The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02% is ______ m.

Question

$(T a k e d e n s i t y o f s e a w a t e r = 10^{3} k g m^{- 3}, B u l k m o d u l u s o f r u b b e r = 9 \times 10^{8} N m^{- 2}, a n d g = 10 m s^{- 2})$ [2024]

Smart Achievers · Accepted Answer

(18) $(T a k e d e n s i t y o f s e a w a t e r = 10^{3} k g m^{- 3}, B u l k m o d u l u s o f r u b b e r = 9 \times 10^{8} N m^{- 2}, a n d g = 10 m s^{- 2})$

$ρ g h = - β \frac{Δ V}{V}$

$10^{3} \times 10 \times h = - 9 \times 10^{8} \times (- \frac{0.02}{100}) \Rightarrow h = 18 m$

Solution

Q.
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02% is ______ m.

$(T a k e d e n s i t y o f s e a w a t e r = 10^{3} k g m^{- 3}, B u l k m o d u l u s o f r u b b e r = 9 \times 10^{8} N m^{- 2}, a n d g = 10 m s^{- 2})$ [2024]

Ans.
(18) $β = \frac{- Δ P}{\frac{Δ V}{V}} \Rightarrow Δ P = - β \frac{Δ V}{V}$

$ρ g h = - β \frac{Δ V}{V}$

$10^{3} \times 10 \times h = - 9 \times 10^{8} \times (- \frac{0.02}{100}) \Rightarrow h = 18 m$

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Solution

Q. The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02% is ______ m. (Take density of sea water=103kgm-3,Bulk modulus of rubber=9×108Nm-2,and g=10ms-2) [2024]

Ans. (18) β=-ΔPΔVV⇒ΔP=-βΔVV ρgh=-βΔVV 103×10×h=-9×108×(-0.02100)⇒h=18m

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Q.
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.02% is ______ m.

$(T a k e d e n s i t y o f s e a w a t e r = 10^{3} k g m^{- 3}, B u l k m o d u l u s o f r u b b e r = 9 \times 10^{8} N m^{- 2}, a n d g = 10 m s^{- 2})$ [2024]

Ans.
(18) $β = \frac{- Δ P}{\frac{Δ V}{V}} \Rightarrow Δ P = - β \frac{Δ V}{V}$

$ρ g h = - β \frac{Δ V}{V}$

$10^{3} \times 10 \times h = - 9 \times 10^{8} \times (- \frac{0.02}{100}) \Rightarrow h = 18 m$