Rotation The Basic Definition(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

ROTATION THE BASIC DEFINITION MCQs

Total Questions : 30 | Page 1 of 3 pages

As a part of a maintenance inspection the compressor of a jet engine is made to spin according to the graph as shown. The number of revolutions made by the compressor during the test is

9000
16570
12750
11250

Discuss Question

Answer: Option D. -> 11250
:
D
Number of revolution = Area between the graph and time axis = Area of trapezium

= \frac{1}{2} \times (2.5 + 5) \times 3000 = 11250

revolution.

Question 2.

Figure below shows a small wheel fixed coaxially on a bigger one of double the radius. The system rotates about the common axis. The strings supporting A and B do not slip on the wheels. If 'x' and 'y' be the distances travelled by A and B in the same time interval, then

x = 2y
x = y
y = 2x
None of these

Discuss Question

Answer: Option C. -> y = 2x
:
C
Linear displacement (S) = Radius (r)

\times

Angular displacement (

θ

)

∴ S \propto r (i f θ = c o n s t a n t)

\frac{D i s t a n c e t r a v e l l e d b y m a s s A (x)}{D i s t a n c e t r a v e l l e d b y m a s s A (y)} = \frac{R a d i u s o f p u l l e y c o n c e r n e d w i t h m a s s A (r)}{R a d i u s o f p u l l e y c o n c e r n e d w i t h m a s s A (2 r)} = \frac{1}{2} \Rightarrow y = 2 x .

Question 3.

If the position vector of a particle is $⃗ r = (3 ˆ i + 4 ˆ j)$ meter and its angular velocity is $⃗ ω = (ˆ j + 2 ˆ k)$ rad/sec then its linear velocity is (in m/s)

$(8 ˆ i - 6 ˆ j + 3 ˆ k)$
$(3 ˆ i + 6 ˆ j + 8 ˆ k)$
- $(3 ˆ i + 6 ˆ j + 6 ˆ k)$
$(6 ˆ i + 8 ˆ j + 3 ˆ k)$

Discuss Question

Answer: Option A. ->

(8 ˆ i - 6 ˆ j + 3 ˆ k)

:
A

⃗ v = ⃗ ω \times ⃗ r = (3 ˆ i + 4 ˆ j + 0 ˆ k) \times (0 ˆ i + ˆ j + 2 ˆ k) = ∣ ∣∣∣ ∣ \begin{matrix} ˆ i & ˆ j & ˆ k 3 & 4 & 0 0 & 1 & 2 \end{matrix} ∣ ∣∣∣ ∣ = 8 ˆ i - 6 ˆ j + 3 ˆ k

Question 4.

A circular disc X of radius 'R' is made from an iron plate of thickness 't', and another disc Y of radius '4R' is made from an iron plate of thickness ' $\frac{t}{4}$ '. Then the relation between the moment of inertia $I_{x}$ and $I_{y}$ is

$I_{y} = 64 I_{x}$
$I_{y} = 32 I_{x}$
$I_{y} = 16 I_{x}$
$I_{y} = I_{x}$

Discuss Question

Answer: Option A. ->

I_{y} = 64 I_{x}

:
A
Moment of Inertia of disc

I = \frac{1}{2} M R^{2} = \frac{1}{2} (π R^{2} t ρ) R^{2} = \frac{1}{2} π t ρ R^{4}

[As

M = V \times ρ = π R^{2} t ρ

where t = thickness,

ρ

= density]

∴ \frac{I_{y}}{I_{x}} = \frac{t_{y}}{t_{x}} {(\frac{R_{y}}{R_{x}})}^{4}

[ If

ρ

= constant]

\Rightarrow \frac{I_{y}}{I_{x}} = \frac{1}{4} (4)^{4} = 64

[G i v e n R_{y} = 4 R_{x}, t_{y} = \frac{t_{x}}{4}]

\Rightarrow I_{y} = 64 I_{x}

Question 5.

Five particles of mass 2 kg are attached to the rim of a circular disc of radius 0.1 m and negligible mass. Moment of inertia of the system about the axis passing through the centre of the disc and perpendicular to its plane is

$1 k g m^{2}$
$0.1 k g m^{2}$
$2 k g m^{2}$
$0.2 k g m^{2}$

Discuss Question

Answer: Option B. ->

0.1 k g m^{2}

:
B
We will not consider the moment of inertia of ring because it doesn't have any mass. So, moment of inertia of five particle system

I = 5 m r^{2} = 5 \times 2 \times (0.1)^{2} = 0.1 k g - m^{2} .

Note: The masses are concentrated at fixed distance from the axis similar to that of ring. That is why you simply need to add all the individual contributions.

Question 6.

Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be

Discuss Question

Answer: Option B. -> 6 l
:
B
Moment of inertia of disc about a diameter

= \frac{1}{4} M R^{2} = I

(given)

∴ M R^{2} = 4 I

Now moment of inertia of disc about an axis perpendicular to its plane and passing through a point on its rim

= \frac{3}{2} M R^{2} = \frac{3}{2} (4 I) = 6 I .

Question 7.

Four thin rods of same mass 'M' and same length 'l', form a square as shown in the figure. Moment of inertia of this system about an axis through centre O and perpendicular to its plane is

$\frac{4}{3} M l^{2} .$
$\frac{M l^{2}}{3}$
$\frac{M l^{2}}{6}$
$\frac{2}{3} M l^{2} .$

Discuss Question

Answer: Option A. ->

\frac{4}{3} M l^{2} .

:
A
Moment of inertia of rod AB about point

P = \frac{1}{12} M l^{2}

M.I. of rod AB about point

O = \frac{M l^{2}}{12} + M {(\frac{1}{2})}^{2} = \frac{1}{3} M l^{2}

[by the theorem of parallel axis]
and the system consists of 4 rods of similar type so by the symmetry

I_{S y s t e m} = \frac{4}{3} M l^{2} .

Question 8.

Three rings each of mass M and radius R are arranged as shown in the figure. The moment of inertia of the system about YY' will be

$3 M R^{2}$
$\frac{3}{2} M R^{2}$
$5 M R^{2}$
$\frac{7}{2} M R^{2}$

Discuss Question

Answer: Option D. ->

\frac{7}{2} M R^{2}

:
D
M.I of system about YY'

I = I_{1} + I_{2} + I_{3}

here

I_{1}

= moment of inertia of ring about diameter,

I_{2} = I_{3} = M . I .

of inertia of ring about a tangent in a plane

∴

I = \frac{1}{2} m R^{2} + \frac{3}{2} m R^{2} + \frac{3}{2} m R^{2} = \frac{7}{2} m R^{2}

Question 9.

Two circular discs A and B are of equal masses and thickness but made of metals with densities $d_{A}$ and $d_{B} (d_{A} > d_{B}) .$ If their moments of inertia about an axis passing through centres and normal to the circular faces be $I_{A}$ and $I_{B}$ , then