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Question: If a rubber ball is taken down to a 100 m deep lake, its volume decreases by 0.1%. If \[g = 10m/{s^2...

If a rubber ball is taken down to a 100 m deep lake, its volume decreases by 0.1%. If g=10m/s2g = 10m/{s^2} then the bulk modulus of elasticity for rubber, in N/m2N/{m^2}, is
A.108{10^8}
B.109{10^9}
C.1011{10^{11}}
D.1010{10^{10}}

Explanation

Solution

In this question we have to find the bulk modulus of elasticity for rubber. For this we will use the formula of bulk modulus. For this first we will find the difference between pressure and volume strain to use in the formula of bulk modulus. We will also be careful that all the variables are in the same unit system.

Complete step by step answer:
Given,
h=100 m
g=10m/s2g = 10m/{s^2}
The volume decreases by 0.1%,
Formula used:
Bulk modulus =ΔPΔVVBulk{\text{ }}modulus{\text{ }} = \dfrac{{\Delta P}}{{\dfrac{{\Delta V}}{V}}}
Where,
ΔP\Delta Pis change in pressure
ΔVV\dfrac{{\Delta V}}{V}is volume strain
Pressure on rubber ball, P=P0+ρghP = {P_0} + \rho gh
Where,
P0{P_0}is the atmospheric pressure which is equal to 1.03×105 m1.03 \times {10^5}{\text{ m}}
ρ\rho is density of water which is equal to   1000 kg/m3\;1000{\text{ }}kg/{m^3}
g is acceleration due to gravity
hh is the distance deep in the lake.
Now, we will put the values of variables in the above equation of pressure.
P=P0+ρghP = {P_0} + \rho gh
P=1.03×105+1000×10×100P = 1.03 \times {10^5} + 1000 \times 10 \times 100
P=1.03×105+10×105P = 1.03 \times {10^5} + 10 \times {10^5}
P=11.03×105 N/m2P = 11.03 \times {10^5}{\text{ N/}}{{\text{m}}^2}
It is given that there is a decrease of 0.1 % in volume,
ΔVV=0.05100\dfrac{{\Delta V}}{V} = \dfrac{{0.05}}{{100}}
ΔVV=5×104\dfrac{{\Delta V}}{V} = 5 \times {10^{ - 4}}
Change in pressure ΔP=PP0\Delta P = P - {P_0}
Putting the values of PPand P0{P_0}in above equation,
ΔP=11.03×1051.03×105\Delta P = 11.03 \times {10^5} - 1.03 \times {10^5}.
ΔP=10.00×105N/m2\Delta P = 10.00 \times {10^5}N/{m^2}
ΔP=106N/m2\Delta P = {10^6}N/{m^2}
Bulk modulus =ΔPΔVVBulk{\text{ }}modulus{\text{ }} = \dfrac{{\Delta P}}{{\dfrac{{\Delta V}}{V}}}
Putting the values of ΔP\Delta Pand ΔVV\dfrac{{\Delta V}}{V}in above formula,
Bulk modulus =1065×104Bulk{\text{ }}modulus{\text{ }} = \dfrac{{{{10}^6}}}{{5 \times {{10}^{ - 4}}}}
Bulk modulus =0.2×1010N/m2Bulk{\text{ }}modulus{\text{ }} = 0.2 \times {10^{10}}N/{m^2}
Result- Hence, from the above calculation we have found the value ofBulk modulus =0.2×1010N/m2Bulk{\text{ }}modulus{\text{ }} = 0.2 \times {10^{10}}N/{m^2}.

Note:
Hence, it is clear from the above explanation that in such types of questions we have to be aware of the formulae and we must know the concept to solve these questions. The units of all the variables must be in the same system. We should be careful while doing the calculation.