET - 2021 Concept

If $I$ is the moment of inertia of a circular ring about a tangent of ring in its plane then the moment of inertia of the same ring about a tangent of ring perpendicular to its plane is:

$\frac{2 I}{3}$

$\frac{3 I}{2}$

$\frac{4 I}{3}$

$\frac{3 I}{4}$

Answer

Correct option is

$\frac{4 I}{3}$

Moment of Inertia of a ring about any axis perpendicular to its plane is $M R^{2}$
By perpendicular axis theorem , moment of Inertia about any diameter is $\frac{M R ^{2}}{2}$ .
Moment of Inertia about any tangent in the plane is $\frac{M R ^{2}}{2} + M R^{2} = \frac{3 M R ^{2}}{2} = I$
Moment of Inertia about any tangent and perpendicular plane is $M R^{2} + M R^{2} = 2 M R^{2}$
$= 4 (\frac{M R ^{2}}{2}) = \frac{4}{3} (\frac{3 M R ^{2}}{2}) = \frac{4 I}{3}$

Question. Solved by 60% From NCERT Derived Question NEET - 2021 Concept

$M$ is mass and $R$ is radius of a circular ring. The moment of inertia of same ring about an axis in the plane of ring at a perpendicular distance $\frac{2 R}{3}$ from centre of ring is:

$\frac{2 M R ^{2}}{3}$

$\frac{4 M R ^{2}}{9}$

$\frac{3 M R ^{2}}{8}$

$\frac{1 7 M R ^{2}}{1 8}$

Answer

Correct option is

$\frac{1 7 M R ^{2}}{1 8}$

Moment of inertia of ring about an axis in the plane $I_{c m} = M R^{2} / 2$

Now using parallel axis theorem $I = I_{c m} + M d^{2}$ where d=2R/3

$I = M R^{2} / 2 + M \times 4 R^{2} / 9 = \frac{1 7 M R ^{2}}{1 8}$

Question. Solved by 90% From NCERT Derived Question NEET - 2021 Concept

A wire of mass $m$ and length $l$ is bent in the form of a circular ring. The moment of inertia of the ring about its axis is :

$m l^{2}$

$\frac{m l ^{2}}{4 π ^{2}}$

$\frac{m l ^{2}}{2 π ^{2}}$

$\frac{m l ^{2}}{8 π ^{2}}$

Answer

Correct option is

$\frac{m l ^{2}}{4 π ^{2}}$

length is $l$
$l = 2 π r$
$r = \frac{l}{2 π}$

$I = m r^{2}$

$= \frac{m l ^{2}}{4 π ^{2}}$

Question. Solved by 60% From NCERT Derived Question NEET - 2021 Concept

A thin rod of mass $M$ and length $L$ is bent into a circular ring.The radius of gyration of this ring is:

$\frac{L}{2 π}$

$\frac{L}{2 2 π}$

$\frac{L}{4 π}$

Answer

Correct option is

$\frac{L}{2 π}$

The length of the rod is the circumference of the formed circle.
Thus $L = 2 π R$
$⟹ R = \frac{L}{2 π}$
Radius of gyration k = $\frac{I}{m}$
The moment of inertia of the thin ring about the axis passing through its geometric center is: I = $m R^{2} = m k^{2}$
hence, the radius of a ring is the radius of gyration of the object. Thus the radius of gyration of the ring is $\frac{L}{2 π}$

Question. Solved by 90% From NCERT Derived Question NEET - 2021 Concept

The moment of inertia of a ring of mass $M$ and radius $R$ about $P Q$ axis will be:

$M R^{2}$

$\frac{M R ^{2}}{2}$

$\frac{3}{2} M R^{2}$

$2 M R^{2}$

Medium

Answer

Correct option is

$M R^{2}$

$I = \int d l = \oint λ d x R^{2} = λ R^{2} (2 π R) = λ (2 π R) R^{2} = M R^{2} i . e I = M R^{2}$
Option $A$ is correct .

Question. Solved by 90% From NCERT Derived Question NEET - 2021 Concept

Three particles (each of mass 10g) are situated at the three corners of equilateral triangle of side 5cm. Determine the moment of inertia of this system about an axis passing through one corner of the triangle and perpendicular to the plane of the triangle

Answer

A is correct

Question.

Answer

Question.

Answer

Question.

Answer

Question.

Answer

Question.

Answer

Question.

Answer

Facebook SDK (Plugin)

Systems of Particles and Rotational Motion MCQs*

From a circular ring of mass M and radius R, an arc corresponding to a 900 sector is removed The moment of inertia of the remaining part of the ring about an axis passing through the center of the ring and perpendicular to the plane of the ring is k times MR2. Then the value of k is

43​

47​

23​

53​

43​

From a complete ring of mass 'm' and radius 'R' a 300 sector is removed.The M.I. of the remaining ring about an axis through centre of ring and normal to its plane is.

43​mR2

1211​mR2

53​mR2

mR2

1211​mR2

The moment of inertia of a ring of mass 10g and radius 1cm about a tangent to the ring and normal to its plane is :

2×10−6kgm2

5×10−7kgm2

5×10−8kgm2

4×10−6kgm2

2×10−6kgm2

The moment of inertia about a tangent to the ring and normal to its plane is given as,I=2MR2=2×10×10−3(1×10−2)2=2×10−6kgm2Thus, the moment of inertia about a tangent to the ring and normal to its plane is 2×10−6kgm2.

