Laws of Motion Test

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Laws of Motion Test - 68

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Question 1
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The vertical section of a road over a canal bridge in the direction of its length is in the form of circle of radius $$8.9\;m$$. Then the greatest speed at which the car can cross this bridge without losing contact with the road at its highest point, the centre of gravity of the car being at a height $$h=1.1\;m$$ from the ground (Take $$g=10m/sec^2$$)

A
$$\;5\;m/sec$$

B
$$\;10\;m/sec$$

C
$$\;15\;m/sec$$

D
$$\;20\;m/sec$$

Solution

Given : Radius of the circular arc $$r =8.9$$ m $$g = 10 m/s^2$$
Heigh of centre of gravity of car above the road $$h =1.1$$ m
Centre of gravity of the car is revolving in a circular motion.
$$\therefore$$ Distance of centre of gravity of car and the centre of circle $$l = r+h = 8.9 + 1.1 =10$$ m
Using $$\dfrac{mv^2}{l} = mg - N$$
For maximum speed of car $$(v_{max})$$, $$\implies N = 0$$
$$\therefore$$ $$\dfrac{mv_{max}^2}{l} = mg $$

OR $$\dfrac{m v^2 _{max}}{10} = m \times 10$$ $$\implies v_{max} =10$$ $$m/s$$
Question 2
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A body crosses the topmost point of a vertical circle with critical speed. What will be its centripetal acceleration when the string is horizontal. :-

A
$$\;g$$

B
$$\;2g$$

C
$$\;3g$$

D
$$\;6g$$

Solution

The critical speed of the body at the topmost point is equal to $$\sqrt{gR}$$ i.e $$v = \sqrt{gR}$$
Let the velocity of the body when the string is horizontal be $$v_2$$
Using work-energy theorem : $$W = \Delta K.E$$
$$\therefore$$ $$mgR = \dfrac{1}{2} m v_2^2 - \dfrac{1}{2}mv^2$$
OR $$mgR = \dfrac{1}{2} m v_2^2 - \dfrac{1}{2}m(gR)$$ $$\implies v_2^2 = 3gR$$
Centripetal acceleration $$a_r = \dfrac{v_2^2}{R} = 3g$$
Question 3
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A metal bar $$70\ cm$$ long and $$4.00\ kg$$ in mass supported on two knife-edges placed $$10\ cm$$ from each end. A $$6.00\ kg$$ load is suspended at $$30\ cm$$ from one end. Find the reactions at the knife-edges. (Assume the bar to be of uniform cross section and homogeneous.)

A
$$45\ N$$ and $$43\ N$$

B
$$50\ N$$ and $$35\ N$$

C
$$55\ N$$ and $$43\ N$$

D
$$54\ N$$ and $$30\ N$$

Solution

Figure shows the rod AB, the positions of the knife edges $$K_1$$ and $$K_2$$, the centre of gravity of the rod at G and the suspended load at P.
Note the weight of the rod W acts as its centre of gravity G. The rod is uniform in cross section and homogenous$$;$$ hence G is at the centre of the rod$$; AB=70$$cm. $$AG=35$$cm, $$AP=30$$cm, $$PG=5$$cm, $$AK_1=BK_2=10$$cm and $$K_1G=K_2G=25$$cm. Also, $$W=$$ weight of the rod $$=4.00$$kg and $$W_1=$$ suspended load$$=6.00$$kg $$;$$ $$R_1$$ and $$R_2$$ are the normal reactions of the support at the knife edges.
For translational equilibrium of the rod, $$R_1+R_2-W_1-W=0$$ ....$$(i)$$
Note $$W_1$$ and $$W$$ act vertically down and $$R_1$$ and $$R_2$$ act vertically up.
For considering rotational equilibrium, we take moments of the forces. A convenient point to take moments of the forces is G. The moments of $$R_2$$ and $$W_1$$ are anticlockwise $$(+ve)$$, whereas the moment of $$R_1$$ is clockwise $$(-ve)$$.
For rotational equilibrium,
$$-R_1(K_1G)+W_1(PG)+R_2(K_2G)=0$$ $$(ii)$$
It is given that $$W=4.00$$g N and $$W_1=6.00$$g N, where $$g=$$ acceleration due to gravity. We take $$g=9.8m/s^2$$.
With numerical values inserted, from $$(i)$$
$$R_1+R_2-4.00g-6.00g=0$$
or $$R_1+R_2=10.00g N$$ $$(iii)$$
$$=98.00N$$
From $$(ii), -0.25 R_1+0.05W_1+0.25R_2=0$$
or $$R_1-R_2=1.2g$$ $$N=11.76N$$ $$(iv)$$
From $$(iii)$$ and $$(iv)$$, $$R_1=54.88N$$,
$$R_2=43.12N$$
Thus the reactions of the support are about $$55N$$ at $$K_1$$ and $$43N$$ at $$K_2$$.
Question 4
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An object is said to be in equilibrium when:
1. there is no net force acting on the object
2. the total clockwise moments about any point is equal to the total anti-clockwise moments about the same point
3. the object moves with constant velocity

A
$$1$$ only

B
$$2$$ only

C
$$1$$ and $$2$$ only

D
$$1, 2$$ and $$3$$

Solution

A body is said to be in equilibrium if it in translational equilibrium as well as rotational equilibrium.
For translational equilibrium, net force acting on the body must be zero i.e. $$F_{net} = 0$$
For rotational equilibrium, net torque acting on the body must also be zero or in other words, total clockwise moments about any point is equal to the total anti-clockwise moments about the same point.
Question 5
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A weightless thread can bear tension upto $$37 N$$. A stone of mass $$500 g$$ is tied to it and revolved in a circular path of radius $$4 m$$ in a vertical plane. If $$g=10 {ms}^{-2}$$, then the maximum angular velocity of the stone will be:

A
$$2\ rad\ {s}^{-1}$$

B
$$4\ rad\ {s}^{-1}$$

C
$$8\ rad\ {s}^{-1}$$

D
$$16\ rad\ {s}^{-1}$$

Solution

Maximum tension in the string will be at the lowest point of the circular path, where the tension will support the weight as well as provide the centripetal force for circular motion.

