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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).…

Heat Transfer | concept booster , study notes for IIT JEE |

HEAT TRANSFER

*. Heat is the form of energy which flows from one body to another body in the form of radiation, molecular vibrations and molecular displacement. The mode of heat flow from this method are collectively known as heat transfer.

*. There are three mode of heat transfer.
1). Conduction 
2). Convection
3). Radiation
The three mode of heat transfer.

*. THERMAL CONDUCTIVITY - It is the ability of the material to conduct heat from them.

*. HEAT CONDUCTION - It is time rate of flow of heat through any material for a given temperature difference.

*. The rate of heat transfer through a metal rod of length L and cross sectional area A and the temperature at the ends are T1 and T2 is -

 Rate of heat transfer = ∆Q/∆t = KA(T1-T2)/L
   H = KA∆T/L and also H = Q/t
So then Q = KA∆T.t/L
Heat transfer study notes for IIT JEE
Where K is called the cofficient of thermal conductivity.
SI unit of K is J/smK or W/mK
The term ∆T/L is called temperature gradient.

*. The rate of flow of heat = heat current or thermal current, and it is denoted by H.
H = ∆Q/∆t = KA(T1-T2)/L = T1-T2/(L/KA)

 therma resistance = L/KA = ∆T/H = T1-T2/H
   R = L/KA

CONNECTION OF ROD WITH DIFFERENT THERMAL CONDUCTIVITIES

SERIES CONNECTION OF TWO THERMAL CONDUCTORS
Heat transfer study notes for IIT JEE.
*. Equivalent resistance = R(eq) = R1 + R2
*. Equivalent thermal conductivity if their length are equal= K(eq) = 2K1K2/K1+K2

PARALLEL CONNECTION OF TWO THERMAL CONDUCTORS

*. Equivalent resistance = 1/R(eq)= 1/R1 + 1/R2
*. Equivalent thermal conductivity if their cross sectional area are equal = K(eq) = K1+K2/2
Heat transfer study material for IIT JEE

CONVECTION

*. It is the mode of heat transfer in which heat transfer occurs due to molecular displacement from high temperature region to the low temperature region. This mode of heat transfer usually takes place in liquids.

*. Application of convection in daily life

1). In boiling of water.
2). Formation of land and sea breeze.
3). The invariable 4°C temperature of the bottom of the deep sea whether it is summer or winter.
4). Illumination of room with candle , but not the interior of a free falling lift or satellite.
5). Heating of atmosphere.

RADIATION

*. It is also a mode of heat transfer in which heat is transferred from one place to another place without heating the intervening medium.

*. In the context of heat transfer, radiation is the process, in which energy is emitted by a body who travelled through space and falls on a body.
But in electromagnetism, radiation is the electromagnetic waves which carry energy by itself.


SOME PROPERTIES OF RADIATION

*. All objects emit radiation simply, whose temperature is just above the obsolute zero and all objects absorbs some of the radiation which falls on them .

*. According to the electromagnetic theory Maxwell shows that all radiation are the electromagnetic waves and their source are the vibrations of the charged particles.

*. If the tempt of a object is high then more radiation emit if tempt is low then less radiation emit.

*. The radiations which emit on the high temperature their wavelength is short in comparison with the wavelength of radiation which emit on the low temperature.

*. All thermal radiation travels with the speed of light.

*. All radiation are the electromagnetic waves and can travel through space or vaccum.

*. Just similar to light, radiation can also be reflected , refracted, diffracted and polarised.

*. All radiation from their point source obeys inverse square law.

PREVOST THEORY OF EXCHANGE

*. According to this theory, all objects radiate thermal radiation at all tempt.

*. The amount of thermal radiation radiated per unit time depends on the nature of the object , surface area and it's temperature.

*. The rate of emitting of thermal radiation is higher in high temperature body.

PERFECTLY BLACK BODY AND BLACK BODY RADIATION

*. PERFECTLY BLACK BODY - It is body which absorbs all the heat radiations of every wavelength , which are incidents on it. This body doesn't reflect and transmit the radiation so it appears black. Whatever be the colour of the radiation. It just appears black.
Heat transfer study notes for IIT JEE.
*. There is no any perfectly black body found naturally. But lamp black,  platinum black and ferry black are some black bodies which are nearly perfectly black. They absorbs about 99% of incidents radiation.

ABSORPTION , REFLECTION , AND EMISSIONS OF RADIATION

*. If Q = Qr + Qt + Qa and after dividing by Q both side we get, 1 = Qr/Q + Qt/Q + Qa/Q

Where r = reflective power
t = transmission power 
a = absorptive power

If r= 0 , t=0 , a = 1 then it will be perfectly black body.
If r= 0 , t= 1 , a= 0 then it will be perfect transmitter
If r=1 , t=0 ,a=0 then it will be perfect reflector

ABSORPTIVE POWER

*. It is the ratio of the energy absorb to the energy incident.
a = energy absorb / energy incident

EMISSIVE POWER

*. It is the energy radiated per unit time and per unit area along the normal of the area.
                       E = Q/∆A.∆t

*. Emissivity - It is the ratio of the emissive power of a body at temperature T to the emissive power of a black body at same temperature. It is denoted by ε.

