(i) \(u=0,\quad g=-9.8\quad m/s^{ 2 },\quad s=-4.9m\) 
\(s=ut+\frac { 1 }{ 2 } { gt }^{ 2 }\) 
\(\therefore -4.9=0-\frac { 1 }{ 2 } \times 9.8\times { t }^{ 2 }\quad or\quad t=\sqrt { \frac { 2\times 4.9 }{ 9.8 } } =1\ s.\) 
(ii) \(u=0,\quad s=-4.9m,\quad g=-9.8\quad m/s^{ 2 },\) 
\({ v }^{ 2 }-{ u }^{ 2 }=2gs\) 
\(\therefore \ { v }^{ 2 }-0=2\times (-9.8)\times (-4.9)\quad or\quad v=\sqrt { 2\times 9.8\times 4.9 } =9.8\ m/s.\) 
(iii) \(u=0,\quad g=-9.8\quad m/s^{ 2 },\quad s--7.9m\)
\(\because { v }^{ 2 }-{ u }^{ 2 }=2gs\) 
\(\\ \therefore { \ v }^{ 2 }-0=2\times (-9.8)\times (-7.9)\) 
\(\\ v=\sqrt { 2\times 9.8\times 7.9 } m/s=10.46\quad m/s\)
(iv) Acceleration after 1 s and after 2 s will be 9.8 m/s² , because acceleration due to gravity remains almost same for small heights.

The practice of naming is one of the greatest inventions of man. Millions of animals and plants are named differently in different languages in different parts of the world. It means the names used for one organism differs from place to place. For example, Argemone a common weed has as many as ten to twele Hindi names alone. Some call it 'Firangi Dhatura', Sial Kanta', 'Pila Dhatura', 'Bharband', 'Kandhari', 'Satya nasi buti', 'Katela', 'Jahriber', 'Ugar Kanta' etc. Same thing is applicable to animals and other plants. Sometimes the common names are misleading. Same name may be used for different organisms in different parts.
To avoid this confusion Linnaeus gave a system in which an organism is given a name with two components-the first part is the generic name and the second part is the specific name. Thus the scientific name of common cat is Felis domestica, and that of tiger is Felis tigris. This system of giving a name with two components to an organism is called 'Binomial system of nomenclature'. The scientific names thus given to the organism are recognized all over the world. Hence, confusion is avoided on this aspect.

(a) 359K-273=\(86°C\)
(b) Boiling point of water = \(100°C\)+273=373K

(i) Ship displaces more water than needle as volume of ship is more than that of needle. Since, upthrust depend on volume of object (U =Vdg), so more the volume of object, more upthrust act on it and object floats.
(ii) Since, pressure act on the body is inversely proportional to the surface area of contact, i.e. \(p\propto \cfrac { 1 }{ A } \)
It means that more the area of contact, less pressure will act on the body. As the broad and the thick handle of our suitcase has large area, due to which less pressure acts on our hand and it is very easy to take from one place to another.

() Difference between thrust and pressure are

The pressure exerted by 1 N of force, acting perpendicular on the s~rface of 1m 2 area is called 1pascal.
1 Pa= 1Nm^-2
(1)
The force required to accelerate 1 kilogram of mass at the rate of 1 metre per second square is called 1newton.
1N = 1 kg ms^-2
Since, pressure is inversely proportional ro the area 0 contact, Le. \(p\propto \cfrac { 1 }{ A } \)
Therefore, pressure will reduce to half, if area of contact is doubled.

(a) Take a tiny crystal of dry blue ink.Ink is a matter.Dissolve it in 100mL water.The colour of water becomes blue.Dilute 1mL of this blue water with 99mL clear water.Water again becomes blue though intensity of colour decreases.Repeat 2-3 times.The solution obtained remains coloured through intensity of colour decreases gradually.This shows that one tiny crystal of link contains millions of ink particles which are very very small
(b) Add 2g salt in a cylinder containing water marked at 100mL.Salt dissolves and disappears but the volume of water remains at 100 mL mark.This means salt particles get into spaces between water particles.Water is a matter and particles of water have spaces between them.
(c) Put a drop of perfume in one corner of the room.Stand at the other end.Stil you will get smell of perfume.This is because of matter, i.e., perfume are moving with air
(d) In the above case particles of perfume and that of air intermix with each other.This intermixing of particles of two different types of matter on their own is called diffusion.
(e) Open a water tap.Try to break the stream of water with your fingers,You will find that the stream of water does not cut.This is because particles of matter attract each other with a force that keeps the particles together.

