The internal dimension of a room is 10m × 10m × 4m (height).

The internal dimension of a room is 10m × 10m × 4m (height). The total area of the doors and windows are 16 m². Keeping the doors and windows closed, the reverberation time of the room becomes 1.2 second. Assume all the interior surfaces including doors and windows have some absorption coefficient. If all doors and windows of the room are kept fully open, the reverberation time will be _____ second (rounded of to one decimal place).

Consider the uniform absorption coefficient = a

Now, RT (Reverberation Time) = Constant ×		Volume
		Absorption

Constant = 0.16 (in metric system)

1.2 = 0.16 ×		400
		360a

Since, Total surface area of room is 360

360a = 400 ×	0.16
	1.2

a = 0.148
If all the doors and windows of the room are kept open,

New RT = 0.16 ×		400
		344a

Since, Surface Area = Total Surface Area – Area under Doors & Windows = 360 – 16

RT =	(0.16 × 400)	= 1.26 seconds
	(344 × 0.148)