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CBSE Questions for Class 11 Engineering Physics Laws Of Motion Quiz 6 - MCQExams.com
CBSE
Class 11 Engineering Physics
Laws Of Motion
Quiz 6
A plane banks its wings $${30}^{o}$$ relative to the horizontal to enter into a circular turn. The circular path has a radius of $$1500 m$$.
Which of these values best represents the velocity of the plane if the lift force exerted on the wings is equal to twice the weight of the plane?
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$$120 {m}/{s}$$
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$$140 {m}/{s}$$
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$$160 {m}/{s}$$
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$$240 {m}/{s}$$
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$$320 {m}/{s}$$
Explanation
In the flight of a plane the angle of banking with vertical is given by ,
$$\tan \theta=v^{2}/rg$$
given $$\theta=90-30=60^{o} , r=1500m$$
therefore $$v^{2}=rg\tan60$$
or $$v^{2}=1500\times9.8\times\sqrt3=25460.4$$
or $$v=160m/s$$
A mass of 2 kg at rest travels for 4 seconds with an acceleration of 1.5 m $$s^{-2}$$. Find the gain of the momentum of the body.
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5 kg m $$s^{-1}$$
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10 kg m $$s^{-1}$$
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12 kg m $$s^{-1}$$
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14 kg m $$s^{-1}$$
Explanation
$$v=u+at=1.5\times 4=6ms^{-1}$$
Gain in momentum=$$mv=2\times 6=12kgms^{-1}$$
To avoid slipping while walking on ice, one should take smaller steps because:
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frictional force of ice is large
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of larger normal reaction
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frictional force of ice is small
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of smaller normal reaction
Explanation
Hint :- To avoid slipping friction should be sufficient.
Explanation
:-
$$\bullet$$To avoid slipping one should take smaller steps because the friction coefficient of ice is small. So we didn’t get enough friction by the ice.
$$\bullet$$Smaller steps will give larger normal force and more the normal force will give more friction and hence we get sufficient friction to avoid slipping.
$$\textbf{Hence option C correct}$$
A cricket ball of mass 500 g is moving with a speed of 36 km $$h^{-1}$$. It is reflected with the same speed. What is the impulse applied on it?
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20 kg m $$s^{-1}$$
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5 kg m $$s^{-1}$$
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25 kg m $$s^{-1}$$
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10 kg m $$s^{-1}$$
Explanation
Initial velocity of the ball $$u = -36 \ km/h =- 36\times \dfrac{5}{18} = -10 \ m/s$$
Initial momentum $$P_i = mu = -0.5\times 10 = -5 \ kg \ m/s$$
Final velocity of the ball $$v = 36 \ km/h = 36\times \dfrac{5}{18} = 10 \ m/s$$
Initial momentum $$P_f = mv = 0.5\times 10 = 5 \ kg \ m/s$$
Impulse on the ball $$I = P_f - P_i = 5-(-5) = 10 \ kg \ m/s$$
Due to an impulse, the change in the momentum of a body is 1.8 kg m $$s^{-1}$$. If the duration of the impulse is 0.2 s, then what is the force produced in it?
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9 N
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8 N
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7 N
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6 N
Explanation
Impulse=$$\dfrac{F}{\Delta t}=\dfrac{1.8}{0.2}=9N$$
For the plank depicted above, a man who weighs $$\dfrac{1}{8}$$ the weight of the plank stands halfway between the pivot and the center of mass of the plank.
If the plank weighs $$500 N$$ and $$\theta={30}^{o}$$, which of the following values best represents the tension in the cable?
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$$500 N$$
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$$471 N$$
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$$531 N$$
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$$1800 N$$
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$$1500 N$$
Explanation
Given : $$F_w = 500$$ N
As the plank is in rotational equilibrium, thus net torque acting on the plank about the point O is zero i.e $$\tau_o = 0$$
$$\therefore$$ $$F_t sin\theta \times L - F_w \times \dfrac{L}{2} - \dfrac{F_w}{8} \times \dfrac{L}{4} = 0$$
$$\implies $$ $$\dfrac{17}{32}F_w =F_t$$ $$sin\theta$$
OR $$\dfrac{17}{32} \times 500 = F_t \times 0.5$$ $$\implies F_t \approx 531$$ N
China wares are wrapped in straw or paper before packing. What is the basis of this application?
