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 \(=\frac { 1 }{ 2 } { I\omega }^{ 2 }\) 
 \(=\frac { 1 }{ 2 } { I\omega }^{ 2 }=\frac { 1 }{ 2 } \times 0.625\times { \left( 100 \right) }^{ 2 }=3125J\).

On reflection from the denser medium, there will be a phase change of 180⁰
Net amplitude = \(\frac { 2 }{ 3 } \times 0.6=0.4\)
Hence, equation of reflected wave will be 
y = 0.4sin\(2\pi \left[ t+\frac { x }{ 2 } +\pi \right] \)
\(\\ =0.4sin2\pi (t+{ x }/{ 2 })\) 

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

Given, T₁= -3⁰C = - 3 + 273 = 270K
T₂ = 27⁰C = 27 + 273 = 300K
Efficiency, 
\(\eta =1-\frac { { T }_{ 1 } }{ { T }_{ 2 } } =1-\frac { 270 }{ 300 } =\frac { 1 }{ 10 } \)%
\(or\quad \frac { W }{ Q } =0.5\quad \eta =\frac { 1 }{ 20 }\)
or Q = 20 W = 20kJ per second

We have , 0.25 x 6 x 10²³ molecules, each of volume
Molecular volume = 2.5 x 10^-7 m³
Supposing, ideal gas law is valid.
Final volume =  \(\frac { { V }_{ in } }{ 100 }\) = \(\frac { { (3) }^{ 3 }\times { 10 }^{ -6 } }{ 100 }\)
\(\approx\) 2.7\(\times\)  10 ^-7 m³ 
Which is about the molecular volume. hence, intermolecular forces cannot be neglected. Therefore, the ideal gas situation does not hold.

Frequency of A, v₀ = 512 Hz
Number of beats/s = 5
Frequency of B = 512\(\pm \) 5 = 517 or 517 Hz
On loading its frequency decreases from 571 to 507, so that number of beats/s remain 5.
Hence, frequency of B when not loaded = 517Hz.

On reflection from the denser medium, there will be a phase change of 180⁰
Net amplitude = \(\frac { 2 }{ 3 } \times 0.6=0.4\)
Hence, equation of reflected wave will be 
y = 0.4sin\(2\pi \left[ t+\frac { x }{ 2 } +\pi \right] \)
\( =0.4sin2\pi (t+{ x }/{ 2 })\)

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.

(a) 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.
\(\frac { 1 }{ x+vt } \)
\(\\ For\ x=0,\ t=0,\ f(x)\rightarrow \infty \)
Though the function is of (x±vt) type still at x=0, it is infinite, hence, it is not a wave.
(b) 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 ± 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.
For x = 0,
If t →0, then (x-vt)²→0 which is finite, hence, it is a wave as it passes the two tests.