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

Solution

At Equilibrium (Rest), Free body diagram of block

$T_0+mg=F_{up}$

$T_o+V\rho_Sg=\;\;V\rho_Lg\\$

$Vg(\;\rho_L-\;\rho_S)\;=T_o$

$\rho_L-\;\rho_S=\frac{T_o}{Vg}\;-----(1)$

Free body diagram of block when it moves upwards.

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.

List of the “top 20 government medical colleges in India for MBBS.” We are just providing you with a list of some renowned government medical colleges in India that have historically been considered among the top. It must be noted that rankings can change, and it’s advisable to check the most recent rankings for accurate information. Here is a list (in no particular order) of some prestigious government medical colleges in India:

All India Institute of Medical Sciences (AIIMS), New Delhi
Armed Forces Medical College (AFMC), Pune
Maulana Azad Medical College (MAMC), New Delhi
JIPMER, Puducherry
Grant Medical College, Mumbai
King George’s Medical University (KGMU), Lucknow
Lady Hardinge Medical College (LHMC), New Delhi
Bangalore Medical College and Research Institute (BMCRI), Bangalore
Institute of Medical Sciences, Banaras Hindu University (IMS-BHU), Varanasi
Osmania Medical College, Hyderabad
Madras Medical College, Chennai
Government Medical College and Hospital (GMCH), Chandigarh
Jawaharlal Nehru Medical College (JNMC), Aligarh
Gandhi Medical College, Hyderabad
Government Medical College, Thiruvananthapuram
Madurai Medical College, Madurai
Byramjee Jeejeebhoy Government Medical College (BJMC), Pune
Stanley Medical College, Chennai
Government Medical College, Kozhikode
Sawai Man Singh Medical College (SMSMC), Jaipur

This list is not exhaustive, and there are many other excellent government medical colleges across India. Rankings may vary, and it’s essential to check the latest rankings and admission criteria before making any decisions. Additionally, new colleges may have gained recognition, and existing ones might have seen changes in their status since my last update.

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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.

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

Solution

At Equilibrium (Rest), Free body diagram of block

$T_0+mg=F_{up}$

$T_o+V\rho_Sg=\;\;V\rho_Lg\\$

$Vg(\;\rho_L-\;\rho_S)\;=T_o$

$\rho_L-\;\rho_S=\frac{T_o}{Vg}\;-----(1)$

Free body diagram of block when it moves upwards.

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.

Top 20 Government Medical Colleges in India for M.B.B.S.

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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)

Solution

At Equilibrium (Rest), Free body diagram of block

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.

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.

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

$T_0+mg=F_{up}$

$T_o+V\rho_Sg=\;\;V\rho_Lg\\$

$Vg(\;\rho_L-\;\rho_S)\;=T_o$

$\rho_L-\;\rho_S=\frac{T_o}{Vg}\;-----(1)$

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

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

$F_{net}=F_{up}^I-mg-T$

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

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