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Impulse
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Momentum
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Acceleration
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Force
Explanation
Because the sharp impulse given to objects may cause damage.
While dusting a carpet, we give it a sudden jerk or beat it with a stick because:
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the inertia of rest keeps the dust in its position and the dust is removed when the carpet moves away.
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the inertia of motion removes dust.
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no inertia involved in process.
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jerk compensates for the force of adhesion between dust and carpet and dust is removed.
Explanation
While dusting we give a sudden serk our beat with stick because when we do that stick because when we do that the particle of carpet move with our help and the dust particle stays at rest by inertia and eventually they separate from the carpet.
A pushing force acts on the trolley as shown. Which arrow shows the frictional force acting on the wheel X?
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A
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B
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C
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D
Explanation
As w.k.t friction always oppose the motion.
Hence, cart move right then friction act on wheel left along -B
Match the entries in Column I with those in Column II.
Column - I
Column - II
a
Force
1
N $$m^{-1}$$
b
Weight
2
kg
c
Pressure
3
N
d
Impulse
4
N s
e
Mass
5
kg wt
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a-3,b-2,c-4, d-1, e-5
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a-5,b-4,c-3, d-1, e-2
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a-3,b-5,c-1, d-4, e-2
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a-4,b-2,c-1, d-3, e-5
Explanation
Si unit of force is Newton denoted by $$N$$, that of weight is $$kg \ wt$$, pressure is $$N m^{-2}$$, impulse is $$N \ s$$ and mass is $$kg$$.
Thus option C is correct.
A cricket ball of mass 500 g is moving with speed of $$36 \,km\,h^{-1}$$. It is reflected back with the same speed. What is the impulse applied on it?
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20 Kg m/s
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10 Kg m/s
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5 Kg m/s
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none of the above
Explanation
Impulse $$I=$$ change in momentum $$=mv_2-mv_1$$
Here, $$v_1=-v_2=36 km/h$$
So, $$I=(500/1000)[-36-36]=-36 kg.m/s$$
The magnitude of impulse $$=36 kg.m/s$$
A $$1\ m$$ long uniform beam is balanced as shown. Calculate the force F.
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$$2\ N$$
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$$3\ N$$
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$$7\ N$$
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$$11\ N$$
Explanation
Moment of force $$ M=$$ force $$\times $$ perpendicular distance from pivot
For the balanced beam, the net moment of force on left side of pivot is equal to the net moment of force on right side.
Thus, $$5N(50-10 )cm= F(60 -50) cm+ (3N)(80-50)cm$$
or $$20=F+9$$ or $$F=11 N$$
The impulse of a body is equal to:
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Rate of change of its momentum.
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Change in its momentum.
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The product of force applied on it and the time of application of the force.
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Both (B) and (C)
Explanation
Impulse $$I=Fdt$$
Also force $$F=m\dfrac{dv}{dt}$$
So, $$I=m\dfrac{dv}{dt}dt=mdv=m(v_2-v_1)=mv_2-mv_1=p_2-p_1=$$ change in momentum
Thus, both B and C are correct.
A cricket ball of mass 120 g is moving with a velocity of 12 m $$s^{-1}$$. It turns back and moves with a velocity of 20 m $$s^{-1}$$ when hit by a bat. Find the impulse if the force acts on the ball for 0.02 s.
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4 N s
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8 N s
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32 N s
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192 N s
Explanation
Initial velocity of ball $$u = -12 \ m/s$$
Final velocity of ball $$v = 20 \ m/s$$
Magnitude of change in momentum $$|\Delta P| = m (v-u) = 0.12\times (20-(-12)) =3.84 \ kg \ m/s $$
Time of action $$\Delta t = 0.02 \ s$$
Thus impulsive force $$F = \dfrac{\Delta P}{\Delta t} = \dfrac{3.84}{0.02} = 192 \ N$$
State whether true or false.
