Physics Mechanical Properties of Solids

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New answer posted

2 weeks ago

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A
Aadit Singh Uppal

Contributor-Level 10

A diamond crystal with young modulus of more than 1050 Gpa is known to be the element which has the highest elastic moduli found in nature. This is possible due to the high electron density as well as strong covalent bonds between the carbon atoms.

New answer posted

3 weeks ago

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A
alok kumar singh

Contributor-Level 10

d m = m L d x

T = m ω 2 2 L L 2 - x 2

? L = 0 L m ω 2 2 L π r 2 Y L 2 - x 2 d x

= ? L = m ω 2 L 2 3 π r 2 Y

New answer posted

3 weeks ago

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A
alok kumar singh

Contributor-Level 10

Deformation energy per unit volume,

U = d U = P 2 2 B A d y

= ρ 2 g 2 y 2 B A d y

U = ρ 2 g 2 2 B h 3 3 A

= ρ 2 g 2 h 2 6 B ( v o l u m e )

= 1.5 μ J

New answer posted

3 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

If μ  is Poisson's ratio,

Y = 3K (1 - 2 μ )      ……… (1)

and Y = 2 n ( 1 + μ )  ……… (2)

With the help of equations (1) and (2), we can write

  3 Y = 1 η + 1 3 k K = η Y 9 η 3 Y

New answer posted

3 weeks ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

If   μ is Poisson's ratio,

Y = 3K (1 - 2  μ )      ……… (1)

and Y = 2  n ( 1 + μ ) ……… (2)

With the help of equations (1) and (2), we can write

3 Y = 1 η + 1 3 k K = η Y 9 η 3 Y            

New answer posted

4 weeks ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

dm = (m/L)dx
∴ T = (mω²/2L) (L² - x²)
∴ ΔL = ∫? (mω²/2Lπr²Y) (L² - x²)dx
= ΔL = mω²L²/3πr²Y

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Stress = F A = T A = W A

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Resistance of P & Q should be approximately equal as it decreases error in experiment.

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Surface energy of bubble = 2 * charge in surface area *  surface tension

= 8 π R 2 * T = 8 * 3.142 * 4 * 10 - 4 * 3 * 10 - 2 J = 30.1 * 10 - 5 J

New answer posted

a month ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

For Wire-A
F / (πr_A²) = Y (l_A / 2) . (1)
For Wire-B
F / (πr_B²) = Y (l_B / 4) . (2)
From equations (1) and (2), we can write
(l_A / (2r_A²) = (l_B / (4r_B²) ⇒ l_B/l_A = 2 (r_B/r_A)² ⇒ x = l_B = 32

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