Laws of Motion Test

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Laws of Motion Test - 67

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Question 1
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The bob of the pendulum shown in figure describes an arc of circle in a vertical plane. If the tension in the cord is $$2.5\ times$$ the weight of the bob for the position shown. Find the velocity and the acceleration of the bob in that position.

A
$$\displaystyle 16.75ms^{-1}, 5.66ms^{-2}$$

B
$$\displaystyle 5.66ms^{-1}, 16.75ms^{-2}$$

C
$$\displaystyle 2.88ms^{-1}, 16.75ms^{-2}$$

D
$$\displaystyle 5.66ms^{-1}, 8.34ms^{-2}$$

Solution

Let the mass of the bob be $$m$$ and the velocity of the at point B be $$v$$
Given : $$r =2$$ m $$T =2.5 mg$$
Using circular motion equation at B : $$\dfrac{mv^2}{r} = T - mg cos30^o$$
$$\therefore$$ $$\dfrac{mv^2}{2} = 2.5 mg - mg \times 0.866$$
OR $$v^2 = 3.27 g =3.27 \times 9.8 = 32 $$
$$\implies v =5.66 $$ m/s
$$\therefore$$ Radial acceleration $$a_r = \dfrac{v^2}{r} = \dfrac{32}{2} =16$$ $$m/s^2$$
Also tangential acceleration $$a_t = g sin 30^o =9.8 \times 0.5 = 4.9$$ $$m/s^2$$
Total acceleration $$a = \sqrt{a_r^2 + a_t^2} = \sqrt{16^2 + (4.9)^2} = 16.75$$ $$m/s^2$$
Question 2
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The simple $$2 kg$$ pendulum is released from rest in the horizontal position. As it reaches the bottom position, the cord wraps around the smooth fixed pin at $$B$$ and continues in the smaller are in the vertical plane. Calculate the magnitude of the force $$R$$ supported by the pin at $$B$$ when the pendulum passes the position $$\displaystyle \theta = 30^{\circ}.\left ( g= 9.8m/s^{2} \right )$$

A
$$15 N$$

B
$$30 N$$

C
$$45 N$$

D
$$60 N$$

Solution

using energy conservation
$$\Rightarrow ({ K.E.) }_{ E }+({ P.E.) }_{ E }=({ K.E.) }_{ C }+({ P.E.) }_{ C }$$
$$\Rightarrow 0+2g\times 800\times { 10 }^{ -3 }=\frac { 1 }{ 2 } \times 2\times { { v }_{ 1 }^{ 2 } }+0$$
$$\Rightarrow v_{ 1 }=4m/s$$
again using energy conservation at point $$C$$ and $$D$$
$$\Rightarrow ({ K.E.) }_{ C }+({ P.E.) }_{ C }=({ K.E.) }_{ D }+({ P.E.) }_{ D }$$
$$\Rightarrow \frac { 1 }{ 2 } \times 2\times { { v }_{ 1 }^{ 2 } }+0=\frac { 1 }{ 2 } \times 2\times { { v }_{ 2 }^{ 2 }+ }2g \times 400(1-\cos { 30° } )\times { 10 }^{ -3 }$$
$$\Rightarrow { v }_{ 2 }^{ 2 }=8+4\sqrt { 3 } $$
From the diagram, at point $$D$$
$$ \Rightarrow T=mg\cos { \theta } +\frac { m{ v }_{ 2 }^{ 2 } }{ R } \\ \Rightarrow T=2\times 10\times \cos { 30° } +\frac { 2\times (8+\sqrt { 3 } ) }{ 400\times { 10 }^{ -3 } } $$
$$\Rightarrow T=90.2N$$
Question 3
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A car is travelling along a circular curve that has a radius of $$50$$ m. If its speed is $$16$$ m/s and is incrcasing uniformly at $$\displaystyle 8m/s^{2}.$$ Determine the magnitude of its acceleration at this instant.

