Question

༒ Bʀᴀɪɴʟʏ Sᴛᴀʀs

༒Mᴏᴅᴇʀᴀᴛᴇʀs

༒Other best user

Air column of 20 cm length in a resonance tube resonates with a certain tuning fork when sounded at its upper open end. The lower end of the tube is closed and adjustable by changing the quantity of mercury filled inside the tube. The temperature of the air is 27°C. The change in length of the air column required, if the temperature falls to 7°C and the same tuning fork is again sounded at the upper open end is nearly
(a) 1mm
(b) 7mm
(c) 5mm
(d) 13mm
Answer with proper explaination
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Answer 1

Hey there,

Here's your Answer- B) 7mm

Explaination-

Frequency of oscillation in organ pipe ∝ rootT/L

=> rootT1/L1 = rootT2/L2

=> L2 = (rootT2/T1*L1 ) = 19.3 cm

=> ΔL = L1 - L2 = 7mm

Hope this helps you buddy.

Answer 2

Solution:-

Frequency of oscillation in organ pipe ∝

$\sqrt{ \frac{t _{}}{1} }$

$\sqrt{ \frac{t _{1} }{1 _{1}} } = \sqrt{ \frac{t _{2} }{1 _{2}} }$

$1 = \sqrt{ \frac{t _{2}}{t _{1} } } 1 _{1} = 19.3cm$

$1 = 1_{1} - 1 _{2} = 7cm$

(b) is correct.