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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).…

what is Roche limit| Roche limit and the formation of the rings |

Have you ever think, why did Earth have no rings? Why did only some planets have rings? Who initiates the rings formation? And how did rings form?
If you want to know the answers of these questions then stay with us till end.
Before giving the explanation of these questions, first of all, we have to know about the Roche limit. Who is responsible for these phenomena.
So let's starts:
ROCHE LIMIT

Then what is Roche limit? And how is it responsible for the formation of the rings?
We all know that, in the starting of the universe , various planets, stars and satellites and other celestial bodies formed by applying the Gravitational force to the nearby small rocks, gaseous matter and particles and get integrated to a bigger one. But sometimes,
This integration of the celestial bodies get hindered when a comparatively larger body came nearer to the integrating celestial body. This hindrance occurred due the Roche limit of the larger celestial body.
So Roche limit(also called Roche radius) is the minimum distance from the larger celestial body within which, when a other smaller celestial body if comes , then it will disintegrates into the smaller pieces due to the tidal force applied by the larger celestial body.
If the orbiting body is inside the Roche limit, then it dispersed into smaller pieces, but if it is outside the Roche limit then it coalesce(integrates). The term Roche is named after the French astronomer Édouard Roche. Who has first calculated the theoretical limit in 1848.

ILLUSTRATION OF ROCHE LIMIT


you can clearly see here, that smaller celestial body is outside of the Roche limit and held together by the Gravitational force. The body is practically spherical.

here you can see the deformation of the body due to tidal force. When it is closer to the Roche limit.

when the body is within the Roche limit, then its own gravity can't longer withstand the tidal force so it starts disintegrating.

the particles which are nearer to the larger celestial body moves fastly than the particles away from it, It is shown by the red arrows.

the different orbital speed resulting to the formation of the rings around the larger celestial body.

EXPLAINATION OF THE ABOVE PHENOMENON

We have seen in illustration that within the Roche limit smaller celestial body starts disintegrating due to the effect of tidal force which is applied by the larger celestial body.
But why this disintegrating occurred? This is the questions which we have to understand.
The parts of the smaller body which is nearer to the larger body get attracted more strongly by the force of gravity other than the parts which is away from it.
This distinction in the force pulls the nearer parts and far parts apart from each other of the smaller body. If the distinction of the force is larger than force of gravity (which hold the smaller body together) then the smaller body disintegrates into the smaller pieces of rocks, and this rocks spread around the larger celestial body and forms rings.

But it is found some satellites(natural and artificial) can orbit within their Roche limit, it is because they are held together by the force which is different from the Gravitational force. 
 If any object are used to be put on the surface of these satellites then it is lifted away or escaped away by the tidal forces.
Within the Roche limit no satellites coalesce gravitationally out of smaller pieces.

CALCULATION OF ROCHE LIMIT

*. If the material mean density of the both bodies (larger and smaller) are same then the Roche limit is 2.5 times of the radius of the larger celestial body.
*. The formula for the calculation of the Roche limit is:
Where Rp is the radius of the primary or larger celestial body , ρp (bar) is mean density of primary body , ρm(bar) is the mean density of the satellites (secondary or smaller body).

*. The Roche limit of the earth with respect to moon is R(lim) ~ 18,237.327KM 

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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 …