SPECTRAL EMISSIVE POWER (Eλ)

*. Spectral emissive power(Eλ) is the emissive power per unit wavelength.
                   Eλ = dE/dλ

KIRCHHOFF'S LAW

*. This laws states that the ratio of the emissive power to the absorptive power of a radiation whose wavelength is given is the same for all objects at the same temperature and it is equal to the emissive power of a perfectly black body for the same temperature and the wavelength.
    E(body)/a(body) = E(perfectly black body)

*. Good absorber of a particular wavelength λ is also a good emitter of the same wavelength.

STEFAN - BOLTZMANN' S LAW

*. This law states that in a black body the amount of radiation radiated per unit time from a area A at absolute temperature T is directly proportional to the fourth power of the temperature.
                         u = σAT^4
where σ is the proportionality constant called Stefan constant and its value is = 
5.67×10^-8W/m^2.K^4

*. If emissivity of a body is given then the amount of radiation radiated per unit time is 
                       u = εσAT^4

*. If the surrounding temperature is T0 then the net amount of energy radiated by area A in per unit time is , 
   ∆u = u - u0 = εσA(T^4 - T0^4)

NEWTON'S  LAW OF COOLING

*. This law states that the rate of cooling of a body is directly proportional to the temperature difference and area exposed.
      dQ/dt = -k(Ti - T0) ,where Ti and T0 is the temperature of the body and the surroundings.

*. Temperature of the body at any instant of time t is given as-
 T = T0 + (Ti -T0)e^-kt , where k is constant.

*. The value of k can be directly find from the Stefan law and it can be given as 
               k = 4εσT0^3A/ms

*. If Qemission < Qabsorption , then the body become hotter.
   If Qemission>Qabsorption, then the body become colder.
   IfQabsorption=Qabsorption , then the body remains constant( thermal equilibrium attained)

WIEN'S DISPLACEMENT LAW

*. This law states that the wavelength λm corresponding to the maximum intensity of the emission decreases inversely with the increase in the temperature of the black body.
  
   Here b is called the wien's constant and its value is b = 0.282cm-K           
Heat transfer study material for IIT JEE.
*. Higher temperatures of the body  , means higher area under the curves.
*. High amount of energy is radiated if the temperature of the body is higher.
*. Energy emitted by the body at temperatures is not uniform , for long and short wavelength energy emission is low.
*. For a given temperature there is a particular wavelength λm, for which energy emission Eλ is maximum.
*. Maxima of the curve shift towards the shorter wavelength if temperature is increased in black body.

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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…

CORONA VIRUS, history of origin , discovery , infection mechanism, symptoms and treatment.

Today we are going to talk about a virus , which is spreading very fastly all over the world. The virus which we are going to talk about is the CORONA VIRUS. So today we will talk about everything of this virus. So let's starts ...

OVERVIEW OF CORONAVIRUS
According to the biological study , Coronavirus is a cluster of viruses that causes diseases in birds and mammals. Therefore humans are also mammals then in human being this viruses cause respiratory infections , and one of the respiratory infections is mild common cold. Coronavirus can lead to diarrhea in cows and pigs but in chicken they can cause upper respiratory infections. Currently there is no vaccine or antiviral drugs for the treatment of diseases caused by Coronavirus.
BIOLOGICAL INTRODUCTION OF coV
The family of Coronavirus is coronaviridae, and it's subfamily is Orthocoronavirinae and order is Nidovirales, Coronavirus is a member of Orthocoronavirinae subfamily. All Coronavirus is coated with positive sense single …

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…

Speed , Distance and Time problems tricks in Hindi | fast track arithmetic formulae for problem solving.

SPEED , DISTANCE AND TIME PROBLEMS TRICKS IN HINDI
1). दूरी = चाल × समय
2). समय = दूरी/चाल
3). चाल = दूरी/समय
4). किलोमीटर को मील बनाने के लिए गुना किया जाता है =       5/8 से
5). मील को किलोमीटर बनाने के लिए गुना किया जाता है =       8/5 से
6). फुट - सेकंड को मील - घंटा बनाने के लिए गुना किया जाता है = 15/22 से
7). मील - घंटा को फुट - सेकंड बनाने के लिए गुना किया जाता है = 22/15 से
8). मी - सेकंड को किमी - घंटा बनाने के लिए गुना किया जाता है = 18/5 से
9). किमी - घंटा को मी - सेकंड बनाने के लिए गुना किया जाता है = 5/18 से
10). यदि एक व्यक्ति दो निश्चित स्थानों के बीच की दूरी a किमी/घंटा की चाल से खत्म करता है, तो t1 घंटे देर से पहुंचता है, तथा जब b किमी/घंटा की चाल से तय करता है, तब वह t2 घण्टे पहले पहुंचता है, तो दोनो स्थानों के बीच की दूरी =     ab(t1+t2)/(b-a) km
11). यदि कोई व्यक्ति a km/h की चाल से चलता है, तो वह अपनी मंजिल पर t1 घंटे लेट पहुंचता है, अगली बार वह अपनी चाल में b km/h की वृद्धि करता है, तो वह t2 घंटे लेट पहुंचता है, तब उसके द्वारा तय की गई दूरी = a(a+b)(t1-t2)/b
12). दो व्यक्ति X …