Question

The tension in a string holding a solid block of density ${\rho }_S$ below the surface of a liquid of density ${\rho }_l({\rho }_l>{\rho }_S)$ is $T_0$ when the beaker is at rest. Tension in the string when beaker moves up with acceleration $a$ is

• A. $T=0$
• B. $T=T_0$
• C. $T=T_0[1+\frac{a}{g}]$
• D. $T=T_0[1-\frac{a}{g}]$

Answer 1

The correct option is C $T=T_0[1+\frac{a}{g}]$

Let $m$ be the mass of block$.$

Initially fore the equilibrium of block$.$

$F=T_0+mg$ ------------------$\left(1\right)$

Here$,$ $F$ is the upthrust on the bl;ock

when the lift is accelerated upwards$,$ get becomes $g+$ $a$ instead of $g.$

Hence,

$F^{\prime }=F\left(\frac{g+a}{g}\right)$--------------------$\left(2\right)$

From newton's second law$,$

$F^{\prime }-T-mg=ma$-----------------$\left(3\right)$

Solving $e{q}^n$ $\left(1\right),\left(2\right)$ and $\left(3\right),$ we get

$T=T_0\left(1+\frac{a}{g}\right)$

Hence,

option $\left(C\right)$ is correct answer.