The mean distance between the atoms of iron is 3 × 10-10 m and interatomic force constant for iron is 7 N / m. The Young's modulus of elasticity for iron is

Question

The mean distance between the atoms of iron is 3 × 10-10 m and interatomic force constant for iron is 7 N / m. The Young's modulus of elasticity for iron is (A) 2.33 × 105 N / m 2 (B) 23.3 × 1010 N / m 2 (C) 2.33 × 109 N / m 2 (D) 2.33 × 1010 N / m 2

Tardigrade Admin · Accepted Answer

Given, r0=3 × 10-10 m , k=7 N / m As k=Y r0 ∴ Y=(k/r0) =(7/3 × 10-10) =2.33 × 1010 N / m 2

Q. The mean distance between the atoms of iron is $3 \times 1 0^{- 10}$ $m$ and interatomic force constant for iron is $7 N / m$ . The Young's modulus of elasticity for iron is

$2.33 \times 1 0^{5} N / m^{2}$

$23.3 \times 1 0^{10} N / m^{2}$

$2.33 \times 1 0^{9} N / m^{2}$

$2.33 \times 1 0^{10} N / m^{2}$

Given, $r_{0} = 3 \times 1 0^{- 10} m, k = 7 N / m$
As $k = Y r_{0}$
$∴ Y = \frac{k}{r _{0}}$
$= \frac{7}{3 \times 1 0 ^{- 10}}$
$= 2.33 \times 1 0^{10} N / m^{2}$

Q. The mean distance between the atoms of iron is 3×10−10 m and interatomic force constant for iron is 7N/m. The Young's modulus of elasticity for iron is

2.33×105N/m2

23.3×1010N/m2

2.33×109N/m2

2.33×1010N/m2