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NCERT MCQ CLASS-12 CHAPTER-3 | BIOLOGY NCERT MCQ | HUMAN REPRODUCTION | EDUGROWN

In This Post we are providing Chapter-3 Human Reproduction NCERT MCQ for Class 12 Biology 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 MCQ ON HUMAN REPRODUCTION

Which of the following characters is seen in females?
(a) Muscles are strong
(b) Mammary gland is well developed
(c) Voice is heavy
(d) Facial hair

Answer (b) Mammary gland is well developed

Which of the following characters is seen in males?
(a) Muscles are comparatively weak
(b) Voice is shrill
(c) Voice is heavy
(d) Beard and mustache is not seen

Answer (c) Voice is heavy

Where testes are situated?
(a) Abdominal cavity
(b) Dorsal side of the abdominal cavity
(c) (a) and (b) both
(d) Scrotal sac

Answer (d) Scrotal sac

Which hormone is released from testes?
(a) Testosterone
(b) Estrogen
(c) Progesterone
(d) Relaxin

Answer (a) Testosterone

Which hormone is released from ovaries?
(a) Testosterone
(b) Estrogen
(c) Progesterone
(d) (b) and (c) both

Answer (d) (b) and (c) both

Which of the following glands is seen in the male reproductive system?
(a) Seminal vesicle
(b) Prostate gland
(c) Bulbourethral gland
(d) All of these

Answer (d) All of these

How much lower is the temperature of the scrotal sac than the body temperature?
(a) 3 degree Celsius
(b) 4-degree Celsius
(c) 5-degree Celsius
(d) 6-degree Celsius

Answer (a) 3 degree Celsius

What is the size of the testis?
(a) 6 cm length and 2.5 cm diameter
(b) 5 cm length and 2.5 cm diameter
(c) 5 cm length and 3.5 cm diameter
(d) 6 cm length and 3.5 cm diameter

Answer (b) 5 cm length and 2.5 cm diameter

Which connective tissue surrounds the testis?
(a) Fibrous tissue
(b) Spongy connective tissue
(c) Tunica albuginea
(d) None of them

Answer (c) Tunica albuginea

Seminiferous tubules in the testis are lined with which type of cells?
(a) Germinal cells
(b) only germinal cells
(c) Sertoli cell
(d) Both a and c

Answer (d) Both a and c

In testis, which cells produce sperms?
(a) Germinal cells
(b) Epithelial cell
(c) Sertoli cell
(d) Both a and c

Answer (a) Germinal cells

Which cells provide nutrition to the sperms?
(a) Germinal cells
(b) Epithelial cell
(c) Sertoli cell
(d) None of them

Answer (c) Sertoli cell

In tests, which cells are present in the interstitial space between seminiferous tubules?
(a) Sertoli cells
(b) Germinal cells
(c) Leydig’s cells
(d) (a)and(b)both

Answer (c) Leydig’s cells

Which cells secrete testosterone?
(a) Sertoli cells
(b) Germinal cells
(c) Interstitial cells
(d) (a)and(b)both

Answer (c) Interstitial cells

Where do the seminiferous tubules of each lobe empty their sperms?
(a) Vas deferens
(b) Vasa efferentia
(c) Epididymis
(d) Seminal vesicles

Answer (b) Vasa efferentia

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NCERT MCQ CLASS-12 CHAPTER-2 | BIOLOGY NCERT MCQ | SEXUAL REPRODUCTION IN FLOWERING PLANTS | EDUGROWN

In This Post we are providing Chapter-2 Sexual Reproduction in Flowering Plants NCERT MCQ for Class 12 Biology 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 MCQ ON SEXUAL REPRODUCTION IN FLOWERING PLANTS

Question 1.
What is the function of filiform apparatus in an angiosperms embryo sac?
(a) Brings about opening of the pollen tube
(b) Guides the pollen tube into a synergid
(c) Prevents entry of more than one pollen tube into a synergid
(d) None of these
Answer:
(b) Guides the pollen tube into a synergid

Question 2.
The female gametophyte of a typical dicot at the time of fertilization is
(a) 8 – celled
(b) 7 – celled
(c) 6 – celled
(d) 5 – celled
Answer:
(b) 7 – celled

Question 3.
Polygonum type of embryo sac is
(a) 8 – nucleate, 7 – celled
(b) 8 – nucleate, 8 – celled
(c) 7 – nucleate, 7 – celled
(d) 4 – nucleate, 3 – celled
Answer:
(a) 8 – nucleate, 7 – celled

Question 4.
Both chasmogamous and cleistogamous flowers are present in
(a) Helianthus
(b) Commelina
(c) Rosa
(d) Gossypium
Answer:
(b) Commelina

Question 5.
Even in absence of pollinating agents seed-setting is assured in
(a) Commelina
(b) Zostera
(c) Salvia
(d) Fig
Answer:
(a) Commelina

Question 6.
Male and female flowers are present on different plants (dioecious) to ensure xenogamy, in
(a) papaya
(b) bottle gourd
(c) maize
(d) all of these.
Answer:
(a) papaya

Question 7.
Feathery stigma occurs in
(a) pea
(b) wheat
(c) Datura
(d) Caesalpinia
Answer:
(b) wheat

Question 8.
Plants with ovaries having only one or a few ovules are generally pollinated by
(a) bees
(b) butterflies
(c) birds
(d) wind
Answer:
(d) wind

Question 9.
Which of the following is not a water pollinated plant ?
(a) Zostera
(b) Vallisneria
(c) Hydrilla
(d) Cannabis
Answer:
(d) Cannabis

Question 10.
Spiny or sticky pollen grains and large, attractively colored flowers are associated with
(a) hydrophily
(b) entomophily
(c) ornithophily
(d) anemophily
Answer:
(b) entomophily

Question 11.
Endospermic seeds are found in
(a) castor
(b) barley
(c) coconut
(d) all of these
Answer:
(d) all of these

Question 12.
In albuminous seeds, food is stored in _______ and in non albuminous seeds, it is stored in _______.
(a) endosperm, cotyledons
(b) cotyledons, endosperm
(c) nucellus, cotyledons
(d) endosperm, radicle
Answer:
(a) endosperm, cotyledons

Question 13.
Persistent nucellus is called as _______ and is found in _______.
(a) perisperm, black pepper
(b) perisperm, groundnut ‘
(c) endosperm, black pepper
(d) endosperm groundnut
Answer:
(a) perisperm, black pepper

Question 14.
Identify the wrong statement regarding post-fertilizations development.
(a) The ovary wall develops into pericarp.
(b) The outer integument of ovule develops into tegmen.
(c) The fusion nucleus (triple nucleus) develops into endosperm.
(d) The ovule develops into seed.
Answer:
(b) The outer integument of ovule develops into tegmen.

Question 15.
Polyembryony commonly occurs in
(a) banana
(b) tomato
(c) potato
(d) citrus.
Answer:
(d) citrus.

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Class 12Th Chapter – 2 Electrostatic Potential and Capacitance |NCERT Physics Solutions | Edugrown

NCERT Solutions for Class 12 Physics Chapter 2 Electrostatic Potential and Capacitance includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 12 exams must go through NCERT Solutions for Class 12 Physics Chapter Chapter 2 Electrostatic Potential and Capacitance. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

Students can also find NCERT intext, exercises and back of chapter questions. Also working on Class 12 Physics Chapter 1 NCERT Solutions will be most helpful to the students to solve their Home works and Assignments on time.

NCERT Solutions for Class 12 Physics Chapter : 2 Electrostatic Potential and Capacitance

NCERT Exercises

Question 1.
Two charges 5 × 10^-88 C and -3 × 10^-8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero. .
Solution:
(i) Let C be the point on the line joining the two charges, where electric potential is zero, then

So, electric potential is zero at distance of 10 cm from charge of 5 × 10^-8 C on line joining the two charges between them. If point C is not between the two charges, then

So, electric potential is also equal to zero a*a distance of 24 cm from charge of -3 × 10^-8 C and at a distance of (24 +16) = 40 cm from charge of 5 × 10^-8 C, on the side of charge of -3 × 10^-8 C.

Question 2.
A regular hexagon of side 10 cm has a charge 5 μC at each of its vertices. Calculate the potential at the center of the hexagon.
Solution:

Question 3.
Two charges 2 μC and -2 μC are placed at points A and B 6 cm apart.

Identify an equipotential surface of the system.
What is the direction of the electric field at every point on this surface?

Solution:

Since it is an electric dipole, so a plane normal to AB and passing through its mid-point has zero potential everywhere.
Normal to the plane in the direction AB.

Question 4.
A spherical conductor of radius 12 cm has a charge of 1.6 × 10^-7C distributed uniformly on its surface. What is the electric field

(a) inside the sphere
(b) just outside the sphere
(c) at a point 18 cm from the centre of the sphere?

Solution:

Question 5.
A parallel plate capacitor with air between the plates has a capacitance of 8 μF. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant 6?
Solution:

Question 6.
Three capacitors each of capacitance 9 μF are connected in series.
(a) What is the total capacitance of the combination?
(b) What is the potential difference across each capacitors if the combination is connected to a 120 V supply?

Solution:

Question 7.
Three capacitors of capacitances 2 μF, 3 μF and 4 μF are connected in parallel.

What is the total capacitance of the combination.
Determine the charge on each capacitor if the combination is connected to a 100 V supply.

Solution:

C = C₁ + C₂ + C₃ = 2 + 3 + 4 or C = 9 μF
Since the capacitors are in parallel, so potential difference across each of them is same i.e.

Question 8.
In a parallel plate capacitor with air between the plates, each plate has an area of 6 × 10^-3m² and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this, capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Solution:

Question 9.
Explain what would happen if in the capacitor given in above question, a 3 mm thick mica sheet (of dielectric constant = 6) were inserted between the plates.

(a) While the voltage supply remained connected.
(b) After the supply was disconnected.

Solution:

Question 10.
A 12 μF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor?
Solution:

Question 11.
A 600 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply and is connected to another uncharged 600 μF capacitor. How much electrostatic energy is lost in the process?
Solution:

C, = 600 μF, V₁ = 200 V, C₂ = 600 μF, V₂ = 0

Question 12.
Acharge of 8 mC is located at theorigin. Calculate the work done in taking a small charge of – 2 × 10^-9 C from a point P(0,0,3 cm) to a point Q (0,4 cm, 0), via a point R (0,6 cm, 9 cm).
Solution:
As electric field is conservative field, so work done in moving a charge in electric field is independent of path chosen to move the charge in electric field and depends only on the electric potential difference between the two end points. So

Question 13.
A cube of side b has a charge q at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.
Solution:
The length of diagonal of the cube of each side b is

E = 0, as electric field at centre due to a charge at any corner of cube is just equal and opposite to that of another charge at diagonally opposite corner of cube.

Question 14.
Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:

(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.

Solution:
(a) At mid point C of line joining two charges, electric potential is V_c = V_CA + V_CB

Question 15.
A spherical conducting shell of inner radius r, and outer radius r₂ has a charge Q.

A charge q is placed at the centre of the shell. What is the surface charge density on the inner and outer surfaces of the shell?
Is the electric field inside a cavity (with no charge) zero, even if the shell is not spherical, but has any irregular shape? Explain.

Solution:

Surface charge density on the inner surface of shell is
The electric flux linked with any closed surface S inside the conductor is zero as electric field inside conductor is zero.
So, by Gauss’s theorem net charge enclosed by closed surface S is also zero i.e., $q_{ net }=0$
So, if there is no charge inside the cavity, then there cannot be any charge on the inner surface of the shell and hence electric field inside the cavity will be zero, even though the shell may not be spherical.

Question 16.
Show that the normal component of electrostatic field has a discontinuity from one side of a charged surface to another given by $\left( \overrightarrow { E } _{ 2 }-\overrightarrow { E } _{ 1 } \right) .\hat { n } =\frac { \sigma }{ \varepsilon _{ 0 } }$ where n is a unit vector normal eo to the surface at a point and o is the surface charge density at that point. (The direction of n is from side 1 to side 2). Hence show that just
outside a conductor, the electric field is $\sigma \frac { \hat { n } }{ \varepsilon _{ 0 } }$ .
(b) Show that the tangential component of electrostatic field is continuous from one side of a charged surface to another.
Solution:
Normal component of eDctric field intensity due to a thin infinite plane sheet of

(b) To show that the tangential component of electrostatic field is continuous from one side of a charged surface to another, we use the fact that work done by electrostatic field on a closed loop is zero.

Question 17.
A long charged cylinder of linear charge density k is surrounded by a hollow co-axial conducting cylinder. What is the electric field in the space between the two cylinders?
Solution:
In Figure, A is a long charged cylinder of linear charge density k, length l and radius a. A hollow co-axial conducting cylinder B of of length l and radius b surrounds A. The charge q = kl spreads uniformly on the outer surface of A. It induces – q charge on the cylinder B, which spreads on the inner surface of B. An electric field E is produced in the space between the two cylinders, which is directed radially outwards. Let us consider a co-axial cylindrical Gaussian surface of radius r. The electric flux through the cylindrical Gaussian surface is

The electric flux through the end faces of the cylindrical Gaussian surface is zero, as E is parallel to them. According to Gauss’s

Question 18.
In a hydrogen atom, the electron and proton are bound at a distance of about 0.53 A:
(a) Estimate the potential energy of the system in eV, taking the zero of the potential energy at infinite separation of the electron from proton.
(b) What is the minimum work required to free the electron, given that its kinetic energy in the orbit is half the magnitude of potential energy obtained in (a)?
(c) What are the answers to (a) and (b) above if the zero of potential energy is taken at 1.06 A separation?
Solution:

Question 19.
If one of the two electrons of a H₂ molecule is removed, we get a hydrogen molecular ion ${ H }_{ 2 }^{ + }$ . In the ground state of an ${ H }_{ 2 }^{ + }$ , the two protons are separated by roughly 1.5 A, and the electron is roughly 1 A from each proton. Determine the potential energy of the system. Specify your choice of the zero of potential energy.
Solution:
It is a system of three point charges and the potential energy stored is this system of charges is

with zero potential energy at infinity.

Question 20.
TWO charged conducting spheres of radii a and b are connected to each other by a wire. What is the ratio of electric fields at the surfaces of the two spheres? Use the result obtained to explain why charge density on the sharp and pointed ends of a conductor is higher than on its flatter portions.
Solution:
As the two spheres are connected to each other by wire, so they have same electric potential i.e.,V_a = V_bif b > a, then E_a > E_b
i.e. sphere with smaller radius produces more electric field on its surface. Hence, the charge density on the sharp and pointed ends of conductor is higher than on its flatter portions.

