Solution

                                       Solutions

Solutions : It is the homogeneous mixture of two or more than two components.

Exam : Mixture of sugar and water.

Solvent : The component of a solution present in the huge quantity is referred  to as solvent which determines the physical state of the solution.

 Solute : The components of a solution present in the lesser amount and get easily dissolved in solvent  is referred  as solute.

Exam : In the case of lemonade Water is the solvent whereas salt/sugar are the solute.

Types of solutions

Solutions can be broadly categorized into two types :-

Homogeneous solutions :   Solutions whose components have uniform composition and properties throughout the solution are known as homogeneous solution.

E.g :  solution of salt or sugar in water, cough syrup, are homogeneous mixture of chemicals and dyes, etc.

                                      

Homogenous solution of sugar dissolved in water.

Heterogeneous solutions :  Solutions whose component  have non-uniform composition and properties throughout the solution are known as heterogeneous solution.

 E.g :  solution of water and sand, solution of oil and water, water and chalk powder, etc.

        

                                                        

Type of solution	Solute	Solvent	Example
Gaseous	Gas	Gas	Hydrogen and nitrogen mixture
	Liquid	Gas	Chloroform with nitrogen gas
	Solid	Gas	Camphor in nitrogen gas
Liquid	Gas	Liquid	Oxygen in water
	Liquid	Liquid	Ethanol in water
	Solid	Liquid	Glucose in water
Solid	Gas	Solid	Hydrogen solution in palladium
	Liquid	Solid	Amalgam of mercury with sodium
	Solid	Solid	Dissolved copper in gold

Percentage compositions

Mass percentage

Mass percentage can be defined as the percentage of mass of the component of the solution present in total mass of the solution.

                                 

Volume percentage

    Volume percentage can be defined as percentage of volume of the component of the solution present in total mass of the solution.

              

Mass by volume percentage

         Mass by volume percentage can be defined as the mass of solute dissolved in 100 mL of the solution.

For instance, mass by volume percent of a solution with 1 g of solute dissolved in 100 mL of solution will be 1% or 1% (mass/volume).

     

        Parts per million

                     Parts per million can be defined as the ratio of number of parts of the component to the total number of parts of all components of the solution multiplied by 10⁶. It is denoted by ppm.

v It is used to express concentration of a solution where solute is present in trace.

v For instance, concentration of pollutants in water or atmosphere is expressed in terms of

g mL^-1 or ppm.

            

    Mole fraction

      Mole fraction can be defined as the ratio of number of moles of the component in the solution to the total number of moles of all components in the solution.

Molality

                                                     

     Molality (m) can be defined as the number of moles of solute dissolved in per Kg of solvent.                                                          

        Unit – molal or m.

Molarity

                   Molarity (M) can be defined as the  number of moles of solute dissolved in per liter of solution.

             It is function of temperature due to the dependence of volume on temperature whereas Mass %,  ppm,  mole fraction and molality are independent of temperature because mass does not depend on temperature.

          Unit – molar or M.

Solubility

                          Solubility is a physical property of a solution. It can be defined as the maximum amount of solute that can be dissolved in a specific amount of solvent at constant temperature and pressure.

         This property of solution depends upon the factors like:

v The nature of solute and solvent and the interaction between them : The stronger the attractions between solute and solvent molecules, the greater will be the solubility.

v Molecular size of solute : Larger molecules are more difficult to dissolve in solvents whereas the smaller molecules dissolve easily and are more soluble in solvents.

v Polarity : Polar solvents (having bonds with different electro negativities) like Water, ethanol, formaldehyde dissolve polar solutes, whereas non-polar solvents (having bonds with similar electro negativities) like pyridine, toluene, and hexane dissolve non-polar solutes.

v Temperature :  Rise in temperature increases the solubility.

v  Pressure :  Rise in pressure increases the solubility.

Solubility of a solid in a liquid

                             Solution of sugar or salt dissolved is a common example of solubility of a solid in liquid. But it is not necessary that all solids will dissolve in liquid.

                 For instance, solids like naphthalene and anthracene do not dissolve in water but dissolves easily in benzene but sugar and salt does not dissolves in benzene. This is because it is property of solution that polar solutes dissolve in polar solvents like Water, ethanol, formaldehyde and chloroform whereas non polar solutes dissolves in non-polar solvents like pyridine, toluene, and hexane. This phenomenon can be easily defined as like dissolves like.

v The phenomenon of increase in the concentration of solution due to dissolving of solid solute to the solvent is known as dissolution.

v The phenomenon of collision of solute particles in a solution resulting in the separation of the solute particles from the solution is known as crystallization.

v According to Le Chateliers Principle in a saturated solution, if the dissolution process is endothermic (Δ_sol H > 0), the solubility should increase with rise in temperature but if the dissolution process is exothermic (Δ_sol H > 0) the solubility should decrease.

v Pressure does not have crucial effect on solubility of solids in liquids because they remain unaffected to pressure due to the high incompressibility of solids and liquids.

