The length of the largest rod that can be kept in a room of length 5 m, breadth 4 m and height 3 m is :

1. $3\sqrt{2}\text{ m}$

2. $5\sqrt{2}\text{ m}$

3. $7\sqrt{2}\text{ m}$

4. $\sqrt{2}\text{ m}$

Given,

Length (l) = 5 m

Breadth (b) = 4 m

Height (h) = 3 m

Length of the largest rod that can fit in the room is the diagonal of the room.

Calculating the length of the diagonal of the room (cuboid),

$\text{Diagonal of room (d)} = \sqrt{l^2 + b^2 + h^2} \\[1em] = \sqrt{5^2 + 4^2 + 3^2} \\[1em] = \sqrt{25 + 16 + 9} \\[1em] = \sqrt{50} \\[1em] = \sqrt{25 × 2} \\[1em] = 5 \sqrt{2} \text{ m}.$

Hence, option 2 is the correct option.