Laws of Motion Test

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Laws of Motion Test - 69

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Question 1
1 / -0

If a particle of mass $$m$$ resting at the top of a fixed smooth hemisphere of radius $$R$$ is displaced, then the angle with vertical at which it looses contact with the surface is:

A
$$\tan^{-1}\left (\dfrac {1}{3}\right )$$

B
$$\cos^{-1}\left (\dfrac {1}{3}\right )$$

C
$$\tan^{-1}\left (\dfrac {2}{3}\right )$$

D
$$\cos^{-1}\left (\dfrac {2}{3}\right )$$

Solution
Question 2
1 / -0

Directions For Questions

Two identical beads $$A$$ and $$B$$ can move along a vertical ring without friction. The mass of each bead is $$m$$. The ring is fixed. The beads are connected by a thin rod of length $$\sqrt {2}R$$ where $$R$$ is radius of the ring. At time $$t = 0$$, the bead $$A$$ is at the lowest position of the ring. The beads are given a sharp impulse to follow a vertical circle, the separation between them remains same in the consequent motion. Neglect the mass of the rod. $$(g =$$ acceleration due to gravity).

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If the beads are released from rest, such that the bead $$A$$ is at the highest position of ring at time $$t = 0$$, the magnitude of acceleration of the bead $$B$$ in the vertical direction at the same instant is

A
Zero

B
$$\dfrac {g}{4}$$

C
$$\dfrac {g}{2}$$

D
$$g$$

Solution

For A that is rest at top
$$T\cos { 45° } =mg\\ T=\sqrt { 2 } mg$$
at B.
$$T\sin { 45° } -mg=m0\\ \sqrt { 2 } mg\times \dfrac { 1 }{ \sqrt { 2 } } -mg=mg\\ 0=mg\\ g=O$$
Question 3
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A mass tied to a string moves in a vertical circle and at the point $$P$$ its speed is $$5m/s$$. At the point $$P$$ the string breaks. The mass will reaches height above $$P$$ of nearly $$\left( g=10m/{ s }^{ 2 } \right) $$

A
$$1m$$

B
$$0.5m$$

C
$$1.27m$$

D
$$1.25m$$

Solution
Question 4
1 / -0

A particle is released on a vertical smooth semicircular track from point X so that OX makes angle $$\theta$$ form the vertical (see figure). The normal reaction of the track on the particle vanishes at point Y where OY makes angle $$\phi$$ with the horizontal. Then:

A
$$\sin {\phi} = \cos {\theta}$$

B
$$\sin{\phi} = \dfrac{1}{2}\cos{\theta}$$

C
$$\sin {\phi} = \dfrac{2}{3}\cos {\theta}$$

D
$$\sin {\phi} = \dfrac {3}{4}\cos {\theta}$$

Solution

When the particle is at $$X$$:
$$N-mg\cos\theta=\cfrac{mV^2}{R}\quad$$ (equation $$1$$)
Since the particle moves along the circle.
When the particle is at $$Y$$ (Just before it leaves the circular track):
$$N-mg\sin\phi=\cfrac{mV^2}{R}\quad$$ (equation $$2$$)
From $$1$$ and $$2$$:
$$\cos\theta=\sin\phi$$
Question 5
1 / -0

A uniform rod of length $$L$$ rests against a smooth wall as shown in figure. Find the friction coefficient between the ground and the lower end if the minimum angle that the rod can make with the horizontal is $$\theta$$.

A
$$\dfrac{L\cos^2\theta \sin\theta}{2h-l\cos\theta \sin^2\theta}$$

B
$$\dfrac{L\cos\theta \sin^2\theta}{2h-l\cos\theta \sin^2\theta}$$

C
$$\dfrac{L\cos^2\theta \sin\theta}{2h-l\cos^2\theta \sin\theta}$$

D
$$\dfrac{L\cos\theta \sin^2\theta}{2h-l\cos^2\theta \sin\theta}$$

Solution

Figure(b) gives the forces .
$$R_2=$$Reactional force on rod by ground
$$R_1=$$ Reactional force on rod by wall
$$f=$$ frictional force $$=\mu R_2$$
$$\theta$$ is the maximum angle the rod can make with horizontal.
Equating the forces on verticle direction we get
$$R_2=mg-R_1cos\theta--------(1)$$
Equating the forces on horizontal direction we get
$$R_1 sin\theta = f =\mu R_2$$
Now the torque about point B should be balanced
$$R_1 cos\theta \times AB + R_1 Sin\theta \times h = mg\cfrac{l}{2}cos\theta$$
$$\implies R_1 h[\cfrac{cos^2\theta +sin^2\theta}{sin\theta}]=\cfrac{mgl}{2}cos\theta$$
$$\implies R_1=\cfrac{mgl}{2h} cos\theta sin\theta$$
$$\implies R_1 cos\theta = \cfrac{mgl}{2h}cos^2\theta sin\theta$$
So, $$R_2 = mg-\cfrac{mgl}{2h}cos^2\theta sin\theta = \cfrac{mg}{2h} (2h-lcos^2\theta sin\theta)$$
And $$\mu = \cfrac{R_1 sin\theta}{R_2} = \cfrac{\cfrac{mgh}{2h} cos\theta sin^2\theta}{\cfrac{mg}{2h} (2h-lcos^2\theta sin^2 \theta)}$$
$$\mu =\cfrac{lcos\theta sin^2 \theta}{2h-lcos^2\theta sin\theta}$$
Question 6
1 / -0