Newton's first law of motion gives the concept of momentum.
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True
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False
Explanation
No the given statement is false as newton first law of motion gives the concept of inertia while the concept of momentum was given by second law.
Momentum is a measure of quantity of _______.
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motion
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friction
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gravitation
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electric force
Explanation
Momentum is a measure of the quantity of motion possessed by an object by virtue of its mass and velocity.
Massive objects have high momentum than lighter objects and fast-moving objects have high momentum than slow moving objects.
So, option $$A$$ is the correct answer.
Which of the following statements are true?
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Impulse is the product of force and time for which the force acts.
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SI unit of impulse is N s.
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When two bodies have the same velocity; the lighter body has more momentum.
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Conservation of linear momentum has no connection with "Newton's third law of motion".
Explanation
If force $$(F)$$ is applied for time $$ t$$, the impulse of the force will be $$I=Ft$$
As the SI unit of force is $$N$$ and unit of time is second (s), so the SI unit of impulse will be N-s.
Momentum of a body of mass m is given by, $$p=mv$$
If velocity $$(v)$$ is constant, $$p\propto m$$
Thus, the heavier body has more momentum.
Newton's second law is the consequence of conservation of linear momentum.
A hammer of mass 5 kg moving with a speed of $$2ms^{−1}$$ strikes the head of a nail driving it 20 cm into the wall. Find the impulse(in N s).
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0
0%
-100
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-10
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-10000
Explanation
Impulse = change in momentum
Impulse = final momentum - initial momentum
= $$0$$ - $$(5kg\times 2ms^{-1})$$
= $$-10kgms^{-1}$$
= $$-10Ns^{-1}$$
An aeroplane executes a horizontal loop at a speed of 720 kmph with its wings banked at $$45^o$$. What is the radius of the loop?
Take g = 10 $$ms^{-2}$$.
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4 km
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4.5 km
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7.2 km
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2 km
Explanation
Given : $$v = 720$$ km/h $$ = 720\times \dfrac{5}{18} = 200$$ $$m/s$$
Angle of banking $$\theta = 45^o$$
Radius of loop $$r = \dfrac{v^2}{g\tan\theta}$$
$$\therefore$$
$$r = \dfrac{200^2}{10\times 1} = 4000$$ m $$(\because \tan45^o =1)$$
$$\implies$$ Radius of loop $$r = 4$$ km
When a body is stationary, then:
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there is no force acting on it.
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the body is in vacuum.
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the forces acting on it are not in contact with it.
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the net forces acting on it balance each other.
Explanation
If the net external force acting on the body is zero, the body either remains at rest or move with constant velocity. Hence, if the given body is stationary, the net forces acting on it balance each other.
The time, in which a force of 2 N produces a change in momentum of $$0.4\, kg\, m\, s^{-1}$$ in a body whose mass is 1 kg is:
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0.2 s
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0.02 s
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0.5 s
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0.05 s
Explanation
Impulse $$I=Ft$$
also $$I=$$change in momentum $$=0.4 kg m/s$$
So, $$0.4=(2)t $$ or $$t=0.2 s$$
A simple pendulum of length $$L$$ carries a bob of mass $$m$$. When the bob is at its lowest position, it is given the minimum horizontal speed necessary for it to move in a vertical in circle about the point of suspension. When the string is horizontal the net force on the bob is
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$$\sqrt {10}mg$$
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$$\sqrt {5}mg$$
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$$4 mg$$
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$$1 mg$$
Explanation
From Newton's third law, Action and reaction____.
Which of the following will not fit in the above sentence.
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acts on the same body
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are equal
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are opposite in direction
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acts on different bodies
Explanation
According to newton's third law of motion, every action has an equal and opposite reaction. The action-reaction pair always acts on different bodies.
So option A is wrong and rest three options are correct.
On which one of the following conservation laws, does a rocket work?
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Mass
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Energy
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Linear momentum
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Angular momentum
Explanation
Rocket works on the principle of conservation of linear momentum.