Question 21.
Two charges -q and +q are located at points (0,0, -a) and (0,0, a), respectively.

(a) What is the electrostatic potential at the points (0,0, z) and (x, y, 0)?
(b) Obtain the dependence of potential on the distance r of a point from the origin when r/a »1.
(c) How much work is done in moving a small test charge from the point (5, 0, 0) to (-7, 0, 0) along the x-axis? Does the

answer change if the path of the test charge between the some points is not along the x-axis?
Solution:
(a) The two point charges form an electric dipole of moment p = q 2_a directed along + z-axis. Point A (0, 0, z) lies on the axis of electric dipole, so electric potential at point A is

Point B (x, y, 0) lies on the equatorial plane of electric dipole, so electric potential at point B is zero i.e. V_B = 0
(b) Electric potential at any point on the axis of electric dipole at distance r from its centre is

(c) As both the points C (5, 0, 0) and D (-7, 0, 0) lie on the perpendicular bisector of electric dipole, so electric potential at both the points is zero. Hence work done in moving the charge from C to D is
W_CD = q₀ [V_D – V_C] × O or W_CD = O
This work done will remain equal to zero even if the path of the test charge between the same points is changed, as electric field is conservative field and work done in moving a charge between the two points in electric field is independent of the path chosen to move the charge.

Question 22.
Figure shows a charge array known as an electric quadrupole. For a point on the axis of quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).

Solution:

Question 23.
An electrical technician requires a capacitance of 2 μF in a circuit across a potential difference of 1 kV. A large number of 1 μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Solution:
Minimum number of capacitors that must be connected in series in a row are

Question 24.
What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm?
Solution:

Question 25.
Obtain the equivalent capacitance of the network in figure. For a 300 V supply, determine the charge and voltage across each capacitor.

Solution:

Question 26.
The plates of a parallel plate capacitor have an area of 90 cm² eacn and are separated by 2.5 mm.The capacitor is charged by connecting it to a 400 V supply
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electro-static field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.
Solution:

Question 27.
A 4 μF capacitor is charged by a 200 V supply. It is then disconnected from the supply, and is connected to another uncharged 2 μF capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?
Solution:

Question 28.
Show that the force on each plate of a parallel plate capacitor has magnitude equal to $\frac { 1 }{ 2 }$ QE, where Q is the charge on the capacitor, and E is the magnitude of electric field between the plates. Explain the origin of the factor $\frac { 1 }{ 2 }$ .
Solution:
Magnitude of electric field between the
plates of charged capacitor is $E\frac { \sigma }{ \varepsilon _{ 0 } }$ However, magnitude of electric field of one plate or! the other plate of charged capacitor is $E_{ 1 }\frac { \sigma }{ 2\varepsilon _{ 0 } }$ So, force on the one plate of charged capacitor due to the other is

The factor $\frac { 1 }{ 2 }$ is because the electric field of one 2 plate on the other plate of charged capacitor is $\frac { 1 }{ 2 }$ of the resultant electric field E between the 2 plates of charged capacitor.

Question 29.
A spherical capacitor consists of two concentric spherical conductors held in position by suitable insulating supports. Show that the capacitance of a spherical capacitor is given by $C=\frac { 4\Pi \varepsilon _{ 0 }r_{ 1 }r_{ 2 } }{ r_{ 1 }-r_{ 2 } }$ Where r₁ and r₂ are the radii of outer and inner spheres, respectively.
Solution:
Let + Q be the charge on outer spherical shell A of radius r, and + Q be the charge on inner spherical shell B of radius r2. Then electric potential on shell A is

This gives the capacitance of the spherical capacitor.

Question 30.
A spherical capacitor has an inner sphere of radius 12 cm and outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 μC. The space between the concentric spheres is filled with liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.
(b) What is the potential of the inner sphere.
(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.
Solution:

Here C > C₀, because a single conductor A can be charged to a electric potential till it reaches the breakdown value of surroundings. But when another earthed metallic conductor B is ; brought near it, negative charge induced on it decreases the electric potential on A, hence more charge can not be stored on A.

Question 31.
Answer carefully:

Two large conducting spheres carrying charges Q₁ and Q₂ are brought close to each other, is the magnitude of electrostatic force between them exactly given by $\frac { Q_{ 1 }Q_{ 2 } }{ 4\pi \varepsilon _{ 0 }r^{ 2 } }$ r is the distance between their centres?
If coulomb’s law involved 1 /r³ dependence (instead of 1/r²), would Gauss’s law be still true?
A small test charge is released at rest at a point in an electrostatic field configuration. Will it travel along the field line passing through that point?
What is the work done by the field of a nucleus in a complete circular orbit of the electron? What if the orbit is elliptical?
We know that electric field is discontinuous across the surface of a charged conductor. Is electric potential also discontinuous there?
What meaning would you give to the capacitance of a single conductor?
Guess a possible reason why water has a much greater dielectric constant {- 80) then say, mica (= 6).

Solution:

No, because coulomb’s law holds good only for point charges. ‘
No, because in that case electric flux linked with the closed surface will also become dependent on V other than charge enclosed by it.
No, it will travel along the field line only if it is a straight line.
Zero, whatever may be the shape of orbit may be. It is because work done in moving a charge in closed path in electric field is zero, as electric field is a conservative field.
No, electric potential is continuous
there. As E = 0, so $\frac { dV }{ dr }$ = 0 or V = constant.
It means that a single conductor is a capacitor whose other plate can be considered to be at infinity.
A water molecule is a polar molecule with non-zero electric dipole moment, however mica does not have polar molecules. So, dielectric constant of water is high.

Question 32.
A cylindrical capacitor has two co-axial cylinders of length 15 cm and radii 1.5 cm and 1.4 cm. The outer cylinder is earthed and the inner cylinder is given a charge of 3.5 μC. Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects (i.e. bending of the field lines at the ends).
Solution:

Question 33.
A parallel plate capacitor is to be designed with a voltage rating 1 kV, using a material of dielectric constant 3 and dielectric strength about 10⁷V m^-1. For safety, we should like the field never to exceed, say 10% of the dielectric strength. What minimum area of the plates is required to have a capacitance of 50 μF?
Solution:

Question 34.
Describe schematically the equipotential surfaces corresponding to

A constant electric field in the z-direction.
a field that uniformly increases in magnitude but remains in a constant (say, z) direction.
a single positive charge at the origin, and
a uniform grid consisting of long equally spaced parallel charged wires in a plane.

Solution:

Planes parallel to x-y plane or normal to the electric field in z-direction.
Planes parallel to x-y plane or normal to the electric field in z-direction, but the planes having different fixed potentials will become closer with increase in electric field intensity.
Concentric spherical surfaces with their centres at origin.
A time dependent changing shape nearer to grid, and at far off distances from the grid, it slowly becomes planar and parallel to the grid.

Question 35.
In a Van-de-Graff type generator, a spherical metal shell is to be a 15 × 106 V electrode. The dielectric strength of the gas surrounding the electrode is 5 × 10⁷V m¹. What is the minimum radius of the spherical shell required?
Solution:

Question 36.
A small sphere of radius r₁ and charge q₁ is enclosed by a spherical shell of radius r₂ and charge q₂. Show that if q: is positive, charge will necessary flow from the sphere to the shell (when the two are connected by a wire) no matter what the charge q² on the shell is.
Solution:
The potential on inner small sphere is VA

higher potential than outer conducting shell B, for any value of charge qv So, when inner sphere A is connected to outer shell B, then charge will flow from inner sphere A to outer shell B, until electric potentials on them is same i.e.
V_A – V_B = 0 or q₁ = 0 [As r₁ # r₂]

So, charge ql given to sphere A will flow on the shell B, no matter what the charge on the shell B is.

Question 37.
Answer the following:

The top of the atmosphere is at about 400 kV with respect to the surface of the earth, corresponding to an electric field that decreases with altitude. Near the surface of the earth, the field is about 100 V m^-1. Why then do we not get an electric shock as we step out of our house into the open? (Assume the house to be a steel cage, so there is no field inside).
A man fixes outside house one evening a two metre high insulating slab carrying on its top a large aluminium sheet of area 1 m₂. Will he get an electric shock if he touches the metal sheet next morning?
The discharging current in the atmosphere due to the small conductivity of air is known to be 1800 A on an average over the globe. Why then does the atmosphere not discharge itself completely in due course and become electrically neutral? In other words, what keeps the atmosphere charged?
What are the forms of energy into which the electrical energy of the atmosphere is dissipated during a lightning?

Solution:

Our body and the ground are at same electric potential. As we step out into the open, the original equipotential surfaces of open air change, keeping our head and the ground at the same potential.
Yes, it is because the aluminium sheet gets charged due to discharging current and raises to the extent depending on the capacitor formed by the sheet and the ground.
The atmosphere of earth gets continuously charged due to lightning, thunderstorms but simultaneously it gets discharged through normal weather zones. This keeps the system balanced.
Light, sound and heat energy. Light energy in lightning and heat and sound energy in the accompanying thunder.

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Class 12Th Chapter – 1 Electric Charges and Fields |NCERT Physics Solutions | Edugrown

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges And Fields includes all the important topics with detailed explanation that aims to help students to understand the concepts better. Students who are preparing for their Class 12 exams must go through NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges And Fields. Going through the solutions provided on this page will help you to know how to approach and solve the problems.

NCERT Solutions for Class 12 Physics Chapter : 1 Electric Charges And Fields

NCERT Exercises

Question 1.
What is the force between two small charged spheres having charges of 2 × 10^-7C and 3 × 10^-7C placed 30 cm apart in air?
Solution:

Question 2.
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8 μC in air is 0.2 N.
(a)What is the distance between the two spheres?
(b)What is the force on the second sphere due to the first?
Solution:

Question 3.
Check that the ratio $\frac { Ke^{ 2 } }{ Gm_{ e }m_{ p } }$ is dimensionless. Look up a table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Solution:
The ratio of electrostatic force to the gravitational force between an electron and a proton separated by a distance r from each other is

This ratio signifies that electrostatic forces are 10³⁹ times stronger than gravitational forces.

Question 4.
(a) Explain the meaning of the statement ‘electric charge of a body is quantised’.
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large scale charges?
Solution:
(a) The net charge possessed by a body is an integral multiple of charge of an electron i.e. q = ± ne, where n = 0,1,2,3,… is the number of electrons lost gained by the body and e = 1.6 × 10^-19 C is charge of an electron. This is called law of quantization of charge.
(b) At macroscopic level, charges are enormously large as compared to the charge of an electron, e = 1.6 × 10^-19C. Even a charge of 1 nC contains nearly 10¹³ electronic charges. So, at this large scale, charge can have a continuous value rather than discrete integral multiple of e, and hence, the quantization of electric charge can be ignored.

Question 5.
When a glass rod is rubbed with a silk cloth, charges appears on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Solution:
When a glass rod is rubbed with a silk cloth, electrons from the glass rod are transferred to the piece of silk cloth. Due to this, the glass rod acquires positive (+) charge whereas the silk cloth acquires negative (-) charge. Before rubbing, both the glass rod and silk cloth are neutral and after rubbing the net charge on both of them is also equal to zero. Such similar phenomenon is observed with many other pairs of bodies. Thus, in an isolated system of bodies, charge is neither created nor destroyed, it is simply transferred from one body to the other. So, it is consistent with the law of conservation of charge.

Question 6.
Four point charges q_A = + 2 μC, q_B = -5μC, q_c = + 2 μC and q_D = -5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?
Solution:

Forces of repulsion on 1 μC charge at O due to 2 μC charge, at A and C are equal and opposite. Therefore, they cancel. Similarly, forces of attraction on 1 μC charge at O, due to -5 μC charges at B and at D are also equal and opposite. Therefore, these also cancel. Hence, the net force on the charge of 1 μC at O is zero.

Question 7.
(a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
(b)Explain why two field lines never cross each other at any point?
Solution:

A field line cannot have sudden breaks because the moving test charge never jumps from one position to the other.
(b)Two field lines never cross each other at any point/because if they do so, we will obtain two tangents pointing in two different directions of electric field at a point, which is not possible.

Question 8.
Two point charges q_A = 3 μC and q_B = -3 μC are located 20 cm apart in vacuum.
(a)What is the electric field at the midpoint O of the line AS joining the two charges?
(b)If a negative test charge of magnitude 1.5 × 10^-9 C is placed at this point, what is the force experienced by the test charge?
Solution:
(a) Electric fields at O due to the charges at A and B are

In direction opposite to that of E i.e. along OA.

Question 9.
A system has two charges q_A = 2.5 × 10^-7C and q_B = -2.5 × 10^-7C located at point A (0,0, -15 cm) and B (0,0, +15 cm), respectively. What are the total charge and electric dipole moment of the system?
Solution:

Question 10.
An electric dipole with dipole moment 4 × 10^-9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ N C^-1. Calculate the magnitude of the torque acting on the dipole.
Solution:

Question 11.
A Polythene piece rubbed with wool is found to have a negative charge of 3 × 10^-7C.
(a) Estimate the number of electrons transferred (from which to which?)
(b) Is there a transfer of mass from wool to polythene?
Solution:
(a) q = ne or n = $\frac { q }{ e } =\frac { 3\times 10^{ -7 } }{ 1.6\times 10^{ -19 } }$
= 1.875 × 10¹²
so, 1.875 × 10¹² electrons have transferred from wool to polythene, as polythene acquires negative charge.
(b) Yes, mass of 1.875 × 10¹² electrons i.e., m = nm_e = 1.875 × 10¹² × 9.1 × 10^-31 kg = 1.71 × 10^-18 kg has transferred from wool to polythene

Question 12.
(a) Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10^-7C ? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Solution:

Question 13.
Suppose the spheres A and B in previous question have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
Solution:

Question 14.
Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Solution:
Particles 1 and 2 are negatively charged as they experience forces in direction opposite to that of electric field E, whereas particle 3 is positively charged as it experience force in the direction of electric field E.

Particle-3 has the highest charge to mass ratio, as it shows maximum deflection in the electric field.

Question 15.
Consider a uniform electric field $\overrightarrow { E } =3\times 10^{ 3 }\hat { i } NC^{ -1 }$ .
(a)What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
(b)What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
Solution:

Question 16.
What is the net flux of the uniform electric field in previous question through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?
Solution:
${ \phi }_{ net }=0,$ As the net electric flux with closed surface like cube in uniform electric
field is equal to zero, because the number of lines entering the cube is the same as the number of lines leaving the cube.

