Author Archives: Priyanka

In This Post we are  providing Chapter-6 WORK, ENERGY AND POWER NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON WORK, ENERGY AND POWER

1. What is the speed of an aircraft if the pilot remains in contact with the seat, even while looping in vertical plane?

Ans. For the pilot to stay at rest, centripetal force should

be provided by his weight, i.e., mv2r=mg

∴v=rg

Q.2 The kinetic energy of a particle moving in a horizontal circle may remain the same everywhere. Is it true for the motion in a vertical circle.?

Ans. No. For the motion in a vertical circle, the speed of the particle at the bottom is 5gr and at the top is gr. Therefore, K.E. of the particle in a vertical circle is maximum at the bottom and minimum at the top.

Q.3 A stone of mass m tied to the ended of a string of length l is whirled in a vertical circle. What is the net force acting on the stone at the bottom and at the top of the vertical circle, if tension in the string at the bottom at the top is Tb and Tt respectively

?

Ans. Net force acting on the stone at the bottom of

the circle =Tb−mg

Net force at the top =Tt+mg

Q.4 Can magnitude of kinetic energy be negative?

Ans. No, kinetic energy can be zero or positive only

 because KE=12mv2, because m and v2 cannot 

be negative.

Q.5 What is the significant of negative mechanical energy?

Ans. Negative value of mechanical energy indicates a bound state electron in an atom and a satellite revolving around a planet both are in bound state having negative mechanical energy.

Q.6 If rain drop move decreasing acceleration, does work done by gravitational force on the drop change.

Ans. It should be clearly understood that whether the drop move with decreasing acceleration or with uniform speed work done by the gravitational force on the drop remains the same.

Q.7 Can a body have momentum without energy? Ans. Yes, when E=k+u=0 either both are zero or k=−u. thus K.E. may or may not be zero As P=2mk

∴P=0 only when k=0

and P≠0 when k=−u

Q.8 Can a body have momentum when its energy is negative?

Ans. Yes, when k<u, total energy E=k+u is negative, the body has the momentum (∵k≠0). For example in an atom, electron has momentum though its energy is negative.

Q.9 Can work be defined at a position or at instant?

Ans. No, work is not defined at a position or at an instant. Work is defined during the time interval within which the force has undergone displacement unlike velocity and acceleration, momentum and kinetic energy it is not the characteristic of moving body it is something which happen an a body which produces or destroys motion of the body.

Q.10 A man rowing boat upstream is at rest with respect to the shore is he doing work.

Ans. The boat is at rest with respect to the shore but itis moving up stream with respect to water. The man is doing work relative to the stream because he is applying force to produce relative motionbetween the boat and stream, but he does no work relative to the shore as displacement relative to shore is zero.

Q.11 Why a metal ball rebounds better than a rubber ball?

Ans. When rubber ball hits a massive object say earth the ball is distorted. A large amount of heat is generated in the ball by the rubbing of the rubber molecules against each other this effect is essentially absent in a band material so metal ball would often loseless energy upon collision than would a rubber ball.

Q.12 The outer casing of a rocket is burnt due to friction between casing and air, who supplies the energy necessary for burning. The rocket or the atmosphere?

Ans. Due to friction with the air the speed hence the kinetic energy of the rocket decreases. This decrease in the kinetic energy appears in the form of heat energy, thus energy necessary for burning is supplied by the rocket itself.

Q.13 An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance- however small, why then does its speed increase progressively as it comes closer and closer to the earth.

Ans. As a satellite comes closer to the earth its potential energy decrease. According to the law of conservation of energy the total energy of the system must remain constant. Now total energy =KE+PE+ heat energy. Because the heat loss due to friction is very small a decrease in potential energy result in a net increase in kinetic energy. Hence speed of a satellite progressively increase as it comes closer to the earth.

Q.14 In a tug-of-war one team is slowly giving way to the other is work being done on the losing team. How about the wining team.

Ans. Work is done on the losing team because that team is being dragged in the direction of the force applied by the wining team. This work done is said to be positive work, work is also done on the wining team because the wining team is being displaced opposite to the direction of the force applied by the losing team. This work done is said to be negative work.

Q.15 In a circus, the diameter of globe of death is 30 m. From what minimum height must a cyclist start in order to roll down the inclined and go round the globe successfully?

Ans. Diameter of globe =30m

Radius of globe, r=15m

Let h be the minimum height from which the

cyclist after rolling down an incline will acquire velocity =2gh

For looping the loop, the minimum velocity at the lowest point should be 5gr.

∴5gr=2gh

⇒h=5r2=5×152=37.5m

In This Post we are  providing Chapter-5 LAWS OF MOTION NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON LAWS OF MOTION

Question 1.
(a) State and prove impulse-momentum Theorem.

Answer:
It states that the impulse of force on a body is equal to the change in momentum of the body.
i.e. J = F_t = P₂ – P₁

Proof: From Newton’s Second law of motion, we know that

Let P₁ and P₂ be the linear momenta of the body at time t = 0 and t respectively.
∴ integrating equation (i) within these limits, we get


Hence proved.

(b) Prove that Newton’s Second law is the real law of motion.

Answer:
Proof: If we can show that Newton’s first and third laws are contained in the second law, then we can say that it is the real law of motion.
1. First law is contained in second law: According to Newton’s second law of motion,
F = ma …(i)
where m = mass of the body on which an external force F is applied and a = acceleration produced in it.

If F = 0, then from equation (1), we get
ma = 0, but as m ≠ 0
∴ a = 0

which means that there will be no acceleration in the body if no external force is applied. This shows that a body at rest will remain at rest and a body in uniform motion will continue to move along the same straight line in the absence of an external force. This is the statement of Newton’s first law of motion. Hence, the First law of motion is contained in the Second law of motion.

2. Third law is contained in second law: Consider an isolated system of two bodies A and B. Let them act and react internally.
Let FAB = force applied on body A by body B
and FBA = force applied on body B by body A

It \frac{\mathrm{d} \mathbf{p}_{\boldsymbol{A}}}{\mathrm{dt}} = rate of change of momentum of body A
and
\frac{\mathrm{d} \mathbf{p}_{\boldsymbol{B}}}{\mathrm{dt}} = rate of change of momentum of body B

Then according to Newton’s second law of motion,


(2) and (3) gives

As no external force acts on the system (∵ it is isolated), therefore according to Newton’s second law of motion,

or
Action = – Reaction,
which means that action and reaction are equal and opposite. It is the statement of Newton’s 3rd law of motion. Thus 3rd law is contained in the second law of motion.

As both First and Third Law is contained in Second law, so Second law is the real law of motion.

Question 2.
Derive the general expression for the velocity of a rocket in flight and obtain the expression for the thrust acting on it.