A
$$\displaystyle 7.5 ms^{-2}$$

B
$$\displaystyle 8.5 ms^{-2}$$

C
$$\displaystyle 9.5 ms^{-2}$$

D
$$\displaystyle 9.8 ms^{-2}$$

Solution

Given : $$r =50$$ m $$a_t =8$$ $$m/s^2$$ $$v =16 $$ $$m/s$$
Centripetal acceleration $$a_r = \dfrac{v^2}{r} = \dfrac{(16)^2}{50} =5.12$$ $$m/s^2$$

$$\therefore$$ Total acceleration $$a =\sqrt{a_r^2 + a_t^2} = \sqrt{(5.12)^2 + 8^2} =\sqrt{90.2} = 9.5$$ $$m/s^2$$
Question 4
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A uniform rod of length L is placed symmetrically on two walls as shown in the figure. The rod is in equilibrium. If $$N_1$$ and $$N_2$$ are the normal forces exerted by the walls on the rod then

A
$$N_1$$ < $$N_2$$

B
$$N_1$$ > $$N_2$$

C
$$N_1$$ = $$N_2$$

D
$$N_1$$ and $$N_2$$ would be in the vertical directions

Solution

$$N_{1} \times \dfrac{l}{4}=N_{2} \times \dfrac{l}{4}$$
$$N_{1}=N_{2}$$
(by torque conservation)
Correct option is (C).
Question 5
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The mass of the bob of a simple pendulum of length $$L$$ is $$m$$. If the bob is left from its horizontal position then the speedof the bob and the tension in the thread in the lowest position of the bob will be respectively:

A
$$\;\sqrt{2gL}\;and\;3\;mg$$

B
$$\;3\;mg\;and \sqrt{2gL}$$

C
$$\;2\;mg\;and\;\sqrt{2gL}$$

D
$$\;2\;gL\;and\;3\;mg$$

Solution

Velocity of the bob at point A is zero, thus its kinetic energy at A is zero.
Let the velocity of the bob at point B be $$v$$ and tension in the string be $$T$$.
Using work-energy theorem from A to B : $$W = \Delta K.E$$
$$\therefore $$ $$mgL = \dfrac{1}{2} mv^2 - 0$$ $$\implies v =\sqrt{ 2gL}$$

Circular motion equation at B : $$\dfrac{mv^2}{L} = T - mg$$
$$\therefore$$ $$T = \dfrac{mv^2}{L} + mg = \dfrac{m (2gL)}{L} + mg = 3mg$$
Question 6
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A weightless thread can bear tension upto $$3.7\;kg$$ wt. A stone of mass $$500\;gm$$ is tied to it and revolved in a circular path of radius $$4m$$ in a vertical plane. If $$g=10\;ms^{-2}$$, then the maximum angular velocity of the stone will be:

A
$$\;16\;rad/s$$

B
$$\;\sqrt{21}\;rad/s$$

C
$$\;2\;rad/s$$

D
$$\;4\;rad/s$$

Solution

Given : $$T_{max} =3.7 kg wt = 37$$ N $$m =500$$ gm $$= 0.5 kg$$ $$r =4$$ m $$g = 10 m/s^2$$
Tension in the string is maximum when the stone is at the lowest position of the vertical motion.
Let the maximum angular velocity be $$w_{max}$$.
Using circular motion equation : $$mrw^2_{max} =T_{max} - mg$$
$$\therefore$$ $$0.5 \times 4 \times w^2_{max} =37 - 0.5 \times 10$$
OR $$w^2_{max} =16$$ $$\implies w_{max} = 4$$ $$rad/s$$
Question 7
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Stone tied at one end of light string is whirled round a vertical circle. If the difference between the maximum and minimum tension experienced by the string wire is $$2\ kg\ wt$$, then the mass of the stone must be

