Question

15. A sphere is taken to the bottom of sea 3 km deep
then the fractional change in radius of sphere is
(Bulk modulus of the sphere is 9.8*108 N/m2)
(1) 0.1
L27 -0.01
(3) -0.03
(4) 3

Answer 1

Given info : A sphere is taken to the bottom of sea 3km deep. Bulk modulus of the sphere is 9.8 × 10^8 N/m².

To find : the fractional change in radius of sphere is ..

solution : pressure at the depth of 3km from the surface of sea, P = P₀ + ρgh

here P₀ is atmospheric pressure.i.e, 1.013 × 10⁵ N/m²

ρ is density of sea water = 1030 kg/m³

g is acceleration due to gravity = 9.8 m/s²

h is depth = 3km = 3000 m

now, P = 1.013 × 10⁵ + 1030 × 9.8 × 3000

= 1.013 × 10⁵ + 302.82 × 10⁵

= 303.833 × 10⁵ N/m²

now, bulk modulus = -P/(∆V/V)

⇒9.8 × 10^8 = -(303.833 × 10^5)/(∆V/V)

⇒∆V/V = -(303.833 × 10^5)/(9.8 × 10^8)

= -31 × 10¯³

= -0.031

we know, ∆V/V = 3 ∆r/r , if change is very small.

so, ∆r/r = (∆V/V)/3 = -0.031/3 ≈ -0.01

Therefore fraction change in radius of sphere is -0.01 [ negative sign indicates sphere has shrunk ]