
Focus \(f=10\ cm\)
\(\frac 1f=(μ−1)(\frac 1R−\frac {1}{−R})\)
\(\frac {1}{10}=\frac {1.5−1}{1}×\frac 2R\)
\(\frac {1}{10}=\frac {0.5×2}{R}\)
\(\frac {1}{10}=\frac 1R\)
\(R=10\ cm\)
So, the answer is \(10\ cm.\)
In the following \(p\text{–}V\) diagram, the equation of state along the curved path is given by \[ (V-2)^2 = 4ap, \] where \(a\) is a constant. The total work done in the closed path is: 
Let \( ABC \) be a triangle. Consider four points \( p_1, p_2, p_3, p_4 \) on the side \( AB \), five points \( p_5, p_6, p_7, p_8, p_9 \) on the side \( BC \), and four points \( p_{10}, p_{11}, p_{12}, p_{13} \) on the side \( AC \). None of these points is a vertex of the triangle \( ABC \). Then the total number of pentagons that can be formed by taking all the vertices from the points \( p_1, p_2, \ldots, p_{13} \) is ___________.
Consider the following two reactions A and B: 
The numerical value of [molar mass of $x$ + molar mass of $y$] is ___.
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