(i) Worms that affect stomach and intestines
(ii) Fluckes damage the liver.

If F be a gravitational force on a body of mass m, then
\(F=G\frac { Mm }{ { r }^{ 2 } } =mg\) (or) \(g=\frac { GM }{ { r }^{ 2 } } \)
Clearly, F ∝ m but g does not depend on m. Hence, all bodies fall with the same rapidness when there is no air resistance.

Here,  \(m=200\quad g=0.2\quad kg,\quad u=10\quad { ms }^{ -1 },v={ -5 }ms^{ -1 }\) 
Change in momentum  \(=m(v-)=0.2(-5-10)\)
\(={ -3\quad kg\quad ms }^{ -1 }.\)

Scalar quantities Temperature,mass,volume,speed,time,distance,electric charge,density ,work and energy.
Vector quantities Displacement,velocity,acceleration,force,momentum and weight.

Advanced organisms will be same as complex organisms. This is because the advanced organisms have acquired their particular body designs relatively recently. There is possibility that advanced or younger organisms undergone increasingly complexity in body design during evolutionary time to complete and survive.

Ordinary elements are composed of a mixture of isotopes. For example, chlorine is a mixture of 75% \(_{17}^{35}Cl\) (35 -17 = 18 neutrons) and 25% \(_{17}^{37}Cl\) (37-17=20 neutrons).The atomic mass of chlorine is the average of these two isotopes and is, therefore, equal to \(\left(35\times{75\over 100}\right)+\left(37\times{25\over100}\right)=35.5\).This accounts for the fractional atomic masses of elements.

(a) The illumination of the beam of light due to scattering on collision with certain particles is called the Tyndall effect. If a beam of light is passed through a true solution, the solution is not illuminated and light passes unaffected.
(b) (i) Milk,
(iv) starch solution,
(vi) Tincture iodine. In these solutions, the size of particles is larger enough to scatter a beam of light.

(i) Potassium chloride crystallises out.
(ii) Sugar remains as residue in the form of a solid mass.
(iii) A black coloured compound is formed.

Common dairy and poultry farming practices are-
(i) Proper cleaning
(ii) Proper shelter with roof and ventilation
(iii) Sanitation and spraying of disinfectants at regular intervals.
(iv) Balanced food with required additives.
(v) Appropriate vaccination.

Mass of bullet, \(m_1=50 \ g = 0.05 kg\)
Mass of rifle,   \(m_2=4\) kg
Initial velocity of bullet, \(u_1=0\)
Initial velocity of rifle, \(u_2=0\)
Final velocity of bullet, \(v_1=35 ms^{-1}\)
Final velocity of rifle, \(v_2 = ?\)
According to the law of conservation of momentum,
total momenta after the fire = Total momenta before the fire
 \(m_1v_1+m_2v_2=m_1u_1+m_2u_2\)
 \(0.05 \times 35 + 4v_2=0+0\)
\({ v }_{ 2 }=-\frac { 0.05\times 35 }{ 4 } =-\frac { 7 }{ 16 } =-0.44{ ms }^{ -1 }\)
The negative sign indicates that the direction in which the rifle would recoil is opposite to that of the bullet.    

For developing a hierarchy of classification, we choose the fundamental characteristic among several other characteristics. For example, plants differ from animals in the absence of locomotion, chloroplasts, cell wall, etc. But, only locomotion is considered as the basic or fundamental feature that is used to distinguish between plants and animals. This is because the absence of locomotion in plants gave rise to many structural changes such as the presence of a cell wall for protection, and the presence of chloroplast for photosynthesis (as they cannot move around in search of food like animals). Thus, all these features are a result of locomotion. Therefore, locomotion is considered to be a fundamental characteristic. By choosing the basic or fundamental characteristic, we can make broad divisions in living organisms as the next level of characteristic is dependent on these. This goes on to form a hierarchy of characteristics.

(a) Carbon dioxide
(b) Elements: Carbon and oxygen.Valency of oxygen is -2 and carbon is +4 in CO
(c) Number of moles in 44 g CO = 1
Number of mole in 360 g CO = \(\frac { 360 }{ 44 } \) = 8.18 moles
(d) Number of molecules CO in 44 g = 6.022 x 10²³
Number of molecxules of CO in 88 g = \(\frac { 88 }{ 44 } \times 6.022\times { 10 }^{ 23 }\)=12.044 x 10²³

Tissue is a group of cells that are similar in structure and are organised together to perform a specific task.