Question

3. A solid ball 300 cm in diameter is submerged in a
lake at such a depth that the pressure exerted by
water is 1.00 kgfcm Find the change in volume of
the ball at this depth. x for material of the ball
= 1.00 x 10-dyne cm
(Ans. 1.385 cm)​

Answer 1

Answer:

1.385 $cm^{3}$

Explanation:

Given Bulk's Modulus = 1 x $10^{13}$ dyne cm

Bulk's Modulus =$\frac{PV}{deltaV}$

Find volume using the volume of sphere formula then V= 1.413 x $10^7$ $cm^3$

We need to find ΔV:

ΔV =$\frac{PV}{B}$     (P = 1000 x 920 dyne/$cm^2$)

ΔΔ =  $\frac{1000 x 920 x 1.413 x 10^7}{10^{13}}$

ΔV = 1.385 $cm^{3}$

Short And Simple

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