AP EAPCET 04-Jul-2022 Shift 2 Paper

Given, volume of block $A = 0.25\ \mathrm{m}^3$Let the density of iron be $\rho\ \mathrm{kg}/\mathrm{m}^3$.Thus, mass of iron block, $A = 0.25 \rho\ \mathrm{kg}$When block $A$ is attached to the spring elongation produced $x = 0.2\ \mathrm{m}$.Thus, force on spring of spring constant $k$ is $F = -k x$(-ve sign denotes force is in opposite direction of elongation)$m g = -k(0.2)$$0.25 \rho g = -0.2 k \Rightarrow k = 1.25 \cdot \rho g\ \mathrm{N/m}$ Now, the same spring is attached to another block $B$.Volume of block $B = 0.75\ \mathrm{m}^3$Mass of block $B = 0.75 \rho\ \mathrm{kg}$ FBD of the spring system on inclined surface is given asForce on block $B,\ F = -(k x)$$M g \sin \theta = -k x$$0.75 \rho g \sin 30^{\circ} = -(-1.25 \rho g) x$$\frac{0.75}{1.25} \sin 30^{\circ} = x \Rightarrow \frac{0.75}{1.25} \times \frac{1}{2} = x$$\Rightarrow x = 0.3\ \mathrm{m}$ Thus, the distance of block from the top $=$ original length of spring + elongation$= 1\ \mathrm{m} + 0.3\ \mathrm{m} = 1.3\ \mathrm{m}$