Question

A metal rope of density $6000kg/{m^3}$ has a breaking stress $9.8 \times {10^8}m{s^{ - 2}}$. This rope is used to measure the depth of a sea. Then the depth of the sea that can be measured without breaking is _________
(A) $10 \times {10^3}m$
(B) $20 \times {10^3}m$
(C) $30 \times {10^3}m$
(D) $40 \times {10^3}m$

Answer 1

In order to solve this problem,we are going to calculate the net force acting on the rope due to its mass and the buoyant force. Using this force, calculate the stress and equate it to the breaking stress.

Complete step by step answer:
So when rope is dipped in water, there are multiple forces acting on it, one is the gravitational force which is due to its own mass and the other one is the buoyant force which is acting due to the water displaced by the rope. So, the net force acting on the rope would be the difference of these two forces.
${F_{mass}} = {m_{rope}}g$
${F_{mass}} = {V_{rope}}{d_{rope}}g$
\left\\{ {{d_{rope}} = \dfrac{{{m_{rope}}}}{{{V_{rope}}}}} \right\\}
${F_{mass}} = {A_{rope}}{l_{rope}}{d_{rope}}g$
${F_{buoyant}} = {V_{displaced}}{d_{water}}g$
${F_{buoyant}} = {V_{rope}}{d_{water}}g$
\left\\{ {{V_{displaced}} = {V_{rope}}} \right\\}
${F_{buoyant}} = {A_{rope}}{l_{rope}}{d_{water}}g$
$\therefore {F_{rope}} = {F_{mass}} - {F_{buoyant}} = {A_{rope}}{l_{rope}}\left( {{d_{rope}} - {d_{water}}} \right)g$
Now the net force acting on the rope will create a stress in it and if this stress exceeds the breaking point of the rope, then it will break. So the deeper the rope will go into the water, the more will be the mass of rope dipped, more will be the gravitational force, more the buoyant force, and more the stress will increase, up to a limit when it will be high enough to break the rope.
$stress = \dfrac{{{F_{rope}}}}{{{A_{rope}}}} = {l_{rope}}\left( {{d_{rope}} - {d_{water}}} \right)g$
So, the maximum length that it would go will be just before the rope breaks.
$stres{s_{breaking}} = 9.8 \times {10^8}N{m^{ - 2}}$
At breaking point,
$\therefore stress = stres{s_{breaking}}$
${l_{rope}}\left( {{d_{rope}} - {d_{water}}} \right)g = 9.8 \times {10^8}$ ${d_{rope}} = 6000kg{m^{ - 3}},{d_{water}} = 1000kg{m^{ - 3}},g = 9.8m{s^{ - 2}}$
${l_{rope}}\left( {6000 - 1000} \right)9.8 = 9.8 \times {10^8}$
${l_{rope}} = \dfrac{{{{10}^8}}}{{5000}} = 20 \times {10^3}m$
Hence, option (B) is the correct answer.

Note: Sometimes students have confusion between mass per unit area and mass per unit volume. So always remember the mass per unit volume is known as the density of the materia and mass per unit area is known as the surface density of the material.