Solubility of a gas in a liquid

           The amount of gas that can be dissolved in a particular amount of liquid at constant temperature and pressure is called solubility of gas in liquid.

Solubility of gas in liquid is highly affected by temperature and pressure.

v Increase in temperature reduces the solubility of gases in liquids because on dissolving gas molecules in liquid the process of dissolution is similar to condensation leading to the evolution of heat. Therefore, dissolution is an exothermic process, due to which the solubility decreases with increase of temperature.

v Increase in pressure by compressing the gas to a smaller volume increase the number of
gaseous particles per unit volume over the solution and also the rate at which the gaseous particles are striking the surface of solution to enter it leading to the increase in the solubility of gases.

          Solubility of gas in liquid can be explained by Henry’s law : -

Henry’s law

                             According to this law, “At constant temperature and pressure, the solubility of gas in liquid will be directly proportional to the pressure applied over it.

                      It can also be stated as - the partial pressure of the gas in vapour phase (p) is proportional to the mole fraction of the gas (x) in the solution.

  Mathematically,

                                                  p = K_H χ

               

                     where,  K_H = Henry’s law constant.         

v At same temperature different gasses have different K_H

v Value of K_H increases with the increase in temperature therefore solubility of gases increases with decreasing temperature. Due to this reason cold water is more sustainable for aquatic life than warm water.

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General Physics Numerical

General Physics Numerical

1)A physical quantity Q is related to four observables x, y, z and t by the relation
Q = 𝑥^2/5𝑧3𝑦 𝑡
The percentage errors of measurement in x, y, z and t are 2.5%, 2%, 0.5% and 1% respectively. Find the percentage error in Q.

2) The length and breadth of a rectangle are (5.7 ± 0.1)cm and (3.4 ± 0.2)cm. Find the error in area of rectangle with limited error.

3) The dimension of 12 Є₀E^2 (Є₀=permittivity of free space, E = Electric field) is.

4) If are A, velocity V and density ρ are base units, then find the dimensional formula of force.

5) To find the value of g using simple pendulum, T =(2 ± 0.01)sec and l = (50 ± 0.1)cm was measured. Find the maximum percentage error in g.

6) A wire has a length l = (6 ± 0.06)cm, radius r =(0.05 ± 0.005)cm and mass m = (0.3 ± 0.003)gm. Find the maximum error in the density of wire.

7) The density of a sphere is measured by measuring its mass and diameter. If it is known that the maximum percentage error in the measurement is 2% and 3%. Find the maximum percentage error in the measurement of density.

8) Find the maximum and minimum value of the functions –
a) y = 25x^2 + 5 -10x                                                     b) y = 9 – (x – 3)2
c) y = (sin 2x – x) where, −𝜋2 < x < 𝜋2                   d) y = x^3 – 6x^2 + 9x +15
e) y = 5 – (x – 1)^2

9) If the error in measurement of radius of a sphere is 1%. Find the error in measurement of volume.

10) If the momentum of particle is increased by 100%.Find the percentage increase in the kinetic energy of the particle.

11) By what percentage should the pressure of a given mass of a gas be increased, so as to decrease its volume by 10% at a constant temperature.

12) Percentage error in the measurement of mass and velocity are 2% and 3% respectively. Find the error in the estimate of kinetic energy obtained by measuring mass and velocity.

13) The radius of a sphere is measured to be (2.1 ± 0.5)cm. Calculate its surface area with limit error.

14) The value of gravitation constant G = 6.67 x 10-11 Nm^2/kg^2 in S.I units. Convert it into CGS unit.

15) A force is given by F= at + bt^2, where t is the time. Find the dimension of a and b.

16) The velocity v of a particle is given by
V = at + 𝑏𝑡+𝑐 , where a, b and c are constant.
Find the dimension of a, b and c.

17) Find the dimension of 𝑎𝑏 in the equation P = 𝑎−𝑡^2𝑏𝑥 , where P is pressure, x is distance and t is time.

18) Find the physical quantity whose dimension is [ M^-1 L^-3 T^3A^2 ].

