A.

${\mathrm{\mu }}_{\mathrm{s}} > {\mathrm{\mu }}_{\mathrm{k}} > {\mathrm{\mu }}_{\mathrm{r}}$

Static friction is a force that keeps an object at rest. It must be overcome to start moving the object. When external force exceeds the maximum limit of static friction body begins to move.

 

Once the body is in motion it is subjected to sliding or kinetic friction which opposes relative motion between the two surfaces in contact. At every instant, there is just one point of contact

between the body and the plane and this point has no motion relative to the plane. In this ideal situation, kinetic or static friction is zero and the body should continue to roll with constant velocity. 

In practice, this will not happen and some resistance to motion (rolling friction) does occur, i.e. to keep the body rolling. Static friction is always greater than kinetic friction and rolling friction is less than kinetic friction.

$\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathbf{\mu }}_{\mathbf{s}}\mathbf{ }\mathbf{>}\mathbf{ }\mathbf{ }{\mathbf{\mu }}_{\mathbf{k}}\mathbf{ }\mathbf{>}\mathbf{ }{\mathbf{\mu }}_{\mathbf{r}}$

A.

$\frac{4 {\mathrm{w}}_1{\mathrm{w}}_2}{{\mathrm{w}}_1 + {\mathrm{w}}_2}$

Equation of motion for first weight

This is a frictionless and inextensible pulley.

for first weight

T - m₁g = m₁ ( a - g)

⇒ T - 2m₁ g = m₁a

For second weight 

m₂g -T = m₂ ( a - g)

2 m₂ g - T = m₂ a

on solving equations (1) & (2)

$$\begin{gathered} \mathrm{T} =\frac{4 {\mathrm{m}}_1 {\mathrm{m}}_2}{\left({\mathrm{m}}_1 + {\mathrm{m}}_2\right)} \\ \mathrm{T} = \frac{4 {\mathrm{m}}_1 {\mathrm{m}}_2}{\left({\mathrm{w}}_1 + {\mathrm{w}}_2\right)} \end{gathered}$$

C.

2:1

Force on charged particle is electric field F = q E

$$\begin{gathered} \mathrm{Acceleration} \mathrm{a} = \frac{\mathrm{F}}{ \mathrm{m}} \\ =\frac{\mathrm{q} \mathrm{E}}{\mathrm{m}} \\ \mathrm{Velocity} \mathrm{v}=\mathrm{u} + \mathrm{a} \mathrm{t} \\ = 0+ \frac{\mathrm{q} \mathrm{E}}{\mathrm{m}} \mathrm{t} \\ \mathrm{u} - \mathrm{initial} \mathrm{velocity} , \mathrm{v} -\mathrm{final} \mathrm{velocity} \\ \mathrm{So} \mathrm{kinetic} \mathrm{energy} \mathrm{K} = \frac{1}{2}{\mathrm{mv}}^2 \\ \mathrm{K} =\frac{1}{2}\mathrm{m} {\left(\frac{\mathrm{q} \mathrm{E}}{\mathrm{m}} \mathrm{t}\right)}^2 \\ \mathrm{Ratio} \mathrm{of} \mathrm{kinetic} \mathrm{energies} \\ =\frac{{\displaystyle \frac{1}{2}{\mathrm{m}}_{1 }{\left(\frac{\mathrm{q} \mathrm{E}}{{\mathrm{m}}_1} \mathrm{t}\right)}^2 }}{{\displaystyle \frac{1}{2}{\mathrm{m}}_2 {\left(\frac{\mathrm{q} \mathrm{E}}{{\mathrm{m}}_2} \mathrm{t}\right)}^2}} \\ = \frac{{\mathrm{m}}_2}{{\mathrm{m}}_1} \\ =\frac{2 \mathrm{m}}{\mathrm{m}} \\ =\frac{2}{1} \\ \mathrm{Ratio} \mathrm{of} \mathrm{kinetic} \mathrm{energies} \\ =\mathbf{ }\mathbf{2}\mathbf{:}\mathbf{1} \end{gathered}$$

B.

FAT²

Let energy E = F^x A^y T³

Writing the dimensions on both sides

$$\begin{gathered} \left[\mathrm{M} {\mathrm{L}}^2 {\mathrm{T}}^{ -2}\right] ={\left[\mathrm{M} \hspace{0.17em}\mathrm{L}\hspace{0.17em}{\mathrm{T}}^{ -2}\right]}^{ \mathrm{x}} {\left[\mathrm{L} {\mathrm{T}}^{ -2}\right]}^{ \mathrm{y}} \left[\mathrm{T}\right]{ }^2 \\ \left[\mathrm{M} {\mathrm{L}}^2 {\mathrm{T}}^{ -2}\right] =\left[{\mathrm{M}}^{\mathrm{x}} {\mathrm{L}}^{\mathrm{x}+\mathrm{y}} {\mathrm{T}}^{ -2\mathrm{x}-2\mathrm{y}+\mathrm{z}}\right] \\ \mathrm{On} \mathrm{comparing} \mathrm{both} \mathrm{sides} \\ \mathrm{x} =1 \\ \mathrm{x} + \mathrm{y}=1 \\ \Rightarrow \mathrm{y} = 2-\mathrm{x} = 2 = 2-1 =1 \\ \mathrm{and} -2\mathrm{x} - 2\mathrm{y} +\mathrm{z} = -2 \\ \Rightarrow \mathrm{z} = -2\times 1-2\times 1+\mathrm{z} \\ \therefore \mathrm{Dimensional} \mathrm{formula} \mathrm{of} \mathrm{energy} \\ =\mathbf{F}\mathbf{ }\mathbf{A}\mathbf{ }{\mathbf{T}}^{\mathbf{2}} \end{gathered}$$

A.

$\frac{\mathrm{a} \mathrm{b} \mathrm{t}}{\left(\mathrm{a} + \mathrm{b} \right)}$

Let car accelerates for time t₁ and decelerates for time t₂ then

t₁ + t₂ = t

From  ν = u + at

v = u + at₁

at₁ = bt₂

$$\begin{gathered} \Rightarrow {\mathrm{t}}_2 =\frac{{\mathrm{at}}_1}{\mathrm{b}} \\ \therefore {\mathrm{t}}_1 + \frac{{\mathrm{at}}_1}{\mathrm{b}} =1 \\ \Rightarrow {\mathrm{t}}_1\left(1 +\frac{\mathrm{a}}{\mathrm{b}}\right) =\mathrm{t} \\ \Rightarrow {\mathrm{t}}_1 = \frac{\mathrm{bt}}{\mathrm{a}+\mathrm{b}} \\ \therefore {\mathrm{t}}_1 =\frac{\mathrm{bt}}{\mathrm{a} + \mathrm{b}} \\ \therefore \mathrm{Maximum} \mathrm{velocity} \mathrm{of} \mathrm{car} \\ \mathrm{v} = {\mathrm{at}}_1 =\frac{\mathrm{a} \mathrm{b} \mathrm{t}}{\mathrm{a}+ \mathrm{b}} \end{gathered}$$