Explain why a long pole is more beneficial to the tight rope walker if the pole has slight bending.

A star mass $M$ (equal to the solar mass) with a planet (much smaller than the star) revolves around the star in a circular orbit. The velocity of the star with respect to the centre of mass of star-planet system is shown below:

The radius of the planet's orbit is closest to ($1 \mathrm{AU}=\mathrm{Distance} \mathrm{Earth} \mathrm{and} \mathrm{Sun}$)

A very long (length $L$) cylindrical galaxy is made of uniformly distributed mass and has radius $R (R<<L)$. A star outside the galaxy is orbiting the galaxy in a plane perpendicular to the galaxy and passing through its centre. If the time period of the star is $T$ and its distance from the galaxy's axis is $r$, then

Six objects are placed at the vertices of $\mathrm{a}$ regular hexagon. The geometric centre of the hexagon is at the origin with objects $1$ and $4$ on the $X$-axis (see figure). The mass of the $\mathrm{k}$ th object is ${\mathrm{m}}_{\mathrm{k}}={\mathrm{k}}^{\mathrm{i}}\mathrm{M}\left|{\mathrm{cos\theta }}_{\mathrm{k}}\right|,$ where $\mathrm{i}$ is an integer, $\mathrm{M}$ is $\mathrm{a}$ constant with dimension of mass and ${\mathrm{\theta }}_{\mathrm{k}}$ is the angular position of the $\mathrm{k}$ th vertex measured from the positive $\mathrm{x}$ -axis in the counter-clockwise sense. If the net gravitational force on $\mathrm{a}$ body at the centroid vanishes, the value of $\mathrm{i}$ is

A uniform metre scale is in equilibrium as shown in the diagram :

Calculate the weight of the metre scale.
Which of the following options is correct to keep the ruler in equilibrium when $40 \mathrm{gf}$ weight is shifted to $0 \mathrm{cm}$ mark?
$F$ is shifted towards $0 \mathrm{cm}$ or $F$ is shifted towards $100 \mathrm{cm}$.