Answer:
The working of a rocket is based upon the principle of conservation of momentum. Consider the flight of the rocket in outer space where no external forces act on it.
Let m_o = initial mass of rocket with fuel.
V_u = initial velocity of the rocket,
m = mass of the rocket at any instant t.
v = velocity of the rocket at that instant.
d_m = mass of the gases ejected by the rocket, in a small-time it.
u =H velocity of exhaust gases,
DV = increase in the velocity of the rocket in a time dt.

∴ Change in the momentum of exhaust gases = dm. u
Change in momentum of rocket = – (m – dm) dv.

A negative sign shows that the rocket is moving in a direction opposite to the motion of exhaust gases.

Applying the law of conservation of linear momentum,
dm.u = – (m – dm) dv …(1)

As dm being very small as compared to m, so it can be neglected, Thus, eqn. (1) reduces to
dm.u = – m dv
or
dv = – u \frac{dm}{m} …(2)

Instantaneous velocity of the rocket:
At t = 0, mass of rocket = m0, velocity of rocket = v_o.
At t = t, mass of rocket = m, velocity of rocket = v.

∴ Integrating Eqn. (1) within these limits, we get

In actual practice, the velocity of exhaust gases nearly remains constant.

equation (3) gives the instantaneous velocity of the rocket. In general v_o = 0 at t = 0,

∴ Eqn. (3) reduces to

From Eqn. (4), we conclude that the velocity of the rocket at any instant depends upon:

1. speed (u) of the exhaust gases.
2. Log of the ratio of initial mass (m0) of the rocket to its mass (m) at that instant of time.

Upthrust on the rocket (F): It is the upward force exerted on the rocket by the expulsion of exhaust gases. It is obtained as follows:
Dividing Eqn. (2) by dt, we get


where F = ma is the instantaneous force (thrust).

From Eqn. (5), we conclude that the thrust (F) on the rocket at any instant is the product of the velocity of exhaust gases and the rate of combustion of fuel at that instant. Here negative sign shows that the thrust and velocity of exhaust gases are in opposite direction.

Question 3.
(a) Define inertia. What are its different types? Give examples.

Answer:
The tendency of bodies to remain in their state of rest or uniform motion along a straight line in the absence of an external force is called inertia.

Inertia is of the following three types:
1. The inertia of rest: When a body continues to lie at the same position with respect to its surrounding, it is said to possess inertia of rest. This situation may be changed only by the application of external force. For example, if a cot or sofa is lying in a particular place in the house, it will remain there even after days or years unless someone removes (by applying force) the same from its position. This is an example of the inertia of rest.

2. The inertia of motion: When a body is moved on a frictionless surface or a body is thrown in a vacuum, it will continue to move along its original path unless acted upon by an external force. In actual situations, air or floor etc. exert friction on the moving bodies so we are unable to visualize a force-free motion. This type of inertia when a body continues to move is called the inertia of direction.

3. In the above examples it is found that the direction of motion of the body or particle also does not change unless an external force acts on it. This tendency to preserve the direction of motion is called the inertia of direction.

(b) Explain Newton’s First law of motion. Why do we call it the law of inertia?
Answer:
According to the First law of motion, “Everybody continues to be in the state of rest or of uniform motion along a straight line until it is acted upon by an external force.”

It means that if a book lying on a table,-it will remain there for days or years together unless force is applied on it from outside to pick it.

Similarly, if a body is moving along a straight line with some speed, it will continue to do so until some external force is applied to it to change its direction of motion.

Thus First law tells us the following:

1. It tells us about the tendency of bodies to remain in the state of rest or of motion and the bodies by themselves can neither change the state of rest nor of uniform motion. This tendency is called inertia. To break the inertia of rest or motion or direction, we need an external force. Thus the definition of the first law matches with the definition of inertia and hence first law is called the law of inertia.
2. The first law of motion also provides the definition of another important physical quantity called force. Thus force is that agency which changes or tends to change the state of rest or of uniform motion of a body along a straight line.

(c) State Newton’s Second law of motion. How does it help to measure force? Also, state the units of force.
Answer:
It states that the time rate of change of momentum of a body is directly proportional to the force applied to it.

where a = \frac{dv}{dt} = acceleration produced in the body of mass m.
k = proportionality constant which depends on the system of units chosen to measure F, m, and a.

In the S.I. system, k = l,
∴ F = ma

The magnitude of the force is given by
F = ma …. (2)

Note: We have assumed that the magnitude of velocity is smaller and much less than the speed of light. Only under this condition Eqns. (1) and (2) hold good.

The definition of the Second law and its mathematical form is given in Eqn. (2) provide us a mean of measuring force.

One can easily find the change in velocity of a body in a certain interval of time. Both velocity and time can be easily measured. Thus by knowing the mass of the body one can determine both change in momentum as well as the acceleration of the body produced by an external force. If the force is increased, the rate of change of momentum is also found to increase. So also is the acceleration. Now with known values of m and we can find F.

Units of force: Force in S.I. units is measured in newton or N. From Eqn. (1) or (2) we can see that a newton of force is that fore? which produces 1 ms^-2 acceleration in the body of mass 1 kg.
1 newton = 1 kilogram × 1 metre/(second)²
or
1 N = 1 kg × 1 ms^-2 = 1 kg ms^-2

In CGS system force is measured in dyne
1 dyne = 1 gram × 1 cm/s2 = 1 g cm s^-2
Since 1 N = 1 kgm s^-2= 1000 g × 100 cm s^-2
= 10⁵ g cm s^-2 = 105 dyne
1 N = 10⁵ dyne
or
1 dyne = 10⁵ N

Gravitational Unit: If a falling mass of 1 kg is accelerated towards the Earth with 9.8 ms^-2, then the force generated is called 1 kg wt (1-kilogram weight) force. It is the S.I. gravitational unit of force.

We know that the earth accelerates the mass with g = 9.8 ms^-2
1 Kg wt = 9.8 N [1 kg × 9.8 ms 2 = 9.8 N]

C.G.S. gravitational unit is gf or g wt.
1 gf = 1g × 980 cms^-2
= 980 dyne

Question 4.
A hunter has a machine gun that can fire 50 g bullets with a velocity of 150 ms^-1. A 60 kg tiger springs at him with a velocity of 10 ms^-1. How many bullets must the hunter fire into the tiger so as to stop him in his track?

Answer:
m = mass of bullet = 50 gm = 0.050 kg
M = mass of tiger = 60 kg
v = velocity of bullet = 150 ms^-1
V = velocity of tiger = – 10 ms^-1
(∵ it is coming from opposite direction) .

Let n = no. of bullets fired per second at the tiger so as to stop it.
∴ p_i = 0 before firing …. (i)
p_f = n(mv) + MV …. (ii)

∴ According to the law of conservation of momentum,

Question 5.
A mass of 200 kg rests on a rough inclined plane of angle 300. If the coefficient of limiting friction is \frac{1}{\sqrt{3}} find the greatest and the least forces in newton, acting parallel to the plane to keep the mass ¡n equilibrium.

