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Oct 21, 2015

Newton's Second Law of Motion For Class 9

Newton's Second law of motion: It states that rate of change of momentum is directly proportional to unbalance external force. And change of momentum is in the direction of unbalance force.
Mathematical Derivation:

Mathematical Derivation of Newton

Initial momentum , Pi = mu
Final momentum , Pf = mv
Change in momentum ,∆p = Pf - Pi
       = mv - mu
But we know that rate of change of momentum is directly proportional to the applied unbalanced force.
∴ F ∝ ∆p/t
or F ∝ (mv - mu ) / t/
Or F ∝ m. ( v -u )/t
Or F ∝ m. a , Where a= Acceleration
Or F = k. m. a ,
Or F = 1. m. a , In SI system K=1
F = m. a




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Oct 20, 2015

Newton's First Law of Motion Class 9 | Law of Inertia

Newton's First Laws of Motion { Law of Inertia }- It states that a body continue in its state of rest or of uniform motion until it is comepelled by any external unbalance forces.

This laws is also known as law of inertia.

Inertia is the proerty of a body which cause it to maintain its state of motion or rest.

Unbalanced Forces : When two unequal forces acts on an object and the object change its state then we say that the forces are unbalance.
Unbalance forces can be understood from example of tug of war.
Watch this video.

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List of Hindi Difficult Words

Learning Hindi is easy if we are going in right way. Many of us are struggling for Hindi learning. We already have created Youtube videos for learning hindi . However We need practice to write correctly.

This is the list of Hindi words in which a student of class 5 have created mistakes are. This is a tested list of words. I think many students do mistakes while writing these Hindi words. I suggest you to practice these words and post your mistakes if possible. There are Videos at the end for learning Hindi


उचित 
पछताएगा
न्याय
गिर
बकरी
पेचिश
पीलिया
त्वचा
योग्य
मानव
इकाई
व्यवस्थित
वर्णमाला
उच्चारण
व्यवस्थापना
उच्चारित 
इकाईयां
तुम्हारे 
विवाह 
धृष्टता
दूँगी
पक्षी
विशेष
समाज
ब्राह्मण
पहरा
कार्य
प्राप्त
खरीद


शीतल
जादू
स्वच्छ
वायु
बीमारी
पढ़ने
अकाल
समस्याए 
प्रयोग
अविष्कार
देवदार
रसायनों
आत्मकथा
परिशान
बचाया
चंगुल
कूदना
भावना
सहभागिता
पढ़ाई
निगरानी
चिंता
मौखिक
संस्कृत
इसलिए
देवनागरी
लिपि
उचित
विमाला
ओतिकड़म
चमत्कार
उपाय
सरोवर
घाटी


मराठी
पहाड़ी
द्रव्यवाचक
परिर्वतन
मात
मातृत्व
व्यक्ति
घाल
छाल
ईट
आदि
सुरक्षित
आविष्कार
इतिहास
छिप
अनुशासन
बढ़ती
बढ़
मुख्यमंत्री
नदी
मंत्रिया
अचानक
दीवार
कर्ज
शिक्षक
फलस्वरूप
निश्चित









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Oct 14, 2015

In triangle ABC, AB > AC. | E is the mid-point of BC | and AD is perpendicular to BC

In triangle ABC, AB > AC. E is the mid-point of BC and AD is perpendicular to BC.
Prove that
AB2 + AC2 = 2.AE2 + 2.BE2

Solution: In ∆ ABD AB2 = AD 2 + BD 2
Or AB2 = AD 2 + (BE + ED ) 2
Or AB2 = AD 2 + BE 2 + 2. BE . ED + ED 2
Or AB2 = AE 2 + BE 2 + 2. BE . ED -----(i)
In ∆ ADC AC2 = AD 2 + CD 2
OR AC2 = AD2 + ( CE - ED) 2
Or AC2 = AD2 + CE 2 - 2. CE . ED + ED 2
Or AC2 = AE2 + CE 2 - 2. CE . ED ------(ii)
Adding equation 1 and situation 2 we have:
AB2 + AC2 = AE2 + AE2 + BE2 + CE2 + 2. BE . ED - + 2. CE . ED
Or AB2 + AC2 = AE2 + AE2 + BE2 + BE2 + 2. BE . ED - 2. BE . ED (As BE = CE )
AB2 + AC2 = 2 AE2 + 2BE2 Proved

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