
A cube is shifted to a depth of 100m in a lake. The change in volume is 0.1%. The bulk modulus of the material is nearly,
(a) $10 Pa$
(b) ${{10}^{4}}$ Pa
(c) ${{10}^{7}}$ Pa
(d) \[{{10}^{9}}\] Pa
Answer
608.1k+ views
Hint: First use the formula \[\nabla P=h\rho g\] to calculate the Bulk stress and then use the formula of bulk modulus I.e. \[B=\dfrac{-\nabla P}{\dfrac{\nabla V}{V}}\] of material given below to get the answer.
Complete step-by-step answer:
We will write the given values first,
Depth of cube in a lake = h = -100m …………………………… (1)
(As depth is measured downward hence negative sign.)
Change in volume = \[\dfrac{\nabla V}{V}\] = 0.1% \[=\dfrac{0.1}{100}\]……………………….. (2)
Now what we have to find is the bulk modulus of the material therefore we should know the formula of bulk modulus as shown below,
Formula:
Bulk Modulus = B = \[\dfrac{-\nabla P}{\dfrac{\nabla V}{V}}\]………………………………… (3)
Where, \[\nabla P\]- Change in pressure (Bulk Stress)
\[\dfrac{\nabla V}{V}\]- Volume strain (change in volume)
As we have the value of change in volume therefore we should find the value of Bulk Stress to find the value of Bulk Modulus.
As we know that the bulk stress is nothing but the change in pressure while shifting the cube at a depth in a lake therefore we should know the formula of change in pressure to find the bulk stress,
Formula:
\[\nabla P=h\rho g\]
Where, h- height of the material.
\[\rho \]- Density of liquid above the material = \[{{10}^{3}}kg/{{m}^{3}}\] for water
g - Acceleration due to gravity = \[10m/{{s}^{2}}\]
If we put the values of equation (1) and given in formula we will get,
\[\therefore \nabla P=\left( -100 \right)\times {{10}^{3}}\times 10\]
\[\therefore \nabla P=-{{10}^{6}}\]……………………………………………… (4)
Now, we will put the values of equation (2) and equation (4) in equation (3) to get,
\[\therefore B=\dfrac{-\left( -{{10}^{6}} \right)}{\dfrac{0.1}{100}}\]
If we multiply by negative sign inside the bracket we will get,
\[\therefore B=\dfrac{{{10}^{6}}}{\dfrac{0.1}{100}}\]
Division can simply be written in the form of multiplication by simply inverting it therefore we will get,
\[\therefore B={{10}^{6}}\times \dfrac{100}{0.1}\]
Can be expressed in the form of exponent as shown below,
\[\therefore B={{10}^{6}}\times \dfrac{100}{{{10}^{-1}}}\]
By multiplying \[{{10}^{6}}\] by 100 we will get,
\[\therefore B=\dfrac{{{10}^{8}}}{{{10}^{-1}}}\]
If we take \[{{10}^{-1}}\] in numerator we will get,
\[\therefore B={{10}^{8}}\times 10\]
\[\therefore B={{10}^{9}}\]
Therefore the value of bulk modulus of the material is \[{{10}^{9}}\] Pa.
Therefore option (d) is correct.
Note: The bulk modulus is a measure of the ability of a substance to withstand changes in volume when under compression on all sides. It is equal to the quotient of the applied pressure divided by the relative deformation.The volumetric strain is the unit change in volume, i.e. the change in volume divided by the original volume.In the above question If you write 0.1% as only the change in volume then you will not be able to get any answer as it is the value of volume strain. Also, take the value of depth as negative as we are going downward, otherwise you will get a negative answer.
Complete step-by-step answer:
We will write the given values first,
Depth of cube in a lake = h = -100m …………………………… (1)
(As depth is measured downward hence negative sign.)
Change in volume = \[\dfrac{\nabla V}{V}\] = 0.1% \[=\dfrac{0.1}{100}\]……………………….. (2)
Now what we have to find is the bulk modulus of the material therefore we should know the formula of bulk modulus as shown below,
Formula:
Bulk Modulus = B = \[\dfrac{-\nabla P}{\dfrac{\nabla V}{V}}\]………………………………… (3)
Where, \[\nabla P\]- Change in pressure (Bulk Stress)
\[\dfrac{\nabla V}{V}\]- Volume strain (change in volume)
As we have the value of change in volume therefore we should find the value of Bulk Stress to find the value of Bulk Modulus.
As we know that the bulk stress is nothing but the change in pressure while shifting the cube at a depth in a lake therefore we should know the formula of change in pressure to find the bulk stress,
Formula:
\[\nabla P=h\rho g\]
Where, h- height of the material.
\[\rho \]- Density of liquid above the material = \[{{10}^{3}}kg/{{m}^{3}}\] for water
g - Acceleration due to gravity = \[10m/{{s}^{2}}\]
If we put the values of equation (1) and given in formula we will get,
\[\therefore \nabla P=\left( -100 \right)\times {{10}^{3}}\times 10\]
\[\therefore \nabla P=-{{10}^{6}}\]……………………………………………… (4)
Now, we will put the values of equation (2) and equation (4) in equation (3) to get,
\[\therefore B=\dfrac{-\left( -{{10}^{6}} \right)}{\dfrac{0.1}{100}}\]
If we multiply by negative sign inside the bracket we will get,
\[\therefore B=\dfrac{{{10}^{6}}}{\dfrac{0.1}{100}}\]
Division can simply be written in the form of multiplication by simply inverting it therefore we will get,
\[\therefore B={{10}^{6}}\times \dfrac{100}{0.1}\]
Can be expressed in the form of exponent as shown below,
\[\therefore B={{10}^{6}}\times \dfrac{100}{{{10}^{-1}}}\]
By multiplying \[{{10}^{6}}\] by 100 we will get,
\[\therefore B=\dfrac{{{10}^{8}}}{{{10}^{-1}}}\]
If we take \[{{10}^{-1}}\] in numerator we will get,
\[\therefore B={{10}^{8}}\times 10\]
\[\therefore B={{10}^{9}}\]
Therefore the value of bulk modulus of the material is \[{{10}^{9}}\] Pa.
Therefore option (d) is correct.
Note: The bulk modulus is a measure of the ability of a substance to withstand changes in volume when under compression on all sides. It is equal to the quotient of the applied pressure divided by the relative deformation.The volumetric strain is the unit change in volume, i.e. the change in volume divided by the original volume.In the above question If you write 0.1% as only the change in volume then you will not be able to get any answer as it is the value of volume strain. Also, take the value of depth as negative as we are going downward, otherwise you will get a negative answer.
Recently Updated Pages
Which cell organelles are present in white blood C class 11 biology CBSE

What is the molecular geometry of BrF4 A square planar class 11 chemistry CBSE

How can you explain that CCl4 has no dipole moment class 11 chemistry CBSE

Which will undergo SN2 reaction fastest among the following class 11 chemistry CBSE

The values of mass m for which the 100 kg block does class 11 physics CBSE

Why are voluntary muscles called striated muscles class 11 biology CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Explain zero factorial class 11 maths CBSE

State the laws of reflection of light

Difference Between Prokaryotic Cells and Eukaryotic Cells

Show that total energy of a freely falling body remains class 11 physics CBSE

