To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by 0.4 % [density of sea water =1000 kg m -3, Bulk modulus of rubber .=9 × 108 Nm -2, g =10 m / s 2]

Question

To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by 0.4 % [density of sea water =1000 kg m -3, Bulk modulus of rubber .=9 × 108 Nm -2, g =10 m / s 2] (A) 5 × 102 m (B) 30 × 106 m (C) 40 × 106 m (D) 36 × 101 m

Tardigrade Admin · Accepted Answer

B=(P/(-Δ V)V) ⇒ B=(h ρ g/0.4 × 10-2) h=(9 × 108 × 0.4 × 10-2/103 × 10) =36 × 101 m

Q. To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by $0.4%$ [density of sea water $= 1000 k g m^{- 3},$ Bulk modulus of rubber $= 9 \times 1 0^{8} N m^{- 2}, g = 10 m / s^{2}]$

$5 \times 1 0^{2} m$

$30 \times 1 0^{6} m$

$40 \times 1 0^{6} m$

$36 \times 1 0^{1} m$

$B = \frac{P}{\frac{- Δ V}{V}}$
$\Rightarrow B = \frac{h ρ g}{0.4 \times 1 0 ^{- 2}}$
$h = \frac{9 \times 1 0 ^{8} \times 0.4 \times 1 0 ^{- 2}}{1 0 ^{3} \times 10}$
$= 36 \times 1 0^{1} m$

Q. To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by 0.4% [density of sea water =1000kgm−3, Bulk modulus of rubber =9×108Nm−2,g=10m/s2]

5×102m

30×106m

40×106m

36×101m