Answer:
Here, m = mass of body = 200 kg
Let angle of inclination = θ
μ_s = coefficient of limiting friction = \frac{1}{\sqrt{3}}

The rectangular components of mg are as shown in fig.

Here mg sin θ acts along the plane in the downward direction and is given by

(i) the least forces in newton, acting parallel to the plane to keep the mass in equilibrium. given by
f₂₁ =mg sinθ – F = 980 – 9800 = 0

(ii) The greatest force to be applied to keep the mass in equilibrium is given by
f₂ = mg sin θ + F = 980 + 980 = 1960 N.

Question 6.
Find the force required to move a train of 2000 quintals up an incline of 1 in 50, with an acceleration of 2 ms^-2, the force of friction being 0.5 newtons per quintal.

Answer:
Here, m = 2000 quintals
= 2000 × 100 kg (v 1 quintal = 100 kg)
sin θ = \frac{1}{50} , acceleration, a = 2 ms^-2

F = force of friction
= 0.5 N per quintal
= 0.5 × 2000 = 1000 N

In moving up an inclined plane, force required against gravity
= mg sin 0
= 2000 × 100 × 9.8 × \frac{1}{50}
= 39200N.
Also if f = force required to produce acce. = 2 ms^-2.
Then f = ma = 200000 × 2 = 400000 N

∴Total force required = F + mg sin θ + f
= 1000 + 39200 + 400000 = 440200 N.

Question 7.
A bullet of mass 0.01 kg is fired horizontally into a 4 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between the block and the surface is 0.25. The bullet remains embedded in the block and the combination moves 20 m before coming to rest. With what speed did the bullet strike the block?

Answer:
Here, m₁ = mass of the bullet = 0.01 kg
m₂ = mass of the wooden block = 4 kg
μ₂ = coefficient of kinetic friction = 0.25
initial velocity ofblock, u₂ = 0, s = distance moved by combination=20 m

Let u₁ = initial velocity of the bullet
If v = velocity of the combination, then according to the principle of conservation of linear momentum,

If F = kinetic force of friction,
Then

Then retardation ‘a’ produced is given by

Question 8.
A force of 100 N gives a mass m₁ an acceleration of 10 ms^-2, and of 20 ms^-2 to a mass m₂. What acceleration would it give if both the masses are tied together?

Answer:
Let a = acceleration produced if m₁ and m₂ are tied together.
F = 100 N, Let a₁ and a₂ be the acceleration produced in m₁ and m₂ respectively.
∴ a₁ = 10 ms^-2, a₂ = 20 ms^-2 (given)

Question 9.
A balloon with mass M is descending down with an acceleration ‘a’ < g. What mass m of its contents must be removed so that it starts moving up with an acceleration ‘a’?

Answer:
Let F = retarding force acting on the balloon in the vertically upward direction.

When the balloon is descending down with an acceleration ‘a’, then the net force acting on the balloon in the downward direction is given
by
Ma = Mg – F
or
F = Mg – Ma ….(i)
When the mass m is taken out of the balloon, then its weight is
= (M – m)g

Now as the balloon is moving upward with acceleration ‘a’, so the net force acting on the balloon in the upward direction is given by:

Question 10.
Three blocks are connected as shown below and are on a horizontal frictionless table. They are pulled to right with a force F = 50 N. If m₁ = 5 kg, m₂ = 10 kg and m₃ = 15 kg, find tensions T₃ and T₂.

Answer:
Here, F = 50 N, m₁ = 5 kg, m₂ = 10 kg m₃ = 15 kg.

As the three blocks move with an acceleration ‘a’

To determine T₂: Consider the free body diagram (1). Here F and T2 act towards the right and left respectively.

As the motion is towards the right side, so according to Newton’s Second law of motion:

To determine T₃: Consider the free body diagram (2)

In This Post we are  providing Chapter-4 MOTION IN A PLANE NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON MOTION IN A PLANE

Question 1.
A body is acted upon by the following, velocities:
(i) 7 ms^-1 due to E,
(ii) 10 ms^-1 due S,
(iii) 52–√ ms^-1 due N.E.
Find the magnitude and direction of the resultant velocity.

Answer:
Let OA, OB and OC represent the velocities given in the statement i.e.
OA = 7 ms^-1
OB = 10 ms^-1
and OC = 52–√ ms^-1
To find their resultant velocity, resolve OC into two rectangular components along east and north.

Hence resultant velocity along east = 7 + 5 = 12 ms^-1 and resultant velocity along south = OB – OF = 10 – 5 = 5 ms^-1.

If R be the resultant velocity, then the magnitude of R is obtained by applying the parallelogram law of vector addition as

When OG = 12ms^-1 and OH 5ms^-1.

The direction of R: Let θ be the angle made by R with the east.
∴ in rt. ∠d ΔOGI,

Question 2.
A projectile is fired horizontally with a velocity of 98 ms^-1 from the top of a hill 490 m high. Find:
(i) the velocity with which it strikes the ground.
(ii) the time is taken to reach the ground.
(iii) the distance of the target from the hill.
Answer:

(i) h = 490 m, a = g = 9.8 ms²
U_y = initial velocity along the y-axis at the top of the tower = 0

(ii) Let v be the velocity along the y-axis with which the projectile hits the ground.

If V be the resultant velocity of hitting the ground

Let θ be the angle made by V with the horizontal

(iii) Let x, be the distance of the target from the hill.
∴ x = horizontal distance covered with u in a time t.
ut = 98 × 10 = 980 m.

Question 3.
A boy stands at 78.4 m from a building and throws a ball which just enters a window 39.2 m above the ground. Calculate the velocity of the projection of the ball.

Answer:
Let the boy standing at A throw a ball with initial velocity u.
θ = angle of the projection made with the horizontal.

As the boy is at 78.4 m from the building and the ball just enters above the ground.

Question 4.
Two particles located at a point begin to move with velocities 4 ms^-1 and 1 ms^-1 horizontally in opposite directions. Determine the time when their velocity vectors become perpendicular. Assuming that the motion takes place in a uniform gravitational field of strength g.

Answer:
Let v₁ and v₂ be the velocities of first and 2nd particles respectively after a time t.
∴ v₁ = 4î – gt ĵ
v₂ = – î – gt ĵ
For v₁ and v₂ to be ⊥ to each other, then their dot product must be zero.

Question 5.
A body is projected with a velocity of 40 ms^-1. After two seconds, it crosses a verticle pole of 20.4 m. Find the angle of projection and the horizontal range.

Answer:
Here, u = 40 ms^-1
height of vertical pole, h = 20.4 m
t = 2 seconds

Let us take vertical motion

∴ The horizontal range is given by the relation,

Question 6.
The greatest and the least resultant of two forces acting at a point are 29 N and 5 N respectively. If each force is increased by 3 N, find the resultant of two new forces when acting at a point at an angle of 90° with each other.

