(a) (i) To get minimum current-Resistance will be connected in series
(ii) To get maximum current the resistance should be connected in parallel.
(b) Strength of current in series
R = R₁ + R₂ = 5 + 10 = 15 \(\Omega \)
V=6V    \(\therefore\) I =\(\overset { V }{ \underset { R }{ \_ \_ } } \quad \overset { 6 }{ \underset { 15 }{ \_ \_ } } =0.4A\)
:Strength of current in parallel
\(\overset { { 1 } }{ \underset { { R }_{ P } }{ \_ \_ } } =\overset { 1 }{ \underset { { R }_{ 1 } }{ \_ \_ } } +\overset { 1 }{ \underset { { R }_{ 2 } }{ \_ \_ } } =\overset { 1 }{ \underset { 5 }{ \_ \_ } } +\overset { 1 }{ \underset { 10 }{ \_ \_ } } =\overset { 2+1 }{ \underset { 10 }{ \_ \_ } } =\overset { 3 }{ \underset { 10 }{ \_ \_ } } \)
\(\therefore { R }_{ P }=\overset { 10 }{ \underset { 3 }{ \_ \_ } } \)  V = 6V 
\(\therefore\)\(I = \overset { V }{ \underset { R }{ \_ \_ } } =\overset { 6 }{ \underset { 10/3 }{ \_ \_ } } =\overset { 18 }{ \underset { 10 }{ \_ \_ } } =1.8 \ A\) 

Given, \(\mu\)_kerosene = 1.44, \(\mu\)_turpentine=1.47, \(\mu\)_water = 1.33
Light travels the fastest in a medium having minimum optical density or lowest refractive index.
Since, \(\mu\)_water(=1.33) is the lowest, speed of light is maximum in water.

(i) Regeneration in Planaria
Planaria: If it is divided into 3 parts a, b and c, each part grows as a new individual

(ii) Budding in Hydra

(a) Concentrated sulphuric acid is called the king of chemicals as it is a very strong acid and is highly corrosive.

(b) The three chemical reactions that take place during the conversion of sulfur dioxide to sulphuric acid in the 'Contact Process' are given below.

Step I: Sulphur dioxide is mixed with air and passed over vanadiumpent (V) oxide catalyst heated to \(450^oC\) to form sulphur trioxide.

\(\underset{Sulphur\ dioxide}{2SO_{2(g)}}+\underset{Oxygen}{O_{2(g)}}\underset{\rightleftharpoons }{V_2O_5}\underset{Sulphur\ trioxide}{2SO_{3(g)}}+Heat\)

Step II : Sulphur trioxide is dissolved in concentrated sulphuric acid to form a thick fuming liquid called oleum.

\(\underset{Sulphuric\ acid}{H_2SO_4(l)}+\underset{Sulphur\ trioxide}{SO_{3(g)}}\begin{matrix} V_2O_5 \\ \rightleftharpoons \\ \end{matrix}\underset{Oleum}{H_2S_2O_{7(l)}}\)

Step III: Oleum is diluted with a calculated quantity of water to obtain sulphuric acid.

\(H_2S_2O_{7(l)}+\underset{Water}{H_2O_{(l)}}\rightarrow\underset{Sulphuric\ acid}{2H_2SO_{4(aq)}}\)

(c) If water is added to sulphuric acid then a large amount of heat is evolved.  This changes water to steam, which can splash the acid.  Hence, it is dangerous.

Hydrogen gas is usually liberated when an acid reacts with metal.
e.g. Zn(s) + H₂SO₄ (aq) \(\rightarrow\) ZnSO₄(aq) + H₂ \(\uparrow\)
Test for H₂ gas Testing of H₂ gas presence can be done by passing the gas through soap solution and then bringing a burning splinter near the soap bubble filled with the gas. If the gas burns with a pop sound, it confirms hydrogen.