Three charged particles are collinear and are in equilibrium, then
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all the charged particle have the same poarity
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the equilibrium is unstable
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all the charged particles cannot have the same polarity
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both (b) and (c) are correct.
Explanation
If a charged particle is in equilibrium under electrostatic force, then that particle never be in stable equilibrium.
Hence option B is correct.
Since all charges are in equilibrium, net force an any one of the charges are in equilibrium,so net force an any one of charges should be zero-for this two equal and opposite forces will act on any charge.
Hence option C is correct.
A small stone tied to an inextensible string of negligible mass is rotated in a circle of radius $$2 m$$ in a vertical plane. Find the speed at a horizontal point on the circle. (in $$m/s$$)
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$$7.67$$
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$$2.36$$
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$$6.32$$
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$$8.32$$
Explanation
$$ v= \sqrt {3gr}$$
$$ v= \sqrt {3 * 9.8* 2} = 7.67 m/s $$
A stone of mass $$10 kg$$ tied with a string of length $$0.5 m$$ is rotated in a vertical circle. Find the total
energy of the stone at the highest position. (in $$J$$)
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$$123.36$$
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$$122.5$$
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$$154.3$$
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$$185.6$$
Explanation
assume that potential energy at lowest point is zero and stone is rotating with critical velocity
so the magnitude of velocity at highest point is $$v=\sqrt { gl } $$ where $$l=$$length of string
total energy $$E=K.E.+P.E.$$
$$\Rightarrow K.E.=\dfrac { 1 }{ 2 } m{ v }^{ 2 }=\dfrac { 1 }{ 2 } 10\times { (\sqrt { 9.8\times 0.5 } ) }^{ 2 }=24.5J$$
$$P.E.=mgh=mg(2l)=10\times 9.8\times \left( 2\times 0.5 \right) =98J$$
so the total energy $$E=98+24.5=122.5J$$
A car executes a turn of radius $$22 m$$ on a banked road while travelling at a speed of $$45 km/h$$. If the height of the outer edge above the inner edge of the road is $$1.1 m$$, what is the breadth of the road? (in $$m$$)
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$$1.745$$
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$$1.569$$
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$$1.875$$
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$$1.236$$
Explanation
given that $$R=22m$$, $$v=45km/h=12.5m/s$$
as the diagram shows, we have to find the value of $$L$$, we can see there is a right triangle $$\Delta OQP$$ so
$$\Rightarrow { L }=\sqrt { \left( { a }^{ 2 }+1.1^{ 2 } \right) }...(1) $$
and $$\tan { \theta } =\dfrac { 1.1 }{ a } \Rightarrow a=\dfrac { 1.1 }{ \tan { \theta } } ...(2)$$
balancing the force in car we gets two equations
$$\Rightarrow N\cos { \theta } =mg...(3)$$
$$\Rightarrow N\sin { \theta } =\dfrac { m{ v }^{ 2 } }{ R } ...(4)$$
by dividing eqn(3) by eqn(4)
$$\Rightarrow \tan { \theta } =\dfrac { { v }^{ 2 } }{ gR } $$ now putting the value of $$\tan{\theta}$$ in eqn(2)
$$\Rightarrow a =\dfrac { 1.1\times gR }{ { v }^{ 2 } } =\dfrac { 1.1\times 10\times 22 }{ { 12.5 }^{ 2 } } =1.548m$$
now again putting the value of $$a$$ in eqn(1)
$${ L }=\sqrt { \left( { 1.5488 }^{ 2 }+1.1^{ 2 } \right) } =1.875m$$
A motorcyclist executes a horizontal loop at a speed of $$65km/h$$ while himself making an angle of $$12^o$$ with the horizontal. What is the radius of the loop? (in $$m$$)
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$$123.6$$
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$$156.6$$
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$$156.3$$
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$$174.3$$
Explanation
as the diagram shows
$$N\cos { \theta } $$ provides it centripetal force and $$N\sin { \theta } $$ balance the weight($$mg$$)
