In the given figure, the block of mass m is dropped from the point ‘A’. The expression for kinetic energy of block when it reaches point ‘B’ is
\(\frac{1}{2} mgy^2_{0}\)
\(\frac{1}{2} mgy^2\)
mg(y - y0)
mgy0
The correct answer is (D) : \(mgy_0\)
Loss is potential energy = gain in kinetic energy
– (mg(y – y0) – mgy) = KE – 0
⇒ KE = mgy0
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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