From Archimedes' principle,
Volume of sphere
= Volume of water displaced by sphere
= (68 - 59) mL = 9 mL
As, density = \(\cfrac { mass }{ volume } \)
=\(\cfrac { 81 }{ 9mL } =9\cfrac { g }{ mL } \)
= 9000 kgm^-3

1. Manures are rich in organic matter but are poor supplier of nutrients. So, the cultivated field which is not supplied with fertilizers may suffer from deficiency of nutrients and not give very good yield.
2. The cultivated field which is supplied with fertilizers only may face drainage problem and water logging due to lack of organic matter in the soil. This may harm the crop and the yield. Use of fertilizers over long periods of time can destroy the soil structure by killing the soil microorganisms that recycle nutrients in the soil.

The object to be cleaned is placed in a cleaning solution and ultrasonic waves are sent into the solution.Due to the high frequency, the particles of dust, grease and dirt get detached and the object gets thoroughly washed.

Here, m = 5 kg, v = 10 ms^-1, g = 10 ms^-2
Initial K.E. of the body = 1/2mv² = 1/2 x 5(10)² = 250J
P.E. at the highest point = K.E. lost by the body = 250J
mgh = P.E.
5 x 10h = 250
Height attained, h = 250/50 = 5m

A sudden movement of the vehicle results in the sudden change in the state of motion of the vehicle and of our feet in contact with it. But the rest of our body oppose this change due to its inertia and tends to remain where it was. Seat belts are provided to protect the passengers from falling backward or forward during such situations.

(a) Coelom is a cavity which lies in between the body wall and gut (alimentary canal) and is lined by mesodermal cells.
(b) The coelom allows greater body flexibility. In it the organs of the body can accommodate in a better way.

1 mole of carbon atoms = 6.022 x 10²³ atoms
Now 6.022 x 10²³ atoms of carbon weigh = 12 g
One atom of carbon weighs = \(\frac { 12 }{ 6.023\times { 10 }^{ 23 } } =1.99\times { 10 }^{ 23 }\)

Brass is a mixture because:
(i) Different samples of brass may have variable compositions.
(ii) Different constituents of brass may have variable compositions.
(iii) The properties of brass depending on the percentage of each pure substance in it.

Given, ml = 3 kg and m₂ = 12 kg
Let the mass, rrt:, = 0.5 kg be placed at a distance of x from m1 as shown in figure

Then, force acting on m₃ due to m₁ is equal and opposite to the force acting on m₃, due to m₂
\(\therefore \) F₃₁ = F₃₂
\(\cfrac { G{ m }_{ 1 }{ m }_{ 2 } }{ { x }^{ 2 } } =\cfrac { G{ m }_{ 3 }{ m }_{ 2 } }{ \left( 12-x \right) ^{ 2 } } \)
\(\Rightarrow \quad \cfrac { 3 }{ { x }^{ 2 } } =\cfrac { 12 }{ \left( 12-x \right) ^{ 2 } } \)
\(\Rightarrow \quad \left( \cfrac { 12-x }{ x } \right) ^{ 2 }=\cfrac { 12 }{ 3 } =4\Rightarrow \cfrac { 12-x }{ x } =2\)
\(\Rightarrow\)12-x = 2x   \(\Rightarrow\) 12 = 3x
\(\Rightarrow\) x=4m
The position of required point is at a distance 4 m from the mass 3 kg.

Important points to remember in dairy and poultry farming are:
(i) Proper housing, hygienic conditions i.e., cleaning, sanitation, spraying disinfectants at regular intervals.
(ii) Vaccination for prevention of infectious disease.
(iii) Selection of good quality breed.
(iv) Enough space, proper light intensity and duration, maintenance of temperature.
(v) Proper feeding management.

Here,   \({ m }_{ 1 }={ m }_{ 2 }=1.5kg,\ { u }_{ 1 }=2.5{ ms }^{ -1 },\ { u }_{ 2 }=-2.5{ ms }^{ -1 }\) 
Let v be the velocity of the combined object after the collision. By conservation of momentum,
Total momenta after collision = Total momenta before collision
 \(\left( { m }_{ 1 }+{ m }_{ 2 } \right) v={ m }_{ 1 }{ u }_{ 1 }+{ m }_{ 2 }{ u }_{ 2 }\)
\(\left( 1.5+1.5 \right) v=1.5\times 2.5+1.5\times \left( -2.5 \right) \)
 \(3.0\ v=0\)
 \(v=0\ { ms }^{ -1 }.\)

(i) Kingdom. Highest category of classification.
(ii) Phylum (for animals) Division (for plants). A group of closely related classes having certain common characters.
(iii) Class. A group of closely related orders having certain common characters.
(iv) Order A group of closely related families having certain on characters.
(v) Family. A group of closely related genus having certain on characters.
(vi) Genus. A group of closely related species have certain common characters.
(vii) Species.  A group of organisms which are similar enough to breed and perpetuate.

Utility of mole concept:
(i) Form the number of moles of a substance, we can calculate the number of elementary particles because the number of moles of a substance is directly proportional to the number of elements particles.
(ii) One mole of a gas occupies 22.4 litres at S.T.P
(iii) One mole of any gas under the same conditions of temperature and pressures occupies the same volume
(iv) One mole is equal to molecular mass in grams which is equal to 6.022 x 10²³ atoms, molecules, ionsetc.Thus, we can calculate absolute masses of atoms and molecules.

It helps in growth, organisation of different organs and performing different functions.

Difference between mass and weight:

Plants and Animals differ as given in the table below:

(i) The relative density of a substance is the ratio of its density to that of water.
Relative density of a substance
= \(\cfrac { Density\quad of\quad the\quad substance }{ Density\quad of\quad water } \)
Relative density of a substance
=\(=\cfrac { Mass\quad of\quad the\quad substance }{ Volume\quad of\quad the\quad substance } \times \cfrac { Volume\quad of\quad water }{ Mass\quad of\quad water } \)  \(\left[ \therefore Density=\cfrac { mass }{ volume } \right] \)
Given that, mass of the cube = 50 g
Side of cube = 1.5 cm
\(\therefore\) Volume of cube = (1.5)³ cm³ = 3.375 cm³
\(\therefore\) Density =\(\cfrac { mass }{ Volume } \)
=\(\cfrac { 50 }{ 3.375 } \) = 14.81 g cm^-3
(iil) Given that, mass, m = 250 g
Volume, V= 400 cc
\(\therefore \) Density=\(\cfrac { mass }{ Volume } \)
= \(\cfrac { 250 }{ 400 } \)g(cc)^-1= 0.625 g (cc)^-1
As we know that, density of water = 1 g (cc)^-1. So, density of tin is less than that of water and hence tin will float.

() 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 = 8 \(ms^{ -2 }\),v = 32 \(ms^{ -1 }\) 
(ii) t = \(2\sqrt { 2\quad s } \)