System of Particles and Rotational Motion Test

Result

System of Particles and Rotational Motion Test - 38

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

A bicycle wheel rolls without slipping on a horizontal floor. Which one of the following is true about the motion of points on the rim of the wheel, relative to the axis at the wheel's centre?

A
Points near the top move faster than points near the bottom

B
Points near the bottom move faster than points near the top

C
All points on the rim move with the same speed

D
All points have the velocity vectors that are pointing in the radial direction towards the centre of the wheel

Solution

We have, $$V_A=2v\sin\theta /2$$
Hence, velocity of point on rim increases with $$\theta$$ for $$0^o < \theta < 180^o$$ and decreases with $$\theta$$ for $$180^o < \theta < 30^o$$.
Question 2
1 / -0

Directions For Questions

Consider a cross made up of two thin cylindrical rods (each of mass m and length l) joined at their middle point. The frame is in the plane of paper and axis (1), (2), (3), (4) are also in the plane of paper whereas axis (5) is passing through the joint and perpendicular to the plane of figure.

...view full instructions

MI of frame about axis '1' will be

A
$$\dfrac{mL^2}{12}$$

B
$$\dfrac{mL^2}{6}$$

C
$$\dfrac{mL^2}{24}$$

D
$$\dfrac{mL^2}{6\sqrt{2}}$$

Solution
Question 3
1 / -0

Two bodies of mass $$10kg$$ and $$2kg$$ are moving with velocities $$2i-7j+3k$$ and $$-10i+35j-3k$$ $$m/s$$ respectively. The velocity of their CM is

A
$$2i$$ m/s

B
$$2k$$ m/s

C
$$(2i+2k)$$ m/s

D
$$(2i+2j+2k)$$ m/s

Solution

Velocity of CM = $$\dfrac{m_1v_1+m_2v_2}{m_1+m_2}=\dfrac{10(2i-7j+3k)+2(-10+35-3k)}{10+2}=2k$$ $$m/s$$
Question 4
1 / -0

Two particles A and B initially at rest, move towards each other under a mutual force of attraction. At the instant when the speed of A is v and speed of B is 2v, the speed of centre of mass of the system is :

A
zero

B
v

C
1.5 v

D
3 v

Solution

Under mutual force of attraction, the centre of mass will be at rest position.
Question 5
1 / -0

A particle moves in a circle in such a way that kinetic energy is directly proportional to $${ R }^{ -n }$$ and the angular momentum of the particle is independent of radius, then the value of n is:

A
1

B
2

C
-1

D
-2

Solution
Question 6
1 / -0

A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see figure). There is negligible friction at the pivot. The free end is held vertically above the pivot and then released. The angular acceleration of the rod when it makes an angle $$\theta$$ with vertical is

A
$$\dfrac { 3g }{ 2l } \cos { \theta }$$

B
$$\dfrac { 2g }{ 3l } \cos { \theta }$$

C
$$\dfrac { 3g }{ 2l } \sin { \theta }$$

D
$$\dfrac { 2g }{ 3l } \sin { \theta }$$

Solution

$$By\quad taking\quad torque\quad equation\quad about\quad A\\ \tau =\vec F\times\vec r ;\\\quad \quad \tau =I\alpha \\ Mgsin\theta \quad \times \quad l/2\quad =M{ l }^{ 2 }/3\quad \times \quad \alpha \\ \alpha =3gsin\theta /2l$$
Question 7
1 / -0

An automobile engine develops $$100$$ $$kW$$ when rotating at a speed of $$1800\ rev/min$$. The torque it delivers is

A
$$3.33\ N-m$$

B
$$200\ N-m$$

C
$$530.5\ N-m$$

D
$$2487\ N-m$$

Solution

$$\textbf{Step 1: Equation of power}$$
Work done by torque is given by$$;$$ $$W=\tau.\theta$$

Differentiating with respect to time
$$\Rightarrow\ \ \text{Power, } P = \dfrac{dW}{dt}=\tau.\dfrac{d\theta}{dt}=\tau.w$$ $$....(1)$$
where $$w$$ is angular velocity

$$\textbf{Step 2: Calculation of angular velocity}$$
$$1$$ RPM $$\Rightarrow \dfrac{2\pi}{60}\,rad/s$$

$$\Rightarrow 1800$$ RPM $$=\dfrac{2\pi}{60}\times 1800=60\,\pi \,rad/s=w$$

$$\textbf{Step 3: Calculation of power}$$
From eq $$....(1):$$
$$P=\tau.w$$
$$\Rightarrow\ \ 100\,KW=\tau(60\pi)$$

$$\Rightarrow\ \ \tau =\dfrac{100000}{60\pi}=530.5\,Nm$$

Hence option $$C$$ is correct.
Question 8
1 / -0

What would happen if the Earth did not rotate on its axis?

A
All winds would blow curved and deflected at equator only.

B
All winds would not blow curved and deflected at poles only.

C
All winds would blow straight from high pressure to low pressure area and die when the pressure equalised.

D
All winds would blow straight from low pressure to high pressure area and die when the pressure is equalised.

Solution

If the earth stopped spinning suddenly, the atmosphere would still be in motion with the earth original 1100 mile per hour rotation speed at the equator. All of the land masses would be scoured clean of anything not attached to bedrock.
Question 9
1 / -0

The moment of inertia of a metre stick of mass $$300\ g$$, about an axis at right angles to the stick and located at $$30\ cm$$ mark, is

A
$$8.3\times10^5 g\ cm^2$$

B
$$5.8 g\ cm^2$$

C
$$3.7\times10^5 g\ cm^2$$

D
$$None\ of\ these$$

Solution

The moment of inertia , I, of a stick along right angle at the mid-point = Moment of inertia of the rod along right angle at the mid-point
$$I=\dfrac {ML^2}{12}$$
In the given situation we will apply parallel axis theorem, as it is about an axis at right angles to the stick,
Therefore, $$I=\dfrac {ML^2}{12}+MR^2$$
And as per the given criteria,
$$M=300g$$
$$L=1m=100cm$$
$$R=\dfrac L2-30=20cm$$
Substituting these values, we get
$$I=\dfrac {300\times 100\times 100}{12}+300\times 20\times 20\\\implies I=370000=3.7\times 10^5 g/cm^2$$
is the required moment of inertia
Question 10
1 / -0

Consider a ring rolling down a smooth inclined plane of vertical height 'h' and inclination $$\theta$$. Then the true statement in the following is?

A
Acceleration along the plane is g $$\sin\theta$$ and the potential energy at the topmost point is mgh

B
Acceleration along the plane is g and the potential energy at the top most point is mgh

C
Acceleration along the plane is g $$\sin\theta $$ and the potential energy at the top most point as mgh $$\sin\theta$$

D
None of these

Solution

Acceleration along the plane is $$g\sin \theta$$ and the potential energy at the topmost point is $$mgh$$.