Question

Suppose a $0.200kg$ mass stretches a spring by $6cm$ when hung, but only stretches it by $4cm$ when the whole system is underwater. The difference is due to a "buoyant force" that acts upwards on the mass.
What is the magnitude of the "buoyant force"?

• A. $0.654N$
• B. $0.981N$
• C. $1.308N$
• D. $1.962N$
• E. $3.270N$

Answer 1

The correct option is A $0.654N$

When the system is in air : $m=0.200$ kg $x=6cm=0.06$ m

Thus spring constant $k=\frac{mg}{x}=\frac{0.200\times 9.8}{0.06}=32.66$ N/m

When the system is in water : $x^{\prime }=4cm=0.04$ m

Buoyant force $\left(B\right)$ acts in the upward direction along with the spring force $\left(k{x}^{\prime }\right)$.

$\therefore$ $mg=B+k{x}^{\prime }$

OR $0.200\times 9.8=B+32.66\times 0.04$ $⟹B=0.654$ N