Answer:
Let A and B be the two forces.
∴ Greatest Resultant = A + B = 29 N ….(1)
least Resultant = A – B = 5 N ….(2)

Let A and B be the new forces such that
A’ = A + 3 = 17 + 3 = 20N and
B’ = B + 3 = 12 + 3 = 15 N

Here, θ = angle between A’ and B’ = 90°
Let R be the resultant of A’ and B’.
∴ according to parallelogram law of vector addition

The direction of R:
Let β be the angle made by R with A’

Question 7.
An aircraft is trying to fly due north at a speed of 100 ms^-1 but is subjected to a crosswind blowing from west to east at 50 ms^-1. What is the actual velocity of the aircraft relative to the surface of the earth?

Answer:
Let V_a and V_w be the velocities of aircraft and wind respectively.
∴ V_a = 100 ms^-1 along N direction
V_w = 50 ms^-1 along E direction

If V be the resultant velocity of the aircraft, then these may be represented as in the figure given below. So the magnitude of V is given by,


Let ∠AOB = θ be the angle which the resultant makes with the north direction.

Question 8.
Calculate the total linear acceleration of a particle moving in a circle of radius 0.4 m at the instant when its angular velocity is 2 rad s^-1 and angular acceleration is 5 rad s^-2.

Answer:
Since the particle possesses angular acceleration, so its total linear acceleration (a) is the vector sum of the tangential acceleration (a,) and the centripetal acceleration (ac). a₁ and a_c, are at right angles to each other.
a = \sqrt{a_{t}^{2}+a_{c}^{2}} …. (1)

Question 9.
An airplane flies 400 km west from city A to city B then 300 km north-east to city C and finally 100 km north to city D. How far is it from city A to city D? In what direction must the airplane go to return directly to the city A from city D?

Answer:
Given, AB = 400 km
BC = 300 km
CD = 100 km
AD =?

Let N₁, N₂ represent north directions.
∠ABC = 45°
Draw CC’ ⊥ AB, And CB’ ⊥ BN₂
Now in ΔBC’ C

From AAC’D, AD is given by

Question 10.
Which of the following quantities are independent of the choice of the orientation of the coordinates axes:
a + b, 3a_x + 2b_y, [a + b – c], angle between b and c, a?

Answer:
a + b, |a + b – c|, angle between b and c, a are the quantities that are independent of the choice of the orientation of the coordinate axes.

But the value of 3a_x + 2b_y depends on the orientation of the axes.

In This Post we are  providing Chapter-3 MOTION IN A STRAIGHT LINE NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON MOTION IN A STRAIGHT LINE

Question 1.
A rocket is fired vertically from the ground. It moves upwards with a constant acceleration of 10 ms^-2 for the 30s after which the fuel is consumed. After what time from the instant of firing, the rocket will attain the maximum height? Take g = 10ms^-2.

Answer:
Here, a = g = 10ms^-2 for t = 30s
∴ u = 0 for first part of the motion
v = velocity attained by the rocket after 30s = ?

∴ using the relation,
v = u + at …(1)
we get v = 0 + 10 × 30
= 300 ms^-1

for second part of motion,
u = v = 300 ms^-1
v₁ = final velocity = 0

Let t1 be the time of journey for this part of the motion,
∴ from equation (1), we get
0 = 300 – 10 × t₁

Let T = total time of flight in which the rocket attains maximum height,
T = t + t₁
= 30 + 30
= 60 second.

Question 18.
On a foggy day, two car drivers spot each other when they are just 80 m apart. They were travelling at 72 km h^-1 and 60 km h^-1 respectively. Both of them simultaneously apply brakes which retard both the cars at 5ms-2. Tell whether they will avert a collision or not?
Answer:
Here, let u, and u2 be the initial velocities of the two cars.

If S1 and S2 be the distances covered by the two cars before coming to rest (i.e. v = 0), then
using the relation,

If S be the total distance covered by two cars before coming to rest, then
S = S₁ + S₂ = 40 + 27.7 = 67.7 m.

Also S₁ = 80 m.
Now clearly the total distance covered by the cars before coming to rest is less than 80 m, so the collision will be averted.

Question 2.
A stone is dropped from a rising balloon at a height of 76 m above the ground and reaches the ground in 6s. What was the velocity of the balloon when the stone was dropped?

Answer:
Let u be the velocity of the balloon in an upward direction at the point
A = initial velocity of stone in an upward direction.

The stone rises to the top, comes to rest and then it starts coming back to the ground to hit B.

Let us take A as origin :
∴ Net vertical distance covered by the stone, y = – 76 m.
a = – g = – 9.8 ms^-2
(Here y and acc. are taken as -ve because they are in a downward direction w.r.t. A)

Question 3.
A car moves at a velocity of 2.24 km h^-1 in the first minute, at 3.60 km h^-1 in the second minute and at 5.18 km h^-1 in the third minute. Calculate the average velocity in these three minutes.

Answer:
Let x₁, x₂ and x₃ be the displacements of car in 1 st, 2nd and 3rd minutes respectively and let v₁, v₂ and v₃ be the velocities of these time intervals,

∴ v₁ = 2.24 kmh^-1, v₂ = 3.60 kmh^-1, v₃ = 5.18 kmh^-1
time intervals = t₁ = t₂ = t₃ = 1 minute = 160 h

If x be the total displacement, then

If v be the average velocity in these 3 minutes, then

Question 4.
A man is walking due East at the rate of 3 km h^-1. Rain appears to fall down vertically at the rate of 3 km h^-1. Find the actual velocity and direction of the rainfall.

Answer:
Let a = angle made by the rainfall with the vertical Velocity of man represented by OA due East = 3 km h^-1
As rain appears to fall vertically downward, so OB = 3 km h^-1 represents the velocity of rain w.r.t. man. The actual velocity of rainfall is represented by OC and is given by

Thus the rain is actually falling at 32–√ km h^-1 at an angle of 45° west of the vertical.

Question 5.
A car starting from rest and moving with uniform acceleration possesses average velocities 5 ms^-1, 10 ms^-1 and 15 ms^-1 in the first, second and third seconds. What is the total r distance covered by the car in these three seconds?

Answer:
Here, let v₁, v₂ and v₃ be the average velocities of 1st, 2nd and 3rd secs, respectively.
∴ v₁ = 5 ms^-1, v₂ = 10 ms^-1, v₃ = 15 ms^-1
t₁ = t₂ = t₃ = 1 s = time intervals

If x₁, x₂ and x₃ be the distances covered in these seconds resp., then

Question 6.
A person is running at a maximum speed of 3 ms-1 along the length of a train to catch hold of the door of a compartment. When he is just 2.5 m from the door, the train steams off with an acceleration of 1 ms^-2. Find how long it takes him to catch the door?
Answer:

Here, let u₁ = 3 m s^-1, be the initial speed of man at point A.
For train x₁ = 2.5 m = distance of the train from man.
u = 0, a = acceleration of train = 1 ms^-2

Let C be their meeting point after a time t.
If BC = x, then usirig the relation,
S = ut + 12 at2, we get

For train, x = 0 + 12 × 1 × t² = 0.5t² … (1)
For man, using the relation,

∴ t = 1, 5 i.e. he will catch the door first after Is and then after 5s.
t = 1s.

