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(a) Assertion is correct, Reasoning

Prokaryotic 

1. Presence of a true nucleus is eukaryotes , 

2. Eg. A nucleus which is covered by a nuclear membrane. 

Prokaryotic 

1. Presence of a true nucleus is eukaryotes , 

2. Eg. A nucleus which is covered by a nuclear membrane.

(a) Assertion is correct, Reasoning

Xylem and phloem.

Invertebrates are animals without a backbone. 

The invertebrates have been classified into various phyla as follows:

The ovules are naked, without ovary. Hence gymnosperms do not produce fruits.

1. Monera 

2. Protista 

3. Fungi 

4. Plantae 

5. Animalia

(i) Gaspard Bauhin jn 1623, introduced naming of organisms with two names which is known as Binomial nomenclature, and it was implemented by Carolus Linnaeas in 1753 

(ii) Binomial nomenclature an universal system of naming organisms. As per this system, each organism has two names – the first is the Genusname and the second is the Speciesname. 

(iii) Genus name begins with a capital letter and Species name begins with a small letter. 

Example The nomenclature for onion is Allium sativam. Genus name is Allium, species name is sativam. 

(iv) Vernacular name is a local name that is familiar for a particular place. Binomial name is an universal name which never changes. 

(v) Binomial nomenclature and classification helps scientists to identify any organisms and to place them at a particular hierarchy

1. Fungi are eukaryotic, and mostly are multicellular. 

2. They secrete enzymes to digest the food and absorb the food after digested by the enzymes. 

3. Fungi saprophytes as decomposes (decaycausing organisms) or as parasites. 

4. Kingdom Fungi includes molds, mildews, mushrooms and yeast.

1. In this system of classification, viruses have not been given a proper place 

2. Multicellular organisms have originated several times from protists.

1. This system of classification is more scientific and natural. 

2. This system of classification clearly indicates the cellular organization, mode of nutrition, and characters for early evolution of life.

Correct answer is (c) Whitakar

In a charged conductor, at any point inside or at the surface, the potential is same. So the surface is equipotential.

1. 1 coulomb

2. 1 J

Given q₁ = 5 × 10^-8 C, r = 16 cm = 0.16 m q₂ = -3 × 10^-8 C Let potential be zero at a distance × metre from positive charge q₁ .

∴ r₁ = x meter

r₂ = (0.16 – x) metre

S0 V = \(\frac{1}{4\pi\varepsilon_0}\) \(\Big[\frac{q_1}{r_1}+\frac{q_2}{r_2}\Big]\)

or = 9 x 10⁹ \(\Big[\frac{5\times10^8}{x}-\frac{3\times10^{-8}}{0.16-x}\Big]\)

or \(\frac{5\times10^{-8}}{x}=\frac{3\times10^{-8}}{0.16-x}\)

or  0.8 – 5x = 3x

or  x = 0.1 m = 10 cm.

1. The plane normal to AB and passing through its mid point has zero potential everywhere hence the plane is equipotential.

2. Normal to the plane is the direction AB.

From the figure, we have

OP = OQ = OR = OS = OT = OU

= r = 10 cm = 0.1m

And given q = 5µC = 5 × 10^-6 C

∴ Potential at O due to all the charges

\(v=6\times \frac{1}{4\pi \varepsilon_0}\frac{q}{r}\) = \(\frac{6\times 9\times10^9\times5\times10^{-6}}{0.1}\)

= 2.7 x 10⁶ volt

1. Electric potential.

2. Work done w = VQ

\(=\frac{1}{4\pi\Sigma_0}\frac{q}{r}\times Q\) = \(\frac{9\times10^9\times5\times10^{-6}\times1}{3\times10^{-3}}\) = 15 J

 3. The potential energy at p,

PE₁ = \(\frac{9\times10^9\times5\times10^{-6}\times1}{3\times10^{-3}}\) = 15 V

The potential energy at Q

PE₂ = CQ = \(\frac{9\times10^9\times5\times10^{-6}\times1}{5\times10^{-3}}\)

PE₂ = 9V

Work done to move IC from P to Q, W = PE₂ – PE₁ 

= 9 – 15 

w = -6 J

1. Electric field – force per unit (+ve) charge -NC^-1 

2. Electric potential – Work done per (+ve) charge – JC^-1

3. Capacitance – charge per unit potential difference - Farad.

1. lev =1.6 × 10^-19J

∴ 1J = \(\frac{1}{1.6\times10^{-19}}\)ev 1 J = 6.25 × 10¹⁸eV.

2. 1 ev is the energy acquired by an electron, when it is accelerated through a potential difference of one volt.

3. Joule is bigger unit, 1J = 6.25 × 10¹⁸eV.

1. capacitor. 

2. Potential difference between two plates

V = Ed

= \(\frac{\sigma}{\varepsilon_0}d\)                   (∴E = \(\frac{\sigma}{\varepsilon_0}\))

V = \(\frac{Q}{A\varepsilon_0}d\)    .......(1)  (σ = \(\frac{Q}{A}\))

Capacitance C of the parallel plate capacitor,

C =\(\frac{Q}{V}\)   .....(2)

Sub .eq. (1) in eq. (2)

C =  \(\frac{Q}{\frac{Q}{A\varepsilon_0}d}\) C = \(\frac{A\varepsilon_0}{d}\)

3. 

• When mica sheet is introduced, capacitance of capacitor increases
• Mica sheet will prevent electric breakdown.

1. Equipotential surface.

2. properties of the surface:

• Direction of electric field is perpendicular to the equi potential surface,
• No work is done to move a charge from one point to another along the equi potential surface.