19) Young’s modulus of steel is 2.0 x 1011 N/m^2. Express it into dyne/cm2.

20) In the expression y = a sin(ωt +θ), y is the displacement and t is time. Write the dimension of a, ω and θ.

21) Charge on the capacitor is given by Q = Iα𝑒−𝐼𝑡Δ𝑣Є°𝛽 , where β and α are constants. t=time, I = current , Δv= potential difference. Find the dimension of 𝛽𝛼 .

22) Find the percentage error in the expression x = 𝑎𝑏^2𝑐^3. If the percentage error in a, b and c are ±1%, ±3% and ±2% respectively.

23) The momentum of an electron in an orbit is h/λ, where h is plank’s constant and λ is the wavelength associated with it. The nuclear magnetron of electron of charge e and mass me is given as
μn = 𝑒ℎ3672𝜋𝑚𝑒.
Find the dimension of μn .

24) Intensity observed in an interference pattern is I = I₀ sin2θ. At θ = 30° intensity I =5 ± 0.0020 W/m2. Find percentage error in angle if I₀ = 20 W/m2.

25) The length of string of a simple pendulum is measured with a meter scale to be 90 cm. The radius of a bob plus the length of hook is calculated to be 2.13 cm using measurement with a side calipers. What is the effective length of the pendulum? {This effective length is defined as the distance between the point of suspension and the centre of the bob}

26) Vernier calipers has 20 divisions on its vernier scale which coincide with 19 divisions on the main scale. Least count of the instrument is 0.1mm. Find the main scale division.

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Projectile Motion Numerical

                   Projectile Motion Numerical

SINGLE CORRECT CHOICE TYPE.

1. A cannon ball has the same range R on a horizontal plane for two angles of projection. If h1 and h2 are the greatest heights in the two paths for which this is possible, then
a) R=h1h2                  b) R=4(h1h2)1/2         c) R=(h1h2)1/3       d) R=(h1h2)1/4

2. A cricket ball is hit at 30ᵒ with the horizontal with kinetic energy K. The kinetic energy at the highest point is
a) 0                                   b) K/4                             c) K/2                      d) 3K/4

3. A particle is projected at an angle of elevation α and after t sec it appears to have an angle of elevation β as seen from the point of projection. The initial velocity will be
a) gt/2sin(α-β)             b) gtcosβ/2sin(α-β)        c) sin(α-β)/gt        d) 2sin(α-β)/gtcosβ

4. Two balls are thrown at the same time with the same speed (10 m/s). One ball is thrown straight up and the other ball is thrown at an angle of 45°. Which ball hits the ground first?
A) The ball thrown straight up                    B) The ball thrown at an angle
C) Both hit at the same time                        D)Not enough information to tell

5. A ball is kicked 10 m/s horizontally off a cliff. After 1 second, its speed is approximately
 a. 10 m/s                   b. 14 m/s                        c. 20 m/s                             d. 28 m/s

6. In the absence of air resistance, a ball of mass m is thrown straight up and reaches a height of 10 m. The net force on the ball half-way up (at 5 m up) is
 a. mg/4                     b. mg/2                            c. mg                                     d. 2mg

7. The greatest range of a particle, projected with a given velocity on an inclined plane, is x times the greatest vertical altitude above the inclined plane. Find the value of x ;
A. 2
B. 4
C. 3
D. 1/2

8. A person throws vertically n balls per second with the same velocity. He throws a ball whenever the previous oneis at its highest point. The height to which the ball rise is :
a) g/n^2
b) 2gn
c) g/2n^2
d) 2gn^2

9. A particle moves in x - y plane according to the equations x = 3cos4t and y = 3(1 - sin4t). The distance travelled bythe particle in 2 sec is (where x and y is in meter)
a) 48m
b) 24m
c) 48√2 m
d) 24√2 m.

10. A particle is projected at an angle of elevation α and after t seconds it appears to have an angle of elevation β asseen from point of projection. The initial velocity will be
A) gt / 2sin(α -β)
B) gt cosβ / 2sin(α - β)
C) Sin(α -β) / 2gt
D) 2sin(α - β) / gtcosβ.

11. A particle is projected at angle 37º with the inclined plane in upward direction with speed 10 m/s. Theangle of inclination is given 53º. Then the maximum height attained by the particle from the incline plane
will be
A) 3m
B) 4m
C) 5m
D) zero

12. For a stone thrown from the tower of unknown height, the maximum range for a projection speed of 10m/s is obtained for a projection angle of 30º. The corresponding distance between the foot of the tower
and the point of landing of the stone is
A) 10m
B) 20m
C) (20/√3)m
D) (10/√3)m.

