(i) Buoyancy and gravitational force are the two forces which act on a body, when it is immersed in a liquid.The conditions for a body required to float or sink in a liquid are as follows:
(a) If weight of the body is more than the upthrust act on it, then body will sink. 
[\(\therefore\)W > U (sink)]
(b) If weight of the body is less than the upthrust act on it, then body will float. 
[\(\therefore\) W < U (floatl]
(ii) Volume of the water displaced by an iron nail is less than that of piece of wood. That is why, more buoyancy force act on piece of wood and it floats on the surface of water.

Height of object A,\({ h }_{ 1 }=\cfrac { 1 }{ 2 } g{ t }_{ 1 }^{ 2 }\)
Height of object B,\({ h }_{ 2 }=\cfrac { 1 }{ 2 } g{ t }_{ 2 }^{ 2 }\)
\(\therefore \) \({ h }_{ 1 }{ h }_{ 2 }={ t }_{ 1 }^{ 2 }:{ t }_{ 2 }^{ 2 }\)
or  \({ t }_{ 1 }:{ { t }_{ 2 }=\sqrt { { h }_{ 1 } } :\sqrt { { h }_{ 2 } } }\)
(i) Acceleration due to gravity is independent of mass of falling body. So, ratio remains the same
(ii) If bodies are hollow, then also ratio remains the  same, i.e.\({ t }_{ 1 }:{ { t }_{ 2 }=\sqrt { { h }_{ 1 } } :\sqrt { { h }_{ 2 } } }\)
 

(t) Let the two bodies have masses m) and mz and they are placed at the same distance R from the centre of the earth. According to the question, if the sameforce  acts on both of them, then
\({ F }_{ 1 }=\cfrac { GM{ m }_{ 1 } }{ { R }^{ 2 } } \)
and \({ F }_{ 2 }=\cfrac { GM{ m }_{ 2 } }{ { R }^{ 2 } } \)
As, F₁=F₂
Hence,\(\cfrac { GM{ m }_{ 1 } }{ { R }^{ 2 } } =\cfrac { GM{ m }_{ 2 } }{ { R }^{ 2 } } \)
So, m₁ = m₂, their masses will be same
(ii) Mathematically,\(g=\cfrac { GM }{ { R }^{ 2 } } \)
where, g= acceleration due to gravity
G = universal gravitational constant
M = mass of the earth
and R = radius of the earth 
(ii) G is known as the universal gravitational constant because its value remains same all the time everywhere in the universe, applicable to all bodies whether celestial or terrestrial.

(a) Gravitational force acting on the 50 kg,
mg= 50x 9.8 = 490N 
(b) Gravitational force acting on the 50 kg mass due to jupiter,
\({ F }_{ jupiter }=\cfrac { G\times { M }_{ jupiter }\times { { M }_{ person } } }{ \left( distance\quad of\quad jupiter\quad from\quad the\quad earth \right) ^{ 2 } } \)
\({ F }_{ jupiter }=\cfrac { 6.67\times { 10 }^{ -11 }\times 2\times { 10 }^{ 27 }\times 50 }{ 6.3\times { 10 }^{ 11 }\times 6.3\times { 10 }^{ 11 } } \)
\({ F }_{ jupiter }=\cfrac { 6.67\times 2\times 50\times { 10 }^{ -11+27-22 } }{ 6.3\times 6.3 } \)
F_Jupiter = 1.68 X 10^-5 N
\({ F }_{ saturn }=\cfrac { G\times M_{ saturn }\times { M }_{ person } }{ \left( distance\quad of\quad saturn\quad from\quad the\quad earth \right) ^{ 2 } } \)
\({ F }_{ saturn }=\cfrac { 6.67\times { 1 }0^{ -11 }\times 6\times { 10 }^{ 26 }\times 50 }{ 1.28\times 10^{ 12 }\times 1.28\times 10^{ 12 } } \)
\({ F }_{ saturn }=\cfrac { 6.67\times 6\times 50 }{ 1.28\times 1.28 } \times { 10 }^{ -11+26-24 }\)
F_saturn = 0.12x 10^-5N
\(\therefore\) Total gravitational force due to the jupiter and the saturn = (1.68x 10^-5+0.12x 10^-5)N
= 1.8x 10^-5 N
Thus, the combined force due to the planets jupiter and saturn (1.8 x10^-5) N is negligible as compared to the gravitational force due to the earth.
(ii) We know that, g at equator is less than g at poles (Antarctica). Thus, weight at equator is less than weight at pole (Antarctica). A bag of sugar weighs w at a certain place on the equator. If this bag is taken to Antarctica, then it will weigh more due to greater value of g.

