एक बलमापी को 6 kg तथा 4 kg द्रव्यमान के दो ब्लॉकों के साथ जोड़ा गया है। 20 N तथा 10 N के बलों को ब्लॉकों पर चित्रानुसार आरोपित किया गया है। बलमापी का पाठ्यांक ज्ञात कीजिये।
Answers
Answer:
We hereby deduct the values of 13/3 &13/4 as under:
We hereby deduct the values of 13/3 &13/4 as under:13/4=3.25. So (13/4)^2=10.5625……..a
We hereby deduct the values of 13/3 &13/4 as under:13/4=3.25. So (13/4)^2=10.5625……..a13/3=4.33. So (13/3)^4=351.5212……b
We hereby deduct the values of 13/3 &13/4 as under:13/4=3.25. So (13/4)^2=10.5625……..a13/3=4.33. So (13/3)^4=351.5212……bSimilarly, (13/4)^4=111.5664……….a^2
We hereby deduct the values of 13/3 &13/4 as under:13/4=3.25. So (13/4)^2=10.5625……..a13/3=4.33. So (13/3)^4=351.5212……bSimilarly, (13/4)^4=111.5664……….a^2The expression can be rewritten as a^2-b/a-a=a(a-1)-b/a after substituting the values of a & b we get 10.5625(10.5625–1)-351.5212/10.5625
We hereby deduct the values of 13/3 &13/4 as under:13/4=3.25. So (13/4)^2=10.5625……..a13/3=4.33. So (13/3)^4=351.5212……bSimilarly, (13/4)^4=111.5664……….a^2The expression can be rewritten as a^2-b/a-a=a(a-1)-b/a after substituting the values of a & b we get 10.5625(10.5625–1)-351.5212/10.5625=10.5625*9.5625–33.2801=101.0039–33.2801=67.7238 Ans.