Question

To what depth must a rubber ball be taken in deep sea so that it’s volume is decreased by 0.1% . Bulk modulus of rubber = 9.8 * 10^8

Answer 1

Heya......!!!

Given in the question :

=> Bulk modulus of Rubber ( B ) = 9.8 × 10^8

=> Density of sea ( water ) d =. 10³ kg/m³

=> Percentage decrease in volume ΔV/V = 0.1 % = 0.1/100 =>> 10^-3

=>. Height = ??

So ,, Let the height be “ h “

Change in pressure = hdg

We know that ⇒ B = ΔP ÷ ( ΔV/V ) ⇒ ΔP = B × ΔV/V

Putting the values we get ,, :

⇒ ΔP = 9.8 × 10^8 × 10^-3 => 9.8 × 10^5 N/m²

⇒ Δ P is also equal to =>> hdg (( h = height ,, d = density ,, g = acceleration due to gravity ))

⇒ h = ΔP/dg. =>> 9.8 × 10^5 ÷ 10³ × 9.8

⇒ depth ( h ). = 10² m

So ,, rubber ball must be taken to a Depth of 100 m .


Hope It Helps You. :-)

Answer 2

Hey friend, Harish here.

Here is your answer:

Given that,

Bulk modulus of rubber (B)= 9.8 × 10⁸.

To  find, 

The depth at which it's volume is decreased by 0.1%.

Solution,

Density of sea water (ρ) = 1000 Kg/m³.

Let volume be (v), And Change in volume (v')

Then,  

$\frac{v'}{v} = 0.1\% = 10^{-3}$

Let the Height be (h). & Change in pressure be (p')

Then, 

$p' = h\rho g$

⇒ $h = \frac{p'}{\rho g} \ \ \ - (i)$

We know that,

$B = \frac{p'}{( \frac{v'}{v} )}$

⇒ $p' = B\times \frac{v'}{v} = 9.8 \times 10^{8} \times 10^{-3} = 9.8 \times 10^{5} \ \ \ \ \ - (ii)$

Now substitute (ii) in (i).

Then,

$h = \frac{p'}{\rho g} = \frac{9.8\times 10^{5}}{10^{3}\times 9.8} = 100\ m.$

Therefore the height is 100 m.
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Hope my  answer is helpful to you.