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Q)An ant is at a corner of a cubical room of side a. The ant can move with a constant speed u.The minimum time taken to reach the farthest corner of the cube is:a)3a/u   b)31/2a/u  c)51/2a/u  d)(21/2+1)a/uPLZ solve it sir..

Saurabh Kumar
6 years ago
Total distance covered = (21/2 + 1)a
speed = u
hence, time min = (21/2 + 1)a/u
Sanskar gupta
19 Points
4 years ago
You would might think the answer to be 3a or (sqrt2 + 1)a.....but the thing is that it`s neither of them.....now first of all open the cube and draw any two of its adjacent sides ....... therefore you will get a rectangle with the sides 2a and a....then for the last step draw the diagonal of the rectangle.......and find it .....the answer would be (sqrt5)a......and thus for the time divide it by `u` (given) so the answer is C) (sqrt5)a / u