Best MCQ -The tension in a string holding a solid block

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.

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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.

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

$T_0+mg=F_{up}$

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

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

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

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

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

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