Given: P_l  = 60 W, V_l = 240 V, R = ?, P₂ = ?, V₂ = 192 V, I₁ = ?
Using,  \({ P }=\overset { { V }^{ 2 } }{ \underset { R }{ \_ \_ } } \)
We get  \(R=\overset { { V }_{ 1 }^{ 2 } }{ \underset { { P }_{ 2 } }{ \_ \_ } } =\overset { (240)^{ 2 } }{ \underset { 60 }{ \_ \_ \_ } } =960\Omega \)
Again    _{\({ P }_{ 2 }=\overset { { V }_{ 2 }^{ 2 } }{ \underset { R }{ \_ \_ } } =\overset { 192*192 }{ \underset { 960 }{ \_ \_ \_ } } =38.4W\)}
Current drawn by bulb at 192 V is 
I = \(\overset { V }{ \underset { R }{ \_ \_ } } =\overset { 192 }{ \underset { 960 }{ \_ \_ \_ } } \) = 0.2A 

Incident ray I enters the glass slab forms an angle of incidence 'i'. Its bends towards the normal and forms an angle of refraction 'r'. The emergent ray is parallel to the incident ray.

Diverging lens / concave lens

Focal length = -20 cm (lens is concave, hence f is -ve)
Power = P = 1/f = 100/ -20 cm = -5D.

(a) (i) Involvement of two different individuals
(ii) Creation of new combination of variants
(b) (i) Pollen/pollen grain
(ii) By pollination/agents of pollination
(iii) It (pollen tube) helps male gamete to reach egg (ovule)
(iv) Converts into embryo.

Chemical formula - CaOCI₂
Chemical equation - Ca(OH)₂ + CI₂ ⟶ CaOCI₂ + H₂O
Uses:
(i) For bleaching cotton and linen in textile industry
(ii) As an oxidising agent in a chemical industry. Y,
(iii) For disinfecting water.

Differences between binary fission and multiple fission are given below

(i) The 5 Ω resistors are connected in series. Therefore, their effective resistance = (5 + 5) = 10 Ω
    The 10 Ω resistors are connected in series. Therefore, their effective resistance = (10 + 10) = 20 Ω
    Now these 10 Ω equivalent and 20 Ω equivalent are connected in parallel. Therefore, the equivalent resistance         (R_eq) will be:
     \(\frac { 1 }{ { R }_{ eq } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \)                   
     \(\frac { 1 }{ { R }_{ eq } } =\frac { 1 }{ 10 } +\frac { 1 }{ 20 } \)
     \(\frac { 1 }{ { R }_{ eq } } =\frac { 2+1 }{ 20 } \)
     \(\frac { 1 }{ { R }_{ eq } } =\frac { 20 }{ 3 } =6.76\Omega \)
(ii) Total Current = Total voltage / Total equivalent resistor 
                          =12 / 6.67
                          = 1.8 A.

(a) Following is the relation between hydrogen ion concentration of an aqueous solution and its pH.

(b) If the pH of the solution decreases from 3 to 2, then

           \({pH_1\over pH_2}={-log[H^+]\over -log[H^+]_2}\)

\(\therefore {-log[H^+]_1\over -log [H^+]_2}={3\over 2}\)

\(\therefore {[H^+]_1\over [H^+]_2}={10^{-3}\over 10^{-2}}\)

or, \([H^+]_2=10[H^+]_1\)

Hence, hydrogen ion concentration will increase 10 times.

Distilled water does not contain any ions and have no electrolytes. Hence, it does not conduct electricity. But in rain water, a small amount of electrolyte is present in the form of acids. These acids are produced due to dissolution of acidic oxides (e.g. SO₂, NO₂) in rain water and make it a better conductor of electricity.

NaOH dissociates into sodium and hydroxyl ions while water dissociates into H⁺ and OH^- ions. Hence, the solution does contain H⁺ ions. H⁺ ions are less than OH^- ions. 