$$\Rightarrow N\cos { \theta } =mg...(1)$$ and $$N\sin { \theta } =\dfrac { m{ v }^{ 2 } }{ R }...(2)$$
dividing eqn(1) by eqn(2)
$$\Rightarrow \tan { \theta } =\dfrac { { v }^{ 2 } }{ gR } $$ or
$$R=\dfrac { { v }^{ 2 } }{ g\tan { \theta } } $$
given $$v=65km/h=18.056m/s$$, $$\tan { \theta } =0.2126$$
$$\Rightarrow R=\dfrac { { 18.056 }^{ 2 } }{ 9.8\times 0.2126 } =156.6m$$
A $$200 g$$ mass is whirled in a vertical circle making $$60$$ revolutions per minute. What is the tension in the string at the top of the circle if the radius of the circle is $$0.8 m$$? (in $$N$$)
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$$6.3$$
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$$2.36$$
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$$4.35$$
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$$4.23$$
Explanation
$$ w = 60 \space \text {revolutions per minute }= 60 * 2\pi / 60 \space\space\text {rad/ sec} = 2\pi \space\space rad/ sec $$
$$ 200 gm = 0.2 kg $$
$$ T = mrw^2 - mg $$
$$ T = 0.2* 0.8 * (2\pi)^2 - 0.2* 9.8 $$
$$ T = 4.35 N$$
The angle of banking of a turn of radius $$75 m$$ on a road is $$30^o$$. What is the speed at which a car can turn along this curve? (in $$m/s$$)
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$$26.3$$
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$$63.32$$
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$$21.3$$
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$$20.6$$
Explanation
$$ v = \sqrt {r gtan\theta }$$
$$ v= \sqrt {75* 9.8* tan 30}$$
$$ v= 20.6 m/s $$
Mark correct option or options.
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The body of greater mass needs more forces to move due to more inertia
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Force versus time graph gives impulse
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Microscopic area of contact is about $$10^{-4}$$ times actual area of the contact
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All of the above
Explanation
When the mass of the body is higher, then it needs more force to move because of inertia.
Impulse is defined as force per unit time.
Microscopic area of contact is about $$10^{-4}$$ times actual area of the contact,
It is known facts.
Therefore all the given statements are correct.
So the correct option is $$D$$
The centre of gravity of a car is $$0.62 m$$ above the ground. It can turn along a track which is $$1.24 m$$ wide and has radius $$r$$. If the greatest speed at which the car can take the turn is $$22.02 m/s$$, What is the value of $$r$$? (in $$m$$)
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$$49.48$$
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$$58.36$$
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$$21.36$$
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$$96.36$$
Explanation
assume that critical velocity is $$v$$, if speed of car is more than $$v$$ then car will topple due to torque of centrifugal force $${ F }_{ c }$$ around point $$P$$ , so balancing the torque at point $$P$$
$${ \Rightarrow F }_{ c }\times 0.62=mg\times 0.62\\ \Rightarrow \dfrac { m{ v }^{ 2 } }{ R } \times 0.62=mg\times 0.62\\ \Rightarrow \dfrac { m{ v }^{ 2 } }{ R } =mg\\ \Rightarrow R=\dfrac { { v }^{ 2 } }{ g } =\dfrac { { 22.02 }^{ 2 } }{ 10 } =49.48m$$
Railway tracks are banked on curves
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necessary centrifugal force may be obtained from the horizontal component weight of the train
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to avoid frictional force between the tracks and wheels
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necessary centripetal force may be obtained from the horizontal component of the weight of the train
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the train may not fly off in the opposite direction
Explanation
The banking is provided to counteract the centrifugal force which acts (away from the center of the curve) on the train. Here component of weight is towards center providing necessary centripetal force as shown in diagram.