Question 17.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux though the surface of the box is 8.0 × 10³ N m² C^-1.
(a)What is the net charge inside the box?
(b)If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?
Solution:

So, the net charge enclosed by that closed surface is zero, although it may have some charges inside it.

Question 18.
A point charge +10 μC is at a distance of 5 cm directly above the centre of a square of side 10 cm, as shown in figure. What is the magnitude of the electric flux though the square?
Solution:
Let us assume that the given square be one face of the cube of edge 10 cm. As charge of +10 μC is at a distance of 5 cm above the centre of a square, so it is enclosed by the cube. Hence by Gauss’s theorem, electric flux linked with the cube is

Question 19.
A point charge of 2.0 pC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Solution:

Question 20.
A point charge causes an electric flux of -1.0 × 10³ Nm² C^-1 to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?
Solution:
(a) On increasing the radius of the Gaussian surface, charge enclosed by it remains the same and hence the electric flux linked with Gaussian surface also remains the same.

Question 21.
A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 10³ N C^-1 and points radially inward. What is the net charge on the sphere?
Solution:

Question 22.
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC m^-2.
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere?
Solution:

Question 23.
An infinite line charge produces a field of 9 × 10⁴NC^-1 at a distance of 2 cm. Calculate the linear charge density.
Solution:

Question 24.
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitudes 17.0 × 10^-22 C m^-2. What is Electric field in
(a) the outer region of the first plate.
(b) the outer region of the second plate, and
(c) between the plates?
Solution:
As both the plates have same surface charge density σ, so

Question 25.
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 10⁴ N C^-1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm^-3. Estimate the radius of the drop.
(g = 9.81 m s^-2; e= 1.6 × 10^-19C)
Solution:
In equilibrium, force due to electric field on the drop balances the weight mg of drop

Question 26.
Which among the curves shown in Figure cannot possibly represent electrostatic field lines?
Solution:
(a) It is wrong, because electric field lines must be normal to the surface of conductor outside it.
(b) It is wrong because field lines cannot start or originate from negative charge, and also cannot end or submerge into positive charge.
(c) It is correct
(d) It is wrong because electric field lines never intersect each other.
(e) It is wrong because electric field lines cannot form closed loops.

Question 27.
In a certain region of space, electric field is along the z-direction throughout. The magnitude of electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ N C^-1 per metre. What are the force and torque experienced by a system having a total dipole moment equal to 10^-7 C m in the negative z-direction?
Solution:
As electric field increases in positive z-direction from A to B, So

directed in direction of FB i.e., from B to A or along -z direction.
As the two forces on charges of electric dipole are collinear and opposite, so net torque on it is equal to zero.

Question 28.
(a) A conductor A with a cavity is shown in figure (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q. (figure (b))
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.
Solution:
(a) Let us consider a closed surface inside the conductor enclosing the cavity. As electric field inside the conductor is zero,

This shows that there cannot be any charge on the inner surface of conductor at the cavity, so any charge Q given to the conductor will appear only on its outer surface.
(b) Let us consider that the charge -q_I is induced on the inner surface of conductor A at the cavity, when conductor B of charge q is kept in its cavity. Due to this charge of Q + qt is induced on the outer surface of conductor. Let us construct a closed Gaussian surface S inside the conductor A enclosing its cavity. As, electric field inside conductor is zero, so

i.e., equal and opposite charge, -q is induced on inner surface of conductor A at the cavity,

when conductor B of charge, + q is kept is its cavity. Hence, the charge on outer surface of conductor is Q + q₁ = Q + q
(c) As we found that electric field inside the cavity of conductor is zero, even on charging the conductor, so the sensitive instrument can be shielded from the strong electrostatic fields in its environment, by covering it with a metallic cover.

Question 29.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left( \frac { \sigma }{ 2\varepsilon _{ 0 } } \hat { n } \right)$ .where $\hat { n }$ is the unit vector in the outward normal direction, and o is the surface charge density near the hole.
Solution:
Inside a charged conductor, the electric field is zero.

But a uniformly charged flat surface provide

If we consider a small flat part on the surface of charged conductor, it certainly provides an
Now if a hole is made in charged conductor, the fifeld due to small flat part is absent but the field due to rest of charged conductor is present i.e., equal to $\frac { \sigma }{ 2\varepsilon _{ 0 } } \hat { n }$ .

Question 30.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density X without using Gauss’s law.
Solution:
Consider a point P, a unit away from the long charged wire. Electric field due to element dy,

vertical components cancel out and horizontal components are added due to symmetry

Question 31.
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + 2/3 e, and the ‘down’ quark (denoted by d) of charge (-1/3)e, together with electrons build up ordinary matter. [Quarks of other types have also been found which give rise to different unusual varieties of matter]. Suggest a possible quark composition of a proton and neutron.
Solution:

Question 32.
(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = o) of the configuration. Show that the equilibrium of the test charge is necessarily unstable, (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed at certain distance apart.
Solution:
(a) Let us assume that a radial electric field is present with origin as centre. The E is 0 at origin and a small test charge ‘q0’ is placed at origin. Force on test charge at origin is zero. $\overrightarrow { F } =q\overrightarrow { E } =0$ But as the test charge is displaced a little, the electrostatic force will drift it further away so charge is in unstable equilibrium.

(b) A test charge placed at mid point of equal charges is in stable equilibrium, for lateral displacement. Initially at mid point position of test charge, two repulsion forces on test charge are equal and opposite.

But on displacing q₀ close to one of the charge, one of the forces become stronger and pushes the charge towards another.

F_A > F_B so q₀ moves towards B.

Question 33.
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed vx.The length of plate is L and an uniform electric field £ is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/ $2m{ v }_{ x }^{ 2 }$ .
Solution:
Let the point at which the charged particle enters the electric field, be origin O (0, 0), then after travelling a horizontal displacement L, it gets deflected by displacement y in vertical direction as it comes out of electric field. So, co-ordinates of its initial position are x₁ = 0 and y₁ = 0 and final position on coming out of electric field are

and of initial velocity are u_x = v_x and u_y = 0 so, by 2^nd equation of motion in horizontal direction,

This gives the vertical deflection of the particle at the far edge of the plate.

Question 34.
Suppose that the particle in question 33 is an electron projected with velocity v_x = 2.0 × 10⁶ m s^-1. If E between the plates separated by 0.5 cm is 9.1 × 10² N C^-1 where will the electron strike the upper plate? (|e| = 1.6 × 10^-19, me = 9.1 × 10^-31 kg)
Solution:
If the electron is released just near the negatively charged plate, then y = 0.5 cm and hence

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Class 12th Chapter -13 Probability | NCERT Maths Solution | NCERT Solution | Edugrown

It is important that you have a strong foundation in Maths Solutions. This can help you immensely in exams. Because there is no paper pattern set for the board exams. Every year there are different twists and turns in questions asked in board exams. While preparing for maths it becomes to go through every bit of detail. Thus, NCERT Solutions Class 12 Maths can help you lay a strong foundation. There solutions of each and every question given in the NCERT 12 Textbook in this article. There are crucial chapters like integration, algebra, and calculus which you can solve easily using these Class 12 Maths solutions. Using these NCERT Maths Class 12 Solutions can give you an edge over the other students.

NCERT Solutions for Class 12 Maths Chapter :13 Probability

Class 12 Maths Chapter 13 Probability Ex 13.1

Ex 13.1 Class 12 Maths Question 1
Given that E and Fare events such that
P (E) = 0.6, P (F) = 0.3 and P(E∩F) = 0.2
find P(E|F) and P (F|E).
Solution:
Given: P (E)=0.6, P (F)=0.3, P (E ∩ F)=0.2
$P(E|F)=\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.3 } =\frac { 2 }{ 3 }$
$P(F|E)=\frac { P(E\cap F) }{ P(E) } =\frac { 0.2 }{ 0.6 } =\frac { 1 }{ 3 }$

Ex 13.1 Class 12 Maths Question 2
Compute P(A|B) if P(B)=0.5 and P (A∩B) = 0.32.
Solution:
Given: P (B)=0.5, P(A∩B)=0.32
$P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { 0.32 }{ 0.50 } =\frac { 32 }{ 50 } =0.64$

Ex 13.1 Class 12 Maths Question 3.
If P (A)=0.8, P (B)=0.5 and P(B/A)=0.4, find
(i) P(A∩B)
(ii) P(A/B)
(iii)P(A∪B)
Solution:
(i) $P(B/A)=\frac { P(A\cap B) }{ P(A) } \Rightarrow 0.4=\frac { P(A\cap B) }{ 0.8 }$
∴ P(A∩B) = 0.4 x 0.8 = 0.32
(ii) P(A/B) = $P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 0.32 }{ 0.5 } =\frac { 32 }{ 50 } =\frac { 16 }{ 25 }$
(iii) P(A∪B) = P(A) + P(B) – P(A∩B)
= 0.8 + 0.5 – 0.32 = 1.30 – 0.32 = 0.98

Ex 13.1 Class 12 Maths Question 4.
Evaluate P(A∪B) if 2P(A) = P(B) = $\frac { 5 }{ 13 }$ and P(A|B) = $\frac { 2 }{ 5 }$ .
Solution:
Given:
2P(A) = P(B) = $\frac { 5 }{ 13 }$ and P(A|B) = $\frac { 2 }{ 5 }$ .
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q4.1

Ex 13.1 Class 12 Maths Question 5.
If P(A) = $\frac { 6 }{ 11 }$ ,P(B) = $\frac { 5 }{ 11 }$ and P(A∪B) = $\frac { 7 }{ 11 }$ ,
Find
(i) P(A∩B)
(ii) P(A|B)
(iii) P(B|A)
Solution:
Given:
P(A) = $\frac { 6 }{ 11 }$ ,P(B) = $\frac { 5 }{ 11 }$ and P(A∪B) = $\frac { 7 }{ 11 }$ ,
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q5.1

Ex 13.1 Class 12 Maths Question 6.
Determine P(E/F) in question 6 to 9:
A coin is tossed three times, where
(i) E: head on third toss F: heads on first two tosses.
(ii) E: at least two heads F : at most two heads
(iii) E: at most two tails F: at least one tail
Solution:
(i) E = Head occurs on third toss as {HHH, HTH, THH, TTH}
F : Heads on first two tosses = {HHH, HHT} E ∩ F = {HHH}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q6.1

Ex 13.1 Class 12 Maths Question 7.
Two coins are tossed once
(i) E: tail appears on one coin F: one coin shows head
(ii) E: no tail appears F: no head appears.
Solution:
S = {HH, TH, HT, TT} n (S) = 4
(i) E : tail appears on one coin
E = {TH,HT}, P{E) = $\frac { 1 }{ 2 }$
F : one coin shows head,
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q7.1

Ex 13.1 Class 12 Maths Question 8.
A die is thrown three times.
E: 4 appears on the third toss
F: 6 and 5 appears respectively on first two tosses.
Solution:
A die is thrown three times E : 4 appears on third toss = {(1,1,4), (1,2,4), (1,3,4), (1,4,4), (1,5,4),
(1.6.4),(2,1,4), (2,2,4), (2,3,4), (2,4,4), (2,5,4),
(2.6.4), (3,1,4), (3,2,4), (3,3,4), (3,4,4), (3,5,4),
(3.6.4), (4,1,4), (4,2,4), (4,3,4), (4,4,4), (4,5,4),
(4.6.4), (5,1,4), (5,2,4), (5,3,4), (5,4,4), (5,5,4),
(5.6.4), (6,1,4), (6,2,4), (6,3,4), (6,4,4), (6,5,4),
(6.6.4)}
These are 36 cases
F: 6 and 5 appears respectively on first two tosses = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6, 5,5), (6,5,6)}
These are six cases E ∩ F = {6,5,4}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q8.1

Ex 13.1 Class 12 Maths Question 9.
Mother, father and son line up at random for a family picture:
E: son on one end, F: father in middle
Solution:
Mother (m), Father (f) and son (s) line up at random E : son on one end: {(s, m, f), (s, f, m), (f, m, s), (m, f, s)}
F: Father in middle: {(m, f,s), (s, f, m), (s, f, m)}
E∩F = {(m, f, s), (s, f, m)}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q9.1

Ex 13.1 Class 12 Maths Question 10.
A Mack and a red die are rolled.
(a) Find the conditional probability of obtaining a sum greater than 9, given that the black die resulted in a 5.
(b) Find the conditional probability of obtaining the sum 8, given that the red die resulted in a number less than 4.
Solution:
(a) n(S) = 6 x 6 = 36
Let A represent obtaining a sum greater than 9 and B represents black die resulted in a 5.
A= {46,64,55,36,63,45,54,65,56,66}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q10.1

Ex 13.1 Class 12 Maths Question 11
A fair die is rolled. Consider events E = {1,3,5} F = {2,3} and G = {2,3,4,5}, Find
(i) P(E|F) and P(F|E)
(ii) P(E|G) and P(G|E)
(iii) P((E ∪ F) |G) and P (E ∩ F)|G)
Solution:
(i) E = {1,3,5}, F = {2,3},E∩F = {3}
P(E) = $\frac { 3 }{ 6 }$ ,P(F) = $\frac { 2 }{ 6 }$ ,P(E∩F) = $\frac { 1 }{ 6 }$ ,
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q11.1

Ex 13.1 Class 12 Maths Question 12.
Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that
(i) the youngest is a girl,
(ii) at least one is girl?
Solution:
Let first and second girls are denoted by G1 and G2 and Boys by B1 and B2.
Sample space
S = {(G1G2),(G1B2),(G2B1),(B1B2)}
Let A = Both the children are girls = {G1G2}
B = youngest child is a girls = {G1G2, B1G2}
C = at least one is a girl = {G1B2, G1G2, B1G2}
A∩B = {G1G2},
A∩C = {G1G2}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q12.1

Ex 13.1 Class 12 Maths Question 13.
An instructor has a question bank consisting of 300 easy True/ False questions, 200 difficult True/ False questions, 500 easy multiple choice questions and 400 difficult multiple choice questions. If a question is selected at random from the question bank, what is the probability that it will be an easy question given that it is a multiple choice question?
Solution:
The given data may be tabulated as
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q13.1

Ex 13.1 Class 12 Maths Question 14.
Given that the two numbers appearing on throwing two dice are different. Find the probability of the event ‘the sum of numbers on the dice is 4’.
Solution:
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q14.1
Let A represents the event “the sum of numbers on the dice is 4”
and B represents the event “the two numbers appearing on throwing two dice are different.”
A = {13,22,31}, n (A) = 3

Ex 13.1 Class 12 Maths Question 15.
Consider the experiment of throwing a die, if a multiple of 3 comes up, throw the die again and if any other number comes, toss a coin. Find the conditional probability of the event ‘the coin shows a tail’ given that ‘at least one die shows a 3’.
Solution:
Let there be n throws in which a multiple of 3 occurs every time.
Probability of getting a multiple of 3 (i.e. 3 or 6)
in one throw = $\frac { 2 }{ 6 }$ = $\frac { 1 }{ 3 }$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.1 Q15.1

In each of the following choose the correct answer:

Ex 13.1 Class 12 Maths Question 16.
If P(A) = $\frac { 1 }{ 2 }$ , P (B) = 0 then P (A | B) is
(a) 0
(b) $\frac { 1 }{ 2 }$
(c) not defined
(d) 1
Solution:
P(A) = P(B) = 0
∴ P(A∩B) = 0
$\therefore P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { 0 }{ 0 }$
Thus option C is correct

Question 17.
If A and B are events such that P(A | B) = P(B | A) then
(a) A⊂B but A≠B
(b) A = B
(c) A∩B = φ
(d) P(A) = P(B)
Solution:
P(A | B) = P(B | A)
Thus option (d) is correct.
$\frac { P(A\cap B) }{ P(B) } =\frac { P(B\cap A) }{ P(A) }$
⇒ P(A) = P(B)

Class 12 Maths Chapter 13 Probability Ex 13.2

Ex 13.2 Class 12 Maths Question 1.

