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Q.

The inter atomic distance for a metal is $3 \times 10^{- 10} m$ . If the inter atomic force constant is $3.6 \times 10^{- 9} N / A^{0}$ , then the Young’s modulus in $N m^{- 2}$ will be

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a

$2.4 \times 10^{10}$

b

$10.8 \times 10^{8}$

c

$4.2 \times 10^{11}$

d

$1.2 \times 10^{11}$

answer is A.

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Detailed Solution

$F o r c e c o n s \tan t, K = Y r \Rightarrow Y = \frac{K}{r}$

Young Modulus $Y = \frac{3.6 \times 10^{- 9} N / A^{0}}{3 \times 10^{- 10} m}$

$= \frac{3.6 \times 10^{- 9} / 10^{- 10}}{3 \times 10^{- 10} \times 10^{- 10}} = 1.2 \times 10^{11} N m^{- 2}$ $(∵ 1 A^{0} = 10^{- 10} m)$

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