When a lift descending with a downward acceleration a, the apparent weight of a body of mass m is given by w' = R = m (g - a)
Given, Mass of the person, m = 50 kg
Descending acceleration, a = 9 m/s²
Acceleration due to gravity, g = 10 m/s² 
Apparent weight of  the person,
R = m ( g - a ) = 50 ( 10 - 9 ) = 50 N
\(\therefore \)   Reading of the weighing scale = \(\frac { R }{ g } \) = \(\frac { 50 }{ 10 } \) = 5 kg

The workdone by a car against gravity is zero because force of gravity is vertical and motion of car is along a straight horizontal road. As angle \(\theta\) between directions of force and displacement is 90⁰ , hence work done is zero.

All mass of a hollow cylinder at a distance R from axis of rotation. Whereas in case of a sphere, most of mass lies at a distance less than R from axis of rotation. As moment of inertia is \(\sum M_{ i }{ R }_{ i }^{ 2 }\), so sphere as a lower value of moment of inertia.

M = 20 Kg, \(\omega =100rad/s,R=0.25m\).
Moment of inertia of cylinder about its own axis
\(=\frac { 1 }{ 2 } MR^{ 2 }=\frac { 1 }{ 2 } \times 20\times { \left( 0.25 \right) }^{ 2 }\)    
Rotational KE \(=\frac { 1 }{ 2 } { I\omega }^{ 2 }\) 
 \(=\frac { 1 }{ 2 } { I\omega }^{ 2 }=\frac { 1 }{ 2 } \times 0.625\times { \left( 100 \right) }^{ 2 }=3125J\).

Here ,
 \(\mu =\frac { 2.5 }{ 20 } kg/m,\quad T=200N\)
\(\\ V=\sqrt { \frac { T }{ \mu } } =\sqrt { \frac { 200 }{ { 2.5 }/{ 20 } } } =\sqrt { \frac { 200\times 20 }{ 2.5 } } \)
\(\\ =\sqrt { \frac { 4\times 10^{ 4 } }{ 25 } } =\frac { 2\times 10^{ 2 } }{ 5 } =\frac { 20\times 10 }{ 5 } =40m/s\)
\(\\ So,\ t=\frac { l }{ v } =\frac { 20 }{ 40 } =0.50s\) 

We know that, speed,  \(v\infty \sqrt { T } \)
By formula v = \(\frac { xRT }{ p } \)
Where T is in kelvin 
\(\frac { v_{ t } }{ v_{ 0 } } =\sqrt { \frac { 273+t }{ 273+0 } } =3\)   
\(\\ \Rightarrow \frac { 273+t }{ 273 } =9\Rightarrow \ t=9\times 273-273=2184^{ 0 }C\) 

Here, m = 60kg, g = 10m/s², h = 10m
In going up and down once, number of kilocalories burnt
= (mgh + mgh/2) =\(\frac { 3 }{ 2 } \) mgh
 \(=\frac { 3 }{ 2 } \times \frac { 60\times 10\times 10 }{ 4.2\times 1000 } =\frac { 15 }{ 7 } kcal\)
Total number of kilocalories to be burnt for losing 5 kg of weight = 5\(\times \)7000 = 35000 kcal
\(\therefore \) Number of times of the person has to go up and down the stairs 
= \(\frac { 35000 }{ 15/7 } =\frac { 35\times 7 }{ 15 } \times { 10 }^{ 3 }\)  = 16.3 x 10³ times

As we know,
\(\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } =\frac { { T }_{ 2 } }{ { T }_{ 1 } } =\frac { 3 }{ 5 } \)
\(\\ \because \ 1-\frac { { Q }_{ 2 } }{ { Q }_{ 1 } } =1-\frac { { T }_{ 2 } }{ { T }_{ 1 } } \)
\(\\ \Rightarrow \frac { { Q }_{ 1 }-{ Q }_{ 2 } }{ { Q }_{ 1 } } =\frac { 500-300 }{ 500 } \)
\(\\ \Rightarrow \frac { W }{ { Q }_{ 1 } } =\frac { 2 }{ 5 } \)
\(\\ \therefore \ { Q }_{ 1 }={ 10 }^{ 3 }\times \frac { 5 }{ 2 } =2500J\)
and heat transferred by the engine to the cold reservoir in one cycle
Q₂ = Q₁ - W = 2.5 kJ- 1 kJ = 1.5 k J