If P(A) = $\frac { 3 }{ 5 }$ and P(B) = $\frac { 1 }{ 5 }$ , find P(A∩B) if A and B are independent events.
Solution:
A and B are independent if P (A ∩ B)
$=P(A)\times P(B)=\frac { 3 }{ 5 } \times \frac { 1 }{ 5 } =\frac { 3 }{ 25 }$

Ex 13.2 Class 12 Maths Question 2.
Two cards are drawn at random and without replacement from a pack of 52 playing cards.Find the probability that both the cards are black.
Solution:
Number of exhaustive cases = 52
Number of black cards = 26
One black card may be drawn in 26 ways
∴ Probability of getting a black card,
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q2.1

Ex 13.2 Class 12 Maths Question 3.
A box of oranges is inspected by examining three randomly selected oranges drawn without replacement. If all the three oranges are good, the box is approved for sale otherwise it is rejected. Find the probability that a box containing 15 oranges out of which 12 are good and 3 are bad ones will be approved for sale.
Solution:
S = {12 good oranges, 3 bad oranges),
n(S) = 15
P (a box is approved) = $\frac { C(12,3) }{ C(15,3) } =\frac { 12\times 11\times 10 }{ 15\times 14\times 13 } =\frac { 44 }{ 91 }$

Ex 13.2 Class 12 Maths Question 4.
A fair coin and an unbiased die are tossed. Let A be the event ‘head appears on the coin’ and B be the event ‘3 on the die’. Check whether A and B are independent events or not
Solution:
When a coin is thrown, head or tail will occur
Probability of getting head P(A) = $\frac { 1 }{ 2 }$
When a die is tossed 1,2,3,4, 5, 6 one of them will appear
∴ Probabillity of getting 3 = P(B) = $\frac { 1 }{ 6 }$
When a die and coin is tosses, total number of cases are
H1,H2,H3,H4,H5,H6
T1,T2,T3,T4,T5,T6
Head and 3 will occur only in 1 way
∴ Probability of getting head and $3=\frac { 1 }{ 12 }$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q4.1

Ex 13.2 Class 12 Maths Question 5.
A die marked 1,2,3 in red and 4,5,6 in green is tossed. Let A be the event, ‘the number is even’, and B be the event, ‘the number is red’. Are A and B independent?
Solution:
Even numbers on die are 2,4,6
∴ Probability of getting even number
P(A) = $\frac { 3 }{ 6 }$ = $\frac { 1 }{ 2 }$
There are two colours of the die – red and green.
Probability of getting red colour, P (B) = $\frac { 1 }{ 2 }$
Even number in red colour is 2
∴ Probability of getting red colour and even number
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q5.1

Ex 13.2 Class 12 Maths Question 6.
Let E and F be the events with P(E) = $\frac { 3 }{ 5 }$ , P (F) = $\frac { 3 }{ 10 }$ and P (E ∩ F) = $\frac { 1 }{ 5 }$ . Are E and F independent?
Solution:
P(E) = $\frac { 3 }{ 5 }$ ,P(F) = $\frac { 3 }{ 10 }$ ,
∴ P (E) x P (F) = $\frac { 3 }{ 5 } \times \frac { 3 }{ 10 } =\frac { 9 }{ 50 }$
P(E∩F)≠P(E)xP(F)
∴ The event A and B are not independent.

Ex 13.2 Class 12 Maths Question 7.
Given that the events A and B are such that P(A) = $\frac { 1 }{ 2 }$ ,P(A∪B) = $\frac { 3 }{ 5 }$ and P(B) = p. Find p if they are
(i) mutually exclusive
(ii) independent.
Sol. Let P(A∩B) = x,Now P(A) = $\frac { 1 }{ 2 }$ ,P(A∪B) = $\frac { 3 }{ 5 }$ , P(B) = P
P(A∪B) = P (A) + P (B) – P (A∩B)
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q7.1

Ex 13.2 Class 12 Maths Question 8.
Let A and B independent events P(A) = 0.3 and P(B) = 0.4. Find
(i) P(A∩B)
(ii) P(A∪B)
(iii) P (A | B)
(iv) P(B | A)
Solution:
P (A) = 0.3,
P (B) = 0.4
A and B are independent events
(i) ∴ P (A∩B) = P (A). P (B) = 0.3 x 0.4 = 0.12.
(ii) P(A∪B) = P(A) + P(B) – P(A).P(B)
= 0.3 + 0.4 – 0.3 x 0.4 = 0.7 – 0.12 = 0.58.
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q8.1

Ex 13.2 Class 12 Maths Question 9.
If A and B are two events, such that P (A) = $\frac { 1 }{ 4 }$ , P(B) = $\frac { 1 }{ 2 }$ ,and P(A∩B) = $\frac { 1 }{ 8 }$ .Find P (not A and not B)
Solution:
Event not A and not B = $\overline { A } \cap \overline { B }$
$P(\overline { A } \cap \overline { B } )=1-P(A\cup B)$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q9.1

Ex 13.2 Class 12 Maths Question 10
Events A and B are such that
P(A) = $\frac { 1 }{ 2 }$ ,P(B) = $\frac { 7 }{ 12 }$ and P (not A or not B) = $\frac { 1 }{ 4 }$ . State whether A and Bare independent
Solution:
$P(\overline { A } \cup \overline { B } )=1-P(A\cap B)$
$\Rightarrow \frac { 1 }{ 4 } =1-P(A\cap B)$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q10.1

Ex 13.2 Class 12 Maths Question 11
Given two independent events A and B such that P (A) = 0.3, P(B) = 0.6. Find
(i) P(A and B)
(ii) P(A and not B)
(iii) P (A or B)
(iv) P (neither A nor B)
Solution:
(i) A and B are independent events
∴ P(A and B) = P(A∩B) = P(A) x P(B)
= 0.3 x 0.6 [ ∵P (A) 0.3), P (B) = 0.6]
∴ P(A and B) = 0.18
(ii) P(A and B) = and
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q11.1

Ex 13.2 Class 12 Maths Question 12
A die is tossed thrice. Find the probability of getting an odd number at least once.
Solution:
S = {1,2,3,4,5,6},n(S) = 6
Let A represents an odd number.
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q12.1

Ex 13.2 Class 12 Maths Question 13
Two balls are drawn at random with replacement from a box containing 10 black and 8 red balls. Find the probability that
(i) both balls are red.
(ii) first ball is “black and second is red.
(iii) one of them is black and other is red.
Solution:
S = {10 black balls, 8 red balls}, n (S) = 18
Let drawing of a red ball be a success.
A = {8 red balls}, n (A) = 8
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q13.1

Ex 13.2 Class 12 Maths Question 14
Probability of solving specific problem independently by A and B are $\frac { 1 }{ 2 }$ and $\frac { 1 }{ 3 }$ respectively. If both try to solve the problem independently, find the probability that
(i) the problem is solved
(ii) exactly one of them solves the problem.
Solution:
Probability that A solves the problem = $\frac { 1 }{ 2 }$
=> Probablility that A does not solve the problem
P (A) = $1-\frac { 1 }{ 2 } =\frac { 1 }{ 2 }$
Probability that B solves the problem = $\frac { 1 }{ 3 }$
=> Probability that B does not solved the
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q14.1

Ex 13.2 Class 12 Maths Question 15
One card is drawn at random from a well shuffled deck of 52 cards. In which of the following cases are the events E and F independent?
(i) E: ‘the card drawn is a spade’
F: ‘the card drawn is an ace ’
(ii) E: ‘the card drawn is black’
F: ‘the card drawn is a king’
(iii) E: ‘the card drawn is a king or queen ’
F: ‘the card drawn is a queen or jack ’.
Solution:
n(S) = 52
(i) E = {13 spades},P(E) = $\frac { 13 }{ 52 }$ = $\frac { 1 }{ 4 }$
F = {4 aces}, P(F) = $\frac { 4 }{ 52 }$ = $\frac { 1 }{ 13 }$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q15.1

Ex 13.2 Class 12 Maths Question 16
In a hostel, 60% of the students read Hindi newspaper, 40% read English newspaper and 20% read both Hindi and English newspapers. A student is selected at random
(a) Find the probability that she reads neither Hindi nor English newspapers.
(b) If she reads Hindi newspaper, find the probabililty that she reads english newspapers.
(c) If she reads English newspaper, find the probability that she reads Hindi newspaper.
Solution:
(a) Let H and E represent the event that a student reads Hindi and English newspaper respectively
P (H) = 0.6, P (E) = 0.4, P (H ∩ E) = 0.2
Probability that the student reads at least one paper
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q16.1

Choose the correct answer in the following Question 17 and 18:

Ex 13.2 Class 12 Maths Question 17
The probability of obtaining an even prime number on each die when a pair of dice is rolled is
(a) 0
(b) $\frac { 1 }{ 3 }$
(c) $\frac { 1 }{ 12 }$
(d) $\frac { 1 }{ 36 }$
Solution:
(d) n(S) = 36
Let A represents an even prime number one each dice.
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.2 Q17.1

Ex 13.2 Class 12 Maths Question 18
Two events A and B are said to be independent, if
(a) A and B are mutually exclusive
(b) P(A’B’) = [1 – P(A)] [1 – P(B)]
(c) P(A) = P(B)
(d) P (A) + P (B) = 1
Solution:
(b) P(A’and B’) = [1 – P(A)]. [1 – P(B) = P(A). P(B’)
Thus option (b) is correct.

Class 12 Maths Chapter 13 Probability Ex 13.3

Ex 13.3 Class 12 Maths Question 1.
An urn contain 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random.What is the probability that the second ball is red?
Solution:
Urn contain 5 red and 5 black balls.
(i) Let a red ball is drawn.
probability of drawing a red ball = $\frac { 5 }{ 10 }$ = $\frac { 1 }{ 2 }$
Now two red balls are added to the urn.
=> The urn contains 7 red and 5 black balls.
Probability of drawing a red ball = $\frac { 7 }{ 12 }$
(ii) Let a black ball is drawn at first attempt
Probability of drawing a black ball = $\frac { 5 }{ 10 }$ = $\frac { 1 }{ 2 }$
Next two black balls are added to the urn
Now urn contains 5 red and 7 black balls
Probability of getting a red ball = $\frac { 5 }{ 12 }$
=> Probability of drawing a second ball as red
$=\frac { 1 }{ 2 } \times \frac { 7 }{ 12 } +\frac { 1 }{ 2 } \times \frac { 5 }{ 12 } =\frac { 7 }{ 24 } +\frac { 5 }{ 24 } =\frac { 12 }{ 24 } =\frac { 1 }{ 2 }$

Ex 13.3 Class 12 Maths Question 2.
A bag contains 4 red and 4 black balls, another bag contains 2 red and 6 black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
Solution:
Let A be the event that ball drawn is red and let E1 and E2 be the events that the ball drawn is from the first bag and second bag
respectively. P(E1) = $\frac { 1 }{ 2 }$ , P(E2) = $\frac { 1 }{ 2 }$ .
P (A|E 1) = Probability of drawing a red ball from bag
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q2.1

Ex 13.3 Class 12 Maths Question 3.
Of the students In a college, it is known that 60% reside In hostel and 40% are day scholars (not residing In hostel). Previous year results report that 30% of all students who reside in hostel attain A grade and 20% of day scholars attain A grade in their annual examination. At the end of the year, one student Is chosen at random from the college and he has an A- grade what Is the probability that the student is a hostlier?
Solution:
Let E1, E2 and A represents the following:
E1 = students residing in the hostel,
E2 day scholars (not residing in the hostel)
and A = students who attain grade A
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q3.1

Ex 13.3 Class 12 Maths Question 4.
In answering a question on a multiple choice test, a student either knows the answer or 3 guesses. Let $\frac { 3 }{ 4 }$ be the probability that he knows the answer and $\frac { 1 }{ 4 }$ be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability $\frac { 1 }{ 4 }$ . What is the probability that the student knows the answer given that he answered it correctly?
Solution:
Let the event E1 = student knows the answer , E2 = He gusses the answer
P(E1) = $\frac { 3 }{ 4 }$ ,P(E2) = $\frac { 1 }{ 4 }$
Let A is the event that answer is correct, if the student knows the answer
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q4.1

Ex 13.3 Class 12 Maths Question 5.
A laboratory blood test is 99% effective in detecting a certain disease when it is, in fact, present. However, the test also yields a false positive result for 0.5% of the healthy person tested (i.e. if a healthy person is tested, then, with probability 0.005, the test will imply he has the disease). If 0.1 percent of the population actually has the disease, what is the probability that a person has the disease given that his test result is positive?
Solution:
Let
E1 = The person selected is suffering from certain disease,
E2 = The person selected is not suffering from certain disease.
A = The doctor diagnoses correctly
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q5.1

Ex 13.3 Class 12 Maths Question 6.
There are three coins. One is a two headed coin, another is a biased coin that conies up heads 75% of the time and third is an unbiased coin. One of the three coins is choosen at random and tossed, it shows head, what is the probability that it was the two headed coin?
Solution:
Let E1, E2, E3 and A denotes the following:
E1 = a two headed coin, E2 = a biased coin,
E3 = an unbiased coin, A=A head is shown
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q6.1

Ex 13.3 Class 12 Maths Question 7.
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of mi accident are 0.01, 0.03, 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?
Solution:
An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. Total number of drivers
=2000 + 4000 + 6000 = 12,000
Probability of selecting a scooter driver
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q7.1

