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CBSE Solutions for Class 11 English

GSEB std 10 science solution for Gujarati check Subject Chapters Wise::

What mil be the order of energy levels 3s, 3p and 3d in case of H-atom?

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જવાબ : All have equal energy.


Discuss the similarities and differences between a 1s and a 2s orbital.

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જવાબ : Similarities:

  • Both have spherical shape.
  • Both have same angular momentum.
Differences:
  • 1s has no node while 2s has one node.
  • Energy of 2s is greater than that of 1s.


An electron beam on hitting a ZnS screen produces a scientillation on it. What do you conclude?

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જવાબ : Electron has particle nature


How many unpaired electrons are present in Pd (Z = 46) ?

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જવાબ : The electronic configuration of the element palladium (Z = 46) is [Kr]36 4d10 5S°.
This means that it has no impaired electron.


Distinguish between a photon and quantum.

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જવાબ : A quantum is a bundle of energy of a definite magnitude (E = hv) and it may be from any source. However, a photon is quantum of energy associated with light only.


When is the energy of electron regarded as zero?

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જવાબ : The energy of the electron is regarded as zero when it is at infinite distance from the nucleus.
At that point force of attraction between the electron and the nucleus is almost nil. Therefore, its energy is regarded as zero.


What type of metals are used in photoelectric cells? Give one example.

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જવાબ : The metals with low ionisation enthalpies are used in photoelectric cells. Caesium (Cs), an alkali metal belonging to group 1 is the most commonly used metal.


If n is equal to 3, what are the values of quantum numbers l and m?

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જવાબ : I = 0,1, 2
m =- 2, — 1, 0, + 1, + 2 and S = +1/2 and-1/2
for each value of m.


What is common between dxy and dx2-y 2 orbitals?

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જવાબ : Both have identical shape, consisting of four lobes.


What is the difference between ground state and excited state?

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જવાબ : Ground state means the lowest energy state. When the electrons absorb energy and jump to outer orbits, this state is called excited state.


Which series of lines of the hydrogen spectrum lie in the visible region?

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જવાબ : Balmer series.


What is difference between the notations l and L?

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જવાબ : ‘l’ signifies the secondary quantum number.
‘L’ signifies second energy level (n = 2).


What are degenerate orbitals ?

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જવાબ : Orbitals having same energy belonging to the same subshell.


What is the most important application of de Broglie concept?

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જવાબ : In the construction of electron microscope used for the measurement of objects of very small size.


Which one Fe3+, Fe2+is more paramagnetic and why?

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જવાબ : As Fe3+ contains 5 impaired electrons while Fe2+ contains only 4 unpaired electrons. Fe3+ is more paramagnetic.


Which element does not have any neutron?

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જવાબ : Hydrogen.


What is value of Planck’s constant in S.I. units?

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જવાબ : 6.62 x 1034 Js.


Arrange X-rays, cosmic rays and radio waves according to frequency.

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જવાબ : Cosmic rays > X-rays > radio waves.


An electron beam after hitting a neutral crystal produces a diffraction pattern? What do you conclude?

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જવાબ : Electron has wave nature.


Calculate the number of protons, neutrons and electrons in   ?

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જવાબ : Z = 35
A = 80
Atomic no. = 35 No. of protons = 35
No. of protons No. of electrons No. of neutrons
= 80 – 35 = 45


Arrange the electron (e), protons (p) and alpha particle (α) in the increasing order for the values of e/m (charge/mass).

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જવાબ : α < p < e.


An anion A3+has 18 electrons. Write the atomic number of A.

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જવાબ : 15


How many electrons in an atom can have n + l = 6?

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જવાબ : 18


What is the difference between notations l and L?

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જવાબ : I represents the subshell and L represents shell.


Which orbital is non-directional?

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જવાબ : s-orbital


What is the minimum product of uncertainty in the position and momentum of an electron?

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જવાબ : h/4π


Write the electronic configuration and number of unpaired electrons in Fe2+ion.

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જવાબ : Fe (Z = 26) : [Ar]18 3d64s2
Fe2+ion : [Ar]18 3d6
No. of unpaired electrons = 4


Give the relation between wavelength and momentum of moving microscopic particle. What is the relation known as?

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જવાબ :
The relation is known as de Broglie’s relationship.


Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

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જવાબ :


(i) The energy associated with first orbit in hydrogen atom is – 2.17 × 10-18 J atom-1. What is the energy associated with the fifth orbit ?
(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

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જવાબ :
(ii) For hydrogen atom ; rn = 0.529 x n2 Å
r5 = 0.529 x (5)2 = 13.225 Å = 1.3225 nm.


The uncertainty in the position of a moving bullet of mass 10 g is 10-5  m. Calculate the uncertainty in its velocity?

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જવાબ : According to uncertainty principle,


The uncertainty in the position and velocity of a particle are 10-10m and 5.27 x 10-24 ms-1 respectively. Calculate the mass of the particle. (Haryana Board 2000)

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જવાબ : According to uncertainty principle,


With what velocity must an electron journey so that its momentum is equal to that of a photon of wavelength = 5200 A?
 

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જવાબ :


Using Aufbau principle, write the ground state electronic configuration of following atoms.
(i)Boron (Z = 5) (ii) Neon (Z = 10), (iii) Aluminium (Z = 13) (iv) Chlorine (Z = 17), (v) Calcium (Z = 20) (vi) Rubidium (Z = 37)

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જવાબ : (i)Boron (Z = 5) ; 1s2 2s2 1p1
(ii)Neon (Z = 10) ; 1s2 2s2 2p6
(iii)Aluminium (Z = 13) ; 1s2 2s2 2p6 3s2 3p1
(iv)Chlorine(Z = 17) ; 1s2 2s2 2p6 3s2 3p5
(v)Calcium (Z = 20) ; 1s2 2s2 2p6 3s2 3p6 4s2
(vi)Rubidium (Z = 37) ; 1s2 2s2 2p3s3p6 3d104s2 4p6 5s1.


Calculate the de Broglie wavelength of an electron moving with 1% of the speed of light?
 

 

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જવાબ :


(i) Calculate the number of electrons which will together weigh one gram.
(ii) Calculate the mass and charge of one mole of electrons.
 

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જવાબ : (i) Mass of an electron = 9.1 × 10-28 g
9.1 × 10-28 g is the mass of = 1 electron

(ii) One mole of electrons = 6.022 × 1023 electrons
Mass of 1 electron = 9.1 × 10-31 kg
Mass of 6.022 × 1023 electrons = (9.1 × 10.31kg) × (6.022 × 1023) = 5.48 × 10-7 kg
Charge on one electron = 1.602 × 10-19 coulomb
Charge on one mole electrons = 1.602 × 10-19 × 6.022 × 1023 = 9.65 × 104 coulombs


(i) Calculate the total number of electrons present in one mole of methane.
(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = 1.675 × 10-27kg).
(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH3 at STP.

Will the answer change if the temperature and pressure are changed ?

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જવાબ : (i) One mole of methane (CH4) has molecules = 6.022 × 1023
No. of electrons present in one molecule of CH4 = 6 + 4 = 10
No. of electrons present in 6.022 × 1023 molecules of CH4 = 6.022 × 1023 × 10
= 6.022 × 1024 electrons (ii) Step I. Calculation of total number of carbon atoms
Gram atomic mass of carbon (C-14) = 14 g = 14 × 103 mg
14 × 103 mg of carbon (C-14) have atoms = 6.022 × 1023

Step II. Calculation of total number and tatal mass of neutrons
No. of neutrons present in one atom (C-14) of carbon = 14 – 6 = 8
No. of neutrons present in 3-011 × 1020 atoms (C-14) of carbon = 3.011 × 1020 × 8
= 2.408 × 1021 neutrons
Mass of one neutron = 1.675 × 10-27 kg
Mass of 2.408 × 1021 neutrons = (1.675X10-27 kg) × 2.408 × 1021
= 4.033 × 10-6 kg. (iii) Step I. Calculation of total number ofNH3 molecules
Gram molecular mass of ammonia (NH3) = 17 g = 17 × 103 mg
17 × 103 mg of NH3 have molecules = 6.022 × 1023

Step II. Calculation of total number and mass of protons
No. of protons present in one molecule of NH3 = 7 + 3 = 10 . No. of protons present in 12.044 × 1020 molecules of NH3 = 12.044 × 1020 × 10
= 1.2044 × 1022 protons
Mass of one proton = 1.67 × 10-27 kg
Mass of 1.2044 × 1022 protons = (1.67 × 10-27 kg) × 1.2044 × 1022
= 2.01 × 10-5 kg.
No, the answer will not change upon changing the temperature and pressure because only the number of protons and mass of protons are involved.