Mass of silver (Ag) ornament = mg
Mass of gold used for polishing = \(\frac{1}{100}\times\) mg = 0.01 mg
Atomic mass of Ag = 108 u
\(\therefore\) 1 mole of Ag = 108 g = 6.022 x 10²³ atoms
Thus, 108 g of Ag have atoms = 6.022 x 10²³
\(\therefore\) mg of Ag have atoms = \(\frac{6.022\times 10^{23}}{108}\)
Similarly, atomic mass of gold (Au) = 197 u
1 mole of Au = 197 g = 6.022 x 10²³ atoms
Thus, 197 g of Au have atoms = 6.022 x 10²³
\(\therefore\) 0.01 mg of Au will have atoms = \(\frac{6.022\times 10^{23}\times 0.01\ m}{197}\)
\(\therefore\) Ratio of the number of atoms of gold and silver = \(\frac{6.022\times 10^{23}}{197}\times 0.01\ m; \frac{6.022\times 10^{23}}{108}m\)
= \(\frac{1}{19700}:\frac{1}{108}\) 
= 108: 19700

(i) First the battery converts chemical energy into electrical energy. Then the bulb converts this electrical energy into heat and light.
(ii) Here, m = 500 kg, v = 0 and
u = 36 kmh^-1  = \(\frac{36\times 1000}{3600}\) = 10 ms^-1
\(\therefore\) Work done = Change in kinetic energy
= \(\frac{1}{2}m(v^2-u^2)=\frac{1}{2}\times 500(0-10^2)\)
= \(\frac{1}{2}\times 500\times 100\) = - 25000 Js^-1
So, it is negative because work is done to stop the car.

\(v_{ av }=\frac { s_{ 1 }+s_{ 2 } }{ t_{ 1 }+t_{ 2 } } =\frac { s }{ \frac { s_{ 1 } }{ v_{ 2 } } +\frac { s_{ 2 } }{ v_{ 2 } } } \)
\(40=\frac { 100 }{ \frac { 30 }{ 30 } +\frac { 70 }{ v_{ 2 } } } \)
\(=1+\frac { 70 }{ v_{ 2 } } =\frac { 100 }{ 40 } =\frac { 5 }{ 2 } \)
\(\frac { 7 }{ v_{ 2 } } =\frac { 5 }{ 2 } -1=\frac { 3 }{ 2 } \)
\(v_{ 2 }=\frac { 7X2 }{ 3 } =\frac { 14 }{ 3 } =4.67\quad km\quad h^{ -1 }\)

(i) Lysosomes are membrane-bound sacs filled with digestive enzymes.These enzymes are made by rough endoplasmic reticulum.
(ii) Lysosomes are a kind of waste disposal system of the cell.
(iii) During the disturbance in cellular metabolism, e.g. when a cell gets damaged, lysosomes present in the cell may burst and the enzymes digest the damaged cell. Hence, lysosomes are called as 'suicidal bags' of a cell.
(iv) Lysosomes break up the foreign materials entering into the cell, such as bacteria or food into smal piece

(i) Mass of 1 mole CO₂ = 12 + 32 = 44 u
Mass of 5 moles CO₂ = 44 x 5 = 220 u
Mass of 1 mole of H₂O = 2 + 16 = 18 u
Mass of 5 moles of H₂O = 5 x 18 = 90 u
(ii) 40 g calcium make = 1 mole
240 g calcium make = \(\frac { 240 }{ 40 } \) = 6 moles
24 g magnesium make = 1 mole
240 g magnesium make = \(\frac { 240 }{ 24 } \) = 10 moles
Ratio of 240 g calcium and 240 g magnesium in moles = \(\frac { 6 }{ 10 } \) or 3 : 5