Ex 13.3 Class 12 Maths Question 8
A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective.All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B.?
Solution:
E₁ and E₂ are the events the percentage of production of items by machine A and machine B respectively.
Let A denotes defective item.
Machine A’s production of items = 60 %
Probability of production of items by machine
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q8.1

Ex 13.3 Class 12 Maths Question 9.
Two groups are competing for the position on the Board of directors of a corporation. The probabilities that the first and the second groups will win are 0.6 and 0.4 respectively. Further, if the first group wins, the probability of introducing a new product is 0.7 and the corresponding probability is 0.3 if the second group wins. Find the probability that the new product introduced was by the second group.
Solution:
Given: P (G1) = 0.6, P (G2) = 0.4
P represents the launching of new product P(P|G1) = 0.7 and P(P|G2) = 0.3
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q9.1

Ex 13.3 Class 12 Maths Question 10.
Suppose a girl throws a die. If she gets a 5 or 6, she tosses a coin three times and notes the number of heads.If she gets 1,2,3 or 4, she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw 1,2,3, or 4 with the die?
Solution:
When a die is thrown there are 6 exhaustive cases.
If she gets 5 or 6 the probability of E1 = $\frac { 2 }{ 6 }$ = $\frac { 1 }{ 3 }$
i.e.,P(E1) = $\frac { 1 }{ 3 }$
If she gets 1,2,3,4 and probability of E2 = $\frac { 4 }{ 6 }$ = $\frac { 2 }{ 3 }$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q10.1

Ex 13.3 Class 12 Maths Question 11.
A manufacturer has three machine operators A, B and C. The first operator A produces 1% defective items, where as the other two operators B and C produce 5% and 7% defective items respectively. A is on the job for 50% of the time, B is on the job for 30% of the time and C is on the job for 20% of the time. A defective item is produced, what is the probability that it was produced by A?
Solution:
Let E1, E2, E3 and A be the events defined as follows:
E1 = the item is manufactured by the operator A
E2 = the item is manufactured by the operator B
E3 = the item is manufactured by the operator C
and A=the item is defective
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q11.1

Ex 13.3 Class 12 Maths Question 12.
A card from a pack of 52 cards is lost From the remaining cards of the pack, two cards are drawn and are found to be both diamonds. Find the probability of the lost card being a diamond?
Solution:
E1 = Event that lost card is diamond,
E2 = Event that lost card is not diamond.
There are 13 diamond cards, out of a pack or 52 cards
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q12.1

Ex 13.3 Class 12 Maths Question 13.
Probability that A speaks truth is 4/5. A coin is tossed. A reports that a head appears. The probability that actually there was head is
(a) $\frac { 4 }{ 5 }$
(b) $\frac { 1 }{ 2 }$
(c) $\frac { 1 }{ 5 }$
(d) $\frac { 2 }{ 5 }$
Solution:
Let A be the event that the man reports that head occurs in tossing a coin and let E1 be the event that head occurs and E2 be the event head does not occurs.
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q13.1

Ex 13.3 Class 12 Maths Question 14.
If A and B are two events such that A⊂B and P (B) ≠ 0, then which of the following is correct:
(a) P(A | B) = $\frac { P(B) }{ P(A) }$
(b) P (A | B) < P (A)
(c) P(A | B) ≥ P(A)
(d) None of these
Solution: (c) A⊂B => A∩B = A and P(B)≠0
$P(A|B)=\frac { P(A\cap B) }{ P(B) } =\frac { P(A) }{ P(B) }$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.3 Q14.1

Class 12 Maths Chapter 13 Probability Ex 13.4

Ex 13.4 Class 12 Maths Question 1.
State which of the following are not the probability distributions of a random variable. Give reasons for your answer.
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q1.1
Solution:
P (0) + P (1) + P (2) = 0.4 + 0.4 + 0.2 = 1
It is a probability distribution.
(ii) P (3) = -0.1 which is not possible.
Thus it is not a probability distribution.
(iii) P(-1)+P(0)+P(1) = 0.6 + 0.1 + 0.2 = 0.9≠1
Thus it is not a probability distribution.
(iv) P (3) + P (2) + P (1) + P (0) + P (-1)
= 0.3 + 0.2 + 0.4 + 0.1 + 0.05 = 1.05≠1
Hence it is not a probability distribution.

Ex 13.4 Class 12 Maths Question 2.
An urn contains 5 red and 2 black balls. Two balls are randomly drawn. Let X represent the number of black balls. What are the possible values of X ?. Is X a random variable?
Solution:
These two balls may be selected as RR, RB, BR, BB, where R represents red and B represents black ball, variable X has the value 0,1,2, i.e., there may be no black balls, may be one black ball, or both the balls are.black. Yes , X is a random variable.

Ex 13.4 Class 12 Maths Question 3.
Let X represent the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are possible values of X?
Solution:
For one coin, S = {H,T}
n (S) = 2, Let A represent Head
∴ A = {H},n(A) = 1
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q3.1

Ex 13.4 Class 12 Maths Question 4.
Find the probability distribution of
(a) number of heads in two tosses of a coin.
(b) number of tails in the simultaneous tosses of three coins.
(c) number of heads in four tosses of a coin.
Solution:
(a) When two tosses of a coin are there sample space
= {TT, TH, HT, HH}
Zero success => No heads => Two tails (TT)
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q4.1

Ex 13.4 Class 12 Maths Question 5.
Find the probability distribution of the number of successes in two tosses of a die where a success Is defined as
(i) number greater than 4
(ii) six appears on at least one die
Solution:
S = (1, 2, 3,4, 5,6},n(S) = 6
(i) Let A be the set of favorable events.
A = {5,6),n(A) = 2
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q5.1

Ex 13.4 Class 12 Maths Question 6.
From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement Find the probability distribution of the number of defective bulbs.
Solution:
There are 30 bulbs which include 6 defective bulbs
Probability of getting a defective bulb = $\frac { 6 }{ 30 }$ = $\frac { 1 }{ 5 }$
Probability of getting a good bulb = $1-\frac { 1 }{ 5 }$ = $\frac { 4 }{ 5 }$
Let X denotes variable of defective bulbs in a sample of 4 bulb
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q6.1

Ex 13.4 Class 12 Maths Question 7.
A coin is biased so that the head is 3 times as likely to occur as tail. If the coin is tossed twice, find the probability distribution of number of tads.
Solution:
Let p represents the appearance of tail.
q represents the appearance of head.
Now q = 3p As p + q = 1 => p + 3p = 1
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q7.1

Ex 13.4 Class 12 Maths Question 8
A random variable X has the following probability distribution:

Determine
(i) k
(ii) P(X<3) (iii) P(X>6)
(iv) P(0<X<3)
Solution:
(i) Sum of probabilities = 1
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q8.2

Ex 13.4 Class 12 Maths Question 9.
The random variable X has a probability distribution P (X) of the following form, where k is some number

(a) Determine the value of k
(b) FindP(X<2),P (X≤2), P(X≥2)
Solution:
(a) Sum of probabilities = 1
k + 2k + 3k = 1 or 6k = 1,k = $\frac { 1 }{ 6 }$
The probability distribution is as given below
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q9.2

Ex 13.4 Class 12 Maths Question 10.
Find the mean number of heads in three tosses of a fair coin.
Solution:
S = {H,T},n(S) = 2
Let A denotes the appearance of head on a toss A = {H}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q10.1

Ex 13.4 Class 12 Maths Question 11.
Two dice are thrown simultaneously. If X denotes the number of sixes, find the expectation of X.
Solution:
Two dice thrown simultaneously is the same the die thrown 2 times.
Let S= {1,2,3,4,5,6},n(S) = 6
Let A denotes the number 6
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q11.1

Ex 13.4 Class 12 Maths Question 12.
Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find E (X)
Solution:
There are six numbers 1,2,3,4,5,6 one of them is selected in 6 ways
When one of the numbers has been selected, 5 numbers are left, one number out of 5 may be select in 5 ways
∴ No. of ways of selecting two numbers without replacement out of 6 positive integers = 6 x 5 = 30
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q12.1

Ex 13.4 Class 12 Maths Question 13.
Let X denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of X.
Solution:
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q13.1
n (S) = 36
Let A denotes the sum of the numbers = 2
B denotes the sum of the numbers = 3
C denotes the sum of the numbers = 4
D denotes the sum of the numbers = 5
E denotes the sum of the numbers = 6
F denotes the sum of the numbers = 7
G denotes the sum of the numbers = 8
H denotes the sum of the numbers = 9
I denotes the sum of the numbers = 10
J denotes the sum of the numbers = 11
K denotes the sum of the numbers = 12
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q13.2

Ex 13.4 Class 12 Maths Question 14.
A class has 15 students whose ages are 14,17, 15,14,21,17,19,20,16,18,20,17,16,19 and 20 years. One student is selected in such a manner that each has the same chance of being chosen and the age X of the selected student is recorded.What is the probability distribution of the random variable X ? Find mean, variance and standard deviation of X?
Solution:
There are 15 students in a class. Each has the same chance of being choosen.
The probability of each student to be selected
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q14.1

Ex 13.4 Class 12 Maths Question 15.
In a meeting, 70% of the members favour and 30% oppose a certain proposal. A member is selected at random and we take X = 0, if he opposed, and X = 1, if he is in favour Find E (X) and Var (X).
Solution:
Here the variable values are 1 and 0 and the probability of occurrence is 70% = 0.7 and 30% = 0.3
Probability distribution is
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q15.1

Choose the correct answer in each of the following:

Ex 13.4 Class 12 Maths Question 16.
The mean of the number obtained on throwing a die having written 1 on three faces, 2 on two faces and 5 on one face is
(a) 1
(b) 2
(c) 5
(d) $\frac { 8 }{ 3 }$
Solution:
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q16.1
Mean 2
Option (b) is correct

Ex 13.4 Class 12 Maths Question 17.
Suppose that two cards are drawn at random from a deck of cards. Let X be the number of aces obtained. What is the value of E (X)?
(a) $\frac { 37 }{ 221 }$
(b) $\frac { 5 }{ 13 }$
(c) $\frac { 1 }{ 13 }$
(d) $\frac { 2 }{ 13 }$
Solution:
n(S) = 52, n(A) = 4
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.4 Q17.1
Now E(X) = $\frac { 2 }{ 13 }$
Option (d) is correct

Class 12 Maths Chapter 13 Probability Ex 13.5

Ex 13.5 Class 12 Maths Question 1.
A die is thrown 6 times. If ‘getting an odd number’ is a success, what is the probability of
(i) 5 successes?
(ii) at least 5 successes?
(iii) at most 5 successes?
Solution:
There are 3 odd numbers on a die
∴ Probability of getting an odd number on a die = $\frac { 3 }{ 6 }$ = $\frac { 1 }{ 2 }$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q1.1

Ex 13.5 Class 12 Maths Question 2.
There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item ?
Solution:
Probability of getting one defective item = 5%
= $\frac { 5 }{ 100 }$
= $\frac { 1 }{ 20 }$
Probability of getting a good item = $1-\frac { 1 }{ 20 }$ = $\frac { 19 }{ 20 }$
A sample of 10 item include not more than one defective item.
=> sample contains at most (me defective item Its probability = P (0) + P (1)
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q2.1

Ex 13.5 Class 12 Maths Question 3.
A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability, of two successes.
Solution:
n(S) = 36, A = {11,22,33,44,55,66}
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q3.1

Ex 13.5 Class 12 Maths Question 4.
Five cards are drawn successively with replacement from a well- shuffled deck of 52 cards. What is the probability that
(i) all the five cards are spades?
(ii) only 3 cards are spades?
(iii) none is spade?
Solution:
S = {52 cards}, n (S) = 52
Let A denotes the favourable events
A= {13 spade}, n(A)= 13
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q4.1

Ex 13.5 Class 12 Maths Question 5.
The probability that a bulb produced by a factory will fuse after 150 days of use is 0.05. Find the probability that out of 5 such bulbs.
(i) none
(ii) not more than one
(iii) more than one
(iv) at least one will fuse after 150 days of use
Solution:
Probability that a bulb gets fuse after 150 days of its use = 0.05
Probability that the bulb will not fuse after 150 days of its use = 1 – 0.05 = 0.95
(i) Probability that no bulb will fuse after 150
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q5.1

Ex 13.5 Class 12 Maths Question 6.
A bag consists of 10 balls each marked with one of the digits 0 to 9. If four bails are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?
Solution:
S = {0,1,2,3,4,5,6,7,8,9},n(S) = 10
Let A represents that the ball is marked with the digit 0.
A = {0}, n(A) = 1
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q6.1

Ex 13.5 Class 12 Maths Question 7.
In an examination, 20 questions of true – false type are asked. Suppose a student tosses fair coin to determine his answer to each question. If the coin falls heads, he answers ‘true,’ if it falls tails, he answers “ false’. Find the probability that he answers at least 12 questions correctly.
Solution:
Probability that student answers a question true = $\frac { 1 }{ 2 }$
i.e., when a coin is thrown, probability that a head is obtained = $\frac { 1 }{ 2 }$
Probability that his answer is false = $1-\frac { 1 }{ 2 }$ = $\frac { 1 }{ 2 }$
Probability that his answer at least 12 questions correctly = P (12) + P (13) + P (14) +…….. P (20)
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q7.1

Ex 13.5 Class 12 Maths Question 8
Suppose X has a binomial distribution $B\left( 6,\frac { 1 }{ 2 } \right)$ . Show that X = 3 is the most likely outcome.
(Hint: P (X = 3) is the maximum among all P (Xi), xi. = 0,1,2,3,4,5,6)
Solution:
${ \left( \frac { 1 }{ 2 } +\frac { 1 }{ 2 } \right) }^{ 6 }$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q8.1

Ex 13.5 Class 12 Maths Question 9.
On a multiple choice examination with three possible answers for each of the five questions, what is the probability that a candidate would get four or more correct answers just by guessing?
Solution:
P = $\frac { 1 }{ 3 }$ . q = 1 – P = $1-\frac { 1 }{ 3 }$ = $\frac { 2 }{ 3 }$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q9.1

Ex 13.5 Class 12 Maths Question 10.
A person buys a lottery ticket in 50 lotteries, in each of which his chance of winning a prize is $\frac { 1 }{ 100 }$ . What is the probability that he will win a prize?
(a) at least once,
(b) exactly once,
(c) at least twice?
Solution:
Probability that the person wins the prize = $\frac { 1 }{ 100 }$
Probability of losing = $1-\frac { 1 }{ 100 }$ = $\frac { 99 }{ 100 }$
(a) Probability that he loses in all the loteries
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q10.1

