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MCQEasyJEE 2024Superposition Principle & Standing Waves

JEE Physics 2024 Question with Solution

In a closed organ pipe, the frequency of the fundamental note is 30 Hz30 \, \text{Hz}. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz110 \, \text{Hz}. If the organ pipe has a cross-sectional area of 2 cm22 \, \text{cm}^2, the amount of water poured in the organ pipe is xx grams. (Take speed of sound in air as 330 m/s330 \, \text{m/s})

  • A

    200 g200 \, \text{g}

  • B

    300 g300 \, \text{g}

  • C

    400 g400 \, \text{g}

  • D

    500 g500 \, \text{g}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: Fundamental frequency of the closed organ pipe initially is f1=30 Hzf_1 = 30 \, \text{Hz}, final fundamental frequency is f2=110 Hzf_2 = 110 \, \text{Hz}, speed of sound is v=330 m/sv = 330 \, \text{m/s}, and cross-sectional area is A=2 cm2A = 2 \, \text{cm}^2.

Find: The mass of water poured into the pipe.

For a closed organ pipe, the fundamental frequency is

f=v4Lf = \frac{v}{4L}

where LL is the length of the air column.

Initially,

L1=v4f1=3304×30=330120=2.75 mL_1 = \frac{v}{4f_1} = \frac{330}{4 \times 30} = \frac{330}{120} = 2.75 \, \text{m}

After pouring water,

L2=v4f2=3304×110=330440=0.75 mL_2 = \frac{v}{4f_2} = \frac{330}{4 \times 110} = \frac{330}{440} = 0.75 \, \text{m}

So the height of the water column is

h=L1−L2=2.75−0.75=2.0 m=200 cmh = L_1 - L_2 = 2.75 - 0.75 = 2.0 \, \text{m} = 200 \, \text{cm}

The volume of water poured is

V=Ah=2 cm2×200 cm=400 cm3V = Ah = 2 \, \text{cm}^2 \times 200 \, \text{cm} = 400 \, \text{cm}^3

Using density of water ρ=1 g/cm3\rho = 1 \, \text{g/cm}^3,

m=ρV=1×400=400 gm = \rho V = 1 \times 400 = 400 \, \text{g}

Therefore, the amount of water poured is 400 g400 \, \text{g}. The correct option is C.

The solution also states the final answer as 400.

Frequency-Length Ratio

Given: In a closed organ pipe, fundamental frequency is inversely proportional to the air-column length.

Find: The mass of water added.

Since

f=v4Lf = \frac{v}{4L}

we have L∝1fL \propto \frac{1}{f}.

Thus,

L1=3304×30=2.75 m,L2=3304×110=0.75 mL_1 = \frac{330}{4 \times 30} = 2.75 \, \text{m}, \qquad L_2 = \frac{330}{4 \times 110} = 0.75 \, \text{m}

So the water height is 2.0 m=200 cm2.0 \, \text{m} = 200 \, \text{cm}.

Now directly compute displaced volume:

V=2 cm2×200 cm=400 cm3V = 2 \, \text{cm}^2 \times 200 \, \text{cm} = 400 \, \text{cm}^3

Since water has density 1 g/cm31 \, \text{g/cm}^3, the mass is numerically equal to the volume in cm3\text{cm}^3.

Therefore, the mass of water is 400 g400 \, \text{g}, so the correct option is C.

Common mistakes

  • Using the formula for an open organ pipe instead of a closed organ pipe is incorrect because the fundamental frequency for a closed organ pipe is f=v4Lf = \frac{v}{4L}, not v2L\frac{v}{2L}. Always identify that one end becomes effectively closed by the water.

  • Subtracting frequencies instead of air-column lengths is wrong because the water level depends on the change in resonating length, not on f2−f1f_2 - f_1 directly. First calculate L1L_1 and L2L_2, then find h=L1−L2h = L_1 - L_2.

  • Mixing units while calculating volume leads to an incorrect answer because the area is given in cm2\text{cm}^2. Convert the water height from 2 m2 \, \text{m} to 200 cm200 \, \text{cm} before multiplying by the area.

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