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A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ>ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)

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A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ > ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)

A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ > ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)

(A) g\left(\frac\sigma\rho-1\right)\;downwards

(B) g\left(\frac\sigma\rho-1\right)\;upwards

(C) g\left(\frac\sigma\rho\right)\;downwards

(D) g\left(\frac\sigma\rho\right)\;upwards

Solution

Let V = Volume of body

\rho = density of body

\sigma = density of lake water

Thus, m=\;\rho\;V

As the density of liquid is more than that of body( σ > ρ), Therefore the body has tendency to float on the liquid because of larger upwards force in comparison to that of weight of body. Hence, net force will be upwards.

W know that F_{up}=V\;\sigma g\;

According to Newton's 2nd law of motion, we can write

F_{net}= F_{up}-mg (upwards)

F_{net}= V\sigma g -V \rho g

ma = V\sigma g -V \rho g

V\rho a = V\sigma g -V \rho g

a=\frac{V(\sigma-\rho)g}{V\rho}

a=\frac{(\sigma-\rho)g}{\rho}

a=g\left(\frac\sigma\rho-1\right)\;upwards

(B) is correct option

A small body of density ρ is dropped from rest at a height h into a lake of density σ , where σ > ρ . What would be of the acceleration of body till it moves inside the lake? (Neglect all dissipative effects)
Mechanics

The tension in a string holding a solid block below the surface of a liquid (of density greater than that of solid) as shown in the figure is [latex]T_o[/latex] (To) when the system is at rest. What will be the tension in the string if the system has upward acceleration a.

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fluid Mechanics upthrust FBD

Q1.The tension in a string holding a solid block below the surface of a liquid (of density greater than that of solid) as shown in the figure is T_o (To) when the system is at rest. What will be the tension in the string if the system has upward acceleration a.

(A) T_o\frac ag

(B) T_o\left(1-\frac ag\right)

(C) T_o\left(1+\frac ag\right)

(D) T_o\left(\frac ag-1\right)

Top 5 Fluid Mechanics problems for IIT-JEE NEET CBSE

Solution

At Equilibrium (Rest), Free body diagram of block

Fluid Mechanics block fbd
T_0+mg=F_{up}
T_o+V\rho_Sg=\;\;V\rho_Lg\\
Vg(\;\rho_L-\;\rho_S)\;=T_o
\rho_L-\;\rho_S=\frac{T_o}{Vg}\;-----(1)

Free body diagram of block when it moves upwards.

fluid Mechanics upthrust FBD

When system moves up with acceleration 'a' then effective weight of fluid displaced also changes which is called apparent weight (of fluid displaced).

Acc. to Archimedes Principle , Upthrust = Weight (effective) of liquid/fluid displaced by object

F_{up}^I=V\;\rho_L\;g_{eff}

Since, the system is moving up with acceleration 'a' .Thus, \\g_{eff}=\;g+a-----(2)

As 'a' be the acceleration of system , Use Free body diagram of the moving block

According to Newton's 2nd Law of Motion
F_{net}=F_{up}^I-mg-T

Therefore, ma=V\;\rho_L\;g_{eff}-mg-T

m(\;a\;+g)=V\;\rho_L\;g_{eff}\;-T

V\;\rho_S\;(\;a\;+g)=V\;\rho_L\;g_{eff}\;-T

Now use (2), V\;\rho_S\;(\;g+a)=V\;\rho_L\;(g+a)\;-T

V\;\rho_S\;(\;g+a)=V\;\rho_L\;(g+a)\;-T

T=\;V(g+a)\;(\;\rho_L-\;\rho_S)

Using (1), T=\;V(g+a)\;({\textstyle\frac{T_o}{V\;g}})\;

Therefore, the tension in a string holding a solid block is

T=\;\;T_o\left(1+\frac ag\right)\\

Ans. (C) is correct option

More fluid mechanics problems will improve your understanding about the concepts.

Mechanics

A partially immersed solid block of density [latex]\rho_s[/latex] is floating in a liquid of density [latex]\rho_L[/latex] as shown in figure. If beaker container moves up with positive acceleration ‘a’ , what is correct statement about the block?

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Top 5 Fluid Mechanics problems for IIT-JEE NEET

Q.A partially immersed solid block of density \rho_s is floating in a liquid of density \rho_L as shown in figure. If beaker container moves up with positive acceleration 'a' , what is correct statement about the block ?

(A) It sinks more

(B) It sinks lesser

(C) It neither sinks more nor lesser

(D) Cannot be predicted as data is insufficient

Q2.A partially immersed solid block of <span class="wp-katex-eq" data-display="false">\rho_s</span> is floating in a liquid of density <span class="wp-katex-eq" data-display="false">\rho_L</span> as shown in figure. If beaker container moves up with positive acceleration &apos;a&apos; , what is correct statement about the block ?

Solution

Let V_{in} = Volume of block immersed in the liquid

For Equilibrium of rest, see FBD as shown aside

F_{up}=V_{in}\;\rho_Lg\;=\;mg

V_{in}\;\rho_Lg\;=\;V\;\rho_S\;g\;

\;\frac{V_{in}}V\;=\frac{\rho_S}{\rho_L}

Therefore, fraction of volume inside the liquid is-

\;\frac{V_{in}}V\;=\frac{\rho_S}{\rho_L}.....(1)

When block is moving up , the FBD is shown in figure

\;\\F_{up}^I=V_{in}^I\;\rho_L\;g_{eff}\\

F_{up}^I=V_{in}^I\;\rho_L\;(g+a).....(2)

According to Newton's 2nd Law of Motion
F_{net}=F_{up}^I-mg
ma =F_{up}^I-mg
F_{up}^I=m(g+a)

F_{up}^I=V\;\rho_S\;(g+a)\\....(3)

From (2) and (3), V_{in}^I\;\rho_L\;(g+a)=V\;\rho_S\;(g+a)

\;\frac{V_{in}^I}V\;=\frac{\rho_S}{\rho_L}

\;\frac{V_{in}}V\;=\frac{\rho_S}{\rho_L}=\;\frac{V_{in}^I}V\;

Hence, fraction of volume immersed is again equal to \frac{\rho_S}{\rho_L}. It means immersed volume remains same .

