### True Discount and Banker's Discount problems tricks in Hindi | fast track formulae for problem solving.

TRUE DISCOUNT AND BANKER'S DISCOUNT TRICKS IN HINDI

TRUE DISCOUNT 1). वास्तविक बट्टा = मिश्रधन - वर्तमान धन
2). यदि ब्याज की दर r% वार्षिक , समय t व वर्तमान धन (pw) है, तो वास्तविक बट्टा = PW×r×t/100
3). यदि r% तथा समय t के बाद देय धन A है, तो तत्काल धन pw = 100×A/(100+r.t)
4). यदि देय धन A पर r% व समय t दिए गए है, तो वास्तविक बट्टा = A.r.t /(100+r.t)
5). यदि किसी निश्चित समय के पश्चात निश्चित वार्षिक दर पर, देय धन पर वास्तविक बट्टा (TD) व समान समय व दर के लिए साधारण ब्याज (SI) हैं, तो देय धन A =      SI × TD/(SI - TD)
6). यदि t समय पश्चात r% वार्षिक दर पर, देय धन पर वास्तविक बट्टा (TD) व साधारण ब्याज (SI) है , तो -  SI - TD = TD×r×t/100
7). t वर्ष बाद r% चक्रवृद्धि दर पर देय धन A का तत्काल धन (PW) = A/(1+r/100)^t   व वास्तविक बट्टा = A - PW
BANKER'S DISCOUNT1). महाजनी बट्टा = शेष समय (समाप्त न हुए समय) के लिए बिल पर ब्याज = बिल की राशि × दर × शेष समय /100
2). महाजनी लाभ = महाजनी बट्टा - वास्तविक बट्टा
3). यदि बिल का मान / अंकित मूल्य A है, समय t व दर r% है, तो महाजनी बट्टा = A×r×t/100
4).…

### Thermal properties of matter.| Concept booster , chapter highlights, study material for IIT JEE , NEET ETC.| LAWS OF NATURE

THERMAL PROPERTIES OF MATTER
HEAT AND TEMPERATURE

*. Temperature is the measure of degree of hotness or coldness.
*. If two body have tempt T1 and T2, if T1 > T2 then T1 is hotter than T2.
*. Heat is the form of energy which flows from hotter body to colder body by virtue of its temperature difference.
*. Transfer of heat is a non mechanical process.
*. It's unit is joule (J) and sometimes uses calories (cal). 1 joule equal to 4.2 calorie.
*. According to calorific theory of heat - heat is invisible , colourless, odourless and weightless fluid called caloric ,which flow from hotter body to colder body.
*. According to dynamic theory of heat-  all substance (solid, liquid and gases) are made up of molecules. Depending upon the nature and tempt of materials they posses three different types of motion.
1). Translatory motion
2). Vibrational motion
3). Rotational motion

*. Temperature is commonly measured by thermometers.
*. It has different units like Kelvin(K) , degree Celsius (°C) , Fahrenheit (F) etc.
*. Relation between Fahrenheit and degree Celsius is ,Tf= (9/5)Tc+32
*. Relation between degree Celsius and Kelvin is-
Tk= Tc+ 273.15
*. All temperature measuring scales are related to other scales as below-
C-0/100 = F-32/180 = R-0/80 = K-273/100
= Ra-460/212

HEAT CAPACITY

*. Heat or thermal capacity is the amount of heat needed to change the temperature by unity ie. 1°C , heat capacity is denoted by S and it's SI unit is J/K.
Heat capacity (S) = ∆Q/∆T
*. Specific heat capacity defined as- the amount of heat needed to raise the temperature of mass m by unity. It is denoted by (s) and it's SI unit is
J/Kg.K.
Specific heat capacity (s) = S/m = ∆Q/m.∆T
*. Molar specific heat capacity is defined as- it is the amount of heat needed to raise the temperature of one mole of substance (gas) by unity. It is denoted by C , it's SI unit is J/mol.K
C = S/μ = ∆Q/μ.∆T
[μ = no. of mole of substance]
*. Relation between heat capacity and molar specific heat capacity is , C = Ms , where M is molecular mass and s is the specific heat.