If I is the moment of inertia of a circular ring about a tangent of ring in its plane then the moment of inertia of the same ring about a tangent of ring perpendicular to its plane is:

32I​

23I​

34I​

43I​

34I​

M is mass and R is radius of a circular ring. The moment of inertia of same ring about an axis in the plane of ring at a perpendicular distance 32R​ from centre of ring is:

32MR2​

94MR2​

83MR2​

1817MR2​

1817MR2​

Moment of inertia of ring about an axis in the plane Icm​=MR2/2Now using parallel axis theorem I=Icm​+Md2 where d=2R/3I=MR2/2+M×4R2/9=1817MR2​

A wire of mass m and length l is bent in the form of a circular ring. The moment of inertia of the ring about its axis is :

ml2

4π2ml2​

2π2ml2​

8π2ml2​

4π2ml2​

length is ll=2πrr=2πl​I=mr2=4π2ml2​

A thin rod of mass M and length L is bent into a circular ring.The radius of gyration of this ring is:

2​πL​

2πL​

22​πL​

4πL​

2πL​

The moment of inertia of a ring of mass M and radius R about PQ axis will be:

MR2

2MR2​

23​MR2

2MR2

MR2

I=∫dl=∮λdxR2=λR2(2πR)​=λ(2πR)R2=MR2i.eI=MR2​Option A is correct .

Three particles (each of mass 10g) are situated at the three corners of equilateral triangle of side 5cm. Determine the moment of inertia of this system about an axis passing through one corner of the triangle and perpendicular to the plane of the triangle

You may like these posts

$\frac{3}{4}$

$\frac{7}{4}$

$\frac{3}{2}$

$\frac{3}{5}$

$\frac{3}{4}$

From a complete ring of mass 'm' and radius 'R' a $3 0^{0}$ sector is removed.The M.I. of the remaining ring about an axis through centre of ring and normal to its plane is.

$\frac{3}{4} m R^{2}$

$\frac{1 1}{1 2} m R^{2}$

$\frac{3}{5} m R^{2}$

$m R^{2}$

$\frac{1 1}{1 2} m R^{2}$

The moment of inertia of a ring of mass $10 g$ and radius $1 c m$ about a tangent to the ring and normal to its plane is :

$2 \times 1 0^{- 6} k g m^{2}$

$5 \times 1 0^{- 7} k g m^{2}$

$5 \times 1 0^{- 8} k g m^{2}$

$4 \times 1 0^{- 6} k g m^{2}$

$2 \times 1 0^{- 6} k g m^{2}$

If $I$ is the moment of inertia of a circular ring about a tangent of ring in its plane then the moment of inertia of the same ring about a tangent of ring perpendicular to its plane is:

$\frac{2 I}{3}$

$\frac{3 I}{2}$

$\frac{4 I}{3}$

$\frac{3 I}{4}$

$\frac{4 I}{3}$

$M$ is mass and $R$ is radius of a circular ring. The moment of inertia of same ring about an axis in the plane of ring at a perpendicular distance $\frac{2 R}{3}$ from centre of ring is:

$\frac{2 M R ^{2}}{3}$

$\frac{4 M R ^{2}}{9}$

$\frac{3 M R ^{2}}{8}$

$\frac{1 7 M R ^{2}}{1 8}$

$\frac{1 7 M R ^{2}}{1 8}$

Moment of inertia of ring about an axis in the plane $I_{c m} = M R^{2} / 2$

Now using parallel axis theorem $I = I_{c m} + M d^{2}$ where d=2R/3

$I = M R^{2} / 2 + M \times 4 R^{2} / 9 = \frac{1 7 M R ^{2}}{1 8}$

A wire of mass $m$ and length $l$ is bent in the form of a circular ring. The moment of inertia of the ring about its axis is :

$m l^{2}$

$\frac{m l ^{2}}{4 π ^{2}}$

$\frac{m l ^{2}}{2 π ^{2}}$

$\frac{m l ^{2}}{8 π ^{2}}$

$\frac{m l ^{2}}{4 π ^{2}}$

length is $l$
$l = 2 π r$
$r = \frac{l}{2 π}$

$I = m r^{2}$

$= \frac{m l ^{2}}{4 π ^{2}}$

A thin rod of mass $M$ and length $L$ is bent into a circular ring.The radius of gyration of this ring is:

$\frac{L}{2 π}$

$\frac{L}{2 π}$

$\frac{L}{2 2 π}$

$\frac{L}{4 π}$

$\frac{L}{2 π}$

The moment of inertia of a ring of mass $M$ and radius $R$ about $P Q$ axis will be:

$M R^{2}$

$\frac{M R ^{2}}{2}$

$\frac{3}{2} M R^{2}$

$2 M R^{2}$

$M R^{2}$

$I = \int d l = \oint λ d x R^{2} = λ R^{2} (2 π R) = λ (2 π R) R^{2} = M R^{2} i . e I = M R^{2}$
Option $A$ is correct .