A light hollow cube of side length 10 cm and mass 10 g , is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y \pi \times 10^{-2} \mathrm{~s}$, where the value of $y$ is (Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$, density of water $=10^3 \mathrm{~kg} / \mathrm{m}^3$ )

<p>We can determine the time period of the oscillations by considering that when the cube is depressed by a small displacement, the additional buoyant force provided by the displaced water acts as a restoring force. Here’s a step‐by‐step explanation:</p> <p>Define the given values:</p> <p><p>Side length of the cube, $L = 10\text{ cm} = 0.1\text{ m}$.</p></p> <p><p>Mass of the cube, $m = 10\text{ g} = 0.01\text{ kg}$.</p></p> <p><p>Density of water, $\rho = 10^3\ \text{kg/m}^3$.</p></p> <p><p>Acceleration due to gravity, $g = 10\ \text{m/s}^2$.</p></p> <p><p>Cross-sectional area of the cube (face area), </p> <p>$A = L^2 = (0.1)^2 = 0.01\text{ m}^2.$</p></p> <p><p>When the cube is depressed by a small distance $x$, the additional volume of water displaced is </p> <p>$\Delta V = A\,x.$</p> <p>Thus, the additional buoyant force is:</p> <p>$F_b = \rho\,g\,\Delta V = \rho\,g\,A\,x.$</p> <p>This force acts in the upward (restoring) direction. Notice that the force is proportional to the displacement $x$, which is the hallmark of simple harmonic motion (SHM).</p></p> <p><p>The effective spring constant, $k$, associated with this SHM is:</p> <p>$k = \rho\,g\,A.$</p></p> <p><p>The period of oscillation for SHM is given by:</p> <p>$T = 2\pi \sqrt{\frac{m}{k}},$</p> <p>which, upon substituting for $k$, becomes:</p> <p>$T = 2\pi \sqrt{\frac{m}{\rho\,g\,A}}.$</p></p> <p><p>Substitute the given values into the expression:</p> <p>$T = 2\pi \sqrt{\frac{0.01}{1000 \times 10 \times 0.01}}.$</p> <p>Calculate the denominator:</p> <p>$1000 \times 10 \times 0.01 = 100,$</p> <p>so,</p> <p>$T = 2\pi \sqrt{\frac{0.01}{100}} = 2\pi \sqrt{0.0001}.$</p> <p>Since,</p> <p>$\sqrt{0.0001} = 0.01,$</p> <p>we have:</p> <p>$T = 2\pi \times 0.01 = 0.02\pi\ \text{s}.$</p></p> <p><p>The problem states that the time period is given by:</p> <p>$y \pi \times 10^{-2}\ \text{s}.$</p> <p>Writing our result in the same form:</p> <p>$0.02\pi\ \text{s} = 2\pi \times 10^{-2}\ \text{s}.$</p> <p>We see that $y = 2.$</p></p> <p>Therefore, the correct answer is:</p> <p>Option A: 2.</p>