When a certain volume of water is subjected to 100kPa pressure, the volume of water decreases by 0.005%. The speed of sound in water must be

Question

When a certain volume of water is subjected to 100kPa pressure, the volume of water decreases by 0.005%. The speed of sound in water must be (A) 140 m/s (B) 300 m/s (C) 1400 m/s (D) 5000 m/s

Tardigrade Admin · Accepted Answer

Bulk modulus K=(Δ p/(Δ V)V)=(100× 103/0.005/100)=2× 109Pa Speed v=√(K/d)=√(2× 109/103)=√2× 103 ⇒ v=1.400× 103=1400 m/s

Q. When a certain volume of water is subjected to 100kPa pressure, the volume of water decreases by 0.005%. The speed of sound in water must be

140 m/s

300 m/s

1400 m/s

5000 m/s

Bulk modulus $K = \frac{Δ p}{\frac{Δ V}{V}} = \frac{100 \times 10 ^{3}}{0.005/100} = 2 \times 10^{9} P a$ Speed $v = \frac{K}{d} = \frac{2 \times 10 ^{9}}{10 ^{3}} = 2 \times 10^{3}$ $\Rightarrow$ $v = 1.400 \times 10^{3} = 1400 m / s$

Q. When a certain volume of water is subjected to 100kPa pressure, the volume of water decreases by 0.005%. The speed of sound in water must be

140 m/s

300 m/s

1400 m/s

5000 m/s

Bulk modulus K=VΔV​Δp​=0.005/100100×103​=2×109Pa Speed v=dK​​=1032×109​​=2​×103 ⇒ v=1.400×103=1400m/s

Bulk modulus $K = \frac{Δ p}{\frac{Δ V}{V}} = \frac{100 \times 10 ^{3}}{0.005/100} = 2 \times 10^{9} P a$ Speed $v = \frac{K}{d} = \frac{2 \times 10 ^{9}}{10 ^{3}} = 2 \times 10^{3}$ $\Rightarrow$ $v = 1.400 \times 10^{3} = 1400 m / s$