Question 7.
The displacement of a body along the x-axis changes according to the relation: x = 20 – 15t + 4t², where x is in metres, t in seconds. Determine its position, velocity and acceleration at t = O.

Answer:
Here, x = 4t² – 1 5t + 20 … (1)
v = dxdt = 4.2t – 15 = 8t – 15 …(2)

Also a = dvdt = 8
At t = 0, the position (x), velocity (y) and acceleration (a) are given by
x = 4.0 – 15 × 0 + 20 = 20 m
v = 8 × 0 – 15 = – 15ms^-1
and a = 8 ms^-2.

Question 25.
A body undergoes a uniformly accelerated motion. Its velocity after 5 seconds is 25 ems and after 8 seconds it is 34 cms’. Calculate the distance covered in the 12th second.
Answer:
Let u and a be the initial speed and acceleration of the body.
Let v₁ and v₂ be its velocities after 5s and 8s respectively

Let S_nth be the distance covered in the 12th second.

Using the relation,

Value-Based Type:

Question 8.
On a two-lane road, there are hoardings hanged on the electric poles “Save Energy”.
If car a is travelling at a speed of 54 km h1 and car ‘B’ moves with 90 km h^-1 from the opposite direction.
Now, answer these questions:
(a) Which values are depicted in the above problem? Write down the four ways by which the energy can be saved.

Answer:
Save energy to save our environment.

The four ways to save energy are as under :

1. By using CFLs, switching off appliances when not in use.
2. Saving water by using it efficiently.
3. Using solar energy.
4. By using public transport.

(b) Write a slogan on “save energy”.
Answer:
Try to switch of the fans and lights when you are not using it. Use limited non renewable sources of energy.

(c) Find the velocity of B with respect to A? Also, find the velocity of the ground with respect to B?
Solution:
V_A = + 54 km h^-1 = 15 ms^-1
V_B = 90 km h^-1 = – 25 ms^-1

(Takingthe velocity of car A positive N v and car B negative)
∴ Relative velocity of B with respect to A
= V_B – V_A = – 25 – 15 = – 40 ms^-1

i.e. the car Boppers to A to move with a speed 40 ms-1 from opposite direction.
Relative velocity of ground w.r.t B = 0 – V_B = 0 – (- 25)
= 25 ms^-1

Question 9.
A candle march was organized by an NGO. The theme of the candle march was “No use of non-biodegradable products”. The organizer made a semicircular track of radius R on which peoples have to run taking a poster in their hands. If Mahesh travelled from A to B in time t with constant speed, then answer the following:

(i) Why should non-biodegradable products not be used? Justify your answer.
Answer:
Biodegradable products are reusable and cause less pollution. So, non-biodegradable products must be avoided.

(ii) Find the
(a) Displacement
Answer:
Displacement = minimum distance between initial and final point = AB = 2 R

(b) Average speed
Answer:

(c) Average velocity
Answer:

(d) Average acceleration
Answer:

Question 10.
In sports day activities of a public school, the lines were drawn with chalk powder. Gagan an athlete runs a distance of 1500 m in the following manner, (i) Starting from rest, he accelerates himself uniformly at 2ms_1, till he covers a distance of 900 m. (ii) He then runs the remaining distance of 600 m at the uniform speed developed.

(i) Which value is depicted in the above problem?
Answer:
Importance of sports activities.

(ii) Do you think that sports activities are important in day-to-day life?

Answer:
Yes, sports activities are very much important in day-to-day life as
(a) It gives break from regular life
(b) It improves physical strength
(c) It relaxes our mind and body
(d) It helps to concentrate on other works
im-90

(iii) Calculate the time taken by the athlete to cover the two parts of the distance covered.
Answer:
Let the athlete run from O to B Here, OB = 1500 m,
OA = 900 m, AB = 600 m

Case II : Taking motion of athelete from A to B u = 60 m/s, S = 600m, a = o (∵ Speed is uniform or constant)
t = t₂ = ?

In This Post we are  providing Chapter-2 UNITS AND MEASUREMENTS NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON UNITS AND MEASUREMENTS

1.E, m, l and G denote energy, mass, angular momentum and gravitational constant respectively. Determine the dimensions of  

Ans:

Thus, it is dimension less.

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2.Two resistances R₁ = 100 and R₂ = 200 are connected in series. Then what is the equivalent resistance?

Ans:

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3.If velocity, time and force were chosen the basic quantities, find the dimensions of mass?  

Ans:

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4.A calorie is a unit of heat or energy and it equals about 4.2 J where . Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals  m, the unit of time is y s. Show that a calorie has a magnitude in terms of the new units.

Ans. Given that,

1 calorie = 4.2 (1 kg) 

New unit of mass = α kg

Hence, in terms of the new unit, 1 kg =

In terms of the new unit of length,

And, in terms of the new unit of time,

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5. A SONAR (sound navigation and ranging) uses ultrasonic waves to detect and locate objects under water. In a submarine equipped with a SONAR the time delay between generation of a probe wave and the reception of its echo after reflection from an enemy submarine is found to be 77.0 s. What is the distance of the enemy submarine? (Speed of sound in water = 1450).

Ans. Let the distance between the ship and the enemy submarine be.
Speed of sound in water = 1450 m/s
Time lag between transmission and reception of Sonar waves = 77 s
In this time lag, sound waves travel a distance which is twice the distance between the ship
and the submarine (2S).
Time taken for the sound to reach the submarine
∴ Distance between the ship and the submarine (S) = 1450 × 38.5 = 55825 m = 55.8 km

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6.One mole of an ideal gas at standard temperature and pressure occupies 22.4 L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of hydrogen molecule to be about 1). Why is this ratio so large?

Ans. Radius of hydrogen atom, r = 0.5 = m

Volume of hydrogen atom =

Now, 1 mole of hydrogen contains  hydrogen atoms.

∴ Volume of 1 mole of hydrogen atoms,

Molar volume of 1 mole of hydrogen atoms at STP,

Hence, the molar volume is  times higher than the atomic volume. For this reason, the inter-atomic separation in hydrogen gas is much larger than the size of a hydrogen atom.

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7.The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

Ans. Distance of the star from the solar system = 4.29 ly

1 light year is the distance travelled by light in one year.

1 light year = Speed of light × 1 year

=  m

∴4.29 ly =  m

1 parsec =  m

∴4.29 ly = = 1.32 parsec

Using the relation,

Where,

Diameter of Earth’s orbit, d=

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8.Estimate the average mass density of a sodium atom assuming its size to be about 2.5. (Use the known values of Avogadro’s number and the atomic mass of sodium). Compare it with the density of sodium in its crystalline phase: 970 kg. Are the two densities of the same order of magnitude? If so, why?

