.02 \%$ is _______. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,KCET,Mechanical Properties of Fluids,Pressure,Mechanical Properties of Solids,Elasticity,NCERT,NEET, JEE Main Physics,Class 11 Physics,Class 12 Phys.">
.02 \%$ is _______. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,KCET,Mechanical Properties of Fluids,Pressure,Mechanical Properties of Solids,Elasticity,NCERT,NEET, JEE Main Physics,Class 11 Physics,Class 12 Phys.">
.02 \%$ is _______. Tags and topics: JEE MAIN,JEE Advanced,JEE Mains,KCET,Mechanical Properties of Fluids,Pressure,Mechanical Properties of Solids,Elasticity,NCERT,NEET, JEE Main Physics,Class 11 Physics,Class 12 Phys.">
The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by $0.02 \%$ is _______ $m$.
(Take density of sea water $=10^3 \mathrm{kgm}^{-3}$, Bulk modulus of rubber $=9 \times 10^8 \mathrm{~Nm}^{-2}$, and $g=10 \mathrm{~ms}^{-2}$)
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Step-by-Step Solution
Step 1: Understand the Concept of Bulk Modulus
The bulk modulus B for a material is defined by the relation:
B = -\frac{\Delta P}{\frac{\Delta V}{V}},
where \Delta P is the change in external pressure, V is the original volume, and \Delta V is the resulting change in volume. The negative sign indicates that as external pressure increases, the volume decreases.
Step 2: Identify the Change in Pressure Below the Sea Surface
When an object is submerged to a depth h beneath the sea surface, the increase in pressure over atmospheric pressure is due to the column of water above it. This increase in pressure is given by:
\Delta P = \rho g h,
where:
\rho is the density of seawater ( 10^3 \,\mathrm{kg\,m^{-3}} ),
g is the acceleration due to gravity ( 10 \,\mathrm{m\,s^{-2}} ),
h is the depth below the surface (in meters).
Step 3: Relate Bulk Modulus to the Pressure and Volume Change
From the definition of the bulk modulus, we have:
\Delta P = -B \times \frac{\Delta V}{V}.
Since the increase in pressure \Delta P is \rho g h , substituting we get:
\rho g h = B \times \frac{\Delta V}{V}.
(The negative sign is treated with absolute values here, focusing on magnitudes.)
Step 4: Substitute the Known Values
We are given:
Density of seawater, \rho = 10^3 \,\mathrm{kg\,m^{-3}}
Acceleration due to gravity, g = 10 \,\mathrm{m\,s^{-2}}
Bulk modulus of rubber, B = 9 \times 10^8 \,\mathrm{N\,m^{-2}}
Decrease in volume, 0.02\% = 0.02/100 = 0.0002
So,
10^3 \times 10 \times h = 9 \times 10^8 \times 0.0002.
Step 5: Calculate the Depth h
First, compute the right-hand side:
9 \times 10^8 \times 0.0002 = 9 \times 10^8 \times 2 \times 10^{-4} = 9 \times 2 \times 10^{8-4} = 18 \times 10^4 = 1.8 \times 10^5.
Thus,
10^3 \times 10 \times h = 1.8 \times 10^5.
This simplifies to:
10^4 \,h = 1.8 \times 10^5 \quad \Longrightarrow \quad h = \frac{1.8 \times 10^5}{10^4} = 18 \,\mathrm{m}.
Final Answer
The depth below the sea surface required for a 0.02\% reduction in the volume of the rubber ball is 18 m.