Let's, Calculate the number of atoms per unit cell in the FCC lattice:
In an FCC unit cell, there are 8 corner atoms, each shared by 8 unit cells and 6 face-centered atoms, each shared by 2 unit cells.
Total number of atoms in FCC unit cell: \(8\times\frac{1}{8}+6\times\frac{1}{2}\) = 1+3 = 4
Calculate the number of tetrahedral voids per unit cell in the FCC lattice:
In an FCC unit cell, there are 8 tetrahedral voids. If only alternate tetrahedral voids are occupied, then 4 tetrahedral voids are occupied per unit cell.
Total number of occupied tetrahedral voids = 4
Calculate the total number of atoms in the modified structure:
Total atoms in FCC lattice = 4 and Total atoms in tetrahedral voids = 4
Total number of atoms per unit cell=4+4=8
The relationship between the edge length a and the atomic radius r, since tetrahedral voids forms at \(\frac{1}{4}th\) of body diagonal
\(\frac{a\sqrt{3}}{4}=2r\)
\(a=\frac{8r}{\sqrt{3}}\)
Packing efficiency (PE) is the ratio of the volume occupied by atoms to the volume of the unit cell, multiplied by 100%:
Packing efficiency = \(\frac{8\times\frac{4}{3}\pi r^3}{a^3}\times100\)
\(\frac{8\times\frac{4}{3}\pi r^3}{(\frac{8r}{\sqrt{3}})^3}\times100\simeq35\%\)
So, The correct option is (B): 35
Let $ a_0, a_1, ..., a_{23} $ be real numbers such that $$ \left(1 + \frac{2}{5}x \right)^{23} = \sum_{i=0}^{23} a_i x^i $$ for every real number $ x $. Let $ a_r $ be the largest among the numbers $ a_j $ for $ 0 \leq j \leq 23 $. Then the value of $ r $ is ________.
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
The center of a disk of radius $ r $ and mass $ m $ is attached to a spring of spring constant $ k $, inside a ring of radius $ R>r $ as shown in the figure. The other end of the spring is attached on the periphery of the ring. Both the ring and the disk are in the same vertical plane. The disk can only roll along the inside periphery of the ring, without slipping. The spring can only be stretched or compressed along the periphery of the ring, following Hooke’s law. In equilibrium, the disk is at the bottom of the ring. Assuming small displacement of the disc, the time period of oscillation of center of mass of the disk is written as $ T = \frac{2\pi}{\omega} $. The correct expression for $ \omega $ is ( $ g $ is the acceleration due to gravity): 
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