B.

10 cm worm has done more work against gravity

The work done against gravity can be calculated from the increased in height of the centre of mass. The centre of mass of a worm folded in two is located at the middle of either half, i.e., at a point one-quarter of the worm's total length from one end.

Thus, the centre of mass of the narrow flat worm travels 5 cm up the wall; while that of the broad worm moves 7.5 cm. Hence, work done by 10 cm worm is more against gravity. Also, the ratio of the work done is 2:3.

A.

at take-off

The rocket has to reach the highest possible total energy. If the zero level of gravitational potential energy is infinitely far away, then the energy of the rocket standing on the surface on the earth is negative. The energy released during the operation of the engines increases the total energy of the rocket, and the rocket can leave the earth's gravitational field if the sum of its potential and kinetic energies becomes positive.

The energy released in the course of the operation of the principal and auxiliary engines increase the total energy of the rocket and its ejected combustion products by a fixed value; this increase is independent of the moment when the engines are scratched on. However, the speed at which the combustion products fall back to the earth does depend on the timing of the rocket's operations. Indeed, if the auxiliary engine starts working when the rocket is at a greater height, the combustion fall further and their speed and total energy are higher when they hit the ground. This means that the sooner the auxiliary engine is switched on. the higher, the energy ultimately acquired by the rocket. The same concept is valid for the principal engine, and if the only energy consideration applies, it is best to operate the engines for the shortest time and at the highest thrust.

B.

p₀A

The volume of the gas is constant i.e., V = constant

∴ p ∝ T i.e pressure will be doubled if the temperature is doubled.

Let F be the tension in the wire. Then, the equilibrium of anyone pipes gives.

$\mathrm{F} = (\mathrm{p}-{\mathrm{p}}_0) \mathrm{A} = 2({\mathrm{p}}_0 - {\mathrm{p}}_0) = {\mathrm{p}}_0\mathrm{A}$

B.

$\frac{\mathrm{g}({\mathrm{t}}_1 +{\mathrm{t}}_2)}{2}$

Let v be the initial velocity of vertical projection and t be time taken by the body to reach a height h from the ground.

Here u = u,

a = -g,

s = h,

t = t

Using, 

$$\begin{gathered} \mathrm{ut} + \frac{1}{2} (-\mathrm{g}){\mathrm{t}}^2 \\ \mathrm{or} \\ {\mathrm{gt}}^2 - 2\mathrm{ut} + 2\mathrm{h} = 0 \\ \therefore \mathrm{t} = \frac{2\mathrm{u}\pm \sqrt{4{\mathrm{u}}^2 - 4\mathrm{g} \mathrm{x} 2\mathrm{h}}}{2\mathrm{g}} = \frac{\mathrm{u}\pm \sqrt{{\mathrm{u}}^2 -2\mathrm{gh}}}{\mathrm{g}} \\ \mathrm{It} \mathrm{means} \mathrm{t} \mathrm{has} \mathrm{two} \mathrm{values}, \mathrm{i}.\mathrm{e}., \\ {\mathrm{t}}_1 = \mathrm{u} + \frac{\sqrt{{\mathrm{u}}^2 -2\mathrm{gh}}}{\mathrm{g}} \\ {\mathrm{t}}_2 = \frac{\mathrm{u}- \sqrt{{\mathrm{u}}^{2-}2\mathrm{gh}}}{\mathrm{g}} \\ {\mathrm{t}}_1 +{\mathrm{t}}_2 = \frac{2\mathrm{u}}{\mathrm{g}} \mathrm{or} \mathrm{u} = \frac{\mathrm{g} ({\mathrm{t}}_1 +{\mathrm{t}}_2)}{2} \end{gathered}$$

D.

Distance travelled by the particle in the time interval between t = 0 and t = π/2p is a

x = a cos (pt), y =b sin (pt)

$$\begin{gathered} \therefore \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2} + \frac{{\mathrm{y}}^2}{{\mathrm{b}}^2} = 1, \mathrm{i}.\mathrm{e}. \mathrm{equation} \mathrm{of} \mathrm{ellipse} \\ \mathrm{Now}, \mathrm{r} = \mathrm{x}\hat{\mathbf{i}} + \mathrm{y}\hat{\mathbf{j}} = \mathrm{a} \cos \left(\mathrm{pt}\right) \hat{\mathbf{i}} + \mathrm{b} \sin \left(\mathrm{pt}\right)\hat{\mathbf{j}} \\ \mathrm{v} = \frac{\mathrm{dr}}{\mathrm{dt}} = -\mathrm{pasin} \left(\mathrm{pt}\right) \hat{\mathbf{i}} + \mathrm{pb} \cos \left(\mathrm{pt}\right) \hat{\mathbf{j}} \\ \mathrm{a} = \frac{\mathrm{dv}}{\mathrm{dt}} = - {\mathrm{p}}^2\mathrm{a} \cos \left(\mathrm{pt}\right) \hat{\mathbf{i}} - {\mathrm{p}}^2 \mathrm{b} \sin \left(\mathrm{pt}\right)\hat{\mathbf{j}} \\ \mathrm{at} \mathrm{t} = \frac{\mathrm{\pi }}{2\mathrm{p}} \\ \mathrm{v} = - \mathrm{pa} \hat{\mathbf{i}} \mathrm{and} \mathrm{a} = - {\mathrm{p}}^2\mathrm{b} \hat{\mathbf{j}} \\ \mathrm{v} \perp \mathrm{a} \end{gathered}$$

also, a = - p²r, i.e directed towards a fixed point.