The parallactic angle, \(\phi =0.76=\frac { 0.76\times \pi }{ 180\times 60\times 60 } \)radians
=\(\frac { 19\pi }{ 1.62\times { 10 }^{ 7 } } \)radians
The orbital diameter, say D = 1.5 x 10¹¹m
\(\therefore\) The required diameter, say d=\(\frac { D }{ \phi } \)
\(=\frac { 1.5\times { 10 }^{ 11 }\times 1.62\times { 10 }^{ 7 } }{ 19\pi } \\ =\frac { 2.43\times { 10 }^{ 18 } }{ 19\times 3.14 } m\\ =\frac { 243\times { 10 }^{ 18 } }{ 19\times 3.14 } m\\ =4.073\times { 10 }^{ 16 }m\)
Since , 1 Light year=9.5\(\times\)10¹⁵
\(=\frac { 4.073\times { 10 }^{ 16 } }{ 9.5\times { 10 }^{ 15 } } \) light year
=4.29 light year.

(a) Point Q is at a height R above the ground. Thus, the difference in height between points P and Q is 4R. Hence, the difference in gravitational potential energy of the block between these points = 4 mgR. Since the block starts from rest at P its kinetic energy at Q is equal to its change in potential energy. By the conservation of energy.

\(\therefore \quad \frac { 1 }{ 2 } { mv }^{ 2 }=4mgR\)
v² = 8gR
At Q, the only forces acting on the block are its weight mg acting downward and the force N of the track on block acting in radial direction. Since the block is moving in a circular path, the normal reaction provides the centripetal force for circular motion.
\(N=\frac { { mv }^{ 2 } }{ R } =\frac { m\times 8gR }{ R } =8\ mg\)
The loop must exert a force on the block equal to eight times the block's weight.
(b) For the block to exert a force equal to its weight against the track at the top of the loop,
\(\frac { { mv }^{ '2 } }{ R } =2mg\)
or v²  = 2gR
\(\therefore \quad mgh=\frac { 1 }{ 2 } { mv }^{ '2 }\)
\(h=\frac { { v }^{ '2 } }{ 2g } =\frac { 2gR }{ 2g } =R\)
The block must be released at a height 3R above the bottom of the loop. 

At t = 0, \(y=\frac{1}{1+x^{2}}\)
∴ \(1+x^{2}=\frac{1}{y}\)
\(x^{2}=\frac{1}{y}-1=\frac{1-y}{y} x= (\frac{1-y}{y})^{\frac{1}{2}}\)
At t = 2s, \(y=\frac{1}{[1+(x+1)^{2}]}\)
\(1+(x-1)^{2} =\frac{1}{y}\)
\((x-1)^{2} = \frac{1}{y}-1 = \frac{1-y}{y}\)
\((x-1)=(\frac{1-y}{y})^{\frac{1}{2}}\)
⇒ \(x=1+(\frac{1-y}{y})^{\frac{1}{2}}\)
Since, \(v=\frac{x_2-x_1}{t_2-t_1}\)
∴ \(v=\frac{1}{2-0}=0.5 ms^{-1}\)

Given equation is \(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)  
Comparing with standard equation,
\(y=2\cos { 2\pi } \left( 10t-0.0080x+3.5 \right) \)   
\(\\ y=a\cos { \left( \omega t-kx+\phi \right) } \)
\(a=2cm,\omega =\frac { 2\pi }{ T } =20\pi ,\ T=0.1s\)
\(k=\frac { 2\pi }{ \lambda } =0.008\times 2\pi \Rightarrow \lambda =\frac { 2\pi }{ 2\pi \times 0.008 } =1.25m\)
\(\\ \phi =2\pi \times 3.5=7\pi rad\)
When x = \(\frac { \lambda }{2 } \)
\({ \phi }_{ 3 }=\frac { 2\pi }{ \lambda } \times \lambda /2=\pi rad\).        

Let T be the temperature of the hot iron in the furnace. Heat radiated per second per unit area, E = \(\sigma \)T⁴ . When the body is placed in the open at temperature T₀ , the heat radiated/Second/unit area,
E' = \(\sigma \)(T₄-T₀4)
Clearly, E' < E. So, the optical pyrometer gives too low a value for the temperature in the open. 

Given, A = 1000 cm² , L = 0.4 cm
\(\Delta T=T_{ 1 }-T_{ 2 }=37-(-5)=42^{ 0 }C\)
\(\\ K=2.2\times 10^{ -3 }cal\ s^{ -1 }cm^{ -1 }K^{ -1 }\)
Rate of loss heat,
\(\\ H=\frac { Q }{ T } =\frac { KA(T_{ 1 }-T_{ 2 }) }{ L } =\frac { 2.2\times 10^{ -3 }\times 1000\times 42 }{ 0.4 } \)
\(\\ H=231\ cal\ s^{ -1 }\)

As the force acting on the  piston compresses the spring .This force will be balanced by the force acting upward by the fluid known as  upthrust force.
\(p=\frac { F }{ A } =\frac { kx }{ A } \)
Given, F = 12 N, A = 3m² and k = 81.75 N/m
From F= kx
\(\Rightarrow \quad x=\frac { F }{ k } =\frac { 12 }{ 81.75 } \)
Hence, compression in the spring is x = 14.6 cm

We know, y = ut + \(\frac { 1 }{ 2 } \) gt²
Now, v = \(\frac { dy }{ dt } \) = u + gt
acceleration, a = \(\frac { dv }{ dt } \) = g
Then, the force is given by
F = ma = mg
Thus, the given equation describes the motion of a particle under acceleration due to gravity and y is the position coordinate in the direction of g.