Ans.  Diameter of sodium atom = Size of sodium atom = 2.5 

Radius of sodium atom, r =

=  m

Volume of sodium atom, V =

According to the Avogadro hypothesis, one mole of sodium contains  atoms and has a mass of 23 g or  kg.

∴ Mass of one atom =

Density of sodium atom, p =

It is given that the density of sodium in crystalline phase is 970 kg.

Hence, the density of sodium atom and the density of sodium in its crystalline phase are not in the same order. This is because in solid phase, atoms are closely packed. Thus, the inter-atomic separation is very small in the crystalline phase.

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9. It is a well known fact that during a total solar eclipse the disk of the moon almost completely covers the disk of the Sun. From this fact and from the information you can gather from examples 2.3 and 2.4, determine the approximate diameter of the moon.

Ans.

The position of the Sun, Moon, and Earth during a lunar eclipse is shown in the given figure.

Distance of the Moon from the Earth =  m

Distance of the Sun from the Earth = m

Diameter of the Sun =  m

It can be observed that ΔTRS and ΔTPQ are similar. Hence, it can be written as:

Hence, the diameter of the Moon is  m.

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10. Just as precise measurements are necessary in science, it is equally important to be able to make rough estimates of quantities using rudimentary ideas and common observations. Think of ways by which you can estimate the following (where an estimate is difficult to obtain, try to get an upper bound on the quantity):

(a) the total mass of rain-bearing clouds over India during the Monsoon

(b) the mass of an elephant

(c) the wind speed during a storm

(d) the number of strands of hair on your head

(e) the number of air molecules in your classroom.

Ans. (a) During monsoons, a metrologist records about 215 cm of rainfall in India i.e., the height of water column, h = 215 cm = 2.15 m

Area of country, A = 

Hence, volume of rain water, V = A × h =

Density of water, p = 

Hence, mass of rain water = p× V =  kg

Hence, the total mass of rain-bearing clouds over India is approximately  kg.

(b) Consider a ship of known base area floating in the sea. Measure its depth in sea (say).

Volume of water displaced by the ship, 

Now, move an elephant on the ship and measure the depth of the ship () in this case.

Volume of water displaced by the ship with the elephant on board, 

Volume of water displaced by the elephant = 

Density of water = D

Mass of elephant = AD 

(c) Wind speed during a storm can be measured by an anemometer. As wind blows, it rotates. The rotation made by the anemometer in one second gives the value of wind speed.

(d) Area of the head surface carrying hair = A

With the help of a screw gauge, the diameter and hence, the radius of a hair can be determined. Let it be r.

∴Area of one hair = 

Number of strands of hair 

(e) Let the volume of the room be V.

One mole of air at NTP occupies 22.4 l i.e., volume.

Number of molecules in one mole = 

∴Number of molecules in room of volume V

== 

= 

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In This Post we are  providing Chapter-1 PHYSICAL WORLD NCERT MOST IMPORTANT QUESTIONS for Class 11 PHYSICS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON PHYSICAL WORLD

Question 1.
Write the physical principle upon which the working of the gadgets mentioned in the above question is based.

Answer:
These are based on the following physical principles:

1. The boiling point rises with the increase in pressure.
2. Light is produced when the current is passed through a given resistor.
3. Light is emitted when an electric discharge is passed through the gas.
4. A rotating magnetic field is produced on passing current which notates the motor.
5. Due to evaporation of water, cooling in the air which is being forced out by the fan is produced.
6. On absorbing heat from the surroundings, compressed volatile liquid on sudden expansion causes cooling.
7. Current produces a rotating magnetic field that operates the motor.
8. Heat is produced due to the burning of L.P.G.
9. It works on the principle of heating effect of electric current.
10. Torque is produced on the coil due to the electric current passed through it, hence it rotates.
11. Current shows the heating effect when passed through the conductor.
12. It rotates due to the torque produced on the coil on passing an electric current through it.

Question 2.
Name one Scientist each from the following countries who have won Nobel Prize.

(a) Japan
Answer:
H. Yukawa

(b) England
Answer:
Janies Chadwick

(c) India
Answer:
C.V. Raman

(d) The U.S.A.
Answer:
K. Feynman

(e) Germany.
Answer:
Max. Plank.

Question 3.
How Darwin showed that scientific themes are at once simple even though phenomena in nature may be complex.

Answer:
Darwin found a simple basis for the origin of species and descent of man which is “Living things change producing descendants with different characteristics in a process that has been going on for as long as there has been life” by taking a large number of observations on the theory of evolution while onboard ship.

Question 4.
In science sometimes we observed certain phenomena experimentally but are unable to give a logical equation or theory for that sometimes, it also happens that we have a scientific theory supported by’ mathematical formulation yet are unable to test it immediately. Site one such example.

Answer:
Einstein worked to establish a relation between the energy and mass of the body. He was of the view that these are the two sides of the same coin or two facts of the same physical quantity. He succeeded when he gave his mass-energy equation E = mc². But its experimental verification came 40 years later in 1945 when the atomic bomb was exploded over Japan.

Question 5.
Why do we call physics an exact science? What is the aim of science?

Answer:
Physics is called exact science because it is based on the measurement of fundamental quantities.
The main aim of science is to find the truth behind the various processes taking place in the universe.

Question 6.
How science has helped in solving the food problem in several countries?

Answer:
Science has helped in solving food problem in the following ways:
(a) It has given improved and new agricultural implements.
(b) Science has improved the quality of seeds by genetic engineering.
(c) High-yielding hybrid varieties of grains have been developed. Some easily reaping varieties have also been developed and grown.
(d) Use of pesticides and insecticides has saved crops from being destroyed by insects and pests.
(e) Some new types of crops are also developed and are being developed to meet the requirement of society.

Question 7.
What is a scientific temperament and scientific way of doing things?

Answer:
A mindset molded in a particular set of thinking called the scientific way is known as scientific temperament. It is not only based on logic, facts but on reliable observations. The ultimate test of truth in science is experimental verification.

A scientific way of doing things involves the following steps:
(a) Identifying the problem or aim.
(b) Collecting all relevant information or data related to the problem.
(c) Hypothesising or proposing a possible theory.
(d) Taking experimental observation yielding consistent results.
(e) Predicting or making statements.

Question 8.
What is the scope of Physics?

Answer:
The scope of Physics is very wide i.e. the domain of Physics covers a very wide variety of natural phenomena.

For example, the range of distances we study in Physics varies from 10¹⁴ m (size of the nucleus) to 10²⁵ m (size of the universe).

Similarly, the range of masses included in the study of Physics varies from 10^-30 kg (mass of an electron) to 10⁵⁵ kg (mass of the universe). Also, the range of time i.e. time intervals of events we come across in the study of Physics varies from 10²² seconds (time taken by light to cross a nuclear distance) to 10^-8 seconds (lifetime of the sun).