13. A cricket ball is hit at 30º with the horizontal with kinetic energy K. The kinetic energy at the highest
point is
A) 0
B) K/4
C) K/2
D) 3K/4.

14. A body of mass m is projected at an angle of 45º with the horizontal with velocity u. If air resistance is
negligible, then total change in momentum when it strikes the ground is
A) 2mu
B) √2 mu
C) mu
D) mu / √2.

15. A particle is projected with speed 10 m/s at an angle 60º with the horizontal. Then the time after which
its speed becomes half of its initial is:
A) 1/2 sec
B) √3 /2 sec
C) 1 sec
D)2/3sec

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STRUCTURE OF ATOM

STRUCTURE OF ATOM

Thomson model of atom :

 J.J Thomson in 1898, proposed that an atom possesses a spherical shape (radius approx. 10-10m) in which the positive charge is uniformly distributed. The electrons are embedded in it in such a manner as to give the most stable electrostatic arrangement. Many different names are given to this model, like plum pudding, raisin pudding and watermelon.

This model can be visualised as a pudding or watermelon of positive charge with plum and seeds embedded into it. An important feature of this model is that the mass of the atom is assumed to be uniformly distributed over the atom. This model was not able to explain the overall neutrality of the atom.

Rutherford α scattering experiment :

Rutherford’s famous α particle scattering experiment is bombarded very thin gold foil with α particle. A stream of high energy α particle from a radioactive source was directed at a very thin gold foil (thickness 100nm). The thin gold foil has a circular fluorescent zinc sulphide screen around it. Whenever, α- particles stuck to the screen, a tiny flash of light was produced at that point.

Observations of α- scattering experiment :

•  Most of the α - particle passed through the gold foil un-deflected.
• A small fraction of α - particle was deflected through a small angle.
•  A very few α - particle (approx. 1 in 20000) bounced back, that is, were deflected by an angle of 180°.

Conclusion of α - scattering experiment :

•  Most of the space in the atom is empty as most of the α - particles passed through the foil un-deflected.
• A few positively charged α - particles were deflected. The deflection must be due to enormous repulsive force showing that the positive charge of the atom is not spread throughout the atom as assumed by Thomson. The positive charge has to be concentrated in a very small volume that repelled and deflects the positive charged α - particles.
• Calculations by Rutherford showed that the volume occupied by the nucleus is negligibly small as compared to the total volume of the atom. The radius of the atom is about 10-10m while the radius of the nucleus is about 10-15m.

Rutherford’s model of an atom :

• The positive charge and most of the mass of an atom is densely concentrated in extremely small region. This small portion of the atom is known as NUCLEUS.
• The nucleus is surrounded by electrons that moves around the nucleus with a very high speed in a circular path called as orbits. Thus, Rutherford’s model f atom resembles the solar system in which the nucleus is sun and the electrons are the revolving planets.
• Electrons and the nucleus are held together by electrostatic force of attraction.

ATOMIC NUMBER : 

Atomic number is the number of protons present in an atom.
Atomic number = number of proton.
Atomic no. = no. of electron in a neutral atom.
Mass no. or atomic mass = no. of protons + no. of neutrons
No. Of neutrons = atomic mass - atomic no.

ISOTOPES : 

The species having same atomic number but different mass number are called Isotopes.

Exam: 1) 1H1, 1H2, 1H3.
2) 6C12, 6C14.

ISOBARS : 

The species having same mass number but different atomic number are called as Isobars.
Exam: 1) 7N14, 6C14.
2) 18Ar40, 20Ca40.

ISOTOMES : 

Species having same no. of neutrons are called as Isotomes.
Exam: 1) 6C14, 8O16.

ISOELECTRIC SPECIES : 

species having same no. of electrons are called as Iso-electric species.
Exam: 1) Na+, Ne.
2) F-, Mg2+.

ELECTROMAGNETIC WAVES : 

The wave which are produced when a particle is accelerated due to varying electric and magnetic field. EMW consist both electric and magnetic field perpendicular to each other as well as perpendicular to the direction of propagation of wave.
Properties of Electromagnetic waves are - - - -

• They do not require any medium to travel.
• When these waves are arranged from low wavelength to high wavelength region, then they forms a spectrum known as EM spectrum.
• Gamma waves are the lowest wavelength wave and radio waves are the highest wavelength wave of EM spectrum.
•  EM waves shows all the characteristics of wave like frequency, amplitude, wavelength etc.
• These waves follows the relation ---

                                                        f = 1/T and
                                                         c = f入
                                   {c=speed of light, f=frequency, 入=wavelength }
NOTE: Wave number = 1/λ.

