Lazy fish

The Bouyant Force is $F_{bouyant}= \rho V g$ where $\rho$ is density, $V$ is volume, and $g$ is gravity.

The sum of all forces is equal to mass times gravity. $F=ma$

The fish is initally at rest so we can set the two forces equal to each other.

$ma=\rho V g$

We can find the change in Force by applying what we know changes.

$\Delta F= \rho \Delta Vg$

$\Delta F= \rho (1.1V-1V)g$

$\Delta F= \rho (0.1V)g$

Setting the two forces equal to each other we get

$ma=\rho (0.1V) g$

We know that $m= \rho V$ by definition.

Substituting this in for $m$ , we get

$\rho V a=\rho (0.1V) g$

Simplifying this equation gives

$a=0.1g$

$g$ is given to be $9.8 m/s^{2}$

$a=(0.1)(9.8)$

$a=0.98$

At rest $\displaystyle$ $V\cdot\rho_w\cdot\,g\;=m\cdot\,g$ ---- $(i)$ .

After increasing it's volume buoyant force would increase but weight remains same since we are given mass doesn't change, so fish acquires a net upward acceleration(say 'a'), we have

$\displaystyle$ $1.1V\cdot\rho_w\cdot\,g-m\cdot\,g\;=m\cdot\,a$ ---- $(ii)$

From $(i)$ & $(ii)$ $a\;=0.1g\;=\boxed{0.98\,ms^{-2}}$

If you observe closely, the change of 1.1V where V is the original volume is the only factor we have to consider in this problem. The change of 1.1V to normal volume (i.e. 1V) multiplied with the acceleration of gravity (g) will yield the final acceleration of the fish. This is due to all the other factors are kept constant

Let its volume before it inflates be $v$ cubic meters. Then the buoyancy force on the fish is $+vg$ N. (This follows from Archimedes' Principle and the fact that water has density $1 \text{ kg/m}^3.$ ) Notice that the force of gravity must then be $-vg$ N, since the fish is at rest. After it inflates, the gravitational force remains as $-vg$ N (since its mass does not change) but the buoyancy force increases to $+1.1vg$ N. Therefore the net force on the fish is $+0.1vg,$ and so it accelerates upwards at $0.1g = \boxed{0.98} \text{m/s}^2.$

We are given that the fish was stationary when it's volume was Vinitial, therefore, d x Vi x g = m x g. m is the mass of the fish, which is a constant d is the density of the surrounding solution, which is also a constant

Solving for the mass of of the fish in the above equation, we get: m = d x Vi We are given that Vfinal = 1.1Vinitial. To calculate the net force on the fish after the volume change, we plug in and solve: net F = (d)(1.1Vi)(g) - (d)(Vi)(g) net F = .1(d)(Vi)(g) Setting this equal to mass x acceleration, we get: .1(d)(Vi)(g) = (m)(a)

Since we solved above that the mass of the fish is d x Vi, we can cancel out the constants and we get that a = .1(g) or .98 m/s^2

m a = 1.1 m g - mg. So a = 0.1 *g = 0.98 m/s^2