To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by 0.1 % ? (Take density of sea water =1000 kg m-3, Bulk modulus of rubber =9 × 108 Nm -2, acceleration due to gravity =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 0.1 % ? (Take density of sea water =1000 kg m-3, Bulk modulus of rubber =9 × 108 Nm -2, acceleration due to gravity =10 m s-2 ) (A) 9 m (B) 18 m (C) 180 m (D) 90 m

Tardigrade Admin · Accepted Answer

Bulk modulus B = ( textnormal stress/ textvolume strain) ∴ B = (p/Δ V/V) Given (Δ V/V) = (0.1/100) , B = 9 × 108 Nm-2 p = hdg = h × 1000 × 10 ∴ (BV/V) = p = hdg ⇒ h = - (B Δ V/Vdg) h = - (9 × 108 × 0.1/1000 × 100 × 10) h = - 90 m

Q. To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by $0.1%$ ? (Take density of sea water $= 1000 k g m^{- 3}$ , Bulk modulus of rubber $= 9 \times 1 0^{8} N m^{- 2}$ , acceleration due to gravity $= 10 m s^{- 2}$ )

9 m

18 m

180 m

90 m

900 m

Q. To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by 0.1% ? (Take density of sea water =1000kgm−3, Bulk modulus of rubber =9×108Nm−2, acceleration due to gravity =10ms−2 )

9 m

18 m

180 m

90 m

900 m

Bulk modulus B=volume strainnormal stress​ ∴B=ΔV/Vp​ Given VΔV​=1000.1​,B=9×108Nm−2 p=hdg=h×1000×10 ∴VBV​=p=hdg ⇒h=−VdgBΔV​ h=−1000×100×109×108×0.1​ h=−90m

Q. To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by $0.1%$ ? (Take density of sea water $= 1000 k g m^{- 3}$ , Bulk modulus of rubber $= 9 \times 1 0^{8} N m^{- 2}$ , acceleration due to gravity $= 10 m s^{- 2}$ )