A $$500 g$$ particle tied to one end of a string is whirled in a vertical circle of circumference $$14 m$$.If the tension at the highest point of its path is $$2 N$$, what is its speed? (in m/s)
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$$4.365$$
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$$5.369$$
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$$5.546$$
0%
$$4.331$$
Explanation
as the diagram shows
balancing the force at point $$P$$
given that $$T=2N$$, $$m=0.5kg$$ and $$R=\dfrac { 14 }{ 2\pi }=2.23m $$
$$\Rightarrow T+mg=m\dfrac { { v }^{ 2 } }{ R } $$
$$\Rightarrow v=\sqrt { R\dfrac { (T+mg) }{ m } } $$
$$\Rightarrow v=\sqrt { 2.23\times \dfrac { (2+0.5\times 10) }{ 0.5 } } =5.546m/s$$
Keeping the banking angle same to increase the maximum speed with which a vehicle can travel on the curved road by $$10\%$$, the radius of curvature of the road has to be changed from $$20\ m$$ to-
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$$16\ m$$
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$$18\ m$$
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$$24.2\ m$$
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$$30.5\ m$$
Explanation
For speed v, we have
$$\displaystyle \tan { \theta } =\frac { { v }^{ 2 } }{ rg } =\frac { { v }^{ 2 } }{ 20g } $$......(i)
If $$\displaystyle { v }^{ \prime }=v+0.1 v=1.1v$$, then
$$\displaystyle \tan { \theta } =\frac { { v }^{ \prime 2 } }{ { r }^{ \prime }g } =\frac { { \left( 1.1v \right) }^{ 2 } }{ { r }^{ \prime }g } $$....(ii)
Equating (1) and (ii), we get $$\displaystyle \quad \frac { 1 }{ 20 } =\frac { 1.21 }{ { r }^{ \prime } } $$
$$\displaystyle \Rightarrow { r }^{ \prime }=20\times 1.21=24.2m$$
A particle is moving in a vertical circle. The tension in the string when passing through two platform at angles $$\displaystyle 30^o$$ and $$\displaystyle 60^o$$ vertical (lowest position) are $$T_1$$ and $$T_2$$ respectively
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$$\displaystyle T_1 = T_2$$
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$$\displaystyle T_2 > T_1$$
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$$\displaystyle T_1 > T_2$$
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Tension in the string always remain the same
Explanation
Tension, $$\displaystyle T= \frac{mv^2}{r} + mg cos \theta$$
For $$\displaystyle \theta =30^o$$
$$\displaystyle T_1= \frac{mv^2}{r} + mg cos 30^o$$
$$\displaystyle = \frac{mv^2}{r} + \frac{\sqrt{3}}{2} mg$$
For $$\displaystyle \theta =60^o$$
$$\displaystyle T_2= \frac{mv^2}{r} + mg cos 60^o$$
$$\therefore$$ $$\displaystyle \frac{mv^2}{r} + \frac{1}{2} mg$$
$$T_1> T_2$$
A stone of mass m is tied to a string and is moved in a vertical circle of radius r making n revolution per minute. The total tension in the string when the stone is at its lowest point is
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$$mg$$
0%
$$m(g+\pi n r^2)$$
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$$m(g+ n r)$$
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$$m(g+ \frac{\pi^2 n^2 r} {900})$$
Explanation
Tension in string when it reach lower-most point
$$\displaystyle T=mg+m \omega^2 r$$
$$\displaystyle =m(g+4 \pi^2 n^2 r)$$ [as $$\omega=2 \pi n$$]
$$\displaystyle = m \left( g + 4 \pi^2 \left( \frac{n}{60} \right)^2 r \right)$$
$$\displaystyle = m \left( g + \left( \frac{\pi^2 n^2 r}{900} \right) \right)$$
A parachutist with total weight $$75kg$$ drops vertically onto a sandy ground with a speed of $$2m{ s }^{ -1 }$$ and comes to a halt over a distance of $$0.25m$$. The average force from the ground on her is close to
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$$600N$$
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$$1200N$$
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$$1350N$$
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$$1950N$$
Explanation
Initial velocity while dropping on the ground u=2m/s.
Parachutist was brought to rest within s=0.25 m
Therefore retardation f : $$u^2=2fs; \; f=\frac{u^2}{2s}=8m/s^2$$
Averarage force from ground on her close is F=mf=600 N
A car moves at a speed of $$20\ ms^{-1}$$ on a banked track and describes an arc of a circle of radius $$40\sqrt {3}$$. The angle of banking is (take, $$g = 10ms^{-2})$$.
