A) \[\frac{10}{9}V/\mu m\]
B) \[x\]
C) \[\frac{mF}{M}\]
D) \[\frac{(M+m)F}{m}\]
Correct Answer: D
Solution :
Solution is isotonic. \[C{{V}^{2}}(K-1)/K\] \[(K-1)C{{V}^{2}}\] \[{{g}_{E}}\] Density of both the solutions are assumed to be equal to\[{{g}_{M}}\]. \[\frac{electronic\text{ }charge\text{ }on\text{ }the\text{ }moon}{electronic\text{ }charge\text{ }on\text{ }the\text{ }earth}\] \[{{g}_{E}}/{{g}_{M}}\] \[{{g}_{M}}/{{g}_{E}}\]In 100 g, 5.25 g of substance is present. \[\frac{\alpha }{R}\]In 1000 g, 52.5 g of substance is found. Hence, \[\frac{1}{3}\] M = molecular mass of the substance \[\frac{1}{2}\] Alternate Solution 5.25% solution means 5.25 g of substance dissolved in 100 g of solution. 1.50% solution means 1.50 g of urea dissolved in 100 g of solution. \[\frac{1}{6}\]Solution is isotonic with solution of urea \[\frac{1}{4}\] \[I,\] \[\frac{g\sin \theta }{1+I/M{{R}^{2}}}\] \[\frac{g\sin \theta }{1+M{{R}^{2}}/I}\] \[\frac{g\sin \theta }{1-I/M{{R}^{2}}}\] \[\frac{g\sin \theta }{1-M{{R}^{2}}/I}\] \[60{}^\circ \]
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