Sphere Problem

[samirpandeyit62]

The answer choices are not very legible to be, but I would solve it as follows

Now min lenght from a vertice to a pt on the sphere will be when moving diagonally from a vertice & the diameter of sphere is 10

diagonal of base of square is rt(2) *10

so diagonal from side is rt(10^2 + 10*rt(2) ^2)

= 10* rt(3)

so min lenght is (10*rt(3) - 10)/2 = 5(rt(3) -1)

[camitava]

Sorry Samir! Probably I am missing something. I am getting some different answer at all.
Refer the figure below -


We get OA = 5sqrt(2)
Now OD = 5
So AD = OA - OD = 5(sqrt(2) - 1)

I am not getting why you are talking about 5(sqrt(3) - 1)

[camitava]

Sorry Samir! Probably I am missing something. I am getting some different answer at all.
Refer the figure below -

We get OA = 5sqrt(2)
Now OD = 5
So AD = OA - OD = 5(sqrt(2) - 1)

I am not getting why you are talking about 5(sqrt(3) - 1)

[samirpandeyit62]

Hey Amitava the figures mentioned in the Q are 3D figures, ie cube, sphere, & diagonal of a cube is rt(3)*s (lenght of side)

[camitava]

Shi...! A miss again ... God pls help meeeeeeeeeeeeeee

[sankruth]

The distance between 2 opposite vertices of the cube =
SqRt(10^2 + 10^2 + 10^2) = 10*SqRt(3)

Since the Sphere is inscribed in the cude, radius = 5 and the centre will be the midpoint of the line joining the two opposite vertices i.e. [10SqRt(3)/2] = 5SqRt(3)

So, Shortest distance between the surface of Sphere and vertex of square = 5SqRt(3) - 5 = 5[SqRt(3)-1]

Hope that helps!