The tension in a string holding a solid block below the surface of a liquid (of density greater than the solid) is T0when the containing vessel (Fig. 15-18) is at rest. Showthat the tension T, when the vessel has an upward vertical acceleration a, is given by T0(1 + a/g).

When the vessel is moving with an upward vertical acceleration a, there is net forces acting on the block. So, there is change in the buoyant force and the tension in the string.
The figure which shows the forces acting on the block when the vessel is accelerating upward is


This represents the tension in the string when the vessel is accelerating upward.

Dear Student,
Please find below the solution to your problem.

Let m be the mass of block.
Initially fore the equilibrium of block.
F=T0​+mg ------------------(1)
Here, F is the upthrust on the bl;ock
when the lift is accelerated upwards, get becomes g+ a instead of g.
Hence,
F′=F(g+a​)/g--------------------(2)
From newton's second law,
F′−T−mg=ma-----------------(3)
Solving eqn (1),(2) and (3), we get
T=T0​(1+ga​)

Thanks and Regards