When a rubber ball is taken to a depth of m in deep sea, its volume decreases by 0.5 % (The bulk modulus of rubber =9.8 × 108 Nm -2 Density of sea water =103 kgm -3 .g =9.8 m / s 2)

Question

When a rubber ball is taken to a depth of m in deep sea, its volume decreases by 0.5 % (The bulk modulus of rubber =9.8 × 108 Nm -2 Density of sea water =103 kgm -3 .g =9.8 m / s 2) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

B =-(Δ P /((Δ V ) V ))=-(ρ gh /((Δ V ) V )) -( B (Δ V / V )/ρ g)= h (9.8 × 108 × 0.5/100 × 103 × 9.8)= h h =500

Q. When a rubber ball is taken to a depth of ___ $m$ in deep sea, its volume decreases by $0.5%$
(The bulk modulus of rubber $= 9.8 \times 1 0^{8} N m^{- 2}$ Density of sea water $= 1 0^{3} k g m^{- 3} g = 9.8 m / s^{2})$

$B = - \frac{Δ P}{( \frac{Δ V}{V} )} = - \frac{ρ g h}{( \frac{Δ V}{V} )}$
$- \frac{B \frac{Δ V}{V}}{ρ g} = h$
$\frac{9.8 \times 1 0 ^{8} \times 0.5}{100 \times 1 0 ^{3} \times 9.8} = h$
$h = 500$

Q. When a rubber ball is taken to a depth of ___m in deep sea, its volume decreases by 0.5% (The bulk modulus of rubber =9.8×108Nm−2 Density of sea water =103kgm−3g=9.8m/s2)

B=−(VΔV​)ΔP​=−(VΔV​)ρgh​ −ρgBVΔV​​=h 100×103×9.89.8×108×0.5​=h h=500

Q. When a rubber ball is taken to a depth of ___ $m$ in deep sea, its volume decreases by $0.5%$
(The bulk modulus of rubber $= 9.8 \times 1 0^{8} N m^{- 2}$ Density of sea water $= 1 0^{3} k g m^{- 3} g = 9.8 m / s^{2})$

$B = - \frac{Δ P}{( \frac{Δ V}{V} )} = - \frac{ρ g h}{( \frac{Δ V}{V} )}$
$- \frac{B \frac{Δ V}{V}}{ρ g} = h$
$\frac{9.8 \times 1 0 ^{8} \times 0.5}{100 \times 1 0 ^{3} \times 9.8} = h$
$h = 500$