Comparing the given equation with the standard equation of the trajectory of an oblique projectile
y = x tan \(\theta\) - \(\frac { { gx }^{ 2 } }{ 2{ v }^{ 2 }{ cos }^{ 2 }\theta } \)
we get tan \(\theta\) = \(\sqrt{3}\)
\(\Rightarrow\)\(\theta\) = 60⁰
Also v² cos²
\(\Rightarrow\) v² =\(\frac { 1 }{ { cos }^{ 2 }\theta } \)
or v² = \(\frac { 1 }{ { cos }^{ 2 }{ 60 }^{ 0 } } \)=4
\(\Rightarrow\) v = 2 m  s^-1.

Three important characteristics of displacement are:
(i) Displacement is a vector quantity having both magnitudes as well as direction.
(ii) Displacement of a particle between two given positions is unique and is the shortest path through which particle may go from its initial to final position.
(iii) Displacement is independent of the choice of origin to the co-ordinate system.

Mg\(\frac { 1 }{ 2 } \) =\(\frac { 1 }{ 2 } \frac { M{ I }^{ 2 } }{ 3 } .{ \omega }^{ 2 }\)
ω = \(\sqrt { \frac { 3gl }{ { I }^{ 2 } } } =\sqrt { \frac { 3g }{ l } } \)
\(\upsilon =l\omega \)
= \(\sqrt { 3gl } \)

Let L₀ be the length of the open pipe. The fundamental frequency of the pipe is given by
\(v_{0}=\frac{v}{\lambda_{f}}= \frac{v}{2L_{0}}\), v = velocity of sound in air
The second overtone of the open pipe has a frequency
\(3v_{0}=\frac{3v}{2L_{0}}Hz\)
The length of the closed pipe L_C = 2 m 
The frequency of the fundamental omitted by the closed pipe 
\(v_{c}=\frac{v}{\lambda}=\frac{v}{4L_{c}}\)
The first overtone of the closed pipe has a frequency
\(3v_{c}=\frac{3v}{4L_{c}}=\frac{3v}{4\times 2}=\frac{3v}{8}Hz\)
Now, 3v₀=3v₀
or 2L₀ = 8 or L₀ = 4 m,

Here, \(\rho\) = 0.09 kg m^-3 
At S.T.P., Pressure P = 1.01 x 10⁵ pa.
According to kinetic theory of gases.
\(p=\frac { 1 }{ 3 } \rho { C }^{ 2 }\quad or\quad C=\sqrt { \frac { 3p }{ \rho } } \)
\(=\sqrt { \frac { 3\times 1.01\times { 10 }^{ 5 } }{ 0.09 } } =1837.5\quad { ms }^{ -1 }\)
Volume occupied by one mole of hydrogen at S.T.P. = 22.4 liters = 22.4 x 10^-3 m³
\(\therefore \) Mass of hydrogen, M = volume x density
= 22.4 x 10^-3 x 0.09
= 2.016 x 10^-3kg
Average K.E/mole =\(\frac { 1 }{ 2 } \)MC2
=\(\frac { 1 }{ 2 } \) x (2.016 x 10^-3) x (1837.5)²
= 3403.4 J

Displacement-time graph for a uniformly accelerated motion has been shown in adjoining Fig. The graph is parabolic in shape.

No. The rate of cooling of a body at a temperature is defined as the fall in temperature per second at that temperature, while the rate of loss of heat is the quantity of heat lost per second from a body at a given temperature.

\(\Upsilon =\frac { Fl }{ A\triangle l } \Rightarrow \triangle l=\frac { Fl }{ { \pi r }^{ 2 }\Upsilon } \)
\(\triangle l^{ ' }=\frac { 4Fl }{ \pi (4r)^{ 2 }\Upsilon } =\frac { Fl }{ 4{ \pi }^{ 2 }\Upsilon } =\frac { \triangle l }{ 4 } \).

(i) The development on the front of genetic engineering and biotechnology will include:
(a) Production of man, animals and plants with specific characteristics.
(b) High yielding variety of plants and specific crops would be sown.
(ii) Multiple use of laser in various fields or even more developed device which will transform the world. Man would treat himself as the king of universe.
(iii) Man may travel in space with unthinkable speeds and transportation would be totally revolutionised.
(iv) Man would travel deeper into the space and may settle on other planets, befriend strange creatures from other worlds or may wage a war with them.
(v) In the field of communication, 22nd century has many surprises in store. Two persons sitting on the globe or on moon would talk on phone face to face.
(vi) Man may partially conquer diseases and slow down ageing.

Speed of the aircraft, v = 720 km/h
\(=720\times \frac { 5 }{ 18 } { m }/{ s }\quad \quad \quad \left[ \because 1km/h=\frac { 5 }{ 18 } { m }/{ s } \right] \)
= 200 m/s
Angel of banking, \(\theta \) = 15^o
Acceleration due to gravity, g = 9.8m/s²
At turn, \(\tan { \theta } \) = \(\frac { { v }^{ 2 } }{ rg } \)
or     \(r=\frac { { v }^{ 2 } }{ g\tan { \theta } } =\frac { (200)^{ 2 } }{ 9.8\times \tan { 15^{ 0 } } } =\frac { 40000 }{ 9.8\times 0.2679 } \) 
\(\\ r=15420m=15.24\times 10^{ 3 }m=15.24km\)

\(\theta ={ 45 }^{ o }\)

To fully cross the bridge, distance to covered by end far away from train is equal to (500+1000) m
\(\text { Average speed } -\frac{1500}{10} =150 \mathrm{~m} / \mathrm{s}\)

(i) \(4.6\times 0.128\) = 0.5888 = 0.59.
The result has been rounded off to have two significant digits (as in 4.6).
(ii) \(\frac { 0.9995\times 1.53 }{ 1.592 } \) = 0.96057 = 0.961
The result has been rounded off to three significant digits (as in 1.53).
(iii) 876 + 0.4382.
876 + 0.4382 = 876.4382 = 876
As, there is no decimal point in 876, therefore, result of addition has been rounded off to no decimal point.

Imagine you along with your friends are in a spaceship which is moving towards Mars. The body of the speceship is made of a specially designed matter which become more harder as its temperature increases. nuclear power plants in spaceship.
Two of them work alternatively and third is for emergency. The speed of the spaceship is very high and all of you are very happy. The energy produced in power plants are converted into electric energy which runs the motors of the spaceship.
You along with your friends reach safely on Mars, collects data, takes photographs and then returns to the Earth. In return journey is working and due to overheating. its efficiency is decreasing continuously.
You and your friends try to reduce the temperature of the plant and try to repair the fuse of the other power plants.
Finally, fuse of one other plant is repaired and start to work before the first plant crosses the danger limit of an excess of temperature. Finally, you and your friends return safely on Earth.