The dependence of time period T on the quantities l, g and m as a product may be written as :
\(T=k{ m }^{ x }{ l }^{ y }{ g }^{ z }\)
where k is dimensionless constant and x, y and z are the exponents.
By considering dimensions on both sides, we have
\(\left[\mathrm{L}^0 \mathrm{M}^0 \mathrm{~T}^1\right]=\left[\mathrm{L}^1\right]^x\left[\mathrm{~L}^1 \mathrm{~T}^{-2}\right]^y\left[\mathrm{M}^1\right]^z\) = L^x+y T^–2y M^z
On equating the dimensions on both sides,
we have x + y = 0; –2y = 1; and z = 0
So, that x =\(\frac { 1 }{ 2 } \), y =\(-\frac { 1 }{ 2 } \), z = 0
Then \( T=km^{ 0 }l^{ { 1 }/{ 2 } }g^{ { -1 }/{ 2 } }\) or, \(T =k\sqrt { \frac { l }{ g } } \)
Note that value of constant k can not be obtained by the method of dimensions. Here it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions.
Actually, k = 2π so tha \(T=2\pi \sqrt { \frac { l }{ g } } \) 

Given, molar volume of one mole of hydrogen
= 22.4 L = \(22.4\times { 10 }^{ -3 }{ m }^{ 3 }\) 
Diameter of hydrogen molecules (d) = 1\(\mathring{A}\) = 10^-10m
\(\therefore \) Radius of hydrogen molecule (r) = \(\frac { d }{ 2 } =\frac { { 10 }^{ -10 } }{ 2 } \) 
 \(=0.5\times { 10 }^{ -10 }m\)  
Volume of one molecule of hydrogen = \(\frac { 4 }{ 3 } \pi { r }^{ 3 }\) 
 \(=\frac { 4 }{ 3 } \times 3.14\times (0.5\times { 10 }^{ -10 })\)  
\(=5.234\times { 10 }^{ -31 }{ m }^{ 3 }\)
Number of molecules in one mole hydrogen = Avogadro's number (N) = \(6.023\times { 10 }^{ 23 }\) 
\(\therefore \) Atomic volume of one mole of hydrogen = Number of molecules in one mole of hydrogen \(\times \) Volume of one molecule of hydrogen
\(=6.023\times { 10 }^{ 23 }\times 5.234\times { 10 }^{ -31 }\)   
\(\\ =3.152\times { 10 }^{ -7 }{ m }^{ 3 }\)
\(\therefore \ \frac { Molar \ Volume }{ Atomic \ Volume } =\frac { 22.4\times { 10 }^{ -3 } }{ 3.152\times { 10 }^{ -7 } } = 7.1\times { 10 }^{ 4 }\) 
This ratio is very large, which shows that the intermolecular separation in a gas is much larger than the size of a molecule.

No. because the moment of inertia depends upon the axis of rotation and distribution of mass.

Yes, when two bodies move in opposite direction then relative velocity of each is greater than the individual velocities.

The value m = 10.23kg is more accurate, being correct upto 2nd place of decimal.

The energy of a polyatomic gas having In' degrees of freedom is given by
E = n x \(\frac { 1 }{ 2 } \) KT x N = \(\frac { n }{ 2 } \) RT
\(\therefore\)  C_v = \(\frac { dE }{ dT } =\frac { n }{ 2 } R\)
\(\therefore\) C_p = C_v + R = \(\frac { n }{ 2 } R\)+R
= \(\left( \frac { n }{ 2 } +1 \right) R\)
\(\therefore\)  \(\gamma\) = \(\frac { { c }_{ p } }{ { C }_{ v } } =\frac { n/2+1 }{ n/2 } =1+\frac { 2 }{ n } \)
In case of a triatomic gas n = 7
\(\therefore\) \(\gamma\) = 1 + \(1+\frac { 2 }{ 7 } =\frac { 9 }{ 7 } \)

The velocity of a particle changes due to either of the following three causes:
(i) Change in magnitude of velocity,
(ii) Change in direction of motion only, and
(iii) Change in magnitude as well as direction of the motion.