Thus we see that the scope of Physics is really very wide. It includes; optics, electricity waves, and oscillations, heat and thermodynamics, magnetism, atomic and nuclear physics, computers, and electronics.

Question 9.
Physics is an exciting subject! Comment.

Answer:
The study of Physics is exciting in many ways, e g.:

1. Journey to the moon with controls from the grounds.
2. Lasers and their ever-increasing applications.
3. Live transmission of events thousands of kilometers away on the T: V.
4. The speed and memory of the fifth generation of computers.
5. Study of various types of forces in nature.
6. Technological advances in health science.
7. The use of robots is quite exciting.
8. Telephone calls over long distances and so on. Thus, Physics is exciting not only to the scientist but also to a layman, children, women, etc. The musical instruments, toy guns, toy trains, etc. all are constructed using simple principles of physics like collision, potential energy, and vibration, etc. Today the situation is that even our thought process and social values are affected by Physics. Thus, it is quite amazing.

Question 10 .
Write a short note on the origin and development of Physics.

Answer:
Physics as a science took roots from the days of Copernicus, i. e., nearly four centuries ago when it was not well understood and it was considered as a part of philosophy, i.e., knowledge. Later on, with the development of knowledge about nature and its various activities, the knowledge was divided into physical and biological sciences.

Some important developments like Newton’s law of gravitation, ideas about light were developed in the 18th century. The 19th century saw some of the great discoveries in Physics and at the end f the century i.e. 1889, the electromagnetic theory was developed, Fouriuatun of Einstein’s and Plank’s ideas were laid down apart from laying the basis for the industrial revolution. Physics progressed very fast in the first quarter of the 20th century.

Atomic structure, the theory of relativity, quantum theory, nuclear physics, basics of laser theory and most of the other developments took place in this period. Then came transistors, semiconductors, television, radar, and few important discoveries during World War II.

Further development in quantum mechanics, thin-film technology, computers, lasers was developed from 1950 onward. Today we have no theoretical development beyond quantum mechanics. A unified theory is not being tried yet. This is the present status with achievements in applied fields.

In This Post we are  providing Chapter-16 PROBABILITY NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON PROBABILITY

Question 1.
A coin is tossed twice, what is the probability that at least one tail occurs?

Solution:
Let S is the sample space, then
S = {(H, H), (H, T), (T, H), (T, T)}
n(S) = 4
Let E be the event of getting at least one tail, then
E = {(H, T), (T, H), (T, T)}
∴ n(E) = 3
Hence, required probability P(E) = n(E)n(S) = 34

Question 2.
There are four men and six women on the city council. If one council member is selected for a committee at random, then how likely is it that it is a woman?

Solution:
Let S is the sample space, then
n (S) = 10
Let E be the event that woman is selected, then
n(E) = 6
Hence, required probability P(E) = n(E)n(S)
P(E) = 610 = 35

Question 3.
If 211 is the probability of an event A, then find what is the probability of the event is ‘not A’.

Solution:
Given that :
P(A) = 211
We know that,
P(A) + P(A¯) = 1
211 + P(A¯) = 1
P(A¯) = 1 – 211 = 11–211
P(A¯) = 911.

Question 4.
A box contain 10 red marbles, 20 blue marbles and 30 green marbles. 5 marbles are drawn from the box. What is the probability that:

1. All marbles will be blue.
2. At least one marble will be green.

Solution:
In the box, there are 10 red, 20 blue and 30 green marbles.
The No. of marbles = 10 + 20 + 30 = 60
1. Total ways of choosing 5 marbles out of 60 marbles,
n(S) = ⁶⁰C₅
Let E is the event of choosing blue marbles, then
n(E) = ²⁰C₅
Probability = n(E)n(S) = 20C560C5

2. P(At least one green marble)
= 1 – P(No green)
= 1 – 30C560C5

Question 5.
4 cards are drawn from a well shuffled deck of 52 cards. What is the probability of obtaining in card of 3 diamond and one spade?

Solution:
Let S is the sample space.
Total number of selecting 4 cards out of 52 cards, n(S) = ⁵²C₄
If E is the event obtaining card of 3 diamond and 2 spade, then
n(E) = ¹³C₃ x ¹³C₁
Probability P(E) = n(E)n(S) = 13C3×13C152C4

Question 6.
Given : P(A) = 35 and P(B) = 15. If A and B are mutually exclusive events, then find P(A or B). (NCERT)

Solution:
A and B are mutually exclusive events.
∴ P (A or B) = P(A ∪ B) = P(A) + P(B)
P(A ∪ B) = 35 + 15 = 3+15 = 45

Question 7.
In a lottery, a person choose six different natural numbers at random from 1 to 20 and if there six numbers match with the six numbers already fixed by lottery committee, he wins the prize, what is the probability of winning the prize in the game? (Hint: Order of the numbers is not important)

Solution:
Let S is the sample space, then
n(S) = ²⁰C₆
= 20×19×18×17×16×156×5×4×3×2×1 = 38760
Only one prize can be win.
∴ n(E) = 1
Hence, required probability P(E) = n(E)n(S) = 13876

Question 8.
Two dice are thrown simultaneously. Find the probability of getting a sum 9 in a single throw.

Solution:
Total number of ways in which two dice may be thrown
= 6 x 6 = 36
∴ n(S) = 36
Event of getting sum 9 is A = {(3, 6), (4, 5), (5, 4), (6, 3)}
∴ n(A) = 4
∴ Required probability P(A) = n(A)n(S) = 19

Question 9.
A bag contains 8 black, 6 white and 5 red balls. Find the probability of drawing a black or a white ball from it

Solution:
Total number of balls = 8 + 6 + 5 = 19
∴ n(S) = 19
Event A of drawing 1 black or 1 white ball
n(A) = 8 + 6 = 14
∴ n(A) = 14
∴ Required probability P(A) = n(A)n(S) = 1419

Question 10.
From a pack of well shuffled cards two cards drawn simultaneously. Find the probability that both the cards are ace.

Solution:
Total number of ways of drawing two cards out of 52
= ⁵²C₂
Number of ways drawing two ace out of 4 ace
= ⁴C₂
∴ Required Probability

Question 11.
A pair of dice are thrown. Find the probability that the sum is 9 or 11.