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$$25^{\circ}$$
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$$60^{\circ}$$
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$$45^{\circ}$$
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$$30^{\circ}$$
0%
$$40^{\circ}$$
Explanation
Given, $$v = 20\ ms^{-1}$$
$$r = 40\sqrt {2} m$$
$$\Rightarrow g = 10\ ms^{-2}$$
We know that, $$\tan \theta = \dfrac {v^{2}}{rg}$$
$$\Rightarrow \tan \theta = \dfrac {(20)^{2}}{40\sqrt {3}\times 10}$$
$$\Rightarrow \tan \theta = \dfrac {1}{\sqrt {3}}$$ or $$\theta = \tan^{-1}\left (\dfrac {1}{\sqrt {3}}\right )$$
$$\Rightarrow \theta = 30^{\circ}$$.
A weightless thread can bear tension up to $$3.7$$kg wt. 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 $$g=10m/s^2$$, then the maximum angular velocity of the stone will be :
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$$4rad/s$$
0%
$$2rad/s$$
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$$6rad/s$$
0%
none of these
Explanation
Maximum tension $$=\displaystyle\frac{mv^2}{r}+mg$$
$$3.7\times 10=\displaystyle\frac{0.5v^2}{4}+0.5\times 10$$
$$\Rightarrow v=16m/s$$
$$\therefore \omega =\displaystyle\frac{v}{r}=\frac{16}{4}=4rad/s$$.
A particle tied to a string describes a vertical circular motion of radius r continually. If it has a velocity $$\sqrt{3gr}$$ at the highest point, then the ratio of the respective tensions in the string holding it at the highest and lowest points is
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4 : 3
0%
5 : 4
0%
1 : 4
0%
3 : 2
0%
1 : 2
Explanation
Tension at highest point
$$T_H=\dfrac{mv^2}{r}-mg=3mg-mg=2mg$$
and tension at lowest point
$$T_L=\dfrac{mv^2}{r}+mg$$
Here, $$V^2_L=3gr+2g.2r=7gr$$
So, $$T_L=7mg+mg=8mg$$
Hence, $$\dfrac{T_H}{T_L}=\dfrac{2mg}{8mg}=\dfrac{1}{4}$$
The one which does not represent a force in any context is:
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friction
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impulse
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tension
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weight
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viscous drag
Explanation
Impulse is equal to the charge in momentum whereas all the other given terms are referred to as force.
So, option (B) is the correct answer.
If a body of mass $$5\ g$$ initially at rest is acted upon by a force of $$50$$ dynes for $$3\ s$$. Then the impulse will be
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$$0.98\times 10^{-3} Ns$$
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$$1.5\times 10^{-3} Ns$$
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$$2.0\times 10^{-3} Ns$$
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$$2.5\times 10^{-3} Ns$$
Explanation
Impulse, $$I = F\times \triangle t$$
$$= 50\times 10^{-5} \times 3 = 1.5\times 10^{-3}Ns$$.
In figure, two identical particles of mass $$m$$ are tied together with an inextensible light string. This is pulled at its centre with a constant force $$F$$. If the whole system lies on a smooth horizontal plane, then the acceleration of approach of particles at the instant shown will be:
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$$\cfrac { \sqrt { 3 } F }{ m } $$
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$$\cfrac { F }{ 2\sqrt { 3 } m } $$
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$$\cfrac { 2F }{ \sqrt { 3 } m } $$
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$$\cfrac { F }{ \sqrt { 3 } m } $$
Explanation
Make free body diagram of given figure.
