Motion

From figure we can write

$\overrightarrow P = b\overrightarrow R = \overrightarrow S $

It is given that $\overrightarrow R = \overrightarrow Q - \overrightarrow P $. Put in above equation we get

$\overrightarrow P + b(\overrightarrow Q - \overrightarrow P ) = \overrightarrow S \Rightarrow \overrightarrow P + b\overrightarrow Q - b\overrightarrow P = \overrightarrow S $

$ \Rightarrow \overrightarrow S = \overrightarrow P (1 - \overrightarrow b ) + b\overrightarrow Q $

Density $\rho = {{mass\,(m)} \over {volume\,(V)}}$

Volume of sphere $ = {4 \over 3}\pi {R^3} \Rightarrow \rho = {m \over {{4 \over 3}\pi {R^3}}} = {{3m} \over {4\pi {R^3}}}$

Rearranging it we get

$\rho {R^2} = {3 \over {4\pi }}m$

It is given that mas m remains constant. Let the constant be k. Therefore,

$\rho {R^3} = k$

Differentiating it w.r.t. time, we get

${d \over {dt}}(\rho {R^3}) = 0$ (Since, differentiation of constant is zero)

${R^3}{{d\rho } \over {dt}} + 3{R^2}\rho {{dR} \over {dt}} = 0$

Now ${{dR} \over {dt}}$, that is, rate of change of radius is equal to velocity. Therefore,

${R^3}{{d\rho } \over {dt}} + 3{R^2}\rho v = 0 \Rightarrow 3{R^2}\rho v = - {R^3}{{d\rho } \over {dt}}$

$ \Rightarrow v = {{ - 1} \over 3}{{{R^3}} \over {{R^2}}}{1 \over \rho }{{d\rho } \over {dt}} = {{ - 1R} \over 3}{1 \over \rho }{{d\rho } \over {dt}}$

It is given that rate of fractional change in density $\left( {{1 \over \rho }{{d\rho } \over {dt}}} \right)$ is constant. Thus,

v $\propto$ R

Thus velocity of any point on surface of expanding sphere is proportional to R.

Statement 1 is True.

For a moving observer, the nearly objects appear to move in the opposite direction at a large speed. This is because the angular speed of the nearby object w.r.t to observer is large. As the object moves away, the angular velocity decreases and therefore its speed seems to be less. The distant object almost remains stationary.

Statement 2 is True but related to relative velocity

${\overline V _{21}} = {\overline V _{2G}} - {\overline V _{1G}}$

Where, G = Laboratory frame

Thus, both the statement are true but statement 2 is not the correct explanation of statement 1.

Also condition for equilibrium of the piston is

Mg = (P$_0$ $-$ P)$\pi$R$^2$

$ \Rightarrow P = {{ - Mg} \over {\pi {R^2}}} + {P_0}$

Cylinder is thermally conducting, the temperature remains the same.

$ \Rightarrow {P_0} \times (2L \times \pi {R^2}) = Py \times \pi {R^2}$

$ \Rightarrow y = {{{P_0}} \over P} \times 2L$

$ = {{{P_0}} \over {\left[ {{P_0} - {{Mg} \over {\pi {R^2}}}} \right]}} \times 2L$

$ = \left[ {{{{P_0} \times \pi {R^2}} \over {{P_0}\pi {R^2} - Mg}}} \right] \times 2L$

When a projectile is fired from the ground, it follows a parabolic path because the force of gravity acts on it, causing it to accelerate downward. At the top of the projectile's path, its vertical velocity becomes zero, while the horizontal velocity remains constant. At this point, the speed of the projectile is minimum because it is only moving horizontally and has no vertical component of velocity.

As the projectile continues to fall, its speed increases because of the acceleration due to gravity. When the projectile reaches the ground, its speed will be at a maximum, assuming there is no air resistance.

Therefore, the statement that the speed of the projectile is minimum at the top of its path is true.

Acceleration due to gravity is independent of the mass of the body.