A) \[I={{I}_{0}}\sin \omega t,\]
B) \[{{I}_{0}}=10\]
C) \[\omega =100\pi \,\text{rad/s}\]
D) None of these
Correct Answer: D
Solution :
Number of ways of selecting three numbers one on each card \[~=100\times 100\times 100={{100}^{3}}\] We know that \[~(2n+1),(2{{n}^{2}}+2n)\] and \[~(2{{n}^{2}}+2n+1)\]are Pythagorean triplets. \[\therefore \]For n = 1, 2, 3, 4, 5, 6, we get the lengths of three sides of a right angled triangle such that its hypotenuse-is less than or equal.to 100 cm. Now, one Pythagorean triplet (e.g. 3, 4, 5; 5,12,13; etc.) can be chosen in 3! ways. \[\therefore \]Number of ways of selecting 6 Pythagorean triplets \[=6\times 3!\] Hence, required probability \[=\frac{6\times 3!}{{{100}^{3}}}\]
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