An object is taken $1.0\,km$ deep in sea. Density of sea $= 1.025 \times 10^3\,kg/m^3$ , Bulk modulus of object $= 1.6 \times 10^6\, KPa$ Find out percentage change in density of object....... $\%$
A$0.36$
B$0.64$
C$0.40$
D$0.60$
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B$0.64$
b $\mathrm{B}=\frac{-\mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}} \quad \begin{array}{l}{\rho \mathrm{V}=\mathrm{const}} \\ {\frac{\Delta \rho}{\rho}=-\frac{\Delta \mathrm{V}}{\mathrm{V}}}\end{array}$
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A string of area of cross-section $4\,mm ^{2}$ and length $0.5$ is connected with a rigid body of mass $2\,kg$. The body is rotated in a vertical circular path of radius $0.5\,m$. The body acquires a speed of $5\,m / s$ at the bottom of the circular path. Strain produced in the string when the body is at the bottom of the circle is $\ldots . . \times 10^{-5}$. (Use Young's modulus $10^{11}\,N / m ^{2}$ and $g =10\,m / s ^{2}$ )
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