How many protons and neutrons are present in the following nuclei

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જવાબ :


Write the complete symbol for the atom (X) with the given atomic number (Z) and atomic mass (A)
(i) Z = 17,A = 35
(ii) Z = 92, A = 233
(in) Z = 4, A = 9.
 

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જવાબ :


Yellow light emitted from a sodium lamp has a wavelength (2) of 580 nm. Calculate the frequency (v) and wave number (v) of yellow light.

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જવાબ :


Calculate the energy of each of the photons which
(i) correspond to light of frequency 3 × 1015 Hz
(ii) have wavelength of 0-50 A.
 

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જવાબ : (i) Energy of photon (E) = hv
h = 6.626 × 10-34 J s ; v = 3 × 1015 Hz = 3 × 1015s-1
∴ E = (6.626 × 10-34 J s) × (3 × 1015 s-1) = 1.986 × 1018 J
Energy of photon (E) = hv = hcλ
h = 6.626 × 10 34 J s; c = 3 × 108 m s-1 ;
λ= 0.50 Å = 0.5 × 10-10 m.
 


Calculate the wavelength, frequency, and wavenumber of lightwave whose period is 2.0 × 10-10 s.

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જવાબ :


What is the number of photons of light with wavelength 4000 pm which provide 1 Joule of energy ?
 

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જવાબ : Energy of photon (E) = hcλ
h = 6.626 × 10-34 Js, c = 3 × 108 m s-1, λ = 4000 pm = 4000 × 10-12 = 4 × 10-9 m


A photon of wavelength 4 × 10-7 m strikes on metal surface ; the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon,
(ii) the kinetic energy of the emission
(iii) the velocity of the photoelectron. (Given 1 eV = 1.6020 × 10-19 J).

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જવાબ :


Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in k-J mol-1.

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જવાબ :


A 25 watt bulb emits monochromatic yellow light of wavelength 0.57 μm. Calculate the rate of emission of quanta per second.

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જવાબ :


Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 A. Calculate threshold frequency (v0) and work function (W0) of the metal.

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જવાબ :


What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from the energy level with n = 4 to energy level n = 2 ? What is the colour corresponding to this wavelength ? (Given RH = 109678 cm-1)

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જવાબ :


How much energy is required to ionise a hydrogen atom if an electron occupies n = 5 orbit ? Compare your answe r with the ionisation energy of H atom (energy required to remove the electron from n = 1 orbit)

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જવાબ :


What is the maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops to the ground state ?


 

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જવાબ : The maximum no. of emission lines = n(n–1)2 = 6(6–1)2 =3 × 5 = 15


Arrange the following type of radiations in increasing order of wavelength :
(a) radiation from microwave oven
(b) amber light from traffic signal
(c) radiation from FM radio
(d) cosmic rays from outer space and
(e) X-rays.

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જવાબ : Cosmic rays < X-rays < amber colour < microwave < FM


The longest wavelength doublet absorption transition is observed at 589 nm and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

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જવાબ :


If the photon of the wavelength 150 pm strikes an atom, one of its inner bound electrons is ejected out with a velocity of 1.5 x 107 m s-1. Calculate the energy with which it is bound to the nucleus.

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જવાબ :


The ejection of the photoelectrons from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

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જવાબ :


Life times of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 x 1015, calculate the energy of the source.

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જવાબ : Time duration (t) = 2 ns = 2 x 10-9 s

Energy of one photon, E = hv = (6.626 x 10-34 Js) x (109/2 s-1) = 3.25 x 10-25J
No. of photons = 2.5 x 105

Energy of source = 3.3125 x 10-25 J x 2.5 x 1015 = 8.28 x 10-10 J


Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

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જવાબ :


Neon gas is generally used in sign boards. If it emits strongly at 616 nm, calculate :
(a) frequency of emission (b) the distance travelled by this radiation in 30s (c) energy of quantum (d) number of quanta present if it produces 2 J of energy.

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જવાબ :


An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign symbol to the ion.

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જવાબ : Let the no. of electrons in the ion = x
∴ the no. of the protons = x + 3 (as the ion has three units positive charge)
and the no. of neutrons = x + x×31.7100 = xc + 0.304 x
Now, mass no. of ion = No. of protons + No. of neutrons
= (x + 3) + (x + 0.304x)
∴ 56 = (x + 3) + (x + 0.304x) or 2.304x = 56 – 3 = 53
x = 532.304 = 23
Atomic no. of the ion (or element) = 23 + 3 = 26
The element with atomic number 26 is iron (Fe) and the corresponding ion is Fe3+.