*. Latent heat - it is amount of heat needed to change the state of solid of mass m to liquid at melting point of solid and from liquid to gas at boiling point of liquid is called latent heat.
Latent heat is given by Q = mL , where L is the latent heat.

TYPES OF MOLAR SPECIFIC HEAT CAPACITY

*. Molar specific heat capacity at constant pressure - It is the amount of heat needed to raise the temperature of one mole of gas by unity at constant pressure. It is denoted by Cp.
Cp = (∆Q/μ∆T) constant pressure = (f/2 +1)R

*. Molar specific heat capacity at constant volume- It is the amount of heat needed to raise the temperature of one mole of gas by unity at constant volume. It is denoted by Cv.
Cv = (∆Q/μ∆T) constant pressure = (f/2)R

MOLAR HEAT CAPACITY OF IDEAL GAS

*. Heat capacity for monoatomic gases , f= 3
Cv = (3/2)R , Cp = (5/2)R , Cp/Cv = λ = 5/3= 1.67

*. Heat capacity for diatomic gases, f= 5
Cv = (5/2)R , Cp = (7/2)R , Cp/Cv = λ = 7/5= 1.4

*. Heat capacity for triatomic gases, f = 6
Cv = 3R , Cp = 4R , Cp/Cv =λ= 4/3 = 1.33

General equation of heat capacity for gases , if f (degree of freedom of molecules) is given.
Cv = (f/2)R , Cp = (f/2 + 1)R , λ = Cp/Cv = [ 1+2/f]

DULONG AND PETIT LAW

*. Average molar specific heat capacity of all metals at room temperature is nearly same and equal to 3R or 6 cal/mol.K
*. Temperature above which metals have constant Cv is called debye temperature.
*. Mayer's equation is given by Cp - Cv = R , it is only for ideal gases.

THERMAL EXPANSIONS

*. Thermal expansion is three types-
1). Linear expansion
2). Superficial expansion
3). Volumetric expansion

*. Linear expansion - when any rod is being heated, it's increase in length ∆L is directly proportional to the original length L0 and change in temperature ∆T.
∆L = αL0∆T , α = ∆L/L0.∆T , α is called the cofficient of linear expansion.
and total increases in length is
L = L0+∆L = L0(1+α∆T) , [ L is length after heating the rod]

*. If rod are free to expand then there will be no stree or strain , but if it is restricted to expand/contract by increasing or decreasing the temperature , then there strain and stree will be produced.
If temperature are used to produce compressive or tensile stress in the rof, then the stress produced is called thermal stress.
Strain = ∆L/L0 = α∆T

*. Superficial expansion - if any solid material is heated then it's area also increases , this increased area is directly proportional to the original area A0 and change in the temperature ∆T.
∆A = βA0∆T , β = ∆A/A0.∆T , and β is called the cofficient of areal expansion.
Area after heating the solid is -
A = A0+∆A = A0(1+β∆T)

*. Volumetric expansion - if any solid is heated then it's volume is also increases , this increased volume is directly proportional to the original volume V0 and change in temperature ∆T.
∆V = γV0∆T , γ = ∆V/V0.∆T , γ is called the cofficient of volumetric expansion.
Volume after heating the solid material
V = V0+∆V = V0(1+γ∆T)

*. α , β and γ is related as-
For isotropic material - α:β:γ = 1:2:3 or α/1 = β/2 = γ/3
For non isotropic material
β = α1 + α2 +α3 , here α1, α2 and α3 is linear expansion in X , Y and Z directions.

*. Variations of density with temperature is given by- , d = d0/(1+γ∆T)
Value of γ gamma is usually small for solid, so we write ; d = d0(1 - γ∆T)

*. Apparent expansion of liquid - if container is full and when temperature is changed by ∆T.
Volumetric expansion of liquid in container is -
V(L) = V0(1+γL∆T) , and here γL is the cofficient of liquid volume expansion.

