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 figure is `T_(0)` when the system is at rest. What will be the tension in the string if the system has an upward acceleration a?
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Options

A. `T_(0)(1+(a)/(g))`
B. `T_(0)(1-(a)/(g))`
C. `T_(0)(1+(g)/(a))`
D. `T_(0)(1-(g)/(a))`

Answer

A

Explanation

Let `m` be the mass of block.
Initially for the equilibrium of block,
`F=T_(0)+mg` ...(i)
Here,` F` is the upthurst on the block.

When the lift is accelerated upward, `g_(eff)` becomes `g+a` instead of g. hence,
`F'=F((g+a)/(g))` ...(ii)
From Newton's second law,
`F'-T-mg=ma` ...(iii)
Solving Eqs. (i),(ii) and (iii), we get
`T=T_(0)(1+(a)/(g))`.