For first wire R₁ = ρ l/A = 4Ω
Now for second wire
R₂ = ρ\(\frac { l/2}{2A}\) = 1/4 ρ l/A
R₂ = 1/4 R₁
R₂ = 1Ω
The resistance of the new wire is 1Ω.

The differences between transport of materials in xylem and phloem are as follows:

(i) (a) No change in colour will take place in case of dry blue litmus.
(b) The moist blue litmus paper will change its colour to red because sulphur is non-metal and non-metal oxides are acidic in nature.
(ii) S(s) + O₂(g) \(\rightarrow\) SO₂(g)
SO₂(g) + H₂O(l) \(\rightarrow\) H₂SO₃ (aq)
                               Sulphurous acid

If magnification is 1, the image size is same as that of object size.
Since, magnification is positive, the image is erect and virtual.

(i) Distinction based on physical properties:
(a) Smell Ethanoic acid has a pungent smell.
Ethanol has a pleasant smell.
(b) Melting point Ethanol has lower melting point (156 K) than ethanoic acid (290 K).
(c) Physical state Ethanoic acid is solid (glacial acetic acid) in winters but ethanol is always a liquid.
(ii) Distinctions based on chemical properties:
(a) Reaction with sodium hydrogen carbonate On adding a small amount of sodium hydrogen carbonate to ethanoic acid, carbon dioxide gas is evolved with brisk effervescence. However, no such reaction noticed in case of ethanol.
CH₃COOH + NaHCO₃ \(\rightarrow\) CH₃CO\(\bar{O}\)\(\overset{+}{N}\)a + CO₂\(\uparrow\) H₂O
Ethanoic acid
C₂H₅OH + NaHCO₃ \(\rightarrow\) No reaction
Ethanol
(b) Reaction with caustic alkalies Ethanoic acid reacts with both sodium hydroxide (NaOH) and potassium hydroxide (KOH) to form corresponding salt and water. Ethanol fails to react with either of these.
CH₃COOH + NaOH \(\rightarrow\) CH₃CO\(\bar{O}\)\(\overset{+}{N}\)a + H₂O
CH₃COOH + KOH \(\rightarrow\) CH₃CO\(\bar{O}\)\(\overset{+}{K}\) + H₂O

pH in increasing order of H⁺ion concentration is C < E < D < A < B

(a) 2KBr(aq) + BaI₂(aq) \(\rightarrow\) 2KI + (aq) + BaBr₂(s)
Type: Double displacement reaction
(b) ZnCO₃(s) \(\overset{\Delta }{\rightarrow}\) ZnO(s)+CO₂(g)
Type: Thermal decomposition reaction
(c) H₂(g)+Cl₂(g)\(\rightarrow\) 2HCl(g)
Type: Combination reaction
(d) Mg(s)+2HCl(aq)\(\rightarrow\)MgCl₂(aq)+H₂(g)
Type: Displacement reaction

An electric fuse is a device which is used and connected in series of an electric circuit as a safety device to prevent the damage caused by short circuiting or overloading of the circuit.
Fuse wire is made up of small thin wire whose melting point is low. So that it can control the amount of current flowing through it, if an extra current passes through it then the fuse wire melts and breaks the circuit thereby preventing the damage of electric appliances. An alloy of Pb + Sn is used for the same.

In parallel
\(\overset { 1 }{ \underset { { R }_{ 1 } }{ \_ \_ } } =\overset { 1 }{ \underset { { R }_{ 1 } }{ \_ \_ } } +\overset { 1 }{ \underset { { R }_{ 2 } }{ \_ \_ } } \)  
hence            
\(\overset { 1 }{ \underset { 5 }{ \_ \_ } } =\overset { 1 }{ \underset { 10 }{ \_ \_ } } +\overset { 1 }{ \underset { { R }_{ 2 } }{ \_ \_ } } \)       
\(\overset { 1 }{ \underset { { R }_{ 2 } }{ \_ \_ } } =\overset { 1 }{ \underset { 5 }{ \_ \_ } } -\overset { 1 }{ \underset { 10 }{ \_ \_ } } \)                              
 \(\therefore { R }_{ 2 \ = \ 10 \ \Omega }\)                   

Phytohorrnones help in control and coordiantion in plants. The plant hormones also help in growth, vegetative growth, ripening of fruit and promotes to the stem cell divistion.