PHOTOELECTRIC EFFECT :

 It is the emission of photo electron from the surface of metal when a light of certain minimum frequency falls over it.

Properties of photoelectric effect :

•  The energy required to just take out electron from the metal surface is called work function (W₀).
• The minimum frequency of light below which the photoelectric effect does not takes place is called THRESHOLD FREQUENCY.
•  Number of photo electron depends on the brightness or intensity of light striking on the metal surface. Brighter the light greater the number of photo electron emitted.
•  As soon as the beam of light strikes the metal surface, photo electrons emitted by the metal surface. It means there is no time gap between striking of light and emission of photo electron.
• The energy of beam of light striking on metal surface is divided into two parts one for the emission of electron (work function) and the remaining as the kinetic energy of the photo electron.

                                           Energy = work function + kinetic energy
                                                     hν = hν₀ + 1/2 mv2
{ hν= energy of light source, hν₀ = work function, 1/2 mv2 = kinetic energy of photo electron}
                                        hν = W₀ + 1/2 mv2
                                        1/2 mv2 = (hν - hν₀)
                                      kinetic energy = h(ν - ν₀)
{m= mass of electron, h=plank’s constant(6.62 X 10-34 J/Sec), ν=frequency of light, ν₀= threshold frequency}

                                     W₀ = hν₀ (where, ν₀ = c/λ₀).

SPECTRUM :  

The collection of waves of different frequency or wavelength in increasing or decreasing order of their wavelength or frequency is called as spectrum.
Depending on nature waves are of two type:

a) Continuous spectrum : A spectrum in which different regions are not separated by dark lines are called as continuous spectrum.

b) Line spectrum : The spectrum in which different regions are separated by dark fine lines are called as line spectrum.
Line spectrum are also of two type:

1) Atom emission spectrum : It is obtained from the radiation emitted from atom in its excited state (higher energy state of atom).
The radiation of excited atom are made to fall on prism so that a line emission spectrum of an atom is obtained.

2) Atom absorption spectrum : It is obtained from an atom in its ground state (lower most energy state).
When a white light is passed through an atom and it is taken up to prism after passing through an atom then it splits up and form a line absorption spectrum of an atom.

HYDROGEN SPECTRUM :

When electric current is passed through hydrogen gas then its molecules splits up into atoms in excited state. The spectrum obtained from these these atoms of hydrogen is called hydrogen spectrum.

In hydrogen spectrum Balmer region was first of all identified because it is found in visible region. After the advancement in spectrum other regions in series are identified, which are as follow -

• Lyman series.
• Balmer series.
• Paschen series.
• Brackket series.
• P-fund series.

Hydrogen spectrum

NOTE :- Wave number for electronic transition in hydrogen spectrum is given by-

109677 = Rydberg constant.(cm-1).

BOHR’S MODEL OF ATOM

• Electron revolve around the nucleus in a circular path called as orbit.
• Energy of electron do not changes while moving in an energy level but it changes when it shows absorption or emission type of transition. In absorption electron absorb energy to move from its lower energy state to higher energy state, whereas, in emission electron moves from its higher energy state to its lower energy state.
• Frequency of radiation during absorption or emission is given by –

                        ΔE =hν
                       ν= ΔE/h

• Energy of an electron in its stationary state (orbit) can be given by the following formula –

                                       En = −2.18 x10−18n2 J

• Electron can move in those orbit in which their angular momentum is quantized. It means the angular momentum of an electron in that orbit must be an integral multiple of h2π.

                                  Angular momentum = mevr.
                                   Quantization of angular momentum –
                                   mevr = nh2π

• Where, n is an orbit in which electron can move, r = radius of orbit, v = velocity of electron, m= mass of electron.
•  This model can be used for all those species which have single electron (He+, Li+, Be+2).

                 For such species energy of electron in nth orbit can be calculated by-
                                         En = −2.18 X 10−18(Z2)n2

•  Radius of nth orbit can be calculated by –

                            Rn = 5.29xn2Z {where Z is an atomic number}

• It is impossible to calculate the velocity of electron moving in those orbit.

Limitation of Bohr’s model

•  It is applicable for single electron species only.
• It could not explain the Zeeman effect and Stark effect. Zeeman effect is the splitting of spectral lines in the presence of magnetic field. Stark effect, it is the splitting of spectral lines in the presence of electric field.

• It could not explain the doublet and triplet of spectral lines.

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