Volumetric expansion of container is-
Vc = V0(1+γc∆T) , γc is the cofficient of container volume expansion.
∆V = VL - Vc = V0(γL - γc)∆T
(γL - γc) is the cofficient of apparent expansion of liquid with respect to container.
γ(app) = γL - γc

Height of the liquid level in the container , if container is initially completely filled.
h = V0(1+γL∆T)/A0(1+2αc∆T)
= h0{1+(γL - 2αc)∆T} then
h = h0{1+(γL - 2αc)∆T}
[h0 is the original height of liquid in the container and αc is the linear cofficient of expansion of container]

YOU MAY ALSO LIKE

*. Calorimetry - if two substance of different mass, different specific heat capacity and with different temperatures are mixed together, then heat starts to flow from high temperature substance to the low temperature substance till a common temperature is reached.
In this heat transfer, heat lost by one body is equal to the heat gain by other body.
m1×s1×∆T1 = m2×s2×∆T2

JOULE'S MECHANICAL EQUIVALENT OF HEAT

*. J=mechanical work done(W)/heat produced(H)
= 4.2J/cal = 4.2×10^7ergs/cal

*. If a object of mass m is falling from height h , and during falling it's temperature increases to T , then , j = mgh/smT or T = gh/js

*. If a block of ice having mass m, melts during falling from height h then
j = mgh/mL or L = jL/g

*. If a bullet having mass m and moving with velocity v , then it suddenly stops then it's whole kinetic energy converted into heat, then
j = 1/2mv^2/smT or T = v^2/2js

VARIATIONS OF FORCE OF BUOYANCY WITH TEMPERATURE

*.  Force of buoyancy on any object which is completely submerged into the liquid of density dL is , FB = V0dLg
V0 is the volume of object inside the liquid
dLis the density of liquid

Volume of object after increase of the temperature is , V = V0(1+γο∆Τ)
Density of the body after increasing temperature
d'L = dL/(1+γL∆T)
Buoyancy force after increasing temperature
F'B = Vd'Lg
F'B/FB = [1+γο∆Τ]/[1+γL∆T]
But if γL< γo then F'B<FB , if this happens then apparent weight of the object increases
[W - F'B > W - FB]

VARIATIONS OF TIME PERIOD OF PENDULUM

*. If T = 2π√L0/g is the time period of pendulum at temperature T0 and T' = 2π√L/g at temperature T. Then
T'/T = √L0/L = √L(1+α∆T)/L = 1+ (1/2)α∆Τ
So change (gain or loss) in time per unit time lapsed is
T'-T/T = (1/2)α∆Τ

### Boat and stream short-tricks in Hindi | Fast track arithmetic formulae for competitive examination.

BOAT AND STREAM (नाव एवं धारा)
1). यदि शांत जल में नाव या तैराक की चाल x किमी/घंटा व धारा की चाल y किमी/घंटा है, तो धारा के अनुकूल नाव अथवा तैराक की चाल = (x+y) किमी/घंटा
2). धारा के प्रतिकूल नाव अथवा तैराक की चाल = (x-y) किमी /घंटा
3). नाव की चाल = (अनुप्रवाह चाल + उद्धर्वप्रवाह चाल)/2
4). धारा की चाल =  (अनुप्रवाह चाल - उद्धर्वप्रवाह चाल)/2
5). यदि धारा की चाल a किमी/घंटा है, तथा किसी नाव अथवा तैराक को उद्धर्वप्रवाह जाने में अनुप्रवाह जाने के समय का n गुना समय लगता है,(समान दूरी के लिए), तो शांत जल में नाव की चाल = a(n+1)/(n-1) किमी/घंटा
6). शांत जल में किसी नाव की चाल x किमी/घंटा व धारा की चाल y किमी/घंटा है, यदि नाव द्वारा एक स्थान से दूसरे स्थान तक आने व जाने में T समय लगता है, तो दोनो स्थानों के बीच की दूरी = T(x^2 - y^2)/2x km
7). कोई नाव अनुप्रवाह में कोई दूरी a घंटे में तय करती है, तथा वापस आने में b घंटे लेती है, यदि नाव कि चाल c किमी/घंटा है, तो शांत जल में नाव की चाल = c(a+b)/(b-a) km/h
8). यदि शांत जल में नाव की चाल a किमी/घंटा है, तथा वह b किमी/घंटा की चाल से बहती हुई नदी में गत…