Vegetarian habit is safer. Being closer to producers, less DDT will accumulate in our body. Bio magnification leads to higher level of DDT in higher trophic levels.

(i)Removal of waste products from the blood
(ii)Maintenance of the water balance of the body
The procedure used in the working of artificial kidney is haemodialysis.

3Fe(s) + 4H₂O(g) →→ Fe₃O₄(s)  + 4H₂(g)

(i) Ventricles have to pump blood out of the heart into various organs of the body, therefore they have thicker muscular walls than atria.
(ii) Plants do not move and have a large proportion of dead cells in many tissues. As a result, plants have low energy needs and therefore transport system in plants is slow.
(iii) Circulation of blood in aquatic vertebrates such as fishes differs from that in terrestrial vertebrates as fishes have two-chambered heart. The blood is pumped to the gills for oxygenation and is then transported to the rest of the body.
On the other hand, most terrestrial vertebrates have lungs for oxygenation of blood. The deoxygenated blood is transported to heart from where it is transported to lungs for oxygenation and is then distributed to all body parts.
(iv) During daytime, due to transpiration pull, more water is absorbed by roots that is why absorption is faster in daytime.
(v) Veins have valves whereas arteries do not because valves have to prevents the back flow of blood as blood flows in them at low pressure.

(i) Given, power of generator, \(P \approx 2 \mathrm{~kW}\)
Time, t=3h
therefore Energy consumed, \(E=P \times t=2 \times 3=6 \mathrm{kWh}\)
Since, cost of 1 kWh of electrical energy is ₹ 3.50.
Therefore Cost of 6 kWh of electrical energy =6 x 350 =₹ 21.0
(ii) As we know that, Heat, \(H=t^2 R t\)
Here, I is constant, so to reduce the boiling time ( f ), R should be increased.
Since, \(R=\rho \frac{l}{A}\) or \(R \propto l\)
So, the length of the heating element should be increased.

(i) When the student mixed sodium sulphate powder in barium chloride in a dry state, no change would be observed. But when he dissolved them in water, barium sulphate precipitates out and sodium chloride remains in solution. This is called a double displacement reaction.
(ii) (a) Copper < iron < zinc < aluminium

(a) Carbon forms compounds mainly by covalent bonding because carbon has small size, so neither it can lose four electrons easily, because very high amount of energy will be required, nor it can gain four electrons as 6 parotons cannot hold 10 electrons. Hence, it shares four electrons forming covalent bonds.
(b) (i) Due to catenation - Self linking property
(ii) Tetravalency of carbon - Forms compounds with other elements.
(c)(i) 'X' is CH₃COOH, it freezes during winter in cold countries.
        'Y' is ester CH₃COOC₂H₅
(ii) CH₃COOH+ C₂H₅OH \(\overset { Cone.H_2SO_4 }{ \leftrightarrows } \)CH₃COOC₂H₅ + H₂O

(a) A - Stigma B - Pollen tube C - Ovary D - Ovule (ovum)
(b) The transfer of pollen grains from anther (male sexual part) of a flower to the stigma part of the pistil (female sexual part) is known as pollination. Signification of pollination. It may be the cause of variation in a species of a plant.
(c) Fertilisation in flower. A pollen grain contains two male gametes. When a pollen grain falls on the stigma of the carpel, it grows a pollen tube downwards into the
style. One of the male gametes reaches the ovary through pollen tube and fuses with female gamete present in the ovule. This fertilisation of the two germ cells forms a zygote. The zygote is capable of growing into a new plant. The other male gamete fuses with two polar nuclei to form endosperm nucleus. This fusion is called triple fusion. The zygote grows into an embryo and an endosperm nucleus grows into endosperm (the food storage tissue of seed).
(i) All the eggs in the ovule present in the ovary of a flower gets fertilised by male gametes from pollen grains and grow to become seeds.
(ii) The ovary of flower develops and becomes fruit.

