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• Credit money – cheques 
• Full bodied money – gold coins 
• Representative full bodied money – paper money

1. What are the defects of paper money? 

2. What are the advantages of paper or polymer currency?

Yes, I think so. But the quality and measurement should be perfect as per norms.

1. The trade transactions in their kingdom will be perfect. This strengthens their treasury. 

2. Minting coins is an industry. The income through this also strengthens their treasury. 

3. They print their favourite designs on the coins. This act enables the future generation to know about the interest of the past people. 

Eg : 

If we find any musical instrument on the coin, it tells us that they were the lovers of music.

(3) auto reduction

(3) Chalcophyrite

(4) Cu + SO₂

C-OH in phenol is stabilized due to resonance and electron pair at oxygen atom in phenol is not readily available to proton ,thus protonation not occurs readily

Side product formed in this reaction is acetone which is another important organic compound.

Detaited Answer

As oxidation off cumene results in the formation of phenol and acetone as by-product which is a commerical product use as chemical.

In phenol the lone pair of electrons on oxygen involves in delocalization which results in their nonavailability for the protonation. Whereas in ethanol, the electrons on oxygen atom are not delocalised which results in their availability for protonation.

On reaction with FeCl₃ phenol gives violet, blue and red colour. while ethanol give no reaction.

Arranging the metals in decreasing order of their reactivity is known as activity series.

To extract metal from enriched ore it is converted into metallic oxide by reduction reaction. Then this metallic oxide is further reduced to get metal with certain impurities.

The only viable method to extract metals like K, Na, Ca, Mg and Al is electrolysis.

The metals at the top of activity series are extracted by electrolysis of their fused compounds.

Salient features: 

1. Any metal which is placed higher up in the series can displace any metal below it in order to from the salt solution of the later metal. 

2. The larger the difference in the position of metals in the series, the more rapidly does the displacement take place. 

3. Metals which are placed above hydrogen in the series have the ability to reduce ions from dilute sulphuric acid to liberate hydrogen gas.

4. Oxides of metals K, Na, Ca and Mg cannot be reduced by H₂, CO or C.

5. Oxides and nitrates of less reactive metals Hg, Ag and Au decompose to give metals on being strongly heated. 

6. Metals below copper such as mercury, silver, platinum and gold do not rust easily. 

7. Hydrogen though a non-metal, has been included in the series. It occupies the position based on its formation of positive ions.

Alcohols are soluble in water due to formation of intermolecular H-bond with water and break H-bonds already existing between water molecules. Therefore they are soluble in water. On the other hand hydrocarbons cannot form H-bond with water molecule and hence are insoluble in water.

Alcohols are more soluble in water than the hydrocarbons of comparable molecular masses because alcohols form hydrogen bonds with water.

i) Good conductors. 

Reason : Metals containing free electrons are very good conductors of electricity whereas non-metals are bad conductors of electricity because they do not have free electrons. 

ii) Non-malleable. 

Reason : Metals are hard. So, they can be beaten into sheets whereas non-metals are soft, so they are non-malleable. 

iii) Negative ions. 

Reason : Metals are electropositive in nature. They easily lose electrons to form positive ions, whereas non-metals are electronegative in nature. So, they gain electrons to form negative ions.

iv) Basic oxides. 

Reason : Non-metallic oxide solutions turn blue litmus into red. They are acidic in nature. So, they are called acidic oxides whereas metallic oxide solutions turn red litmus into blue. They are basic in nature. So, they are called basic oxides.

1. Highly active metals like potassium, sodium, calcium,  magnesium and aluminium are obtained by the electrolysis of their fused halides or oxides, that is, by electrolytic reduction because their oxides cannot be reduced by common reducing agents like carbon, carbon monoxide and hydrogen.

2. Zinc is obtained only by heating its oxide with carbon. 

3. Iron, lead and copper are obtained by reduction of their oxides with carbon, carbon monoxide and hydrogen. 

4. Copper is obtained by reducing black copper oxide with carbon or by air reduction. 

5. Mercury and silver are obtained by heating their respective oxides to temperature above 300°C when they lose oxygen and are reduced to free metals. 