An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11 -1% more neutrons than the electrons, find the symbol of the ion.
 

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જવાબ : Let the no. of electron in the ion = x
∴ the no. of protons = x – 1 (as the ion has one unit negative charge)
and the no. of neutrons = x + x×11.1100 = 1.111 x
Mass no. or mass of the ion = No. of protons + No. of neutrons
(x – 1 + 1.111 x)
Given mass of the ion = 37
∴ x- 1 + 1.111 x = 37 or 2.111 x = 37 + 1 = 38
x = 382.111 = 18
No. of electrons = 18 ; No. of protons = 18 – 1 = 17
Atomic no. of the ion = 17 ; Atom corresponding to ion = Cl
Symbol of the ion = 3717Cl


An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the symbol to the element.
 

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જવાબ : An element can be identified by its atomic number only. Let us find the atomic number.
Let the number of protons = x
Number of neutrons = x + [x×31.7100 = (x × 0.317x)
Now, Mass no. of element = no. of protons =no. neutrons
81 = x + x + 0-317 x = 2.317 x or x = 812.317 = 35
∴ No. of protons = 35, No. of neutrons = 81 – 35 =46
Atomic number of element (Z) = No. of protons = 35
The element with atomic number (Z) 35 is bromine 8135Br


Symbols 7935Br and 79Br can be written whereas symbols 3579Br and 35Br are not accepted. Answer in brief.
 

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જવાબ : In the symbol BAX of an element :
A denotes the atomic number of the element
B denotes the mass number of the element.
The atomic number of the element can be identified from its symbol because no two elements can have the atomic number. However, the mass numbers have to be mentioned in order to identify the elements. Thus,
Symbols 7935Br and 79Br are accepted because atomic number of Br will remain 35 even if not mentioned. Symbol 3579Br is not accepted because atomic number of Br cannot be 79 (more than the mass number = 35). Similarly, symbol 35Br cannot be accepted because mass number has to be mentioned. This is needed to differentiate the isotopes of an element.


In Rutherford experiment, generally the thin foil of heavy atoms like gold, platinum etc. have been used to be bombarded by the a-particles. If a thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?
 

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જવાબ : We have studied that in Rutherford’s experiment by using heavy metals like gold and platinum, a large number of a-particles sufferred deflection while a very few had to retrace their path.

If a thin foil of lighter atoms like aluminium etc. be used in the Rutherford experiment, this means that the obstruction offered to the path of the fast moving a-particles will be comparatively quite less.

As a result, the number of a-particles deflected will be quite less and the particles which are deflected back will be negligible


The diameter of zinc atom is 2.6 Å. Calculate :
(a) the radius of zinc atom in pm
(b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side length wise.

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જવાબ : (a) Radius of zinc atom =Undefined control sequence \AA= 1.3 Å = 1.3 × 10-10m = 130 × 10-12m = 130 pm
(b) Length of the scale = 1.6 cm = 1.6 × 1010 pm
Diameter of zinc atom = 260 pm


In Millikan’s experiment, the charge on the oil droplets was found to be – 1.282 x 10-18C. Calculate the number of electrons present in it.

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જવાબ :


A certain particle carries 2.5 x 10-16 C of static electric charge. Calculate the number of electrons present in it.

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જવાબ :


Calculate the energy required for the process :
He+fe) → He2+(g) + e
The ionisation energy’ for the H atom in the ground state is 2.18 × 10-18  J atom-1

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જવાબ :

For H atom (Z = 1), En =2.18 × 10-18 × (l)2 J atom-1 (given)
For He+ ion (Z = 2), En =2.18 × 10-18 × (2)2 = 8.72 × 10-18 J atom-1 (one electron species)


If the diameter of carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across a length of a scale of length 20 cm long.

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જવાબ :


2 × 108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.
 

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જવાબ : The length of the arrangement = 2.4 cm
Total number of carbon atoms present = 2 ×108


Calculate the number of atoms present in :
(i) 52 moles of He
(ii) 52 u of He
(iii) 52 g of He.

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જવાબ :


Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

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જવાબ :
Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength.