(i) Snell's law states that, "the ratio of sine of angle of incidence to the sine of angle of refraction is constant for a given pair of media." 
\(\frac { sin\quad i }{ sin\quad r } \)=Constant
The value of constant for a ray of light passing from air into a particular medium is called the refractive index of the medium. 
Refractive index (of a medium) = \(\frac { Speed\quad of\quad light\quad in\quad vacuum }{ Speed\quad of\quad light\quad in\quad medium } \)
(ii) Refractive index of medium 'a' with respect to 'b'=\(_{ b }{ { n }_{ a } }=\frac { 2 }{ 3 } \)
\(_{ b }{ { n }_{ a } }=\frac { Speed\quad of\quad light\quad in\quad medium\quad 'b' }{ Speed\quad of\quad light\quad in\quad medium\quad 'a' } \)
\(=\frac { { v }_{ b } }{ { v }_{ a } } =\frac { 2 }{ 3 } \)
Refractive index of medium 'b' with respect to 'c' =\(_{ c }{ { n }_{ b } }=\frac { 4 }{ 3 } \)
\(_{ c }{ { n }_{ b } }=\frac { Speed\quad of\quad light\quad in\quad medium\quad 'c' }{ Speed\quad of\quad light\quad in\quad medium\quad 'b' } \)
\(=\frac { { v }_{ c } }{ { v }_{ b } } =\frac { 4 }{ 3 } \)
Refractive index of medium 'c' with respect to 'a' =\(_{ a }{ { n }_{ c } }=?\)
\(_{ a }{ { n }_{ c } }=\frac { Speed\quad of\quad light\quad in\quad medium\quad 'a' }{ Speed\quad of\quad light\quad in\quad medium\quad 'c' } \)
\(=\frac { { v }_{ a } }{ { v }_{ c } } =?\)
\(\because \quad \frac { { v }_{ b } }{ { v }_{ a } } =\frac { 2 }{ 3 } \Rightarrow { v }_{ a }=\frac { 3 }{ 2 } { v }_{ b }\)
\(\because \quad \frac { { v }_{ c } }{ { v }_{ b } } =\frac { 4 }{ 3 } \Rightarrow { v }_{ c }=\frac { 4 }{ 3 } { v }_{ b }\)

Observation: As shown in the figure-the current flows from +ve to -ve and the current flows from X to Y. The needle deflects to right.

Compass needle deflects as current passes through a metallic conductor. It means that the electric current through the wire has produced a magnetic effect.

Aim: To show that carbon dioxide is necessary for the starch formation.
Procedure: Take two healthy potted plants which are nearly the same size. Keep them in a dark room for three days. Now place each plant on separate glass plates. Place a watch-glass containing potassium hydroxide by the side of one of the plants. The potassium hydroxide is used to absorb carbon dioxide.
Material Required: Two potted plants, two bell shaped glass jars, watch glass, KOH, vaseline and glass plates.
Cover both plants with separate bell-jars as shown in figure below.
Use vaseline to seal the bottom of the jars to the glass plates so that the set up is air tight. Keep the plants in sunlight for about two hours. Pluck a leaf from each plant and check for the presence of starch.
Observation: The leaf of the pot with KOH will not turn blue-black because KOH absorbs carbon dioxide required for photosynthesis. The leaf of pot without KOH will show blue-black colour due to presence of CO₂ which is essential for starch formation in plants.
Conclusion: Carbon dioxide is essential for starch formation.

Experimental set-up (a) with potassium hydroxide, (b) without potassium hydroxide

Take a spatula full of sodium carbonate in a test tube and add 2 mL of dilute ethanoic acid. Pass the gas produced through freshly prepared lime water. Repeat the above procedure with sodium hydrogen carbonate 
Following reactions take place
2CH₃COOH + Na₂CO₃ \(\to \) 2CH₃COONa + H₂O+ CO₂
CH₃COOH + NaHCO₃ \(\to \) CH₃COONa + H₂O + CO₂
The brisk effervescence of CO₂ gas is obtained which when passed through lime water turns it milky

 

(b): The resistance, \(\begin{equation} R=\frac{V^{2}}{P}, \text { i.e., } R \propto 1 / P \end{equation}\) i.e., higher is the wattage of a bulb, lesser is the resistance and so it will glow bright.

\({ (d) }: \mathrm{Zn}_{(s)}+2 \mathrm{H}_{(a q)}^{+} \rightarrow \mathrm{Zn}_{(a q)}^{2+}+\mathrm{H}_{2(g)}\)
In this reaction, zinc loses electrons and so it is a reducing agent, while H⁺ gains electrons and so it is an oxidising agent.

(a) If both assertion and reason are true and the reason is correct explanation of assertion.

(a) If both assertion and reason are true and the reason is correct explanation of assertion.