Question

The tension in a string holding a solid block $({\rho }_b=700{\text{kg/m}}^3)$ below the surface of a liquid $({\rho }_l=1000{\text{kg/m}}^3)$ as shown in figure is $10\text{N}$ when the system is at rest. What will be the tension in the string if the system has an upward acceleration of $5{\text{m/s}}^2$?
$[\text{Take}g=10{\text{m/s}}^2]$

• A. $10\text{N}$
• B. $20\text{N}$
• C. $15\text{N}$
• D. $5\text{N}$

Answer 1

The correct option is C $15\text{N}$

Let $m$ be the mass of block.
Initially for the equilibrium of block,
$F=T_0+mg$


Here, $F$ is the upthrust on the block.
When the lift is accelerated upwards,
$g_{eff}=g+a$
Hence, $F^{\prime }=\frac{F{g}_{eff}}{g}$


Form Newton's $2^{nd}$ law
$F^{\prime }-(T+mg)=ma$
$\Rightarrow \frac{F{g}_{eff}}{g}-(T+mg)=ma$
$\Rightarrow (T_0+mg)\frac{g_{eff}}{g}=T+mg+ma$
$\Rightarrow T_0\frac{g_{eff}}{g}+m{g}_{eff}=T+m(g+a)$
$\Rightarrow T=T_0\frac{g_{eff}}{g}$
$[\because g_{eff}=g+a]$
$\Rightarrow T=10\times \frac{(10+5)}{10}=15\text{N}$
So, $T=15\text{N}$ will be the tension in the string.