So, we have:

$$T=mg+ml{ \omega }^{ 2 }$$

$$\Rightarrow ml{ \omega }^{ 2 }=T-5$$

For maximum angular speed, the tension must be $$37N$$

So, the maximum angular speed is $$\sqrt { \dfrac { T-5 }{ ml } } =\sqrt { \dfrac { 37-5 }{ 0.5\times 4 } } = 4 rad/s$$

Thus, $${ \omega }_{ max }= 4 rad/s $$
Question 6
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Directions For Questions

These questions consist of two statement, each printed as assertion and reason. While answering these question you are required to choose anyone of the following five responses.

...view full instructions

Assertion: A quick collision between two bodies is more violent than a slow collision; even when the initial and final velocities are identical.
Reason: The momentum is greater in first case

A
If both assertion and reason are true but the reason is the correct explanation of assertion.

B
If both assertion and reason are true but the reason is not the correct explanation of assertion.

C
If assertion is true but reason is false.

D
If both the assertion and reason are false.

E
If reason is true but assertion is false.

Solution

Force = Change in momentum/time taken.
Here, as mentioned, the initial and final velocities are same in both cases. Hence, change in momentum is same. The only deciding factor it time taken.
When lesser time is taken, the force is larger (though the impulse remains same). Thus, quiker collisions seem more violent as more force is exerted.
Assertion is true, but reason is false.(Force in first case is greater, not momentum)
Question 7
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A massless meter stick is in equilibrium about its center point.
If a mass, $${m}_{1}=50 g$$, is placed at the $$30 cm$$ mark and a mass, $${m}_{2}=75 g$$, is placed at the $$70 cm$$ mark, where must a mass, $${m}_{3}=50 g$$, be placed to keep the system in equilibrium?

A
$$10 cm$$

B
$$40 cm$$

C
$$50 cm$$

D
$$60 cm$$

E
$$90 cm$$

Solution

Given : $$m_1 = 50$$ gram $$m_2= 75$$ gram $$m_3 = 50$$ gram
As the meter stick is in rotational equilibrium, thus net torque about point C must be equal to zero i.e $$\tau_c = 0$$
$$\therefore$$ $$m_1 g (20) + m_3 g (x) - m_2 g (20) =0$$

OR $$50 \times 20 + 50 x - 75 \times 20 = 0$$ $$\implies x =10$$ cm
Hence the mass $$m_3$$ must be palced at position $$X_3 = 50 - 10 = 40$$ cm
Question 8
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A car taken a turn along a curved road of radius $$40$$ m and the angle of banking of the road is $$\displaystyle {45} ^ {o} $$. At what maximum speed can the car be driven so that it may not skid. $$\displaystyle (g\ = {10\ ms} ^{-2} $$).

A
$$\displaystyle {20\ ms} ^{-1} $$

B
$$\displaystyle {10\ ms} ^{-1} $$

C
$$\displaystyle {5\ ms} ^{-1} $$

D
$$\displaystyle {1\ ms} ^{-1} $$

Solution

For banking of road:
$$V^2\le rg\tan \theta$$
$$\implies V \le \sqrt{rg\tan \theta}$$
$$\implies V\le \sqrt{40\times 10\times \tan 45^o}$$
$$\implies V\le 20$$
$$\implies V_{max}=20$$ $${m/s}$$
Question 9
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A small sphere is given vertical velocity of magnitude $$v_{0} = 5m/s$$ and it swings in a vertical plane about the end of massless string. The angle $$\theta$$ with the vertical at which string will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere, is $$[g = 10 m/s^{2}]$$.

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

B
$$\cos^{-1}\left (\dfrac {1}{4}\right )$$

C
$$60^{\circ}$$

D
$$30^{\circ}$$

Solution
Question 10
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A uniform meter sticks of mass 200 g is suspended from the ceiling through two verticle strings of equal lengths fixed at the ends. A small object of 20 g is placed on the stick at a distance of 70 cm from the left end. Find the tension in the two strings.

A
$$T_1=1.06N, T_2=1.14N$$

B
$$T_1=1.03N, T_2=1.17N$$

C
$$T_1=1.10N, T_2=1.10N$$

D
$$T_1=1.00N, T_2=1.20N$$

Solution

C is the centre of mass of the stick.
For translation equilibrium
$$T_1 + T_2 = 220g--------(1)$$
Here,g is acceleration due to gravity $$=1000cm/s^2$$
For rotation equilibrium torque about centre of mass shall be zero or sum of clockwise torque is equal to sum anti clockwise torque.
So, AC=CB=50cm
$$T_1 \times 50 + 20g \times (70-50)=T_2\times 50$$
$$\implies T_2 -T_1 = 8g---------(2)$$
So,from (1) & (2) we get
$$T_2=114g=1.14\times 10^5 dyne =1.14N$$
$$T_1=106g=1.06\times 10^5 dyne = 1.06N$$