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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)

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).

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

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

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