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Chapter 7: Problem 973

A rubber ball when taken to the bottom of a \(100 \mathrm{~m}\) deep take decrease in volume by \(1 \%\) Hence, the bulk modulus of rubber is $\ldots \ldots \ldots . . .\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]$ (A) \(10^{6} \mathrm{~Pa}\) (B) \(10^{8} \mathrm{~Pa}\) (C) \(10^{7} \mathrm{~Pa}\) (D) \(10^{9} \mathrm{~Pa}\)

Short Answer

Expert verified
The bulk modulus of the rubber ball is \(10^8 \mathrm{~Pa}\).

Step by step solution

01

Find the pressure at the bottom of the tank

First, we need to determine the pressure at the bottom of the tank using the depth and gravity acting on the water. We can do that using the formula: \(P = \rho \cdot g \cdot h\) where: \(P\) is the pressure at the bottom, \(\rho\) is the density of water (approximately \(1000 \mathrm{~kg/m^3}\)), \(g\) is the acceleration due to gravity (\(10 \mathrm{~m/s^2}\)), and \(h\) is the depth (\(100 \mathrm{~m}\)). By substituting the values, we get: \(P = 1000 \cdot 10 \cdot 100\)
02

Calculate the pressure at the bottom of the tank

Now, let's calculate the pressure at the bottom of the tank: \(P = 1000 \cdot 10 \cdot 100 = 10^6 \mathrm{~Pa}\) The pressure at the bottom of the tank is \(10^6 \mathrm{~Pa}\).
03

Calculate the change in volume percentage

We know that the volume of the rubber ball decreases by \(1 \%\) when submerged at the bottom of the tank, which means the change in volume as a fraction is \(0.01\). Let the original volume be \(V\), then the change in volume is \(0.01V\).
04

Use the formula for bulk modulus and solve for it

We can now use the formula for bulk modulus to calculate the bulk modulus of the rubber ball: Bulk Modulus = \(\dfrac{\Delta P}{\dfrac{\Delta V}{V_0}}\) Here, \(\Delta P = 10^6 \mathrm{~Pa}\), and \(\dfrac{\Delta V}{V_0} = 0.01\). Now, we can solve for the bulk modulus: Bulk Modulus = \(\dfrac{10^6}{0.01} = 10^8 \mathrm{~Pa}\) The bulk modulus of the rubber ball is \(10^8 \mathrm{~Pa}\). Therefore, the correct answer is (B) \(10^8 \mathrm{~Pa}\).

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Most popular questions from this chapter

A capillary tube of radius \(\mathrm{R}\) is immersed in water and water rises in it to a height \(\mathrm{H}\). Mass of water in the capillary tube is M. If the radius of the tube is doubled. Mass of water that will rise in the capillary tube will now be (A) \(\mathrm{M}\) (B) \(2 \mathrm{M}\) (C) \((\mathrm{M} / 2)\) (D) \(4 \mathrm{M}\)

Radius of a soap bubble is \(\mathrm{r}^{\prime}\), surface tension of soap solution is \(\mathrm{T}\). Then without increasing the temperature how much energy will be needed to double its radius. (A) \(4 \pi r^{2} T\) (B) \(2 \pi r^{2} T\) (C) \(12 \pi r^{2} T\) (D) \(24 \pi r^{2} T\)

A soap bubble of radius \(r\) is blown up to form a bubble of radius $2 \mathrm{r}\( under isothermal conditions if the \)\mathrm{T}$ is the surface tension of soap solution the energy spent in the slowing is. (A) \(3 \pi \mathrm{Tr}^{2}\) (B) \(6 \pi \mathrm{Tr}^{2}\) (C) \(12 \pi \mathrm{Tr}^{2}\) (D) \(24 \pi \mathrm{Tr}^{2}\)

A triangular lamina of area \(\mathrm{A}\) and, height \(\mathrm{h}\) is immersed in a liquid of density \(\mathrm{S}\) in a vertical plane with its base on the surface of the liquid. The thrust on lamina is (A) \((1 / 2)\) Apgh (B) \((1 / 3)\) Apgh (C) \((1 / 6)\) Apgh (D) \((1 / 3)\) Apgh

To what depth below the surface of sea should a rubber ball be taken as to decrease its volume by \(0.1 \%\) [Take : density of sea water \(=1000\left(\mathrm{~kg} / \mathrm{m}^{3}\right)\) Bulk modulus of rubber $=9 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$, acceleration due to gravity \(\left.=10\left(\mathrm{~m} / \mathrm{s}^{2}\right)\right]\) (A) \(9 \mathrm{~m}\) (B) \(18 \mathrm{~m}\) (C) \(180 \mathrm{~m}\) (D) \(90 \mathrm{~m}\)
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