How many electrons in an atom may have the following quantum numbers?
(a) n = 4 ; ms = -1/2
(b) n = 3, l = 0.

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જવાબ : (a) For n = 4
Total number of electrons = 2n2 = 2 × 16 = 32
Half out of these will have ms = —1/2
∴ Total electrons with ms (-1/2) = 16
(b) For n = 3
l= 0 ; ml = 0, ms +1/2, -1/2 (two e)


From the following sets of quantum numbers, state which are possible. Explain why the others are not possible.
(i) n = 0, l = 0, ml = 0, ms = +1/2
(ii) n = 1, l = 0, ml = 0, ms – – 1/2
(iii) n = 1, l = 1, ml = 0, ms= +1/2
(iv) n = 1, l = 0, ml = +1, ms= +1/2
(v) n = 3, l = 3, ml = -3, ms = +1/2
(vi) n = 3, l = 1, ml = 0, ms= +1/2

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જવાબ : (i) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(ii) The set of quantum numbers is possible.
(iii) The set of quantum numbers is not possible because, for n = 1, l can not be equal to 1. It can have 0 value.
(iv) The set of quantum numbers is not possible because for l = 0. mt cannot be + 1. It must be zero.
(v) The set of quantum numbers is not possible because, for n = 3, l ≠ 3.
(vi) The set of quantum numbers is possible.


Using s, p and d notations, describe the orbitals with follow ing quantum numbers :
(a) n = 1, l = 0
(b) n = 4, l = 3
(c) n = 3, l = 1
(d) n = 4, l = 2

 

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જવાબ : (a) 1s orbital
(b) 4f orbital
(c) 3p orbital
(d) 4d orbital


(i) An atomic orbital has n = 3. What are the possible values of l and ml ?
(ii) List the quantum numbers ml and l of electron in 3rd orbital.
(iii) Which of the following orbitals are possible ?
1p, 2s, 2p and 3f.
 

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જવાબ : (i) For n = 3; l = 0, 1 and 2.
For l = 0 ; ml = 0
For l = 1; ml = +1, 0, -1
For l = 2 ; ml = +2, +1,0, +1, + 2
(ii) For an electron in 3rd orbital ; n = 3; l = 2 ; ml can have any of the values -2, -1, 0,
+ 1, +2.
(iii) 1p and 3f orbitals are not possible.


(i) Write the electronic configuration of the following ions : (a) H (b) Na+ (c) 02~ (d) F.
(ii) What are the atomic numbers of the elements whose outermost electronic configurations are represented by :
(a) 3s1 (b) Ip3 and (c) 3d6 ?
(iii) Which atoms are indicated by the following configurations ?
(a) [He]2s1 (b) [Ne] 3s2 3p3 (c) [Ar] 4s2 3d1.
 

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જવાબ : (i) (a) 1s2
(b) 1s2 2s2 2p6
(c) 1s22s22p6
(d) 1s22s22p6.
(ii) (a) Na (Z = 11) has outermost electronic configuration = 3s1
(b) N (Z = 7) has outermost electronic configuration = 2p3
(c) Fe (Z = 26) has outermost electronic configuration = 3d6
(iii) (a) Li
(b) P
(c) Sc


Which of the following are iso-electronic species ?
Na+, K+, Mg2+, Ca2+, S2-, Ar.

 

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જવાબ : Na+ and Mg2+ are iso-electronic species (have 10 electrons) K+, Ca2+ , S2- are iso-electronic species (have 18 electrons)


What is the energy in joules required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of light emitted when the electron returns to the ground state ? The ground state electronic energy is – 2.18 × 11-11 ergs.

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જવાબ :


The electronic energy in hydrogen atom is given by En (-2.18 × 10-18 s) / n2J. Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition?

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જવાબ :


Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s-1.

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જવાબ :


The mass of an electron is 9.1 × 10-31 kg. If its kinetic energy is 3.0 × 10-25 J, calculate its wavelength.

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જવાબ :


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Browse & Download CBSE Books For Class 11 All Subjects

The GSEB Books for class 10 are designed as per the syllabus followed Gujarat Secondary and Higher Secondary Education Board provides key detailed, and a through solutions to all the questions relating to the GSEB textbooks.

The purpose is to provide help to the students with their homework, preparing for the examinations and personal learning. These books are very helpful for the preparation of examination.

For more details about the GSEB books for Class 10, you can access the PDF which is as in the above given links for the same.