System of Particles and Rotational Motion Test

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System of Particles and Rotational Motion Test - 48

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Question 1
1 / -0

Four uniform thin rods each of mass 1 kg and length 1 m are joined in the form of a square. If the square is rotated about axis AB, then it moment of inertia is

A
$$\frac{5}{3}{\text{kg}}\;{{\text{m}}^{\text{2}}}$$

B
$$\frac{3}{4}{\text{kg}}\;{{\text{m}}^{\text{2}}}$$

C
$$\frac{5}{4}{\text{kg}}\;{{\text{m}}^{\text{2}}}$$

D
$$\frac{7}{4}{\text{kg}}\;{{\text{m}}^{\text{2}}}$$

Solution
Question 2
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Mass of thin long metal rod is $$2$$ kg and its moment of inertia about an axis perpendicular to the length of the rod and passing through its one end is $$0.5kg{m^2}$$. Its radius of gyration is:

A
$$25$$ cm

B
$$40$$ cm

C
$$50$$ cm

D
$$1$$ m

Solution

Solution we know that,
$$ I_{end} = \dfrac{ML^{2}}{3} = \dfrac{1}{2} $$ [given]
$$ I_{com} = \dfrac{ML^{2}}{12} = \dfrac{1}{4}(\dfrac{ML^{2}}{3}) = \dfrac{1}{8} $$
Also,
radius of gyration $$ = \sqrt{\dfrac{I_{com}}{m}} $$
$$ = \sqrt{\dfrac{1}{8\times 2}} = \dfrac{1}{\sqrt{16}} = 25cm $$
Question 3
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The linear density of a thin rod of length $$1.0$$m varies as $$\lambda =2$$kg/m$$+\left(\dfrac{2kg}{m^2}\right)x$$, where x is the distance from its one end. The distance of its centre of mass from its end is?

A
$$\dfrac{2}{3}$$m

B
$$\dfrac{5}{9}$$m

C
$$\dfrac{4}{3}$$m

D
$$\dfrac{1}{2}$$m

Solution

REF.Image
$$\lambda =2\dfrac{kg}{meter}+\dfrac{2kg}{meter^{2}}.x$$
$$\lambda =(2+2x) kg/m$$ ; x is in meter
suppose there is an element of length dx
& at a distance x from left must end.
mass of this element
is $$dm = \lambda dx = (2+2x)dx$$
$$\Rightarrow $$ mass of rod = sum of marvels of these elements
$$\displaystyle M= \int_{x=0}^{1}dm = \int_{0}^{1}(2+2x)dx= [2x+\frac{2x}{2}]_{0}^{1}$$
$$m= 3 kg$$
$$\displaystyle x_{ c.m} = \frac{\displaystyle \int_{0}^{1}x\, dm}{\displaystyle \int dm}= \frac{\displaystyle \int_{0}^{1}(2+2x)xdx}{\displaystyle \int_{0}^{1}(2+2x)dx}$$
$$\displaystyle x_{ cm}= \frac{\displaystyle \int_{0}^{1}(2x+2x^{2})}{3}dx=\frac{[\dfrac{2x^{2}}{2}+\dfrac{2x^{3}}{3}]_{0}^{1}}{3}$$
$$x_{cm}=\dfrac{1+\dfrac{2}{3}}{3}$$ meter = $$\dfrac{5}{9}$$ meter
from are end (having less density)
Question 4
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A body of mass $$3$$ kg is thrown from ground with a speed $$10$$ m/s at an angle $$53^0$$ with horizontal. At the highest point of its path it is fragmented into three identical pieces such that one of them comes to rest. Velocity of the centre of mass just after fragmentation is

A
$$10$$ m/s

B
$$5$$ m/s

C
$$6$$ m/s

D
$$8$$ m/s

Solution
Question 5
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Moment of a inertia of a sphere about its diameter is $$2/5\ MR^{2}$$. What its moment of inertia about an axis perpendicular to its two diameter and passing through their point of intersection.?