3. Work done = Potential difference × charge = 0 × Charge = 0.

1. 2 v

2. potential at A, V_A = 10 v potential at B, V_B = 12 v.

1. one use of capacitor:

• Capacitor is used to store electric charges
• It is used to prevent dc current.

2. Expression for capacitance of a capacitor: Potential difference between two plates V = Ed

= \(\frac{\sigma}{\varepsilon_0}d\)      (∴\(E=\frac{\sigma}{\varepsilon_0}\))

V = \(\frac{Q}{A\varepsilon_0}d\) ..........(1)  \((σ=\frac{Q}{A})\)

Capacitance C of the parallel plate capacitor,

C = \(\frac{Q}{V}\) ...........(2)

Sub eq. (1) in eq. (2)

C = \(\frac{Q}{\frac{Q}{A\varepsilon_0}d}\)

C = \(\frac{A\varepsilon_0}{d}\)

3. C = 10µF

When dielectric slab is placed, New capacitance

C¹ = KC

10 × 10^-6 = 2 × C, C = 5 × 10^-6 F.

Answer is (a) 80 V

The capacitance of capacitor with air as dielectric is given by

C = \(\frac{\varepsilon_0A}{d}\)

Given C = 8pF = 8 × 10^-12F……(1)

If C is new capacitance when d₁ = \(\frac{d}{2}\)and space is filled with a substance of dielectric constant k = 6. Then

C₁ = \(\frac{\varepsilon_0kA}{d_1}=\frac{\varepsilon_0kA}{d/2}\)

or C₁ = \(\frac{2k\varepsilon_0A}{d}\)

Using Eq.(1)

C₁ = 12 × 8 × 10^-12 F

or C₁ = 96pF.

1. When electric field on air exceeds the 3 × 10⁶ v/m, air becomes conductor and conduct electricity.

2. Insulator shows conducting property at high voltage.

(c) decreases k times

Explanation : F_m = \(\frac{F_0}{k}\) i.e., Decrease K times.

Correct answer is (a) Acidic

(a) Dipole-dipole forces : The force of attraction which act between two molecules having permanent dipoles.

(b) Avogadro Law : Equal volumes of all gases under the similar condition of temperature and pressure contain equal number of molecules . V α n.

(c) Intermolecular Forces: The forces of attractions and repulsion between interacting particles.

Correct answer is (d) 4

• Benzene (C₆H₆) is nonpolar molecule and has zero dipole moment.
• In benzene, only London forces exist due to momentary dipoles.
• The strength of London forces increases with increase in molecular size, molecular mass and number of electrons present in an atom or molecule.
• Hence, due to the presence of London forces, benzene exists in liquid state.

The negative logarithm of hydrogen ion concentration in a solution is calledpH of that solution. 

Let us assume that concentration of hydrogen ion in a solution is 1 x 10^{- 4} gm mole L^-1 . 

The pH of this solution can be calculated as follows: 

pH = -log [1 x 10^-4 ] 

pH = 41og₁₀10 

pH = 4 

pH range is between 0 to 14. The pH of a neutral solution is 7.  

All basic solutions have pH more than 7. 

All acidic solutions have pH less than 7.

(i) Both A and R are true and R is the correct explanation of A.

H₂, N₂, O₂, Ne, He etc.,

For the given binary solution of A and B, it would be ideal if A-B interactions are equal to A-A and B-B interactions and it would be non-ideal if they are different to each other.

The deviation from ideal behaviour will be positive if A-B interactions are weaker as compared to A-A and B-B. The deviation will be negative if A-B interactions are stronger as compared to A-A and B-B.

Dipole-Dipole attractive forces.

(a) Dipole-Dipole interactions. 

(b)H-Bond. 

(c)Dispersion forces. 

(d)Dispersion forces. 

(e)Dispersion forces. 

(f)Induced dipole-dipole interactions.

(i) When we spray the can, the liquid spray of the can is released. Hence, pressure inside decreases a bit. Since the can‘s volume does not change, the temp. falls. 

(ii) We place the dented ping pong ball in warm water. As temp.inside increases, the pressure inside increases. As a result, dent gets removed. 

(iii) The reason is that these cans have a lot of artificial pressure stored in them. When exposed to direct heat, the pressure inside the cans rises because vol. remains constant. When the pressure exceeds the certain limit, the can explodes. 

(iv) When the ball is brought outside, the temp. dropped. As a result, the pressure of the air inside the football also dropped making the ball look deflated. 

(v) Air expands on heating. Hence, its density decreases. Thus, hot air is lighter than atmospheric air. This helps the balloon rise up.

When intermolecular forces are very weak , molecules do not cling together to make liquid and solid.

(i) Dipole-dipole interactions (because H₂S is polar) 

(ii) Hydrogen bonding 

(iii) London dispersion force (because both are non-polar)

(iv) London dispersion forces (because SiH₄ is non-polar)

(v) London dispersion forces (because He atoms have symmetrical electron clouds) 

(vi) Dipole-induced dipole forces (because HCl is polar while He atom has symmetrical r electron cloud)

Correct Answer is: C. SO₃ 

The sulphur trioxide formed is added to sulfuric acid which gives rise to oleum (disulphuric acid).

Answer: (b) NaOH

Answer: (d) KOH

1. True.

2. False. Water exists in three states as solid, liquid, gas.

3. False. Water into steam on heating is called vaporisation.

4. True.

5. False. Chemical change is a permanent change, (or) Physical change is a temporary change.

6. True.

7. False. Construction of building is a human made change.

(a) 1° < 2° < 3°

Correct option is (A) KO₂ 

Correct option is D) All of these

(d) all the above