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 \((g=10\,\text{m/s}^2)\):
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 \((g=10\,\text{m/s}^2)\):
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For bodies immersed in liquids:
\[
\text{Net force}=\text{Weight}-\text{Buoyant force}.
\]
Buoyant force equals the weight of displaced liquid.
Updated On: Jun 24, 2026
- \(5\,\text{kg}\)
- \(6\,\text{kg}\)
- \(3\,\text{kg}\)
- \(4\,\text{kg}\)
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The Correct Option is B
Solution and Explanation
Step 1: Find the acceleration of the object.
The object starts from rest, so \[ u=0 \] Distance travelled: \[ s=2\,\text{m} \] Time taken: \[ t=1\,\text{s} \] Using the equation of motion, \[ s=ut+\frac{1}{2}at^2 \] Substituting the values, \[ 2=0+\frac{1}{2}a(1)^2 \] \[ 2=\frac{a}{2} \] \[ a=4\,\text{m/s}^2 \]
Step 2: Identify the forces acting on the object.
Weight acting downward: \[ W=mg \] Buoyant force acting upward: \[ F_b=m_lg \] where \[ m_l \] is the mass of displaced liquid.
Step 3: Apply Newton’s second law.
Net downward force: \[ mg-m_lg=ma \] Substitute the given values: \[ 10(10)-m_l(10)=10(4) \] \[ 100-10m_l=40 \]
Step 4: Solve for displaced mass.
\[ 10m_l=60 \] \[ m_l=6\,\text{kg} \]
Step 5: Final conclusion.
Hence, the mass of the displaced liquid is \[ \boxed{6\,\text{kg}} \]
The object starts from rest, so \[ u=0 \] Distance travelled: \[ s=2\,\text{m} \] Time taken: \[ t=1\,\text{s} \] Using the equation of motion, \[ s=ut+\frac{1}{2}at^2 \] Substituting the values, \[ 2=0+\frac{1}{2}a(1)^2 \] \[ 2=\frac{a}{2} \] \[ a=4\,\text{m/s}^2 \]
Step 2: Identify the forces acting on the object.
Weight acting downward: \[ W=mg \] Buoyant force acting upward: \[ F_b=m_lg \] where \[ m_l \] is the mass of displaced liquid.
Step 3: Apply Newton’s second law.
Net downward force: \[ mg-m_lg=ma \] Substitute the given values: \[ 10(10)-m_l(10)=10(4) \] \[ 100-10m_l=40 \]
Step 4: Solve for displaced mass.
\[ 10m_l=60 \] \[ m_l=6\,\text{kg} \]
Step 5: Final conclusion.
Hence, the mass of the displaced liquid is \[ \boxed{6\,\text{kg}} \]
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