$$2T\cos30°=F$$
$$T=\dfrac F{\sqrt{3}}$$
acceleration of approach of $$A$$
$$a_A=\dfrac{T\sin\theta}{m}$$
$$a_A=\dfrac{F}{2m\times \sqrt{3}}$$
Similarly,
$$a_B=\dfrac{F}{2\sqrt{3}m}$$
Acceleration of approach $$=\dfrac{F}{\sqrt{3}m}$$
A motor car is travelling at speed of $$30m/s$$ on a circular road of radius $$500m$$. If the speed of the car is increasing at the rate of $$2.0m/{ s }^{ 2 }$$, then the total acceleration will be
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$$1.8m/{ s }^{ 2 }$$
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$$2m/{ s }^{ 2 }$$
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$$2.7m/{ s }^{ 2 }$$
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$$3.8m/{ s }^{ 2 }$$
Explanation
Radial acceleration, $${ a }_{ r }=\cfrac { { v }^{ 2 } }{ r } =\cfrac { { (30) }^{ 2 } }{ 500 } =1.8m/{ s }^{ 2 }\quad $$
Tangential acceleration, $${ a }_{ r }=2m/{ s }^{ 2 }\quad $$
$$\therefore$$ Resultant acceleration, $$a=\sqrt { { a }_{ r }^{ 2 }+{ a }_{ r }^{ 2 }2{ a }_{ n }{ a }_{ r }\cos { \theta } } $$
$$\quad \therefore a=\sqrt { { (1.8) }^{ 2 }+{ (2) }^{ 2 } } =\sqrt { 7.24 } $$
$$\Rightarrow a=2.7m/{ s }^{ 2 }$$
A block of mass $$M$$ at the end of the string is whirled round a vertical circle of radius $$R$$. The critical speed of the block at the top of the swing is
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$${ \left( R/g \right) }^{ 1/2 }$$
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$$g/R$$
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$$M/Rg$$
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$${ \left( Rg \right) }^{ 1/2 }$$
Explanation
Using conservation of energy
Total mechanical energy at lowest point = Total mechanical energy at top
$$\dfrac{1}{2}m{v_{lowest}^2} = \dfrac{1}{2}m{v_{High}^2} + 2mgR$$
we know that $$ {v_{lowest}} = \sqrt{5Rg}$$
$$\therefore \dfrac{1}{2}(5Rg) = \dfrac{1}{2}{v_{high}^2} + 2gR$$
$${v_{high}} = \sqrt{Rg}$$
A railway carriage has its centre of gravity at a height of $$ 1 m $$ above the rails , which are $$ 1.5 m $$ apart.The maximum safe speed at which it could travel round an unbanked curve of radius $$ 100 m $$ is
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$$ 12 m/s$$
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$$ 18 m/s$$
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$$ 22 m/s$$
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$$ 27 m/s$$
Explanation
Here, $$ h=1 m , r=100 m , and 2x=1.5 m$$
For no skidding
$$\dfrac{mv^{2}}{r}\times h=mgx$$
$$v=\sqrt{\dfrac{grx}{h}}$$
$$=\sqrt{\dfrac{9.8 \times 100 \times 0.75}{1}}$$
$$=v=27.1 m/s$$
A pendulum string of length $$l$$ is moved upto a horizontal position and released as shown in figure. If the mass of pendulum is $$m$$, then what is the minimum strength of the string that can withstand the tension as the pendulum passes through the position of equilibrium?
(neglect mass of the string and air resistance)
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$$mg$$
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$$2mg$$
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$$3mg$$
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$$5mg$$
Explanation
We have, $$\quad T-mg=\cfrac { m{ v }^{ 2 } }{ I } $$
$$\quad { v }^{ 2 }=2gI\Rightarrow T=mg+\cfrac { m2gI }{ I } =3mg$$
A batsman deflects a ball by an angle of $$45^o$$ without changing its initial speed which is equal to $$54$$km/hr. Mass of the ball is $$0.15$$kg. What is the impulse imparted to the ball?
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$$4.2$$kg m/s in the direction of the final velocity
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$$4.2$$kg m/s in the direction of the initial velocity
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$$4.2$$kg m/s in the direction opposite to the initial velocity
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$$4.2$$kg m/s directed along the bisector of the initial and the final directions of the velocity
Explanation
Initial momentum of ball $$ = mucos\theta$$ along NO
Final momentum of ball $$ = mucos\theta$$ along ON
Impulse = Change in momentum
=$$2 mu Cos\theta$$
On subsituitng values we get,
$$\therefore Impulse = 4.2 kg/s$$
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Practice Class 11 Engineering Physics Quiz Questions and Answers
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