Properties of Matter

The pressure experienced by a submarine at a certain depth in the sea is given by the formula:

$P = \rho g h$

where:

• $P$ is the pressure
• $\rho$ is the density of the fluid (sea water in this case)
• $g$ is the acceleration due to gravity
• $h$ is the height (or depth in this case)

Given:

• $\rho = 10^3 \, kg/m^3$
• $g = 10 \, m/s^2$

We are looking for the difference in depth, $d_2 - d_1$, which corresponds to the difference in pressure $\Delta P$:

$\Delta P = P_2 - P_1 = \rho g (d_2 - d_1)$

Rearranging the above equation, we get:

$d_2 - d_1 = \frac{\Delta P}{\rho g}$

Given:

• $P_2 = 8.08 \times 10^6 \, Pa$
• $P_1 = 5.05 \times 10^6 \, Pa$

$\Delta P = P_2 - P_1 = 8.08 \times 10^6 \, Pa - 5.05 \times 10^6 \, Pa = 3.03 \times 10^6 \, Pa$

So,

$d_2 - d_1 = \frac{3.03 \times 10^6 \, Pa}{10^3 \, kg/m^3 \times 10 \, m/s^2} = 303 \, m$

The closest answer among the options provided is Option C, 300 m.

When rubber cord is stretched, then it stores potential energy and when released, this potential energy is given to the stone as kinetic energy.

So, potential energy of stretched cord = kinetic energy of stone

$ \Rightarrow {1 \over 2}Y{\left( {{{\Delta L} \over L}} \right)^2}A\,.\,L = {1 \over 2}m{v^2}$

Here, $\Delta$L = 20 cm = 0.2 m, L = 42 cm = 0.42 m, v = 20 ms^$-$1, m = 0.02 kg, d = 6 mm = 6 $\times$ 10^$-$3 m

$\therefore$ $A = \pi {r^2} = \pi {\left( {{d \over 2}} \right)^2} = \pi {\left( {{{6 \times {{10}^{ - 3}}} \over 2}} \right)^2}$

$ = \pi {(3 \times {10^{ - 3}})^2} = 9\pi \times {10^{ - 6}}{m^2}$

On substituting values, we get

$Y = {{m{v^2}L} \over {A{{(\Delta L)}^2}}} = {{0.02 \times {{(20)}^2} \times 0.42} \over {9\pi \times {{10}^{ - 6}} \times {{(0.2)}^2}}}$

$ \approx 3.0 \times {10^6}N{m^{ - 2}}$

So, the closest value of Young's modulus is 10⁶ Nm^$-$2.

When soap bubble is being inflated and its temperature remains constant, then it follows Boyle's law, so

pV = constant (k)

$ \Rightarrow p = {k \over V}$

Differentiating above equation with time, we get

${{dp} \over {dt}} = k.\,{d \over {dt}}\left( {{1 \over V}} \right)$

$ \Rightarrow {{dp} \over {dt}} = k\left( {{{ - 1} \over {{V^2}}}} \right).\,{{dV} \over {dt}}$

It is given that, ${{dV} \over {dt}} = c$ (a constant)

So, ${{dp} \over {dt}} = {{ - kc} \over {{V^2}}}$ ..... (i)

Now, from ${{dV} \over {dt}} = c$, we get

$dV = cdt$

or, $\int {dV = \int {cdt} } $ or, $V = ct$ ..... (ii)

From Eqs. (i) and (ii), we get

${{dp} \over {dt}} = {{ - kc} \over {{c^2}{t^2}}}$

or, ${{dp} \over {dt}} = - {\left( {{k \over c}} \right)^{{t^{ - 2}}}}$

$ \Rightarrow dp = - {k \over c}.\,{t^{ - 2}}dt$

Integrating both sides, we get

$\int {dp = - {k \over c}\int {{t^{ - 2}}dt} } $

$p = - {k \over c}.\,\left( {{{{t^{ - 2 + 1}}} \over { - 2 + 1}}} \right)$

$ = - {k \over c}.\,{{ - 1} \over t} = {k \over {ct}}$

or, $p \propto {1 \over t}$

Hence, p versus ${1 \over t}$ graph is a straight line, which is correctly represented in option (b).