Solution:
Let the sample space be S, then
∴ n(S) = 36
Let E be the event that sum is 9 or 11, then
E = {(5, 4), (4, 5), (6, 3), (3, 6), (6, 5), (5, 6)}
∴ n (E) = 6
The probability of getting sum 9 or 11 is
P(E) = n(E)n(S) = 636 = 16
Hence, probability that the sum is not 9 or 11 is
P(E¯) = 1 – P(E)
= 1 – 16 = 56

Question 12

A letter is selected at random from the word ‘ASSASSINATION’. Find the probability that letter is

1. a vowel
2. a consonant

Solution:
Number of letters is 13 in which there are 6 vowels and 7 consonants.
1. Let sample space is S, then
∴ n(S) = 13
E₁ is the event of choosing a vowel, then

n(E₁) = 6
Hence, required probability P(E₁) = n(E1)n(S)
P(E₁) = 613

2. E₂ is the event of choosing a consonant, then
n(E₁) = 7
Hence, required probability P(E₂) = n(E2)n(S)

Question 13.
In class XI of a school, 40% of the students study Mathematics and 30% study Biology. 10% of the class study both Mathematics and Biology. If a student is selected at random from the class, find the probability that he will be studying Mathematics or Biology.

Solution:
Let M and B denote the students of Mathematics and Biology respectively. Then, as given:
P(M) = 40% = 40100
= P(B) = 30% = 30100
P(M ∩ 5) = 10% = 10100
∴ P(M ∪ B) = P(M) + P(B) – P(M ∩ B)
40100 + 30100 – 10100
= 60100 = 60% = 0.6

Question 14.
In an entrance test that is graded on the basis of two examinations, the probability of a randomly chosen student passing the first examination is 0.8 and the probability of passing the second examination is 0.7. The probability of passing at least one of them is 0.95. What is the probability of passing both? (NCERT)

Solution:
Let probability of passing in first examination is A and passing in the second examination is B.
P(A) = 0.8, P(B) = 0.7, P(A ∪ B) = 0.95, P(A ∩ B) = ?
We know that,
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
0.95 = 0.8 + 0.7 – P(A ∩ B)
⇒ P(A ∩ B) = 1.5 – 0.95
∴ P(A ∩ B) = 0.55.

Question 15.
Check whether the following probabilities P{A) and P(B) are consistently defined:

1. P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
2. P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8.

Solution:
1. Given : P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
If P(A ∩ B) ≤ P(A) and P(A ∩ B) ≤ P(B)
Then, P(A) and P(B) are consistent.
Here

P(A) and P(B) are not consistent.

In This Post we are  providing Chapter-15 STATISTICS NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON STATISTICS

1. In a test with a maximum marks 25, eleven students scored 3,9,5,3,12,10,17,4,7,19,21 marks respectively. Calculate the range.

Ans. The marks can be arranged in ascending order as 3,3,4,5,7,9,10,12,17,19,21.

Range = maximum value – minimum value

=21-3

= 18

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2. Coefficient of variation of two distributions is 70 and 75, and their standard deviations are 28 and 27 respectively what are their arithmetic mean?

Ans. Given C.V (first distribution) = 70

Standard deviation = = 28

C.V 

= 

Similarly for second distribution

C.V 

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3. Write the formula for mean deviation.

Ans.MD

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4. Write the formula for variance

Ans. Variance 

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5. Find the median for the following data.

579101215

862226

Ans.

	5	7	9	10	12	15
	8	6	2	2	2	6
	8	14	16	18	20	26

Median is the average of 13^th and 14^th item, both of which lie in the c.f 14

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6. Write the formula of mean deviation about the median

Ans.

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7. Find the rang of the following series 6,7,10,12,13,4,8,12

Ans. Range = maximum value – minimum value

= 113-4

=9

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8. Find the mean of the following data 3,6,11,12,18

Ans. Mean = 

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9. Express in the form of a + ib (3i-7) + (7-4i) – (6+3i) + i²³

Ans. Let

Z = 

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10. Find the conjugate of 

Ans.

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11. Solve for x and y, 3x + (2x-y) i= 6 – 3i

Ans.3x = 6

x = 2

2x – y = – 3

2 × 2 – y = – 3

– y = – 3 – 4

y = 7

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12. Find the value of 1+i² + i⁴ + i⁶ + i⁸ + —- + i²⁰

Ans.

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13. Multiply 3-2i by its conjugate.

Ans. Let z = 3 – 2i

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14. Find the multiplicative inverse 4 – 3i.

Ans. Let z = 4 – 3i

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15. Express in term of a + ib 

Ans.

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In This Post we are  providing Chapter- 14 MATHEMATICAL REASONING NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS ON MATHEMATICAL REASONING

Question 1.

Give three examples of sentences which are not statements. Give reasons for the answers.

Ans. (i) The sentence “Rani is a beautiful girl” is not a statement. To some Rani may look beautiful and to other she may not look beautiful. We cannot say on logic whether or not this sentence is true.

(ii) The sentence ‘shut the door’ is not a statement. It is only an imperative sentence giving a direction to someone. There is no question of it being true or false.

(iii) The sentence ‘yesterday was Friday’ is not a statement. It is an ambiguous sentence which is true if spoken on Saturday and false if spoken on other days. Truth or false hood of the sentence depends on the time at which it is spoken and not on mathematical reasoning.

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Question 2.

Write the negation of the following statements

(i) Chennai is the capital of Tamil Nadu.

(ii) Every natural number is an integer.

Ans. (i) Chennai is not the capital of Tamil Nadu.

(ii) Every natural number is not an integer.

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Question-3

Ans.

 Question-4

Ans.

Question-5

Ans.

 Question-6

Ans.

 Question-7

Ans.

NCERT Solutions for Class 11 Maths Chapter 14 Exercise.14.3

Question-6

Ans.

 Question-7

Ans.

 Question-8

Ans.

 Question-9

Ans.

Question-10

Ans.

 Question-11

Ans.

Question-12

Ans.

 Question-11

Ans.

Question-12

Ans.

 Question-13

Ans.

 Question-14

Ans.

 Question-15

Ans.

In This Post we are  providing Chapter-13 LIMITS AND DERIVATIVES NCERT MOST IMPORTANT QUESTIONS for Class 11 MATHS which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These MCQS  can be really helpful in the preparation of Board exams and will provide you with a brief knowledge of the chapter.

NCERT MOST IMPORTANT QUESTIONS LIMITS AND DERIVATIVES

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1. Find the derivative of at 

Ans. 

At 

=

=

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2. Find the derivative of with respect to using product rule

Ans. let

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3. Find the derivative of with respect to 

Ans. let

=

=

=

=

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4. Find 

when 

Ans. 

We know that 

L. H. L. 

R. H. L. 

L. H. L. R. H. L does not exist

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5. Find the derivative of the function at Also show that 

Ans. 

At 

 Hence proved

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6. Evaluate 

Ans. 

=

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7. Find derivative of  by first principle

Ans. 

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8. Evaluate 

Ans. 

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9. Evaluate (if it exist)

Ans. 

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Ans.

For all real number exist

For 

 all integral values of exist

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10. Differentiate the function with respect to 

Ans. 

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11. Find 

Ans. 

12. Find 

Ans. 

13. Find derivative of by first principle

Ans. 

14. Find derivative of 

Ans. 

15. Find derivative of 
Ans. 

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