Question

An object of mass \(10\,\text{kg}\) is released from rest in a liquid. If the object moves a distance of \(2\,\text{m}\) while sinking in a time duration of \(1\,\text{s}\), then the mass of the liquid displaced by the submerged object is \((g=10\,\text{m/s}^2)\):

• A. \(5\,\text{kg}\)
• B. \(6\,\text{kg}\)
• C. \(3\,\text{kg}\)
• D. \(4\,\text{kg}\)

Hint:

For bodies immersed in liquids: \[ \text{Net force}=\text{Weight}-\text{Buoyant force}. \] Buoyant force equals the weight of displaced liquid.

Answer 1

The Correct Option is B

Step 1: Find the acceleration using kinematics.
The object starts from rest ($u = 0$) and travels $s = 2\,\text{m}$ in $t = 1\,\text{s}$.
Using $s = ut + \frac{1}{2}at^2$:
\[ 2 = 0 + \frac{1}{2}a(1)^2 \implies a = 4\,\text{m/s}^2 \quad (\text{downward}) \]

Step 2: Identify the forces acting on the sinking object.
Two forces act vertically:
(i) Weight $W = mg = 10 \times 10 = 100\,\text{N}$ (downward)
(ii) Buoyant force $F_b = m_l g$ (upward), where $m_l$ is the mass of liquid displaced.

Step 3: Compute the net upward buoyant force from Newton's second law.
Net downward force = $ma$:
\[ mg - F_b = ma \] \[ F_b = mg - ma = m(g - a) = 10(10 - 4) = 60\,\text{N} \]

Step 4: Relate buoyant force to displaced mass.
By Archimedes' principle, buoyant force equals the weight of displaced liquid:
\[ F_b = m_l g \] \[ 60 = m_l \times 10 \]

Step 5: Solve for $m_l$.
\[ m_l = \frac{60}{10} = 6\,\text{kg} \]

Step 6: State the answer.
\[ \boxed{6\,\text{kg}} \]