A
$$1\ kg$$

B
$$6\ kg$$

C
$$1/3\ kg$$

D
$$2\ kg$$

Solution

$$\textbf{Step 1: FBD [Ref. Fig.]}$$
Tension will be maximum at bottom and minimum at top

$$\textbf{Step 2: Velocities at highest and lowest points}$$
Minimum velocity required at bottom point to complete circle $$= v$$
$$v = \sqrt{5gR}$$ $$....(1)$$

$$\textbf{Step 3: Calculations }$$
For revolving in complete circle tension at highest point $$T\geq 0$$
$$\Rightarrow T_{\min} = 0$$

Applying Newton's second law on the stone along centripetal direction (Considering direction towards center as positive)
$$\Sigma F_c = ma_c$$ $$=m\left (\cfrac{v^2}{R} \right)$$
Using equation $$(1)$$ in the following

$$\Rightarrow T_{max} - mg = \dfrac{mv^2}{R} = 5 mg$$

$$\Rightarrow T_{\max} = 6 mg$$ [T is maximum at bottom]

So, $$ T_{\max} - T_{\min} = 6mg$$
$$\Rightarrow 6mg = 2g$$ Given: $$ T_{\max} - T_{\min} = 2g$$

$$\Rightarrow m = \dfrac{2}{6} = \dfrac{1}{3} kg$$

Hence $$C$$ is the correct option.
Question 8
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If a particle of mass $$M$$ is tied to a light inextensible string fixed at point $$P$$ and particle is projected at A with velocity $$V_A\, =\, \sqrt{4 gL}$$ as shown. Find tension in the string at $$B$$. (Assume particle is projected in the vertical plane.)

A
$$Mg$$

B
$$3Mg$$

C
$$2Mg$$

D
$$7Mg$$

Solution

Taking point A as reference where gravitational potential energy is zero,

Total energy of particle at point A is, $$E_A = \dfrac{1}{2}MV_A^2 = 2MgL$$

Let velocity at point B as $$V_B$$.
So, Total energy at point B is, $$E_B = mgL + \dfrac{1}{2}MV_B^2$$
By conservation of energy, $$E_A=E_B$$
$$\implies \dfrac{1}{2}MV_B^2 = mgL\implies \dfrac{MV_B^2}{L}=2Mg$$

So centripetal force point B is $$2Mg$$
At point B, $$T+Mg=\dfrac{MV_B^2}{L}=2Mg\implies T=Mg$$

So Tension in the string is $$Mg$$, at point B.
Question 9
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A uniform rod of mass m and length L is suspended with two massless strings as shown in the figure. If the rod is at rest in a horizontal position the ratio of tension in the two strings $$T_1$$/$$T_2$$ is

A
1 : 1

B
1 : 2

C
2 : 1

D
4 : 3

Solution

Vertical equilibrium $$\sum F_Y=0$$,
$$T_1 + T_2 = w$$ .........1
Taking Moments about point 'O',
Clockwise moments = Anticlockwise Moments;
$$T_1 \times \dfrac{3L}{4}= w \times \dfrac{L}{2}$$
$$\therefore T_1= \dfrac{2w}{3}$$ .......2
Solving equations 1 and 2,
$$ \Rightarrow T_2= \dfrac {w}{3}$$,
$$\therefore \dfrac{T_1}{T_2}= \dfrac{2}{1}$$
Question 10
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Two uniform rods of equal length but different masses are rigidly joined to form an L-shaped body, which is then pivoted about O as shown in the figure. If the system is in equilibrium in the shown configuration. Then the ratio M/m will be:

A
$$2$$

B
$$3$$

C
$$\sqrt{2}$$

D
$$\sqrt{3}$$

Solution

Taking Moments about point 'O'
clockwise moments = Anticlockwise moments,
$$Mg \times \dfrac{l}{2} sin(30^o)$$ = $$mg \times \dfrac{l}{2} sin(60^o)$$ ,

$$\Rightarrow \dfrac{M}{2}= m \dfrac{\sqrt3}{2}$$

$$\therefore \dfrac{M}{m}= \sqrt{3}$$