The correct option is (B) : 60 J
vi = 3(0)2 + 4 = 4 ≅ x = 0
vF = 3(2)2 + 4 = 16 ≅ x = 2
\(W=ΔK=\frac{1}{2}m(16^2-4^2)\)
\(=\frac{1}{2}\times\frac{1}{2}(256-16)=\frac{240}{4}\)
\(=60\ J\)
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)
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