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2 years ago

Calculate area moment of inertia for a circular cross section with 6 mm diameter

1 answer:

3 0

The area moment of inertia for a circular cross-section with a 6 mm diameter will be 64 mm⁴. Option B is correct.

<h3>What is a moment of inertia?</h3>

The moment of inertia is defined as the resistance offered in the rotation of the moment.

Given data;

The diameter of circular section ,d = 6 mm

The moment of inertia is, I

The moment of inertia for the circular cross-section;

$\rm I = \frac{\pi d^4}{64} \\\\ \rm I = \frac{3.14 \times 6^4 }{64} \\\\ I = 63.585 mm^4$

The area moment of inertia for a circular cross-section with a 6 mm diameter will be 64 mm⁴

Hence option B is correct.

To learn more about the moment of inertia refer;

brainly.com/question/15246709

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Find values of the intrinsic carrier concentration n for silicon at –70° 0° 20° C, 100° C, and C. At 125° each temperature, what

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Part (i) at –70° C, intrinsic carrier concentration of silicon is 2.865 x 10⁵ carriers/cm³ and fraction of the atoms ionized is 5.37 x 10⁻¹⁸

Part (ii) at 0° C, intrinsic carrier concentration of silicon is 1.533 x 10⁹ carriers/cm³ and fraction of the atoms ionized is 3.067 x 10⁻¹⁴

Part (iii) at 20° C, intrinsic carrier concentration of silicon is 8.652 x 10⁹ carriers/cm³ and fraction of the atoms ionized is 1.731 x 10⁻¹³

Part (iv) at 100° C, intrinsic carrier concentration of silicon is 1.444 x 10¹² carriers/cm³ and fraction of the atoms ionized is 2.889 x 10⁻¹¹

Part (iv) at 125° C, intrinsic carrier concentration of silicon is 4.754 x 10¹² carriers/cm³ and fraction of the atoms ionized is 9.508 x 10⁻¹¹

Explanation:

$ni^2 = BT^3(e^{\frac{-E_g}{KT}})\\\\ni = \sqrt{ BT^3(e^{\frac{-E_g}{KT}})}$

where;

B = 5.4 x 10⁻³¹

Eg = 1.12 ev

K = 8.62 x 10⁻⁵ eV/K

T = (273 + ⁰C) K

Number of atoms in silicon crystal = 5 x 10²² atoms/cm³

Part (i) For –70° C, T = (273 -70 ⁰C)K = 203 K

$ni = \sqrt{ 5.4*10^{31}*203^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*203}})}} \ =2.685*10^5 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{2.685*10^5}{5 *10^{22}} = 5.370 *10^{-18}$

Part (ii) For 0° C, T = (273 +0 ⁰C)K = 273 K

$ni = \sqrt{ 5.4*10^{31}*273^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*273}})}} \ =1.533*10^9 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{1.533*10^9}{5 *10^{22}} = 3.067 *10^{-14}$

Part (iii) For 20° C, T = (273 + 20 ⁰C)K = 293 K

$ni = \sqrt{ 5.4*10^{31}*293^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*293}})}} \ =8.652*10^9 \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{8.652*10^9}{5 *10^{22}} = 1.731 *10^{-13}$

Part (iv) For 100° C, T = (273 + 100 ⁰C)K = 373 K

$ni = \sqrt{ 5.4*10^{31}*373^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*373}})}} \ =1.444*10^{12} \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{1.444*10^{12}}{5 *10^{22}} = 2.889 *10^{-11}$

Part (v) For 125° C, T = (273 + 125 ⁰C)K = 398 K

$ni = \sqrt{ 5.4*10^{31}*398^3(e^ \ {\frac{-1.12}{8.62*10^{-5}*398}})}} \ =4.754*10^{12} \ carriers/cm^3$

$Fraction \ of \ atoms \ ionized = \frac{4.754*10^{12}}{5 *10^{22}} = 9.508 *10^{-11}$

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