A particle of mass 'm' strikes a smooth stationary wedge of mass M with a velocity $$v_0$$, at an angle $$\theta$$ with horizontal if the collision is perfectly inelastic, the impulse on the wedge is:

A
$$\dfrac{Mmv_0}{M+m}$$

B
$$Mv_0$$

C
$$\dfrac{Mmv_0cos\theta}{M+m}$$

D
$$\dfrac{Mmv_0sin\theta}{M+m}$$

Solution

Initial momentum along x axis $$= mv_0 \cos\theta$$

Final momentum $$= (M + M)v$$

$$v = \cfrac{mv_0\cos\theta}{M + m}$$

Impulse = Change in momentum
$$ = \cfrac{Mmv_0\cos\theta}{M + m}$$
Question 7
1 / -0

Two blocks are connected by a massless string that passes over a frictionless peg as shown in fig. One end of the string is attached to a mass $$m_1 = 3kg$$, i.e. a distance $$R = 1.20 m$$ from the peg. The other end of the string is connected to a block of mass $$m_2 = 6 kg$$ resting on a table. When the 3 kg block be released at $$\displaystyle \theta = \frac{\pi}{k}$$, the 6 kg block just lift off the table? Find the value of k.

A
$$2$$

B
$$3$$

C
$$4$$

D
$$6$$

Solution

Tension in the string will be maximum when $$m_1$$ is at lowest position. For $$m_2$$ to just lift, tension in the string at this position should equal the weight of $$m_2$$ at this position.

Using conservation of energy,
$$\dfrac{m_1 R^2 \omega^2}{2} = m_1 g R (1-\cos \theta) $$
$$m_1 \omega^2 R = 2m_1 g (1 - \cos \theta) \quad ........(i) $$

From the FBD of $$m_1$$,
$$T_1 - m_1g = m_1 \omega^2 R \quad ...........(ii)$$

From $$(i)\ \& \ (ii),$$
$$T_1 = m_1g + 2m_1g (1-\cos \theta)$$
$$T_1 = m_1g (3-2\cos\theta) \quad .........(iii)$$

From the FBD of $$m_2$$,
$$T_2 + N_2 = m_2 g \quad ........(iv)$$
When the block just leaves the surface,
$$N_2 = 0 \quad .....(v)$$
Also, tension throughout the string is constant,
$$T_2=T_1 \quad ........(vi) $$

Using $$(iii), \ (iv), \ (v) \ \& \ (vi),$$
$$m_2g = m_1g(3-2cos\theta)$$
$$\cos \theta = \dfrac{3 - m_2/m_1}{2}$$
Substituting values,
$$\cos \theta = 1/2$$
$$\theta = \pi/6$$

$$\implies k = 6$$
Question 8
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Two masses $$A$$ and $$B$$ of $$10Kg$$ and $$5Kg$$ respectively are connected with a string passing over a frictionless pulley fixed at the corner of a table as in figure. The coefficient of friction of $$A$$ with the table is $$0.2$$. The minimum mass of $$C$$ that may be placed on $$A$$ to prevent it from moving is equal to:

A
$$20Kg$$

B
$$15Kg$$

C
$$10Kg$$

D
$$5Kg$$

E
$$0Kg$$

Solution
Question 9
1 / -0

A $$"V"$$ shaped rigid body has two identical uniform arms. What must be the angle between the two arms so that when the body is hung from one end, the other arm is horizontal?

A
$$\cos^{-1}(1/3)$$

B
$$\cos^{-1}(1/2)$$

C
$$\cos^{-1}(1/4)$$

D
$$\cos^{-1}(1/6)$$

Solution
Question 10
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A stone of mass $$1\ kg$$ tied to a light inextensible string of length $$L = \dfrac{10}{3}\ m$$, whirling in a circular path in a vertical plane. The ratio of maximum tension in the string to the minimum tension in the string is $$4$$. If g is taken to be $$10\ m/s^2$$ the speed of the stone at the highest point of the circle is

A
$$10\ m/s$$

B
$$5\sqrt{2}\ m/s$$

C
$$10\sqrt{3}\ m/s$$

D
$$20\ m/s$$

Solution

$$The\quad ratio\quad to\quad the\quad mamimum\quad { T }_{ 2 }{ \quad to\quad the\quad minimum\quad tension\quad T }_{ 1 }\quad is\quad -\\ \frac { { T }_{ 2 } }{ { T }_{ 1 } } =4\\ { T }_{ 2 }=4{ T }_{ 1 }\\ Now\quad the\quad difference\quad between\quad the\quad two\quad 6mg;\\ Hence\quad we\quad have\quad \\ { T }_{ 2 }-{ T }_{ 1 }=6mg\\ \therefore 4{ T }_{ 1 }-{ T }_{ 1 }=6mg\\ 3{ T }_{ 1 }=6mg\\ { T }_{ 1 }=2mg\\ Now\quad tension\quad at\quad the\quad top\quad of\quad the\quad circle\quad is-\\ { T }_{ 1+mg }=\frac { { mv }_{ 1 }^{ 2 } }{ r } =\frac { { mv }_{ 1 }^{ 2 } }{ L } \\ 2mg+mg=\frac { { mv }_{ 1 }^{ 2 } }{ \frac { 10 }{ 3 } } \\ { v }_{ 1 }^{ 2 }=3g\frac { 10 }{ 3 } =10g\\ { v }_{ 1 }=\sqrt { 10g } \\ =\sqrt { 10\times 10 } =10m/s$$