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A particle moves along a straight line with constant acceleration, changing speed from $10\text{ m/s}$ to $20\text{ m/s}$ while covering $135\text{ m}$. The time taken is:
Using $v^2 - u^2 = 2as$: $(20)^2 - (10)^2 = 2a(135) \Rightarrow 300 = 270a \Rightarrow a = \dfrac{10}{9}\text{ m/s}^2$
Using $v = u + at$: $20 = 10 + \dfrac{10}{9}t \Rightarrow t = \mathbf{9\text{ s}}$
1. Find Cube Side ($a$): The body diagonal of the cube equals the diameter of the sphere. $\sqrt{3}a = 2R \implies a = \frac{2R}{\sqrt{3}}$.
2. Find Cube Mass ($m$): Density $\rho = \frac{M}{\frac{4}{3}\pi R^3}$. Mass of cube $m = \rho \times a^3 = \left(\frac{3M}{4\pi R^3}\right) \times \left(\frac{8R^3}{3\sqrt{3}}\right) = \frac{2M}{\pi \sqrt{3}}$.
3. Moment of Inertia ($I$): For a cube about the center-face axis, $I = \frac{ma^2}{6}$.
$I = \frac{1}{6} \left( \frac{2M}{\pi \sqrt{3}} \right) \left( \frac{4R^2}{3} \right) = \frac{8MR^2}{18\pi \sqrt{3}} = \frac{4MR^2}{9\sqrt{3}\pi}$.
\(3\% \times x = 30 \Rightarrow x = \dfrac{30 \times 100}{3} = 1000\)
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