Science, asked by sheelathomas304, 6 months ago

calculate the maximum possible error y=4 mgL/πd^2L

Answers

Answered by kulkarninishant346

Explanation:

Y= πD

Y= πD 2

Y= πD 2 l

Y= πD 2 l 4MgL

Y= πD 2 l 4MgL Y

Y= πD 2 l 4MgL Y ΔY

Y= πD 2 l 4MgL Y ΔY =

Y= πD 2 l 4MgL Y ΔY = L

Y= πD 2 l 4MgL Y ΔY = L ΔL

Y= πD 2 l 4MgL Y ΔY = L ΔL +

Y= πD 2 l 4MgL Y ΔY = L ΔL + D

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D +

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2×

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 +

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y ΔY

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y ΔY %=0.03024×100

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y ΔY %=0.03024×100 =3.024%

Answered by Anonymous

Answer:

Y= πD

Y= πD 2

Y= πD 2 l

Y= πD 2 l 4MgL

Y= πD 2 l 4MgL Y

Y= πD 2 l 4MgL Y ΔY

Y= πD 2 l 4MgL Y ΔY =

Y= πD 2 l 4MgL Y ΔY = L

Y= πD 2 l 4MgL Y ΔY = L ΔL

Y= πD 2 l 4MgL Y ΔY = L ΔL +

Y= πD 2 l 4MgL Y ΔY = L ΔL + D

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D +

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2×

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 +

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y ΔY

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y ΔY %=0.03024×100

Y= πD 2 l 4MgL Y ΔY = L ΔL + D 2D + l Δl 200 0.1 +2× 0.075 0.001 + .325 0.001 Y ΔY =0.03024 Y ΔY %=0.03024×100 =3.024%

Explanation:

Previous Question

Next Question

Similar questions

Hindi, 2 months ago

ज़िम्मेवारियाँ translate in English...

Physics, 2 months ago

(a) चित्र-6-20 29, संलग्न परिपथ के बिन्दुओ A व B के बीच तुल्य प्रतिरोध, धारा तथा विभवान्तर ज्ञात कीजिए जबकि प्रत्येक सेल का विचाबल 1.5 वोल्ट तथा आन्तरिक प्रतिरोध 0.5 ओम है। 90 www 90 Aman B B wM w चित्र-6-21 उत्तर:90.0.5 ऐम्पियर, 1.5 वोल्ट...

Science, 2 months ago

cover ----- cheek soil erosion...

Social Sciences, 6 months ago

1.Great sages of ancient India have gifted heritage of __________________ to the world of science. 1 point Meditation art science 2. The statue of ____________________ a masterpiece in sculpture, preserved in a museum at chennai is famous all over the world. 1 point Ram Natraja Buddha 3. Acharya Nagarjuna initiated the use of _____________ as ash a medicine. 1 point blackpepper...

Math, 6 months ago

In APQR, P = 60°, ZQ = 90° and PR = 15 cm. Find PQ?

Math, 10 months ago

Draw two parallel lines at a distance 4 cm apart.

Geography, 10 months ago

how are indian climate and agriculture season related...

Science, 10 months ago

Which Route is taken by raw materials to reach various parts of the plant.

calculate the maximum possible error y=4 mgL/πd^2L​

Answers

calculate the maximum possible error y=4 mgL/πd^2L