Question

If t is the length of diagonal of a cube of volume V, Then
A. \[3V={{t}^{3}}\]
B. \[\sqrt{3}V={{t}^{3}}\]
C. \[3\sqrt{3}V=2{{t}^{3}}\]
D.\[3\sqrt{3}V={{t}^{3}}\]

Answer 1

Hint: We know the formula for diagonal of cube in terms of length of cube similarly we know the volume of cube in terms of length of cube so we will substitute the value of length of cube and find a relationship between volume and diagonal length a cube.

Complete step-by-step answer:
Given t is the length of diagonal of a cube and volume of the cube is V we have to find the relationship between them, so. Assume side length of a cube to be ‘a’. now we know that the value of diagonal of a cube is root 3 times of side length so it should be equals to \[t=\sqrt{3}a.....(1)\]
Similarly, we know that the volume of a cube is a cube of side length so it should be \[V={{a}^{3}}....(2)\]
Getting the value of a from equation (1) as\[\dfrac{t}{\sqrt{3}}=a\]
Putting this value of a in equation (2) we get expression as
\[V={{(\dfrac{t}{\sqrt{3}})}^{3}}\] on solving we get \[V=\dfrac{{{t}^{3}}}{3\sqrt{3}}\]
So, we get it as \[3\sqrt{3}V={{t}^{3}}\]

So, the correct answer is “Option D”.

Note: It should be noted that the cube has two diagonals, face diagonal and body centred diagonal. If it is not mentioned in the question then consider it as body centred diagonal a in this question .Body centre diagonal is the line segment joining two farthest vertex of cube , its value is \[\sqrt{3}a\] and face diagonal is line segment joining farthest vertex of a face ,its value is \[\sqrt{2}a\], here a is the side length of cube .