MCQ
The interatomic distance for a metal is $3 \times {10^{ - 10}}\,m$. If the interatomic force constant is $3.6 \times {10^{ - 9}}\,N/{{\buildrel _{\circ} \over {\mathrm{A}}}}$, then the Young's modulus in $N/{m^2}$ will be
  • $1.2 \times {10^{11}}$
  • B
    $4.2 \times {10^{11}}$
  • C
    $10.8 \times {10^{ - 19}}$
  • D
    $2.4 \times {10^{10}}$

Answer

Correct option: A.
$1.2 \times {10^{11}}$
a
(a) $Y = \frac{{3.6 \times {{10}^{ - 9}}N/\mathop A\limits^o }}{{3 \times {{10}^{ - 10}}m}}$$ = 1.2 \times {10^{11}}N/{m^2}$

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