The average KE will be samea conditions of temperatue and pressure are same.
v_rms \(\propto\)  \(\frac{1}{\sqrt{m}}\)
\(\because\) m_A > m_B  > m_C
\(\Rightarrow\) v_C > v_B > v_A

l = 12m, M = 2.10 kg, T = 2.06\(\times \)10⁴N, v =?
\(\mu =\frac { M }{ l } =\frac { 2.10 }{ 12 } kg/m\)
\(\\ v=\sqrt { \frac { T }{ \mu } } =\sqrt { \frac { 2.06\times { 10 }^{ 4 } }{ 2.10/12 } } =3.43\times { 10 }^{ 2 }m/s\)

Length of pipe(l) = 20 cm = 20 x 10^-2 m
\({ v }_{ funda }=\frac { v }{ 4L } =\frac { 330 }{ 4\times 20\times { 10 }^{ -2 } }\)
\( \\ { v }_{ funda }=\frac { 330\times 100 }{ 80 } =412.5\quad Hz\)
\(\\ \frac { { v }_{ given } }{ { v }_{ funda } } =\frac { 1237.5 }{ 412.5 } =3\)
Hence, 3rd harmonic mode of the pipe is resonantly excited by the source of given frequency.

We know that, speed,  \(v\infty \sqrt { T } \)
By formula v = \(\frac { xRT }{ p } \)
Where T is in kelvin 
\(\frac { v_{ t } }{ v_{ 0 } } =\sqrt { \frac { 273+t }{ 273+0 } } =3\)
\(\\ \Rightarrow \frac { 273+t }{ 273 } =9\Rightarrow \ t=9\times 273-273=2184^{ 0 }C\)

\({ v }_{ 1 }=\frac { 1 }{ { l }_{ 1 }{ D }_{ 1 } } \sqrt { \frac { { T }_{ 1 } }{ { \pi \rho }_{ 1 } } } \)
Where, D = diameter of wire
\( { v }_{ 2 }=\frac { 1 }{ { l }_{ 2 }{ D }_{ 2 } } \sqrt { \frac { { T }_{ 2 } }{ { \pi \rho }_{ 2 } } } \)  
\( { l }_{ 1 }={ l }_{ 2 },\ { \rho }_{ 2 }={ \rho }_{ 1 }\)
\( { T }_{ 2 }={ T }_{ 1 },\ { D }_{ 2 }={ 3D }_{ 1 }\)
\(\\ \Rightarrow { V }_{ 2 }=\frac { { V }_{ 1 } }{ 3 } \)
New frequency is \(\frac { 1 }{ 3 } \) rd of the original frequency.

Conceptual question based on fundamentals of characteristics of travelling wave.
The converse is not true means if the function can be represented in the form  y = f( x \(\pm \) vt ), it does not necessarily express a travelling wave. As the essential condition for a travelling wave is that the vibrating particle must have finite displacement value for all x and t.
(i) For x = 0
If t \(\rightarrow \)0, then (x - vt)²\(\rightarrow \)0 which is finite, hence, it is a wave as it passes the two tests.
(ii) log \(\left( \frac { x+vt }{ x_{ 0 } } \right)\)
l\( \\ At\ x=0\ and\ t=0,\)
\( f(x,t)=log\left( \frac { 0+0 }{ x_{ 0 } } \right) \)
= log 0 \(\rightarrow\) not defined
Hence, it is not a wave. 
(iii) \(\frac { 1 }{ x+vt } \)
\( \\ For\ x=0,\ t=0,\ f(x)\rightarrow \infty\)
Though the function is of (x\(\pm\) vt) type still at x = 0, it is infinite, hence, it is not a wave.