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 \green{Spammer \: Go \: Away}SpammerGoAway​

Answer 1

Answer:

The correct answer is (b) 7mm

Step-by-step explanation:

We know that,

Frequency of oscillation in organ pipe ∝ $\frac{\sqrt{T} }{l}$

Where T is the temperature and l is the length of the air column.

For first organ pipe we have T₁=27°C and l₁=20cm.

For the second organ pipe we have T₂=7°C and we have to find l₂.

Using the above formula we have,

√T₁/l₁=√T₂/l₂

⇒ l₂=√(T₂/T₁) × l₁=√(7/27)×20=19.3 cm.

Δl= l₁ - l₂ = 20 - 19.3= 0.7cm= 7mm.

∴ Δl=7mm. which is option (b)

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Answer 2

Answer:

Step-by-step explanation:

Answer:

The correct answer is (b) 7mm

Step-by-step explanation:

We know that,

Frequency of oscillation in organ pipe

$∝ \frac{\sqrt{T} }{l} </p><p></p><p>$

Where T is the temperature and l is the length of the air column.

For first organ pipe we have T₁=27°C and l₁=20cm.

For the second organ pipe we have T₂=7°C and we have to find l₂.

Using the above formula we have,

√T₁/l₁=√T₂/l₂

⇒ l₂=√(T₂/T₁) × l₁=√(7/27)×20=19.3 cm.

Δl= l₁ - l₂ = 20 - 19.3= 0.7cm= 7mm.

∴ Δl=7mm. which is option (b)

Hope it helps, please mark as Brainliest