Question

As shown in the figure, an iron block \(A\) of volume \(0.25\;m^3\) is attached to a spring \(S\) of unstretched length \(1.0\;m\) and hanging to the ceiling of a roof. The spring gets stretched by \(0.2\;m\). This block is removed and another block \(B\) of iron of volume \(0.75\;m^3\) is now attached to the same spring and kept on a frictionless inclined plane of \(30^\circ\) inclination. The distance of the block from the top along the incline at equilibrium is

• A. \(1.1\;m\)
• B. \(1.3\;m\)
• C. \(1.6\;m\)
• D. \(1.9\;m\)

Hint:

For a block attached to a spring on a smooth inclined plane, balance the spring force with the component of weight along the plane: \[ kx=mg\sin\theta \]

Answer 1

The Correct Option is B

Step 1: Set up the first hanging case.
Block $A$ has volume $V_A = 0.25\ m^3$ and density $\rho$, so $m_A = 0.25\rho$. Hanging vertically, the full weight stretches the spring by $x_A = 0.2\ m$. By Hooke's law at equilibrium, \[ k x_A = m_A g \] \[ k(0.2) = 0.25\rho g \quad (\text{i}) \]
Step 2: Solve for the spring constant grouping.
From (i), \[ k = \frac{0.25\rho g}{0.2} = 1.25\rho g \]
Step 3: Set up the second case on the incline.
Block $B$ has volume $V_B = 0.75\ m^3$, so $m_B = 0.75\rho$. On a frictionless $30^\circ$ incline, only the component of weight along the incline stretches the spring: \[ k x_B = m_B g\sin 30^\circ \]
Step 4: Substitute the known values.
With $\sin 30^\circ = \tfrac{1}{2}$, \[ (1.25\rho g)x_B = 0.75\rho g \times \tfrac{1}{2} \] The factor $\rho g$ cancels: \[ 1.25\,x_B = 0.375 \]
Step 5: Find the spring extension on the incline.
\[ x_B = \frac{0.375}{1.25} = 0.3\ m \]
Step 6: Add the natural length to locate the block.
The unstretched spring length is $1.0\ m$, so the block sits at \[ L = 1.0 + 0.3 = 1.3\ m \] from the top along the incline, matching option (2). \[ \boxed{1.3\ m} \]