A
$$I=\dfrac{2}{5}Mr^{2}$$

B
$$I=\dfrac{3}{5}Mr^{2}$$

C
$$I=\dfrac{4}{5}Mr^{2}$$

D
$$I=\dfrac{5}{5}Mr^{2}$$

Solution

moment of inertia of sphere about its diameter:$$I=\dfrac{2MR^{2}}{5}$$
two diameter always inetersect at the centre.
however,a sphere in symmetric from all directions,
so taking any other diameter axis will have moment,
$$I=\dfrac{2}{5}MR^{2}$$ (i.e. same)
Question 6
1 / -0

The wheel of radius $$r= 300 mm$$ rolls to the right without slipping and has a velocity $$v_0= 3 m/s$$ of its center O. The speed of the point A on the wheel for the instant represented in the figure is:-

A
$$4.36 $$ m/s

B
$$5 $$ m/s

C
$$3 $$ m/s

D
$$1.5$$ m/s

Solution

$$V_0=\omega r_0$$
$$3m/s=\omega\times 300\,mm$$
$$3m/s=\omega\times 0.3m$$
$$10\,rad/s=\omega$$
$$v=\omega r=2m/s$$
$$v_0=3m/s$$
$$V_{net}=\sqrt{v^2+v^2_0+2v\,v_0\cos\theta}$$
$$=\sqrt{4+9+2\times 2\times 3\times \dfrac{1}{2}}$$
$$=\sqrt{13+6}$$
$$=\sqrt{19}\approx 4.36$$
Question 7
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Find the M.I of rod about (i) an axis perpendicular to the rod and passing through left end (ii)An axis through its centre of mass and perpendicular to the length whose linear density varies as $$\lambda=ax$$ where a is a positive constant and $$'x'$$ is the position of an element of the rod relative to its left end.The length of the rod is $$l$$

A
$$\dfrac{al^4}{4}$$

B
$$\dfrac{al^3}{4}$$

C
$$\dfrac{al^4}{2}$$

D
$$\dfrac{al^3}{6}$$
Question 8
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Find the torque of a force $$\vec{F}=-3\hat{i}+\hat{j}+5 \hat{k}$$ acting at the point $$\vec{r}=7\hat{i}+3\hat{j}+\hat{k}$$ with respect to origin:-

A
$$14 \hat{i}-32 \hat{j}-2 \hat{k}$$

B
$$4 \hat{i}+4 \hat{j}+6 \hat{k}$$

C
$$-14 \hat{i}+38 \hat{j}-16 \hat{k}$$

D
$$-21\hat{i}+3 \hat{j}+5\hat{k}$$

Solution

$$\vec{\tau}=\vec{r}\times \vec{F}=(7\hat{i}+3\hat{j}+\hat{k})(-3\hat{i}+\hat{j}+5\hat{k})$$

$$\vec{\tau}=\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 7 & 3 & 1 \\ -3 & 1 & 5 \end{vmatrix} =14\hat{i}-38\hat{j}+16\hat{k}$$
Question 9
1 / -0

A bullet of mass m is fired at a certain angle $$\theta$$b with the horizontal. The bullet returns to ground after time t. Then the change in momentum of the bullet is

A
zero

B
$$ 2mu sin \theta$$

C
$$\dfrac{mgt cos \theta}{2}$$

D
mgt

Solution

$$\Delta D=mV_y-mv_y\\ \quad=m(\mu\sin\theta-(-u\sin\theta))\\ \quad=m(2u\sin\theta)\\ \quad=2mu\sin\theta$$
Question 10
1 / -0

In Figure a spherical part of radius $$\frac { R } { 2 }$$ is removed from a bigger solid sphere of radius $$R$$ . Assuming uniform mass distribution, a shift in the centre of mass will be:

A
$$\frac {- R } { 6 }$$

B
$$-\frac { R } { 14 }$$

C
$$\frac { R } { 9 }$$

D
$$\frac { R } { 3 }$$

Solution

Let the mass of bigger sphere be $$M$$ with radius $$R$$. Then the mass of smaller sphere with radius $$\left( \dfrac{R}{2} \right) $$ is $$\dfrac{M}{4}$$
Centre of the big sphere which is $$C.O.M$$ be the origin $$O. \ C.O.M$$ of small sphere is at distance $$\dfrac{R}{2}$$
cut out $$\rightarrow C.M$$ of the remaining portion shift to $$P$$. mass of remaining portion $$=\dfrac{3m}{4}$$ from conservation of $$C.O.M$$
$$C.M$$ of big sphere $$+C.M$$ of small $$\dfrac{3M}{4} \times (-OP) = M \times OO+ \dfrac{M}{4} \times \dfrac{R}{2}$$
$$\Rightarrow OP= -\dfrac{R}{6}$$