### Emergence of British East India Company as an Imperialist Political Power in India

EMERGENCE OF BRITISH EAST INDIA COMPANY AS AN IMPERIALIST POLITICAL POWER IN INDIA
Dynamically changing India during early eighteenth century had a substantially growing economy under the authority of Mughal emperor Aurangzeb. But after his demise in 1707, several Mughal governors established their control over many regional kingdoms by exerting their authority. By the second half of eighteen century, British East India Company emerged as a political power in India after deposing regional powers and dominating over Mughal rulers. The present article attempts to analyze the reasons for emergence of British East India Company as an imperial political power in India and their diplomatic policies of territorial expansion. In addition to this, I briefly highlighted the Charter Acts (1773, 1793, 1813, 1833 and 1853) to trace its impact on the working process of Company.Establishment of East India Company in India

In 1600, British East India Company received royal charter or exclusive license…

### Three dimensional geometry (part-1) | study material for IIT JEE | concept booster , chapter highlights.

THREE DIMENSIONAL GEOMETRY

ORIGIN
In the following diagram X'OX , Y'OY and Z'OZ are three mutually perpendicular lines , which intersect at point O. Then the point O is called origin.
COORDINATE AXES
In the above diagram X'OX is called the X axes, Y'OY is called the Y axes and Z'OZ is called the Z axes.
COORDINATE PLANES
1). XOY is called the XY plane. 2). YOZ is called the YZ plane. 3). ZOX is called the ZX plane.
If all these three are taken together then it is called the coordinate planes. These coordinates planes divides the space into 8 parts and these parts are called octants.
COORDINATES  Let's take a any point P in the space. Draw PL , PM and PN perpendicularly to the XY, YZ and ZX planes, then
1). LP is called the X - coordinate of point P. 2). MP is called the Y - coordinate of point P. 3). NP is called the Z - coordinate of the point P.
When these three coordinates are taken together, then it is called coordinates of the point P.
SIGN CONVENTI…

### A detailed unit conversion table in Hindi.

UNITS CONVERSION TABLE
CENTIMETRE GRAM SECOND SYSTEM (CGS)1). MEASUREMENT OF LENGTH (लंबाई के माप) 10 millimeter = 1 centimetres10 centimetre = 1 decimetres  10 decimetre = 1 metres 10 metre = 1 decametres 10 decametres = 1 hectometres 10 hectometres = 1 kilometres 10 kilometres = 1 miriametresMEASUREMENTS OF AREAS ( क्षेत्रफल की माप )  100 millimetre sq. = 1 centimetre sq.
100 centimetre sq. = 1 decimetres sq. 100 decimetres sq. = 1 metre sq. 100 metre sq. = 1 decametres sq  100 decametres sq. = 1 hectometres sq. 100 hectometres sq. = 1 kilometres sq. 100 kilometres sq. = 1 miriametres sq.
MEASUREMENTS OF VOLUME ( आयतन की माप) 1000 millimetre cube. = 1 centimetre cube.
1000 centimetre cube. = 1 decimetres cube. 1000 decimetres cube. = 1 metre cube. 1000 metre cube. = 1 decametres cube. 1000 decametres cube. = 1 hectometres cube. 1000 hectometres cube. = 1 kilometres cube. 1000 kilometres cube. = 1 miriametres cube.
MEASUREMENTS OF VOLUME OF LIQUIDS  (द्रव्य के आयतन का माप) 10 millilitre=…

### THE GENERAL THEORY OF RELATIVITY | A Unique way to explain gravitational phenomenon.

Today we are going to talk about a very important and revolutionary concept that is THE GENERAL THEORY OF RELATIVITY.
This theory came into existence after 10 years of special theory of relativity (1905), and published by Albert Einstein in 1915.
This theory generalise the special theory of relativity and refines the Newton's laws of universal gravitation.
After coming this theory people's perspective about space and time has been changed completely. And this theory give a new vision to understand the spacetime geometry.
This theory gives a unified description of gravity as a geometrical properties of space and time.
This theory helps us to explain some cosmological phenomenon that is ,

* why small planets revolve around the big stars?
* Why everything in this universe is keep moving?
* Why mostly planets and stars are spherical in shape?
* Why does gravity create?
* Why does time become slow near the higher gravitating mass. Ie. Gravitational time dilation.
And gravitational…