Ex 13.5 Class 12 Maths Question 11.
Find the probability of getting 5 exactly twice in 7 throws of a die.
Solution:
S = {1,2,3,4,5,6},n(S) = 6
A = {5} => n(A) = 1
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q11.1

Ex 13.5 Class 12 Maths Question 12.
Find the probability of throwing at most 2 sixes in 6 throws of a single die.
Solution:
When a die is thrown,
Probabiltiy of getting a six = $\frac { 1 }{ 6 }$
Probabiltiy of not getting a six = $1-\frac { 1 }{ 6 }$ = $\frac { 5 }{ 6 }$
Probabiltiy of getting at most 2 sixes in 6 throws of a single die = P (0) + P (1) + P (2)
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q12.1

Ex 13.5 Class 12 Maths Question 13.
It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles 9 are defective?
Solution:
p = $\frac { 1 }{ 100 }$ = $\frac { 1 }{ 10 }$
q = 1 – p = $1-\frac { 1 }{ 10 }$ = $\frac { 9 }{ 10 }$
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q13.1

Ex 13.5 Class 12 Maths Question 14.
In a box containing 100 bulbs, 10 are defective. The probability that out of a sample of 5 bulbs, none is defective is
(a) ${ 10 }^{ -1 }$
(b) ${ \left( \frac { 1 }{ 2 } \right) }^{ 5 }$
(c) ${ \left( \frac { 9 }{ 10 } \right) }^{ 5 }$
(d) $\frac { 9 }{ 10 }$
Solution:
p = $\frac { 1 }{ 10 }$
q = $\frac { 9 }{ 10 }$ n = 5, r = 0, P(X=0) = ${ \left( \frac { 9 }{ 10 } \right) }^{ 5 }$
Option (c) is correct

Ex 13.5 Class 12 Maths Question 15.
The probability that a student is not a swimmer is $\frac { 1 }{ 5 }$ . Then the probability that out of five students, four are swimmers is:
vedantu class 12 maths Chapter 13 Probability 15
Solution:
p = $\frac { 4 }{ 5 }$ , q = $\frac { 1 }{ 5 }$ , n = 5,r = 4
NCERT Solutions for Class 12 Maths Chapter 13 Probability Ex 13.5 Q15.1

Option (a) is true

Rohit0 comments

Class 12th Chapter -12 Linear Programming | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter :12 Linear Programming

Class 12 Maths Chapter 12 Linear Programming Ex 12.1

olve the following Linear Programming Problems graphically:

Ex 12.1 Class 12 Maths Question 1.
Maximize Z = 3x + 4y
subject to the constraints:
x + y ≤ 4,x ≥ 0,y ≥ 0.
Solution:
As x ≥ 0, y ≥ 0, therefore we shall shade the other inequalities in the first quadrant only. Now consider x + y ≤ 4.
Let x + y = 4 => $\frac { x }{ 4 } +\frac { y }{ 4 } =1$
Thus the line has 4 and 4 as intercepts along the axes. Now (0, 0) satisfies the inequation i.e., 0 + 0 ≤ 4. Now shaded region OAB is the feasible solution.
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Q1.1

Ex 12.1 Class 12 Maths Question 2.
Minimize Z = -3x+4y
subject to x + 2y ≤ 8,3x + 2y ≤ 12, x ≥ 0, y ≥ 0
Solution:
Objective function Z = -3x + 4y
constraints are x+2y ≤ 8,
3x + 2y ≤ 12, x ≥ 0,y ≥ 0
(i) Consider the line x+2y = 8. It pass through A (8,0) and B (0,4), putting x = 0, y = 0 in x + 2y ≤ 8,0 ≤ 8 which is true.
=> region x + 2y ≤ 8 lies on and below AB.
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Q2.1

Ex 12.1 Class 12 Maths Question 3.
Maximize Z = 5x+3y
subject to 3x + 5y ≤ 15,5x + 2y ≤ 10, x≥0, y≥0
Solution:
The objective function is Z = 5x + 3y constraints
are 3x + 5y≤15, 5x + 2y≤10,x≥0,y≥0
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Q3.1

Ex 12.1 Class 12 Maths Question 4.
Minimize Z = 3x + 5y such that x + 3y ≥ 3, x + y ≥ 2,x,y ≥ 0.
Solution:
For plotting the graph of x + 3y = 3, we have the following table:
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Q4.1

Ex 12.1 Class 12 Maths Question 5.
Maximize Z=3x+2y subject to x+2y ≤ 10, 3x+y ≤ 15, x, y ≥ 0.
Solution:
Consider x + 2y ≤ 10
Let x + 2y = 10
=> $\frac { x }{ 10 } +\frac { y }{ 5 } =1$
Now (0,0) satisfies the inequation, therefore the half plane containing (0,0) is the required plane.
Again 3x+2y ≤ 15
Let 3x + y = 15
=> $\frac { x }{ 5 } +\frac { y }{ 15 } =1$
It is also satisfies by (0,0) and its required half plane contains (0,0).
Now double shaded region in the first quadrant contains the solution.
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Q5.1

Ex 12.1 Class 12 Maths Question 6.
Minimize Z = x + 2y subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.
Solution:
Consider 2x + y ≥ 3
Let 2x + y = 3
⇒ y = 3 – 2x
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Q6.1
(0,0) is not contained in the required half plane as (0, 0) does not satisfy the inequation 2x + y ≥ 3.
Again consider x+2y≥6
Let x + 2y = 6
=> $\frac { x }{ 6 } +\frac { y }{ 3 } =1$
Here also (0,0) does not contain the required half plane. The double-shaded region XABY’ is the solution set. Its comers are A (6,0) and B (0,3). At A, Z = 6 + 0 = 6
At B, Z = 0 + 2 × 3 = 6
We see that at both points the value of Z = 6 which is minimum. In fact at every point on the line AB makes Z=6 which is also minimum.

Show that the minimum of z occurs at more than two points.

Ex 12.1 Class 12 Maths Question 7.
Minimise and Maximise Z = 5x + 10y
subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x,y≥0
Solution:
The objective function is Z = 5x + 10y constraints are x + 2y≤120,x+y≥60, x-2y≥0, x,y≥0
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Q7.1

Ex 12.1 Class 12 Maths Question 8.
Minimize and maximize Z = x + 2y subject to x + 2y ≥ 100,2x – y ≤ 0,2x + y ≤ 200;x,y ≥ 0.
Solution:
Consider x + 2y ≥ 100
Let x + 2y = 100
=> $\frac { x }{ 100 } +\frac { y }{ 50 } =1$
Now x + 2y ≥ 100 represents which does not include (0,0) as it does not made it true.
Again consider 2x – y ≤ 0
Let 2x – y = 0 or y = 2x
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Q8.1

Ex 12.1 Class 12 Maths Question 9.
Maximize Z = -x + 2y, subject to the constraints: x≥3, x + y ≥ 5, x + 2y ≥ 6,y ≥ 0
Solution:
The objective function is Z = – x + 2y.
The constraints are x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Q9.1

Ex 12.1 Class 12 Maths Question 10.
Maximize Z = x + y subject to x – y≤ -1, -x + y≤0,x,y≥0
Solution:
Objective function Z = x + y, constraints x – y≤ -1, -x + y≤0,x,y≥0
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Q10.1

Class 12 Maths Chapter 12 Linear Programming Ex 12.2

Ex 12.2 Class 12 Maths Question 1.
Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs Rs 60/kg and Food Q costs Rs 80/kg. Food P contains 3 units/ kg of Vitamin A and 5 units/ kg of Vitamin B while food Q contains 4 units/kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
Solution:
Let x kg of food P and y kg of food Q are mixed,
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Q1.1

Ex 12.2 Class 12 Maths Question 2.
One kind of cake requires 200 g of flour and 25g of fat, and another kind of cake requires 100 g of flour and 50 g of fat Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients, used in making the cakes.
Solution:
Let number of cakes made of first kind are x and that of second kind is y.
∴ maximize Z = x + y
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Q2.1

Ex 12.2 Class 12 Maths Question 3.
A factory makes tennis rackets and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hour of machine time and 1 hour of craftman’s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman’s time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity?
(ii) If the profit on a racket and on a bat is Rs 20 and Rs 10 respectively, find the maximum profit of the factory when it works at full capacity. .
Solution:
Let x tennis rackets and y cricket bats are produced in one day in the factory.
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Q3.1

Ex 12.2 Class 12 Maths Question 4.
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts. It takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of Rs 17.50 per package on nuts and Rs 7.00 per package on bolts. How many packages of each should be produced each day so as to maximise his profit, if he operates his machines for at the most 12 hours
Solution:
Let x nuts and y bolts are produced.
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Q4.1

Ex 12.2 Class 12 Maths Question 5.
A factory manufactures two types of screws, A and B, Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machines to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machines to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of Rs 7 and screws B at a profit of Rs 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit? Determine the maximum profit.
Solution:
Let the manufacturer produces x packages of screws A and y packages of screw B, then time taken by x packages of screw A and y packages of screw B on automatic machine = (4x + 6y) minutes.
And hand operated machine = (6x + 3y) minutes
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Q5.1

Ex 12.2 Class 12 Maths Question 6.
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/ cutting machine and a sprayer. It takes 2 hours on grinding/ cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp, while it takes 1 hour on the grinding/ cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 20 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is Rs 5 and that from a shade is Rs 3. Assuming that the manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit?
Solution:
Let the manufacturer produces x pedestal lamps and y wooden shades; then the time taken by x pedestal lamps and y wooden shades on grinding/ cutting machines = (2x + y) hours and time taken by x pedestal lamps and y shades on the sprayer = (3x + 2y) hours.
Since grinding/ cutting machine is available for at the most 12 hours, 2x + y ≤ 12 and sprayer is available for at the most 20 hours.
We have: 3x + 2y ≤ 20.
Profit from the sale of x lamps and y shades.
Z = 5x + 3y
So, our problem is to maximize Z = 5x + 3y subject to constraints 3x + 2y ≤ 20,2x + y ≤ 12, x, y ≥ 0.
Consider 3x + 2y ≤ 20
Let 3x + 2y = 20
$y=\frac { 20-3x }{ 2 }$
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Q6.1

Ex 12.2 Class 12 Maths Question 7.
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is Rs 5 each for type A and Rs 6 each for type B souvenirs. How many souvenirs of each type should be company manufacture in order to maximise the profit?
Solution:
Let the company manufactures x souvenirs of type A and y souvenirs of type B, then time taken for cutting x souvenirs of type A and y souvenirs of type A and y souvenirs of type B = (5x + 8y) minutes.
Since 3 hours 20 minutes i.e 200 minutes are available for cutting, so we should have 5x + 8y ≤ 200
Also as 4 hours i.e., 240 minutes are available for assembling, so we have 10x + 8y ≤ 240 i. e., 5x + 4y ≤ 120
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Q7.1

Ex 12.2 Class 12 Maths Question 8.
A merchant plans to sell two types of personal computers – a desktop model and a portable model that will cost Rs 25000 and Rs 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which die merchant should stock to get maximum profit if he does not want to invest more than Rs 70 lakhs and if his profit on the desktop model is Rs 4500 and on portable model is Rs 5000.
Solution:
Let there be x desktop and y portable computers. Total monthly demand of computer does not exceed 250. => x+y ≤ 250, cost of 1 desktop computer if Rs 25000 and 1 portable computer is Rs 40000
∴ cost of x desktop and y portable computer = Rs (25000 x+40000 y)
Maximum investment = Rs 70 lakhs = Rs 70,00,000
=> 25000 x+40000 y≤ 7000000 or 5x+8y ≤ 1400
profit on 1 desktop computer is Rs 4500 and on 1 portable is Rs 5000, total profit,
Z = 4500 x+5000 y
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Q8.1

Ex 12.2 Class 12 Maths Question 9.
A diet is to contain at least 80 units of vitamin A and 100 units of minerals. Two foods Ft and F2 are available. Food F1 costs Rs 4 per unit food and F2 costs Rs 6 per unit.One unit of food F2 contains 3 units of vitamin A and 4 units of minerals. One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Solution:
Let there be x units of food F1 and y units of food F2.
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Q9.1

Ex 12.2 Class 12 Maths Question 10.
There are two types of fertilisers F1, and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a fanner finds that she needs atleast 14 kg of nitrogen and 14 kg of phsophoric acid for her crop. If F2 costs Rs 6/kg and F2 costs Rs 5/ kg, determine how much of each type of fertiliser should be used so that nutrient requirements are met at a minimum cost What is the minimum cost?
Solution:
Let x kg of festiliser and y kg of fertilised F2. be required.
NCERT Solutions for Class 12 Maths Chapter 12 Linear Programming Ex 12.2 Q10.1

Ex 12.2 Class 12 Maths Question 11.
The corner points of the feasible region determined by the following system of linear inequalities:
2x+y≤10,x+3y≤15, x,y≥0 are (0,0),(5,0), (3,4) and (0,5).
Let Z=px+qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3,4) and (0,5) is
(a) p=q
(b) p=2q
(c) p=3q
(d) q=3p
Solution:
Maximum value of Z = px + qy occurs at (3,4) and (0,5),
At (3,4), Z = px + qy = 3p + 4q
At (0,5), Z = 0 + q . 5 = 5q
Both are the maximum values
=> 3p + 4q = 5q or q = 3p
Option (d) is correct.

