Question

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Answer 1

Answer:

$1.7$ × $10^1^0$

Explanation:

Knowing that, the volume of the sphere is given by, $v=\frac{4}{3}\pi ^3$

Thus, the fractional change in volume is given by,

$=3$ × $\frac{0.02}{100}=\frac{0.06}{100}$

Pressure at the bottom of the sea is,

Δp =pgh = $10^3$ × $10$ × $10^3=10^7$ pa.

Knowing that,

Bulk modulus: $\frac{10^7 * 100}{0.06}=\frac{10^9}{6*10^-^2}=\frac{10^1^1}{6}$

$B =\frac{10}{6}*10^1^0$

$B = 1.7*10^1^0N/M^2$

Answer = $1.7$ × $10^1^0$

[RevyBreeze]

Answer 2

Answer:

Explanation:

Sphere contracts in volume by 0.02%

Change in Depth = 1 km = 1000m

Acceleration due to gravity = 9.8 m/s^2

Density of water = 1000 kg/m^3

Assume the volume of sphere is V

V=4/3*pi*R^3

So, dV=4*pi*R^2*dR

dV/V = 4*pi*R^2*dR / 4/3*pi*R^3

= 3*dR/R

Given dR/R = 0.02%

dV/V=3*0.02/100

= 6*10^(-4)

Water pressure at 1km (1000m) deep water

P=ρgH

= 1000*9.8*1000

= 9.8*10^6 N/m^2

Bulk modulus, K, is given by

= P / (dV/V)

= ​9.8*10^6 / 6*10^(-4)

= 1.63*10^10 N/m^2