Two persons pull a rope towards each other. Each person exerts a force of $100 \mathrm{N}$ on the rope. The Young modulus of the material of the rope if it extends in length by $1 \mathrm{cm}$ is $1\times {10}^{\mathrm{y} }\mathrm{N} {\mathrm{m}}^{-2}$ The original length of the rope is$2 \mathrm{m}$, and the area of cross-section is $2 {\mathrm{cm}}^2$. The value of $\mathrm{y}$ is:

Consider the situation, shown in the figure. Force $F$ is equal to $\frac{m_{2}g}{2}$. If the area of the cross-section of the string is $A$ and its Young modulus $Y$, The string is light, and there is no friction anywhere The strain developed in it is $\frac{m_{2}g\left(2m_1+m_2\right)}{xAY\left(m_1+m_2\right)}$. Find $x$.

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The increase in the pressure required to decrease the volume of a water sample by $0.01\%$ is $z\times {10}^5 \mathrm{N} {\mathrm{m}}^{-2}$. Find the value of $z$.

It is given that the bulk modulus of water is $2.1\times {10}^9 \mathrm{N} {\mathrm{m}}^{-2}$.

The change in the density of water in the ocean at a depth of $400 \mathrm{m}$ below the surface is $x \mathrm{kg} {\mathrm{m}}^{-3}$. Find the value of $x$.

It is given that the density of water at the surface is $1030 \mathrm{kg} {\mathrm{m}}^{-3}$, and the bulk modulus of water is $2\times {10}^{9 }\mathrm{N} {\mathrm{m}}^{-2}$. (Take $g=10 \mathrm{m} {\mathrm{s}}^{-2}$.)

The excess pressure inside

(a) a drop of mercury of radius $2 \mathrm{mm}$. say as $x N m^{-2}$

(b) a soap bubble of radius $4 \mathrm{mm}$.say as $y N m^{-2}$

(c) an air bubble of radius $4 \mathrm{mm}$, formed inside a tank of water say as $z N m^{-2}$

The surface tension of mercury, soap solution, and water are $0.465 \mathrm{N} {\mathrm{m}}^{-1}$, $0.03 \mathrm{N} {\mathrm{m}}^{-1}$ and $0.076 \mathrm{N} {\mathrm{m}}^{-1}$, respectively.

Find the value of $\frac{x}{yz}$.