Ans. (C) is correct option

Top 5 Fluid Mechanics problems for IIT-JEE NEET

Actually, this question is all about the Archimedes Principle which focus on the concept of Upthrust/Buoyant force acting on completely or partially immersed solid block or any other solid/hollow object. It helps in calculating the loss of weight when any solid substance is partially or completely immersed into the liquid. The liquid container may be stationary or moving with constant velocity/ variable velocity.

When completely or partially immersed solid block lies in stationary beaker container :- 

When beaker container is stationary then , according to Archimedes Principle , it loses its weight to equal to the weight of liquid displaced by the object (like completely or partially immersed solid block) . Therefore,

F_{up}=V_{in}\;\rho_Lg\;=\;mg

When completely or partially immersed solid block lies in beaker container moving up with constant acceleration :- 

When beaker container is moving up with uniform acceleration, then the loss of weight of object is now equal to apparent weight of liquid displaced by it.Thus, buoyant force/ upthrust changes so as to balance the increased effective weight of solid block when it observed from frame of reference of accelerated beaker. 

Top 5 Fluid Mechanics problems for IIT-JEE NEET

F_{up}^I=F_{pseu}+W

F_{up}^I=ma\;+\;mg

F_{up}^I=m(g\;+\;a)

F_{up}^I=mg_{eff} where g_{eff}=g+a

Therefore, we can write

\;\\F_{up}^I=V_{in}\;\rho_L\;g_{eff}\\

F_{up}^I=V_{in}\;\rho_L\;(g+a)

Conclusion-

From the above discussion, we understand that the volume immersed does not change when the container accelerates uniformly upwards or downwards along vertical. Immersed part remains same. Block neither sinks more nor lesser. This problems is best combine of Newton’s laws of motion, Archimedes’s Principle and Law of Flotation. This is how we reach to the conclusion that the knowledge of Free Body Diagram and Pseudo force  is necessary to get into problems solving process. In fact the basic idea of pseudo force can be summarized into one line that a problem which is solved by application Newton’s 2nd law (from ground frame) gets converted into problem of Equilibrium of rest (from non-inertial frame of reference)  by introducing concept of pseudo force. 
Pseudo force can be written as:

{\overrightarrow{F\;\;}}_{pseu}=-M_{object}\;\;{\overrightarrow{a\;\;}}_{frame}
The negative sign shows that the peudo force is taken opposite to direction of acceleration of non-inertial frame.
Law of flotation simply gives us the condition under which an object floats partially or completely immersed inside the liquid. It suggest that
(i) if the density of solid is lesser than  density of liquid then it floats partially immersed in the liquid
(ii) if the density of solid is equal to the density of liquid then it floats completely immersed in the liquid
(iii) if the density of solid is more than the density of liquid then it sinks completely unitl it settles down to bottom of container.
It is very interesting to note that this law also helps to find out fraction of immersed volume of object inside the liquid. We have to understand the relationships between different concepts of mechanics to solve such type of fluid mechanics problems.

Education: NEET Medical

NEET (UG)

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NEET, which stands for the National Eligibility cum Entrance Test, is a national-level entrance examination in India for admission to undergraduate medical courses (MBBS/BDS) in government or private medical colleges across the country. NEET is the sole entrance test for admission to medical and dental programs in India.

Key features of the NEET exam include:

  1. Eligibility:
    • Candidates must have completed 10+2 or its equivalent with Physics, Chemistry, Biology/Biotechnology, and English as mandatory subjects.
    • The minimum age limit for candidates is generally 17 years, and there is an upper age limit (with some relaxations for certain categories).
  2. Subjects and Exam Pattern:
    • NEET primarily assesses the candidates’ knowledge in Physics, Chemistry, and Biology (Botany and Zoology).
    • The exam consists of multiple-choice questions (MCQs), and each correct answer is awarded four marks, while one mark is deducted for each incorrect answer.
  3. Mode of Examination:
    • NEET is conducted in a pen-and-paper-based (offline) mode. However, there have been discussions about transitioning to a computer-based mode in the future.
  4. Syllabus:
    • The syllabus for NEET is based on the topics covered in the 11th and 12th standards, following the NCERT curriculum. It includes subjects like Physics, Chemistry, and Biology.
  5. National Testing Agency (NTA):
    • NEET is conducted by the National Testing Agency (NTA), which is responsible for the conduct of various entrance examinations in India.
  6. Admission Process:
    • NEET scores are used by medical colleges across India to admit students to MBBS and BDS programs.
    • Counseling for NEET is conducted by the Directorate General of Health Services (DGHS) and the Medical Counseling Committee (MCC), which oversee the allocation of seats in government and private medical colleges.
  7. Attempts:
    • Candidates are allowed a maximum of three attempts to appear for the NEET examination, subject to the upper age limit.

NEET replaced several individual state and university-level medical entrance exams to bring uniformity in the admission process for medical courses. It is considered one of the most challenging and competitive medical entrance exams in the country. NEET plays a crucial role in determining the eligibility of students to pursue undergraduate medical education, contributing to the standardization and transparency of the medical admissions process in India.