Reasoning: $\mathbf{Work = Force \times Displacement}$
If force becomes $\mathbf{5\ }\mathbf{times}$ and displacement also becomes $\mathbf{5\ t}\mathbf{imes}$, then, $\mathbf{\Longrightarrow}$ $\mathbf{Work = 5 \times 5 = 25\ times}$ So, unit of work increases by $\mathbf{25\ times}$.

Reasoning: Centripetal force acts towards the center of circular motion, while displacement is along the tangent. Since force and displacement are perpendicular. $\mathbf{\Longrightarrow}$ $\mathbf{Work\ done = 0}$
So, centripetal force does no work.

Reasoning: Gravitational force, elastic force, and electric force are generally conservative forces. Frictional force is a non-conservative force because it wastes mechanical energy as heat. So, the odd one is frictional force.

Reasoning: Work done on a body is equal to the change in its energy: $\mathbf{W = \Delta E}$
Given: $\mathbf{W = 500\ J}$ $\mathbf{\Longrightarrow}$ Therefore, $\mathbf{\Delta E = 500\ J}$ So, the body’s energy increases by $\mathbf{500\ J}$.

Reasoning: Kinetic energy depends on the square of velocity:
$\mathbf{K}\mathbf{\ }\mathbf{.E\ =}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{mv}^{\mathbf{2}}$
Mass is same. If object $\mathbf{A}$ has twice the speed of object $\mathbf{B}$:
${\mathbf{K}\mathbf{\ }\mathbf{.E}}_{\mathbf{\ }\mathbf{A}}\mathbf{=}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{m}\mathbf{(2v)}^{\mathbf{2}}$ $\mathbf{\Longrightarrow}$ ${\mathbf{K}\mathbf{\ }\mathbf{.E}}_{\mathbf{\ }\mathbf{A}}\mathbf{= 4 \times}$ $\frac{\mathbf{1}}{\mathbf{2}}\mathbf{mv}^{\mathbf{2}}$ $\mathbf{\Longrightarrow}$ ${\mathbf{K}\mathbf{\ }\mathbf{.E}}_{\mathbf{\ }\mathbf{A}}\mathbf{=}\mathbf{4\ }{\mathbf{K}\mathbf{\ }\mathbf{.E}}_{\mathbf{\ }\mathbf{B}}$

Reasoning: Kinetic energy is proportional to the square of velocity:
$\mathbf{K}\mathbf{\ }\mathbf{.E \propto}$ $\mathbf{v}^{\mathbf{2}}$
If velocity becomes half: $\mathbf{New\ }\mathbf{\ }\mathbf{K}\mathbf{\ }\mathbf{.E}\mathbf{\ }\mathbf{=}\mathbf{\ }$ $\left( \frac{\mathbf{1}}{\mathbf{2}} \right)^{\mathbf{2}}\mathbf{=}\mathbf{\ }\frac{\mathbf{1}}{\mathbf{4}}$ So, kinetic energy becomes one fourth of the initial value.

Reasoning: Kinetic energy depends on velocity squared:
$\mathbf{K}\mathbf{\ }\mathbf{.E \propto}$ $\mathbf{v}^{\mathbf{2}}$
If speed doubles: $\mathbf{New\ K}\mathbf{\ }\mathbf{.E =}\mathbf{(2)}^{\mathbf{2}}\mathbf{= 4\ times}$ So, kinetic energy becomes quadruple.

Reasoning: $\mathbf{Efficiency =}$ $\frac{\mathbf{Useful\ output\ }}{\mathbf{Input}}\mathbf{\times 100}$
$\mathbf{Efficiency =}$ $\frac{\mathbf{150}}{\mathbf{600}}\mathbf{\times 100}$ $\mathbf{\Longrightarrow}$ $\mathbf{Efficiency = 25\%}$
So, efficiency of the engine is $\mathbf{25\%}$.

Reasoning: Efficiency means how much input energy is converted into useful output energy. Both engines give the same power output, but Engine $\mathbf{A}$ is more efficient. So, Engine $\mathbf{A}$ needs less input energy or fuel to produce the same power.

Reasoning: $\mathbf{Useful\ work = Efficiency \times Input\ energy}$
For Machine A: $\mathbf{40\%\ }\mathbf{\ }\mathbf{of}\mathbf{\ }\mathbf{\ 500\ J =}$ $\frac{\mathbf{40}}{\mathbf{100}}\mathbf{\times 500 = 200\ J}$ For Machine B: $\mathbf{60\%\ }\mathbf{\ }\mathbf{of}\mathbf{\ }\mathbf{\ 500\ J =}$ $\frac{\mathbf{60}}{\mathbf{100}}\mathbf{\times 500\ J = 300\ J}$ Difference: $\mathbf{ }\mathbf{300\ J - 200\ J = 100\ J}$ So, Machine $\mathbf{B}$ performs $\mathbf{100\ J}$ more useful work.

Reasoning: $\mathbf{Electrical\ energy = 30\%\ of\ wind\ kinetic\ energy}$
$\mathbf{Electrical\ energy =}$ $\frac{\mathbf{30}}{\mathbf{100}}\mathbf{\times 900}$
$\mathbf{Electrical\ energy = 270\ J}$
So, the turbine produces $\mathbf{270\ J\ }$of electrical energy.