TS EAMCET · Physics · Mechanical Properties of Solids

A swimming pool has a depth of $22 m$ and area $700 m^{2}$ . Calculate fractional change $\frac{Δ v}{v}$ of water at the bottom of the swimming pool. Given that the bulk modulus of water is $2.2 \times 10^{9} {Nm}^{- 2}, g = 10 m s^{- 2}$ , and density of water $1000 kg m^{- 3}$ .

A $2.2 \times 10^{- 4}$
B $0.7 \times 10^{- 4}$
C $0.31 \times 10^{- 4}$
D $10^{- 4}$

Verified Solution

Answer & Solution

Correct Answer

(D) $10^{- 4}$

Step-by-step Solution

Detailed explanation

An increase in the pressure at the bottom due to water will be, ∆P=ρWgh. Using the concept of bulk modulus, B=-∆PV∆V⇒∆VV=∆PB=ρWghB=1000×10×222.2×109=10-4

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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A swimming pool has a depth of 22 m and area 700 m2. Calculate fractional change Δvv of water at the bottom of the swimming pool. Given that the bulk modulus of water is 2.2×109Nm-2, g=10 m s-2, and density of water 1000 kg m-3.

Step-by-step Solution

A swimming pool has a depth of $22 m$ and area $700 m^{2}$ . Calculate fractional change $\frac{Δ v}{v}$ of water at the bottom of the swimming pool. Given that the bulk modulus of water is $2.2 \times 10^{9} {Nm}^{- 2}, g = 10 m s^{- 2}$ , and density of water $1000 kg m^{- 3}$ .