Given
\(n_g={3\over 2},\ \ \ \ n_w={4\over 3}\)
Speed of light in glass=2 x 10⁸ m/s
Speed of light water = v = ?
\(n_g={Speed\ of\ light\ in\ air\over Speed\ of\ light\ in\ glass}\)
\(\Rightarrow\ \ {3\over2}={Speed\ of\ light\ in\ air\over 2\times10^8\ m/s}\)
Speed of light in air=\({3\over 2}\times2\times10^8=3\times10^8\ m/s\)
\(n_w={Speed\ of\ light\ in\ air\over Speed\ of\ light\ in\ water}\)
\(\Rightarrow\ \ {4\over 3}={3\times10^8\ m/s\over V}\)
\(3\times10^8\times{3\over4}=v\)
⇒ v=2.25 x 10⁸ m/s

We know that \(n=\frac { c }{ v } \), where 'c' is velocity of light in vaccum and 'v' is the velocity of the light in the medium. If v₁ is velocity of light in a medium of refractive index n₁ and v₂ is velocity of light in another medium of refractive index n₂, then
\({ n }_{ 1 }=\frac { c }{ { v }_{ 1 } } \quad and\quad { n }_{ 2 }=\frac { c }{ { v }_{ 2 } } \\ ^{ 1 }{ n }_{ 2 }=\frac { { n }_{ 2 } }{ { n }_{ 1 } } =\frac { { c/v }_{ 2 } }{ { c/v }_{ 1 } } =\frac { { v }_{ 1 } }{ { v }_{ 2 } } \)

Rhizopus reproduces b spore formation. Conditions favourable for spore formation are:
(i) Cool place,
(ii) Moist place and
(iii) Dark place.

Differences between pollination and fertilisation are

P = 120 W
t = 20 minutes      E = P\(\times\)t
E = ?  = 120 \(\times\) 20 \(\times\) 60 = 120 \(\times\) 1200
E = 144 \(\times\) 10³ J

Electric cell or battery is a device that helps to maintain a potential difference across a conductor.

pH will decrease from 6 because during curd formation, lactic acid is produced which makes it acidic. When milk changes into curd, its pH changes and becomes less than 6.

(i) (b): The electrical wire should have low resistance.
(ii) (b): More is the temperature, more will be the resistance. The resistance of A is more, so temperature of A is more.
(iii) (c):  \(\begin{equation} R_{1}=\rho \frac{L}{A}, R_{2}=\rho \frac{2 L}{A / 2} \end{equation}\)
(iv) (a): In summers, temperature is more, so resistance is more.
(v) (a): \(\begin{equation} R^{\prime}=\frac{R}{5}=\frac{20}{5}=4 \Omega \end{equation}\)

(i) (b): Plant Y is a dioecious plant with only male flowers. As such flowers lack carpel, fruit formation cannot take place in them.
(ii) (c): X could be any monoecious bisexual plant,Y could be a dioecious male plant and Z could be a monoecious plant with separate male and female flowers.
(iii) (d): Plants X and Z have both types of sex organs present in same flower and different flowers on the same plant respectively. Plant X can be self or cross pollinated. Plant Z could undergo geitonogamy (a type of self pollination) or cross pollination.
(iv) (b): When all the four whorls of a flower are present, it is called complete flower. A bisexual flower is generally complete as it has sepals, petals, stamen and pistil.
(v) (d): Fertilisation takes place inside ovary i.e., female reproductive organ which is present inside female flowers present in female plant.

(i) (a): The compound of sodium that is a constituent of baking powder and is used in antacids, is sodium hydrogen carbonate (NaHC03).

\(\text { (iii) }(\mathrm{a}): \mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{H}_{2} \mathrm{O} \ \ \longrightarrow 2 \mathrm{NaOH}+ \mathrm{H}_{2} \mathrm{CO}_{3}\)
                                                  Strong base       (Z)Weak acid
NaOH ionises completely to give a large amount of OH^- ions whereas H₂CO₃ ionises partially to give a small amount ofH⁺ ions. Hence, the solution is overall alkaline.
(iv) (b): Z is carbonic acid, a weak acid formed when Na₂CO₃ is dissolved in water.
(v) (d)    

(d): The size of the image does not affect the nature of the image except that a bigger image as it gathers more light rays due to wider aperture.

(a) Both A and R are true, and R is correct explanation of the assertion.