9.21 The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 108 Pa. A steel ball of initial volume 0.32 m3 is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Ask & Answer Question

physics ncert solutions class 11th Physics Class 11th Mechanical Properties of Solids

9.21 The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 10⁸ Pa. A steel ball of initial volume 0.32 m³ is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

4 Views | Posted 5 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

5 months ago

Water pressure at the bottom, p = 1.1 $\times 10^{8}$ Pa

Initial volume of the steel ball, V = 0.32 $m^{3}$

Bulk modulus of the steel, B = 1.6 $\times 10^{11}$ N/ $m^{2}$

The ball falls at the bottom of the Pacific ocean which is 11 km beneath the surface

Let the change in volume of the ball on reaching the bottom of the trench be ΔV

We know, bulk modulus, B = $\frac{p}{\frac{Δ V}{V}}$ or ΔV = $\frac{p V}{B}$

ΔV = $\frac{1.1 \times 10^{8} \times 0.32}{1.6 \times 10^{11}}$ = 2.2 $\times 10^{- 4} m^{3}$

0 Upvotes 0 Downvotes

Similar Questions for you

If y, K and $η$ are the value of Young’s modulus, bulk modulus of rigidity of any material respectively. Choose

Vishal Baghel

If $μ$ is Poisson’s ratio,

Y = 3K (1 - 2 $μ$ ) ……… (1)

and Y = 2 $n (1 + μ)$ ……… (2)

With the help of equations (1) and (2), we can write

$\frac{3}{Y} = \frac{1}{η} + \frac{1}{3 k} \Rightarrow K = \frac{η Y}{9 η - 3 Y}$

A uniform rod, of mass m, length l and r radius of cross section, is rotated about an axis passing through one of its ends and perpendicular to its length with constant angular velocity ω in horizontal plane. If Y is the Young's modulus of the material of rod, the increase in its length due to rotation of rod is

Vishal Baghel

dm = (m/L)dx
∴ T = (mω²/2L) (L² - x²)
∴ ΔL = ∫? (mω²/2Lπr²Y) (L² - x²)dx
= ΔL = mω²L²/3πr²Y

A hollow spherical shell at outer radius $R$ floats just submerged under the water surface. The inner radius of the shell is r. If the specific gravity of the shell material is $\frac{27}{8}$ w.r.t water, the value of r is.

Vishal Baghel

Initially S? L = 2m
S? L = √2² + (3/2)²
S? L = 5/2 = 2.5 m
? x = S? L - S? L = 0.5 m
So since λ = 1 m. ∴? x = λ/2
So white listener moves away from S? Then? x (= S? L − S? L) increases and hence, at? x = λ first maxima will appear.? x = λ = S? L − S? L.
1 = d - 2 ⇒ d = 3 m.

A stone of mass 20g is projected from a rubber catapult of length 0.1m and area of cross section 10?? m² stretched by an amount 0.04 m. The velocity of the projected stone is _______ m/s. (Young's modulus of rubber = 0.5 × 10? N/m²)

Vishal Baghel

Loss in elastic potential energy = Gain in KE
½ (YA/L)x² = ½mv²
0.5 × (0.5×10? × 10? / 0.1) × (0.04)² = 20×10? ³ v²
0.5 × (5×10²) × 1.6×10? ³ = 20×10? ³ v²
0.4 = 20×10? ³ v²
v² = 20 => v = √20 ≈ 4.47 m/s
(Re-checking calculations)
0.5 * ( (0.5e9 * 1e-6) / 0.1) * (0.04)^2 = 0.5 * (5e2) * 1.6e-3 = 4.
0.5 * 20e-3 * v^2 = 10e-3 v^2
4 = 10e-3 v^2
v^2 = 400 => v = 20 m/s

A uniform metallic wire is elongated by 0.04m when subjected to a linear force F. The elongation, if its length and diameter is doubled and subjected to the same force will be……………….cm.

Vishal Baghel

As we know that

$Y = \frac{F L}{A ? L}$

$? L = 0.04 m = \frac{F L}{A Y} . . . . . . . . . . . . . . . (i)$

If length and diameter both are doubled

$? L' = \frac{F . 2 L}{4 A . Y} = \frac{F L}{2 Y} = 0.02 m = 2 c m$

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
686k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question