Priyanka0 comments

NCERT MCQ CLASS-12 CHAPTER-1 | BIOLOGY NCERT MCQ | REPRODUCTION IN ORGANISM | EDUGROWN

In This Post we are providing Chapter-1 Reproduction in Organism NCERT MCQ for Class 12 Biology 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 MCQ ON REPRODUCTION IN ORGANISM

Question 1: Offspring formed by sexual reproduction exhibit more variation than those formed by Asexual reproduction because

a) gametes of parents have qualitatively different genetic composition
b) gametes of parents have qualitatively different genetic composition
c) greater amount of DNA is involved in sexual reproduction
d) sexual reproduction is a lengthy process

Answer: gametes of parents have qualitatively different genetic composition

Question 2: The property of an undifferentiated cell that has the potential to develop into an entire plant is called

a) Totipotency
b) Sub potency
c) Cloning
d) Budding

Answer: Totipotency

Question 3: The development of root and shoot in tissue culture is determined by ______

a) Auxin and cytokinin ratio
b) None of the above
c) Nutrients
d) Temperature

Answer: Auxin and cytokinin ratio

Question 4: There is no natural death in single celled organisms like Amoeba and bacteria because

a) parental body is distributed among the offspring
b) they are microscopic
c) they reproduce by binary fission
d) they cannot reproduce sexually

Answer: parental body is distributed among the offspring

Question 5: There are various types of reproduction. The type of reproduction adopted by an organism depends on

a) the organism’s habitat, physiology, and genetic makeup
b) morphology and physiology of the organism
c) morphology of the organism
d) None of these

Answer: the organism’s habitat, physiology, and genetic makeup

Question 6: A few statements describing certain features of reproduction are given below:

i. Gametic fusion takes place

ii. Transfer of genetic material takes place

iii. Reduction division takes place

iv. Progeny have some resemblance with parents

Select the options that are true for both asexual and sexual reproduction from the options given below:

a) ii and iv
b) i and iii
c) ii and iii
d) i and ii

Answer: ii and iv

Question 7: Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because

a) nodes have meristematic cells
b) nodes have meristematic cells
c) nodes have meristematic cells
d) nodes are shorter than intenodes

Answer: nodes have meristematic cells

Question 8: Identify the correct sequence of events

a) Gametogenesis → Syngamy → Zygote
b) Gametogenesis → Embryogenesis → Zygote → Syngamy
c) Gametogenesis → Zygote → Syngamy → Embryogenesis
d) Gametogenesis → Syngamy Embryogenesis → Zygote

Answer: Gametogenesis → Syngamy → Zygote

Question 9: The term ‘clone’ cannot be applied to offspring formed by sexual reproduction because

a) Offspring do not possess exact copies of parental DNA
b) DNA of only one parent is copied and passed on to the offspring
c) Offspring are formed at different times
d) NA of parent and offspring are completely different

Answer: Offspring do not possess exact copies of parental DNA

Question 10: Amoeba and Yeast reproduce asexually by fission and budding respectively, because they are

a) Unicellular organisms
b) Uninucleate organisms
c) Heterotrophic organisms
d) Microscopic organisms

Answer: Unicellular organisms

Question 11: A few statements with regard to sexual reproduction are given below:

i. Sexual reproduction does not always require two individuals

ii. Sexual reproduction generally involves gametic fusion

iii. Meiosis never occurs during sexual reproduction

iv. External fertilisation is a rule during sexual reproduction

Choose the correct statements from the options below:

a) i and ii
b) ii and iii
c) i and iv
d) i and iv

Answer: i and ii

Question 12: A multicellular, filamentous alga exhibits a type of sexual life cycle in which the meiotic division occurs after the formation of zygote. The adult filament of this alga has

a) haploid vegetative cells and haploid gametangia
b) diploid vegetative cells and haploid gametangia
c) diploid vegetative cells and diploid gametangia
d) haploid vegetative cells and diploid gametangia

Answer: haploid vegetative cells and haploid gametangia

Question 13: The male gametes of rice plant have 12 chromosomes in their nucleus. The chromosome number in the female gamete, zygote and the cells of the seedling will be, respectively,

a) 12, 24, 24
b) 24, 12, 24
c) 24, 12, 12
d) 12, 24, 12

Answer: 12, 24, 24

Question 14: Given below are a few statements related to external fertilization. Choose the correct statements.

i. The male and female gametes are formed and released simultaneously

ii. Only a few gametes are released into the medium

iii. Water is the medium in a majority of organisms exhibiting external fertilization

iv. Offspring formed as a result of external fertilization have better chance of survival than those formed inside an organism

a) i and iii
b) ii and iv
c) i and iv
d) iii and iv

Answer: i and iii

Question 15 : The statements given below describe certain features that are observed in the pistil of flowers.

i. Pistil may have many carpels

ii. Each carpel may have more than one ovule

iii. Each carpel has only one ovule

iv. Pistil have only one carpel

Choose the statements that are true from the options below:

a) i and ii
b) i and iii
c) ii and iv
d) iii and iv

Answer: i and ii

Rohit0 comments

Class 12th Chapter -11 Three Dimensional Geometry | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter :11 Three Dimensional Geometry

Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1

Ex 11.1 Class 12 Maths Question 1.
If a line makes angles 90°, 135°, 45° with the and z axes respectively, find its direction cosines.
Solution:
Direction angles are 90°, 135°, 45°
Direction cosines are
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1 Q1.1

Ex 11.1 Class 12 Maths Question 2.
Find the direction cosines of a line which makes equal angles with coordinate axes.
Solution:
Let direction angle be α each
∴ Direction cosines are cos α, cos α, cos α
But l² + m² + n² = 1
∴cos² a + cos² a + cos² a = 1
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1 Q2.1

Ex 11.1 Class 12 Maths Question 3.
If a line has the direction ratios – 18,12, -4 then what are its direction cosines?
Solution:
Now given direction ratios of a line are -8,12,-4
∴ a = -18,b = 12,c = -4
Direction cosines are
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1 Q3.1

Ex 11.1 Class 12 Maths Question 4.
Show that the points (2,3,4) (-1,-2,1), (5,8,7) are collinear.
Solution:
Let the points be A(2,3,4), B (-1, -2,1), C (5,8,7).
Let direction ratios of AB be
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1 Q4.1

Ex 11.1 Class 12 Maths Question 5.
Find the direction cosines of the sides of the triangle whose vertices are (3,5, -4), (-1,1,2) and (-5,-5,-2).
Solution:
The vertices of triangle ABC are A (3, 5, -4), B (-1,1,2), C (-5, -5, -2)
(i) Direction ratios of AB are (-4,-4,6)
Direction cosines are
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.1 Q5.1

Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2

Ex 11.2 Class 12 Maths Question 1.
Show that the three lines with direction cosines:
$\frac { 12 }{ 13 } ,\frac { -3 }{ 13 } ,\frac { -4 }{ 13 } ,\frac { 4 }{ 13 } ,\frac { 12 }{ 13 } ,\frac { 3 }{ 13 } ,\frac { 3 }{ 13 } ,\frac { -4 }{ 13 } ,\frac { 12 }{ 13 }$
are mutually perpendicular.
Solution:
Let the lines be L₁,L₂ and L₃.
∴ For lines L1 and L2
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q17.2

Ex 11.2 Class 12 Maths Question 2.
Show that the line through the points (1,-1,2) (3,4, -2) is perpendicular to the line through the points (0,3,2) and (3,5,6).
Solution:
Let A, B be the points (1, -1, 2), (3, 4, -2) respectively Direction ratios of AB are 2,5, -4
Let C, D be the points (0, 3, 2) and (3, 5, 6) respectively Direction ratios of CD are 3, 2,4 AB is Perpendicular to CD if
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q17.3

Ex 11.2 Class 12 Maths Question 3.
Show that the line through the points (4,7,8) (2,3,4) is parallel to the line through the points (-1,-2,1) and (1,2,5).
Solution:
Let the points be A(4,7,8), B (2,3,4), C (-1,-2,1) andD(1,2,5).
Now direction ratios of AB are
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q3.1

Ex 11.2 Class 12 Maths Question 4.
Find the equation of the line which passes through the point (1,2,3) and is parallel to the vector $3\hat { i } +2\hat { j } -2\hat { k }$
Solution:
Equation of the line passing through the point
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q17.4

Ex 11.2 Class 12 Maths Question 5.
Find the equation of the line in vector and in cartesian form that passes through the point with position vector $2\hat { i } -\hat { j } +4\hat { k }$ and is in the direction $\hat { i } +2\hat { j } -\hat { k }$ .
Solution:
The vector equation of a line passing through a point with position vector $\overrightarrow { a }$ and parallel to the
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q17.5

Ex 11.2 Class 12 Maths Question 6.
Find the cartesian equation of the line which passes through the point (-2,4, -5) and parallel to the line is given by $\frac { x+3 }{ 3 } =\frac { y-4 }{ 5 } =\frac { z+8 }{ 6 }$
Solution:
The cartesian equation of the line passing through the point (-2,4, -5) and parallel to the
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q6.1

Ex 11.2 Class 12 Maths Question 7.
The cartesian equation of a line is
$\frac { x-5 }{ 3 } =\frac { y+4 }{ 7 } =\frac { z-6 }{ 2 }$
write its vector form.
Solution:
The cartesian equation of the line is
$\frac { x-5 }{ 3 } =\frac { y+4 }{ 7 } =\frac { z-6 }{ 2 }$
Clearly (i) passes through the point (5, – 4, 6) and has 3,7,2 as its direction ratios.
=> Line (i) passes through the point A with
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q7.1

Ex 11.2 Class 12 Maths Question 8.
Find the vector and the cartesian equations of the lines that passes through the origin and (5,-2,3).
Solution:
The line passes through point
$\therefore \overrightarrow { a } =\overrightarrow { 0 }$
Direction ratios of the line passing through the
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q8.1

Ex 11.2 Class 12 Maths Question 9.
Find the vector and cartesian equations of the line that passes through the points (3, -2, -5), (3,-2,6).
Solution:
The PQ passes through the point P(3, -2, -5)
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q9.1

Ex 11.2 Class 12 Maths Question 10.
Find the angle between the following pair of lines
(i) $\overrightarrow { r } =2\hat { i } -5\hat { j } +\hat { k } +\lambda (3\hat { i } +2\hat { j } +6\hat { k } )$
$and\quad \overrightarrow { r } =7\hat { i } -6\hat { j } +\mu (\hat { i } +2\hat { j } +2\hat { k } )$
(ii) $\overrightarrow { r } =3\hat { i } +\hat { j } -2\hat { k } +\lambda (\hat { i } -\hat { j } -2\hat { k } )$
$\overrightarrow { r } =2\hat { i } -\hat { j } -56\hat { k } +\mu (3\hat { i } -5\hat { j } -4\hat { k } )$
Solution:
(i) Let θ be the angle between the given lines.
The given lines are parallel to the vectors
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q10.1

Ex 11.2 Class 12 Maths Question 11.
Find the angle between the following pair of lines
(i) $\frac { x-2 }{ 2 } =\frac { y-1 }{ 5 } =\frac { z+3 }{ -3 } and\frac { x+2 }{ -1 } =\frac { y-4 }{ 8 } =\frac { z-5 }{ 4 }$
(ii) $\frac { x }{ 2 } =\frac { y }{ 2 } =\frac { z }{ 1 } and\frac { x-5 }{ 4 } =\frac { y-2 }{ 1 } =\frac { z-3 }{ 8 }$
Solution:
Given
(i) $\frac { x-2 }{ 2 } =\frac { y-1 }{ 5 } =\frac { z+3 }{ -3 } and\frac { x+2 }{ -1 } =\frac { y-4 }{ 8 } =\frac { z-5 }{ 4 }$
(ii) $\frac { x }{ 2 } =\frac { y }{ 2 } =\frac { z }{ 1 } and\frac { x-5 }{ 4 } =\frac { y-2 }{ 1 } =\frac { z-3 }{ 8 }$
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q11.1

Ex 11.2 Class 12 Maths Question 12.
Find the values of p so that the lines
$\frac { 1-x }{ 3 } =\frac { 7y-14 }{ 2p } =\frac { z-3 }{ 2 } and\frac { 7-7x }{ 3p } =\frac { y-5 }{ 1 } =\frac { 6-z }{ 5 }$
are at right angles
Solution:
The given equation are not in the standard form
The equation of given lines is
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q12.1

Ex 11.2 Class 12 Maths Question 13.
Show that the lines $\frac { x-5 }{ 7 } =\frac { y+2 }{ -5 } =\frac { z }{ 1 } and\frac { x }{ 1 } =\frac { y }{ 2 } =\frac { z }{ 3 }$ are perpendicular to each other
Solution:
Given lines
$\frac { x-5 }{ 7 } =\frac { y+2 }{ -5 } =\frac { z }{ 1 }$ …(i)
$\frac { x }{ 1 } =\frac { y }{ 2 } =\frac { z }{ 3 }$ …(ii)
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q13.1

Ex 11.2 Class 12 Maths Question 14.
Find the shortest distance between the lines
$\overrightarrow { r } =(\hat { i } +2\hat { j } +\hat { k } )+\lambda (\hat { i } -\hat { j } +\hat { k } )$ and
$\overrightarrow { r } =(2\hat { i } -\hat { j } -\hat { k } )+\mu (2\hat { i } +\hat { j } +2\hat { k } )$
Solution:
The shortest distance between the lines
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q14.1

Ex 11.2 Class 12 Maths Question 15.
Find the shortest distance between the lines
$\frac { x+1 }{ 7 } =\frac { y+1 }{ -6 } =\frac { z+1 }{ 1 } and\frac { x-3 }{ 1 } =\frac { y-5 }{ -2 } =\frac { z-7 }{ 1 }$
Solution:
Shortest distance between the lines
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q15.1

Ex 11.2 Class 12 Maths Question 16.
Find the distance between die lines whose vector equations are:
$\overrightarrow { r } =(\hat { i } +2\hat { j } +3\hat { k) } +\lambda (\hat { i } -3\hat { j } +2\hat { k } )$ and
$\overrightarrow { r } =(4\hat { i } +5\hat { j } +6\hat { k) } +\mu (2\hat { i } +3\hat { j } +\hat { k } )$
Solution:
Comparing the given equations with
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q16.1

Ex 11.2 Class 12 Maths Question 17.
Find the shortest distance between the lines whose vector equations are
$\overrightarrow { r } =(1-t)\hat { i } +(t-2)\hat { j } +(3-2t)\hat { k }$ and
$\overrightarrow { r } =(s+1)\hat { i } +(2s-1)\hat { j } -(2s+1)\hat { k }$
Solution:
Comparing these equation with
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Q17.1

Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3

Ex 11.3 Class 12 Maths Question 1.
In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
(a) z = 2
(b) x+y+z = 1
(c) 2x + 3y – z = 5
(d) 5y+8 = 0
Solution:
(a) Direction ratios of the normal to the plane are 0,0,1
=> a = 0, b = 0, c = 1
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q1.1

Ex 11.3 Class 12 Maths Question 2.
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector $3\hat { i } +5\hat { j } -6\hat { k }$
Solution:
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q2.1

Ex 11.3 Class 12 Maths Question 3.
Find the Cartesian equation of the following planes.
(a) $\overrightarrow { r } \cdot (\hat { i } +\hat { j } -\hat { k) } =2$
(b) $\overrightarrow { r } \cdot (\hat { 2i } +3\hat { j } -4\hat { k) } =1$
(c) $\overrightarrow { r } \cdot [(s-2t)\hat { i } +(3-t)\hat { j } +(2s+t)\hat { k) } =15$
Solution:
(a) $\overrightarrow { r }$ is the position vector of any arbitrary point P (x, y, z) on the plane.
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q3.1

Ex 11.3 Class 12 Maths Question 4.
In the following cases find the coordinates of the foot of perpendicular drawn from the origin
(a) 2x + 3y + 4z – 12 = 0
(b) 3y + 4z – 6 = 0
(c) x + y + z = 1
(d) 5y + 8 = 0
Solution:
(a) Let N (x1, y1, z1) be the foot of the perpendicular from the origin to the plane 2x+3y+4z-12 = 0
∴ Direction ratios of the normal are 2, 3, 4.
Also the direction ratios of ON are (x1,y1,z1)
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q4.1

Ex 11.3 Class 12 Maths Question 5.
Find the vector and cartesian equation of the planes
(a) that passes through the point (1,0, -2) and the normal to the plane is $\hat { i } +\hat { j } -\hat { k }$
(b) that passes through the point (1,4,6) and the normal vector to the plane is $\hat { i } -2\hat { j } +\hat { k }$
Solution:
(a) Normal to the plane is i + j – k and passes through (1,0,-2)
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q5.1

Ex 11.3 Class 12 Maths Question 6.
Find the equations of the planes that passes through three points
(a) (1,1,-1) (6,4,-5), (-4, -2,3)
(b) (1,1,0), (1,2,1), (-2,2,-1)
Solution:
(a) The plane passes through the points (1,1,-1) (6,4,-5), (-4,-2,3)
Let the equation of the plane passing through(1,1,-1)be
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q6.1

Ex 11.3 Class 12 Maths Question 7.
Find the intercepts cut off by the plane 2x+y-z = 5.
Solution:
Equation of the plane is 2x + y- z = 5 x y z
Dividing by 5: $\Rightarrow \frac { x }{ \frac { 5 }{ 2 } } +\frac { y }{ 5 } -\frac { z }{ -5 } =1$
∴ The intercepts on the axes OX, OY, OZ are $\frac { 5 }{ 2 }$ , 5, -5 respectively

Ex 11.3 Class 12 Maths Question 8.
Find the equation of the plane with intercept 3 on the y- axis and parallel to ZOX plane.
Solution:
Any plane parallel to ZOX plane is y=b where b is the intercept on y-axis.
∴ b = 3.
Hence equation of the required plane is y = 3.