The frequency of vibration of string is
given by \(n=\frac{1}{2l}\sqrt{\frac{T}{m}}\)
m (mass per unit length) = \(\frac{mass}{length}\)
=\(\frac{volume \times density}{l} = \frac{\pi r^{2}l \times \rho}{l}=\pi r^{2} \rho\)
= \(\pi (\frac{D}{2})^{2} \ \ \rho=\frac{\pi D^{2}}{4} \rho\)
n =\(\frac{1}{2l}\sqrt{\frac{4T}{\pi D^{2} \rho}}=\frac{1}{Dl}\sqrt{\frac{T}{\pi \rho}}\)
∴ \(n \propto \frac{1}{D}\)      (when l,T,ρ are same)
or nD = constant or n₂D₁ = n₁D₂.
Hence the two strings may vibrate with different frequencies when they have different diameters.

Let M₁, M₂ by the masses and R₁, R₂ be, the radii of the planets.
\(\Rightarrow\frac{M_1}{M_2}=\frac{1}{10}\ and\ \frac{R_1}{R_2}=\frac{2}{5}\)
(a) Ratio of densities \(=\frac{d_1}{d_2}\ or\ \frac{d_1}{d_2}=[\frac{M_1}{\frac{4}{3}\pi R_1^3}][\frac{\frac{4}{3}\pi R_2^3}{M_2}]\)
or, \(\frac{d_1}{d_2}=\frac{M_1}{M_2}[\frac{R_2}{R_1}]^3\Rightarrow \frac{d_1}{d_2}=[\frac{1}{10}][\frac{5}{2}]^3=\frac{25}{16}\)
(b) Acceleration due to gravity at the surface \(=g=\frac{GM}{R^2}\)
\(\therefore\frac{g_1}{g_2}=\frac{M_1}{M_2}[\frac{R_2}{R_1}]^2\)
\(=\frac{1}{10}[\frac{5}{2}]^2=\frac{5}{8}\)
(c) Escape velocity \(=\sqrt {\frac{2GM}{R}}\)
\(=\frac{v_1}{v_2}=\sqrt {\frac{M_1}{M_2}}\sqrt {\frac{R_2}{R_1}}=\sqrt {\frac{1}{10}\times \frac{5}{2}}=\frac{1}{2}\)
(d) Time period of a satellite near the surface (orbit radius = R) \(=\frac{2\pi}{\sqrt {GM}}R\sqrt R\)
\(\Rightarrow \frac{T_1}{T_2}=\sqrt {\frac{M_2}{M_1}}[\frac{R_1}{R_2}][\sqrt {\frac{R_1}{R_2}}]\)
\(=\sqrt {\frac{10}{1}}[\frac{2}{5}][\sqrt {\frac{2}{5}}]=\frac{4}{5}\)

Given, h = 300m, g = 9.8m/s^2 , v = 340ms^-1
t₁ = time taken by stone to strike the water surface 
 \(t_{ 1 }=\sqrt { \frac { 2h }{ g } } =\sqrt { \frac { 300 }{ 49 } } =7.82s\left( as\quad h=0+\frac { 1 }{ 2 } gt^{ 2 }_{ 1 } \right) \)
t₂ = time taken by the splash's sound to reach top of the tower
\(t_{ 2 }=\frac { h }{ v } =\frac { 300 }{ 340 } =0.882\quad \left[ v=\frac { h }{ t_{ 2 } } \right] \)
Total time, t = time to hear splash of sound 
= t₁ + t₂ =7.82 + 0.882
= 8.702

574.6^o

One-eighth the present year
Since \(T^2 \propto r^3 \therefore\left(\frac{T}{T}\right)^2=\left(\frac{1}{4}\right)^3 \Rightarrow T^{\prime}=\frac{1}{8} T\)

Observing the figure,
 
Coordinates of ml, x₁ = 0, y₁ = 0
Coordinates ofmz' x₂ =1, y₂ =0
Coordinates of m3, x₃ =1, y₃ =1
Coordinates of m₄, x₄ =0, y₄ =1
\(\therefore \quad x_{\mathrm{CM}}=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}+m_{4} x_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}\)
\(=\frac{(1)(0)+(2)(1)+(3)(1)+(4)(0)}{1+2+3+4}\)
\(=\frac{5}{10}=0.5\)
Now, \(y_{\mathrm{CM}}=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}+m_{4} y_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}\)
\(=\frac{(1)(0)+(2)(0)+(3)(1)+(4)(1)}{1+2+3+4}\)
\(=\frac{7}{10}=0.7 \mathrm{~m}\)
Thus, centre of mass is located at (0.5m, 0.7m).

Any object can be considered as a point object if the distance travelled by it is very large in comparison to its dimensions.
(i) A railway carriage is moving without jerks between two stations, it means stations are at large distance, therefore railway carriage can be taken as a point object.
(ii) Man along with monkey is cycling smoothly which indicates that the distance travelled by the man is very large, therefore monkey can be taken as a point object.
(iii) The distance travelled by the ball is not so large therefore, spinning cricket ball cannot be taken as a point object.
(iv) Again a tumbling beaker slipped off the edge of a table cannot be considered as a point object because distance covered is not much larger.