Time taken to cool from 60$^\circ$C to 50$^\circ$C = 10 minutes

Temperature of surroundings = 25$^\circ$C

Temperature of body in next 10 minutes = T

Therefore, ${{60 - 50} \over {10\,\min }} = {k_B}\left[ {{{60 + 50} \over 2} - 25} \right] \Rightarrow {k_B}30 = 1$ ...... (1)

and ${{60 - T} \over {20\,\min }} = {k_B}\left[ {{{60 + T} \over 2} - 25} \right] = {k_B}\left[ {{{60 + T - 50} \over 2}} \right]$ ..... (2)

Taking ratio of Eqs. (1) and (2), we get

${{20} \over {60 - T}} = {{30{k_B}} \over {{k_B}\left( {{{10 + T} \over 2}} \right)}} \Rightarrow {{20} \over {60 - T}} = {{30} \over {5 + T/2}}$

$ \Rightarrow 20\left( {5 + {T \over 2}} \right) = 30(60 - T) \Rightarrow 100 + T10 = 1800 - 30T$

$ \Rightarrow 1800 - 100 = 30T + 10T$

$ \Rightarrow 1700 = 40T \Rightarrow T = {{1700} \over {40}} = 42.5^\circ C \sim 43^\circ C$

Given : Force applied F = 10⁵ N; side of cube x = 10 m = 10 $\times$ 10^$-$2 m; shift dx = 0.5 cm = 0.5 $\times$ = 10^$-$2 m

On cube of side x = 10 cm;

Stress = ${F \over {{x^2}}} = {{{{10}^5}} \over {{{(10 \times {{10}^{ - 2}})}^2}}} = {10^7}$ N/m²

Strain = ${{dx} \over x} = {{0.5 \times {{10}^{ - 2}}} \over {10 \times {{10}^{ - 2}}}} = 0.5$

On cube of side x' = 20 cm;

Stress' = ${F \over {x{'^2}}} = {{{{10}^5}} \over {{{(20 \times {{10}^{ - 2}})}^2}}}$

Strain' = ${{dx'} \over {x'}} = {{dx'} \over {20 \times {{10}^{ - 2}}}}$

Now, it is given that the material is same for both cubes, therefore,

${{Stress} \over {Strain}} = {{Stress'} \over {Strain'}} \Rightarrow {{{{10}^7}} \over {0.05}} = {{{{10}^5}} \over {(20 \times {{10}^{ - 2}})}} \times {{20 \times {{10}^{ - 2}}} \over {dx'}}$

$ \Rightarrow {{{{10}^7} \times {{10}^2}} \over 5} = {{{{10}^5}} \over {20 \times {{10}^{ - 2}}}}{1 \over {dx'}}$

$ \Rightarrow dx' = {{{{10}^5}} \over {20 \times {{10}^{ - 2}}}} \times {5 \over {{{10}^7} \times {{10}^2}}} = {{5 \times {{10}^{ - 2}}} \over {20}} = 0.25 \times {10^{ - 2}}$ m

$ \Rightarrow dx' = 0.25$ cm

Given, bubble of radius r rises from bottom of lake. The radius of bubble at top becomes 5r/4. Therefore, pressure in bubble at bottom is ${P_1} = {P_0} + \rho gh + {{4T} \over r}$

Pressure in bubble at top is ${P_2} = {P_0} + {{4T} \over {5r/4}}$

Now, we know ${P_1}{V_1} = {P_2}{V_2}$, therefore,

$\left( {{P_0} + \rho gh + {{4T} \over 4}} \right){{4\pi } \over 3}{r^3} = \left( {{P_0} + {{4T} \over {5r/4}}} \right){{4\pi } \over 3}{\left( {{{5r} \over 4}} \right)^3}$

${P_0} + \rho hg + {{4T} \over r} = \left( {{P_0} + {{4T \times 4} \over {5r}}} \right){{125} \over {64}}$

Now given P₀ = 10$\rho$g, therefore,

$10\rho g + \rho gh + {{4T} \over r} = {{125} \over {64}} \times 10\rho g + {{16T} \over {5r}} \times {{125} \over {64}}$

Neglecting effect of temperature, we get

$10\rho g + \rho gh = {{125} \over {64}} \times 10\rho g$

$ \Rightarrow 10 + h = {{125} \over {64}} \times 10 \Rightarrow h = {{1250} \over {64}} - 10 = 9.53125\,m$

$\Rightarrow$ h $\sim$ 9.5 m