Mechanical Properties of Solids

JEE Main level, numerical answer, medium

90 of 118

Question 90 of 118

The depth below the surface of sea to which a rubber ball be taken so as to decrease its volume by 0.020.02% is _____ mm.
(Take density of sea water =103kgm−3={10}^{3}kg{m}^{-3}, Bulk modulus of rubber =9×108Nm−2=9\times {10}^{8}N{m}^{-2}, and g=10ms−2g=10m{s}^{-2})

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Answer: 18

Bulk modulus is given by, B=−∆P∆VVB=\dfrac{-∆P}{\dfrac{∆V}{V}}
⇒∆P=−B∆VV\Rightarrow ∆P=-B\dfrac{∆V}{V}
⇒ρgh=−B∆VV\Rightarrow \rho gh=-B\dfrac{∆V}{V}
⇒103×10×h=−9×108×(−0.02100)\Rightarrow {10}^{3}\times 10\times h=-9\times {10}^{8}\times \left(-\dfrac{0.02}{100}\right)
⇒h=18m\Rightarrow h=18m

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