Ex 11.3 Class 12 Maths Question 9.
Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2,2,1).
Solution:
Given planes are:
3x – y + 2z – 4 = 0 and x + y + z – 2 = 0
Any plane through their intersection is
3x – y + 2z – 4 + λ(x + y + z – 2) = 0
point (2,2,1) lies on it,
∴3 x 2 – 2 + 2 x 1 – 4 +λ(2+2+1-2)=0
=>λ = $\frac { -2 }{ 3 }$
Now required equation is 7x – 5y + 4z – 8 = 0

Ex 11.3 Class 12 Maths Question 10.
Find the vector equation of the plane passing through the intersection of the planes $\overrightarrow { r } \cdot \left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =7,\overrightarrow { r } \cdot \left( 2\hat { i } +5\hat { j } +3\hat { k } \right) =9$ and through the point (2,1,3).
Solution:
Equation of the plane passing through the line of intersection of the planes
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q10.1

Ex 11.3 Class 12 Maths Question 11.
Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.
Solution:
Given planes are
x + y + z – 1 = 0 …(i)
2x + 3y + 4z – 5 = 0 …(ii)
x – y + z = 0 ….(iii)
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q11.1

Ex 11.3 Class 12 Maths Question 12.
Find the angle between the planes whose vector equations are $\overrightarrow { r } \cdot \left( 2\hat { i } +2\hat { j } -3\hat { k } \right) =5,\overrightarrow { r } \cdot \left( 3\hat { i } -3\hat { j } +5\hat { k } \right) =3$
Solution:
The angle θ between the given planes is
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q12.1

Ex 11.3 Class 12 Maths Question 13.
In the following determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angle between them.
(a) 7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0
(b) 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0
(c) 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0
(d) 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0
(e) 4x + 8y + z – 8 = 0 and y + z – 4 = 0.
Solution:
(a) Direction ratios of the normal of the planes 7x + 5y + 6z + 30 = 0 are 7,5,6
Direction ratios of the normal of the plane 3x – y – 10z + 4 = 0 are 3,-1,-10
The plane 7x + 5y + 6z + 30 = 0 …(i)
3x – y – 10z + y = 0 …(ii)
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q13.1

Ex 11.3 Class 12 Maths Question 14.
In the following cases, find the distance of each of the given points from the corresponding given plane.
Point Plane
(a) (0, 0,0) 3x – 4y + 12z = 3
(b) (3,-2,1) 2x – y + 2z + 3 = 0.
(c) (2,3,-5) x + 2y – 2z = 9
(d) (-6,0,0) 2x – 3y + 6z – 2 = 0
Solution:
(a) Given plane: 3x – 4y + 12z – 3 = 0
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.3 Q14.1 q

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NCERT MCQ CLASS-12 CHAPTER-16 | CHEMISTRY NCERT MCQ | CHEMISTRY IN EVERYDAY LIFE | EDUGROWN

In This Post we are providing Chapter-16 Chemistry in Everyday Life NCERT MCQ for Class 12 Chemistry 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 MCQ ON CHEMISTRY IN EVERYDAY LIFE

Question 1: Drugs can be classified on the basis of

a) Pharmacological effect
b) Drug action
c) Chemical structure
d) All of these

Answer: All of these

Question 2: Antacids include

a) Omeprazole
b) Lansoprazole
c) Sodium bicarbonate
d) All of these

Answer: All of these

Question 3: Drawback of excess of hydrogen carbonate taking as antacid is

a) It is insoluble
b) It can make stomach alkaline and trigger the production of even more acid
c) It causes ulcer
d) It causes pain and irritation

Answer: It can make stomach alkaline and trigger the production of even more acid

Question 4: Tranquilizers are prescribed for curing

a) Anxiety, stress, irritability
b) The growth of microorganism
c) Pain, Fever
d) All of these

Answer: Anxiety, stress, irritability

Question 5: Tincture of iodine is

a) Iodoform
b) 100% Iodine
c) 2-3% Iodine solution in alcohol-water
d) Iodobenzene

Answer: 2-3% Iodine solution in alcohol-water

Question 6: 0.2% of solution of phenol and 0.2–0.4 ppm chlorine in aqueous solution respectively behave as

a) Antiseptic, Disinfectant
b) Disinfectant, Antiseptic
c) Disinfectant, Disinfectant
d) Antiseptic, Antiseptic

Answer: Antiseptic, Disinfectant

Question 7: Birth control pills essentially contains

a) Synthetic estrogen
b) Synthetic progesterone
c) Both (1) & (2)
d) Neither (1) nor (2)

Answer: Both (1) & (2)

Question 8: Which is mismatched, regarding the examples?

a) Broad spectrum Antibiotic – Chloramphenicol
b) Narrow spectrum antibiotic – Ampicillin
c) Antiseptic – Furacine
d) Antifertility – Novestrol

Answer: Narrow spectrum antibiotic – Ampicillin

Question 9: Chemicals are added to food for

a) Preservation
b) Enhancing the appeal
c) Adding nutritive value
d) All of these

Answer: All of these

Question 10: The first popular artificial sweetening agent is

a) Saccharin
b) Aspartame
c) Alitame
d) Both (2) & (3)

Answer: Saccharin

Question 11: The compound with structure

NEET Chemistry Chemistry in Everyday Life Online Test Set A-Q29

is used as

a) Food preservative
b) Artificial sweetener
c) Medicine
d) Edible color

Answer: Artificial sweetener

Question 12: The main disadvantage associated with use of aspartame is

a) Its sweetening power is less
b) It is unstable at cooking temperature
c) It provide calories
d) It is difficult to control its sweetness

Answer: It is unstable at cooking temperature

Question 13: Which of the following can be used as food preservative?

a) Vegetable oil
b) Table salt
c) Sodium benzoate
d) All of these

Answer: All of these

Question 14: Among the following, the maximum high potency sugar is

a) Saccharin
b) Alitame
c) Sucralose
d) Aspartame

Answer: Alitame

Question 15: Soaps are sodium or potassium salt of long chain fatty acids like

a) Palmitic acid
b) Oleic acid
c) Stearic acid
d) All of these

Answer: All of these

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NCERT MCQ CLASS-12 CHAPTER-15 | CHEMISTRY NCERT MCQ | POLYMERS | EDUGROWN

In This Post we are providing Chapter-15 Polymers NCERT MCQ for Class 12 Chemistry 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 MCQ ON POLYMERS

Question 1.
Bakelite is an example of
(a) elastomer
(b) fibre
(c) thermoplastic
(d) thermosetting

Answer: (d) thermosetting

Question 2.
Arrange the following polymers is an increasing order of intermolecular forces: fibre, plastic, elastomer
(a) Elastomer < Fibre < Plastic
(b) Elastomer < Plastic < Fibre
(c) Plastic < Elastomer < Fibre
(d) Fibre < Elastomer < Plastic

Answer: (b) Elastomer < Plastic < Fiber

Question 3.
The correct structure of monomers of Buna-S is
MCQ Questions for Class 12 Chemistry Chapter 15 Polymers with Answers

Answer: (c)

Question 4.
The S in Buna-S refers to
(a) Sulphur
(b) Styrene
(c) Sodium
(d) Salicylate

Answer: (b) Styrene

Question 5.
Identify the type of polymer
(i) -A-A-A-A-A-A-
(ii) -A-B-B-A-A-A-B-A-
(a) (i) Homopolymer, (ii) Copolymer
(b) (i) Natural polymer, (ii) Synthetic polymer
(c) (i) Linear polymer, (ii) Branched polymer
(d) (i) Fiber, (ii) Elastomer

Answer: (a) (i) Homopolymer, (ii) Copolymer

Question 6.
Which of the following are thermoplastic polymers?
(a) Polythene, urea-formaldehyde, polyvinyl
(b) Bakelite, polythene, polystyrene
(c) Polythene, polystyrene, polyvinyl
(d) Urea-formaldehyde, polystyrene, Bakelite

Answer: (c) Polythene, polystyrene, polyvinyl

Question 7.
Glycogen, a naturally occurring polymer stored in animals is a
(a) monosaccharide
(b) disaccharide
(c) trisaccharide
(d) polysaccharide

Answer: (d) polysaccharide

Question 8.
Nylon 6, 6 is obtained by condensation polymerization of
(a) adipic acid and ethylene glycol
(b) adipic acid and hexamethylene diamine
(c) terephthalic acid and ethylene glycol
(d) adipic acid and phenol

Answer: (b) adipic acid and hexamethylene diamine

Question 9.
Teralitre is a condensation polymer of ethylene glycol and
(a) benzoic acid
(b) phthalic acid
(c) dimethyl terephthalate
(d) salicylic acid

Answer: (c) dimethyl terephthalate

Question 10.
Natural rubber is a polymer of
(a) 1, 1-dimethylbufadiene
(b) 2-methyl-1, 3-rbutadiene
(c) 2-chlorobuta-1, 3-diene
(d) 2-chlorobut-2-ene

Answer: (b) 2-methyl-1, 3rbutadiene

Question 11.
Heating rubber with Sulphur is known as
(a) galvanization
(b) bessemerization
(c) vulcanization
(d) sulphonation

Answer: (c) vulcanization

Question 12.
Dacron is an example of
(a) polyamides
(b) polypropenes
(c) polyacrylonitrile
(d) polyesters

Answer: (d) polyesters

Question 13.
Which of the following is a condensation polymer?
(a) Teflon
(b) PVC
(c) Polyester
(d) Neoprene

Answer: (c) Polyester

Question 14.
Which of the following polymers does not involve cross-linkages?
(a) Vulcanized rubber
(b) Bakelite
(c) Melamine
(d) Teflon

Answer: (d) Teflon

Question 15.
Composition of Ziegler-Natta catalyst is
(a) (Et₃)₃Al.TiCl₂
(b) (Me)₃Al.TiCl₂
(c) (Et)₃Al.TiCl₄
(d) (Et)₃Al.PtCl₄

Answer: (c) (Et)₃Al.TiCl₄

NCERT Solutions for Class 12 Physics Chapter : 2 Electrostatic Potential and Capacitance

NCERT Solutions for Class 12 Physics Chapter : 1 Electric Charges And Fields

NCERT Solutions for Class 12 Maths Chapter :13 Probability

NCERT Solutions for Class 12 Maths Chapter :12 Linear Programming

Question 1: Offspring formed by sexual reproduction exhibit more variation than those formed by Asexual reproduction because

Question 2: The property of an undifferentiated cell that has the potential to develop into an entire plant is called

Question 3: The development of root and shoot in tissue culture is determined by ______

Question 4: There is no natural death in single celled organisms like Amoeba and bacteria because

Question 5: There are various types of reproduction. The type of reproduction adopted by an organism depends on

Question 6: A few statements describing certain features of reproduction are given below:

Question 7: Appearance of vegetative propagules from the nodes of plants such as sugarcane and ginger is mainly because

Question 8: Identify the correct sequence of events

Question 9: The term ‘clone’ cannot be applied to offspring formed by sexual reproduction because

Question 10: Amoeba and Yeast reproduce asexually by fission and budding respectively, because they are

Question 11: A few statements with regard to sexual reproduction are given below:

Question 12: A multicellular, filamentous alga exhibits a type of sexual life cycle in which the meiotic division occurs after the formation of zygote. The adult filament of this alga has

Question 13: The male gametes of rice plant have 12 chromosomes in their nucleus. The chromosome number in the female gamete, zygote and the cells of the seedling will be, respectively,

Question 14: Given below are a few statements related to external fertilization. Choose the correct statements.

Question 15 : The statements given below describe certain features that are observed in the pistil of flowers.

NCERT Solutions for Class 12 Maths Chapter :11 Three Dimensional Geometry

Question 1: Drugs can be classified on the basis of

Question 2: Antacids include

Question 3: Drawback of excess of hydrogen carbonate taking as antacid is

Question 4: Tranquilizers are prescribed for curing

Question 5: Tincture of iodine is

Question 6: 0.2% of solution of phenol and 0.2–0.4 ppm chlorine in aqueous solution respectively behave as

Question 7: Birth control pills essentially contains

Question 8: Which is mismatched, regarding the examples?

Question 9: Chemicals are added to food for

Question 10: The first popular artificial sweetening agent is

Question 11: The compound with structure

Question 12: The main disadvantage associated with use of aspartame is

Question 13: Which of the following can be used as food preservative?

Question 14: Among the following, the maximum high potency sugar is

Question 15: Soaps are sodium or potassium salt of long chain fatty acids like

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