System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 44

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Question 1
1 / -0

Three identical masses each of mass $$1kg$$, are placed at the corners of an equilateral triangle of side $$l$$. Then the moment of inertia of this system about an axis along one side of the triangle is

A
3$$Ml^2$$

B
$$l^2$$

C
$$\dfrac{3}{4}$$$$Ml^2$$

D
$$\dfrac{3}{2}$$ $$Ml^2$$

Solution
Question 2
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A thin rod of linear pass density $$ \lambda $$ at right angle at its mid point $$(C)$$ and fixed to points $$A$$ and $$B$$ such that it can rotate about an axis passing through $$AB$$ .The moment of inertia about an axis passing through $$AB$$ is :

A
$$ \dfrac { \lambda l^3} {6 \sqrt2} $$

B
$$ \dfrac { \lambda l^3} {2 \sqrt2} $$

C
$$ \dfrac { \lambda l^3}{4} $$

D
$$ \dfrac { \lambda l^3}{ \sqrt2} $$

Solution

We have length of each rod $$=\dfrac{1}{\sqrt 2}$$
Let a small part $$dx$$ be at a distance $$x$$ from the of first rod it's small mass $$dm=\lambda dx$$ (Linear density of rod $$= \lambda$$)
distance of dm from axis $$=x \cos 45^o=\dfrac{x}{\sqrt 2}$$
So moment of interia of $$dm$$ about the axis is :
$$dI=dm \left(\dfrac{x}{\sqrt 2}\right)^2=\dfrac{\lambda x^2dx}{2}$$
So $$I=\displaystyle \int^{\dfrac{L}{\sqrt {2}}}_{0}\dfrac{\lambda x^2dx}{2}=\dfrac{\lambda}{2}\dfrac{x^3}{3}^{\dfrac{L}{\sqrt 2}}_0=\dfrac{\lambda L ^3}{3 \times 4\sqrt{2}}$$
Now the second rod will also have same moment of inertia about the axis
Hence net moment of inertia $$=2 \dfrac{\lambda L^3}{3 \times 4 \sqrt 2}=\dfrac{\lambda L^3}{6 \sqrt 2}$$
Question 3
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A smooth horizontal rod passes through two identical rings, each of mass $$m$$. Rings are connected to a block of same mass through two strings of length $$2l$$ and $$l$$ as shown in figure. Block is released, when block is crossing lowest point its velocity is $$2\ m/s$$, velocity of left ring is

A
$$4 \ m/s$$

B
$$2 \ m/s$$

C
$$3 \ m/s$$

D
$$None\ of\ these$$
Question 4
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In applying the equation for motion with uniform angular acceleration $$\omega =\omega _0+\alpha t$$ the radian measure

A
Must be used for both $$\omega $$and $$\alpha$$

B
May be used for both $$\omega $$ and $$\alpha$$ `

C
may be used for $$\omega $$ but not $$\alpha$$

D
cannot be used for both $$\omega $$ and $$\alpha$$

Solution
Question 5
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A solid sphere and a hollow sphere have same mass and radius. Ratio of radius of gyration of the solid sphere to that of the hollow sphere about a tangent axis is:

A
$$\sqrt {\frac{7}{3}} $$

B
$$\sqrt {\frac{3}{5}} $$

C
$$\sqrt {\frac{5}{3}} $$

D
$$\sqrt {\frac{{21}}{5}} $$

Solution

For hollow sphere, movement of inertia $$ = \frac{2}{3}M{R^2}$$

$$M{K^{2\,}}\, = \frac{2}{3}\,M{K^{2\,}}$$

or $$K = \,\sqrt {\frac{2}{3}R} $$

For solid sphere, movement of inertia $$ = \frac{2}{5}M{R^2}$$

$$M{K_1}^{2\,}\, = \frac{2}{5}\,M{K^{2\,}}$$

or $${K_1} = \,\sqrt {\frac{2}{5}R} $$

Now, $$\frac{{{K_1}}}{K}\, = \sqrt {\frac{3}{5}} $$
Question 6
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Find the moment of inertia of the triangular lamina of mass M about the axis of rotation AA' shown in the figure:

A
$$\frac{Ml^2}{3}$$

B
$$\frac{Ml^2}{6}$$

C
$$\frac{2}{3} Ml^2$$

D
$$\frac{1}{3} Ml^2$$

Solution

mass per unit area = $$ M \times \dfrac{al}{2}$$ where a= base which is AA' =l(given)
M.O.I $$I = \dfrac{1}{2}\times \dfrac{M}{2}al\times \dfrac{1}{2}axdx$$
where $$x$$ is te small strip along 'l' from the base a(AA')
so $$M.O.I = \dfrac{1}{2}\times \dfrac{M}{l} \int x^2dx$$ =$$ \dfrac{M}{2l}[\dfrac{x^3}{3}] = \dfrac{M}{2l}[\dfrac{l^3}{3}] = \dfrac{Ml^2}{6}$$
Question 7
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A body of mass $$5\ kg$$ is acted on by a net force $$F$$ which varies with time $$t$$ as shown in graph. Then the net momentum in $$SI$$ units gained by the body at the end of $$10$$ seconds is

A
$$0$$

B
$$100$$

C
$$140$$

D
$$200$$

Solution

We know that the Newton's second law of motion is $$F=\dfrac{\Delta p}{\Delta t}$$ or change in momentum
or the $$gain$$ in momentum is $$\Delta p=F\Delta t=\text{Area under the F-t graph}$$
The net area till the end of $$t=10s$$ is consists of a trapezium with parallel side as $$10second$$ and $$8-4=4second$$
and the separation being $$20N$$ so area or momentum is $$\dfrac{sum\times separation}{2}=\dfrac{(10+4)second\times 20N}{2}=140Newtonsecond$$
Option c is correct.
Question 8
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Two blocks of masses $$2kg$$ and $$1kg$$ respectively are tied to the ends of a strings which passes over a light frictionless pulley. The masses are held at rest at the same horizontal level and then released. The distance traversed by centre of mass in $$2 $$ seconds is:($$g=10m/s^2$$)

A
$$1.42m$$

B
$$2.22m$$

C
$$3.12m$$

D
$$3.33m$$

Solution

Given,

Two blocks having masses $$2kg,1kg,t=2s$$

So, $$a=\dfrac{(2-1)mg}{(2+1)m}=\dfrac{g}{3} a_{CM}=\dfrac{2\times\dfrac{g}{3}-1\dfrac{g}{3}}{2+1}=\dfrac{g}{9}$$

So, $$S=ut +\dfrac{1}{2}at^2=\dfrac{1}{2}\times\dfrac{g}{2}\times4=\dfrac{20}{9}=2.22m$$
Question 9
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A triangular loop of side $$l$$ carries a current $$I$$. It is placed in a magnetic field $$B$$ such that the plane of the loop in the direction of $$B$$. The torque on the loop is:

A
Zero

B
$$IBl$$

C
$$\dfrac{\sqrt 3}{2} Il^2B^2$$

D
$$\dfrac{\sqrt 3}{4} B l^2$$

Solution

Since $$q = {90^0}$$
So$$,$$ $$\tau = NIAB = 1 \times I \times \left( {\frac{{\sqrt 3 }}{4}{l^2}} \right)B = \frac{{\sqrt 3 }}{4}{l^2}B$$
Hence,
option $$(D)$$ is correct answer.
Question 10
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A ballet dancer, dancing on a smooth floor is spinning about a vertical axis with her arms folded with angular velocity of $$20\ rad/s$$. When the stretches her arms fully, the spinning speed decreases in $$10\ rad/s$$/. If $$I$$ is the initial moment of inertia of the dancer, the new moment of inertia is

A
$$21$$

B
$$31$$

C
$$I/2$$

D
$$I/3$$

Solution

Here, angular momentum is conserved.

Initial angular momentum = Final angular momentum

$$I\times 20= I'\times 10$$

where, $$I'$$ is the new moment of inertia.

Then,

$$I'=2I$$