6. However, less active mercury can also be obtained by merely heating its sulphide in air. 

7. Silver and Gold are obtained by displacement from solutions containing their ions by more electropositive metal zinc.

C) Sulphide, Oxide

i) If X is a metal, then the litmus turns into blue because metal reacts with oxygen and forms metallic oxide and aqueous solution of metallic oxide ore basic in nature. 

ii) Mf X is a non-metal, then the litmus turns into red because non-metal reacts with oxygen and forms non-metallic oxide and aqueous solution of nonmetallic oxide ore acidic in nature.

iii) If X is a reactive metal, then it releases hydrogen gas from sulphuric acid because more reactive metal displaces hydrogen from acid.

a) Potassium and sodium form oxides of type M₂O in limited supply of oxygen. 

4 K + O₂ → 2 K₂O 

4 Na + O₂ → 2 Na₂O 

b) In excess of oxygen they form peroxides of type M₂O₂. 

2 Na + O₂ → Na₂O₂ 

2 K + O₂ → K₂O₂

1. The metals which are at the bottom of activity series have very low reactivity and do not burn or oxidase even on surface. 

Eg : Ag, Pt, Au. 

2. The metals like Pb, Cu and Hg do not burn but only form a surface layer of oxide, i.e., PbO, CuO, HgO. 

3. The metals like Al, Zn, Fe react with oxygen to form respective oxides.

4. The metals like Ca and Mg burn with decreasing vigorousity to form oxides. 

5. The metals like K, Na burn vigorously to form Na₂O, K₂O in limited supply of oxygen but form peroxides in excess of oxygen.

Correct option is  B) Roasting

Chaleo means ore. We know that most of ores of many metals are oxides and sulphides. That’s why oxygen – sulphur group is called chalcogen family.

 Data : N = 300, A = 0.05 m²,

M = 4.5 A ∙ m², B 

= 0.2 T, θ = 30° 

(i) M = NIA 

∴ The current in the coil,

I = \(\cfrac{M}{NA}\) = \(\cfrac{4.5}{300\times0.05}\) = 0.3 A

(ii) The magnitude of the torque, 

τ = MB sin θ = 4.5 × 0.2 × sin 30°

= 0.9 × \(\cfrac12\) = 0.45 N∙m

Data: N = 10, A = 0.05 m² , 

B = 0.01 T, 

I = 30 μA = 3 × 10^-5 A

(i) The magnetic moment, 

μ = NIA = 10(3 × 10^-5 A)(0.05 m²) 

=1.5 × 10^-5 A∙m² = 15 μA∙m²

(ii) The maximum torque experienced by the coil (when its plane is parallel to \(\vec B\)) is τ_max = MB 

= (1.5 × 10^-5 A∙m²)(0.01 T) 

= 1.5 × 10^-7 N∙m

(iii) The minimum torque experienced by the coil (when its plane is perpendicular to\(\vec B\)) is τ_min = 0

1. Advantage of radial magnetic field in a moving- coil galvanometer:

• As the coil rotates, its plane is always parallel to the field. That way, the deflecting torque is always a maximum depending only on the current in the coil, but not on the position of the coil.
• The restoring torque is proportional to the deflection so that a radial field makes the deflection proportional to the current. The instrument then has a linear scale, i.e., the divisions of the scale are evenly spaced. This makes it particularly straight forward to calibrate and to read.

2. Producing radial magnetic field : 

• The pole pieces of the permanent magnet are made cylindrically concave, concentric with the axis of the coil.
• A soft iron cylinder is centred between the pole pieces so that it forms a narrow cylindrical gap in which the sides of the coil can move. Together, they produce a radial magnetic, field; that is, the magnetic lines of force in the gap are along radii to the central axis.

Suppose the magnetic field is uniform but not radial. Then, when the coil comes to rest after rotation through an angle θ, NIAB cos θ = Cθ (in usual notations).

∴ I ∝ \(\cfrac{\theta}{cos\,\theta}\)

as N, A, B and C are constants in a particular case. Thus, the current is not directly proportional to the deflection. Hence, we cannot have a linear scale for measurement.

Data : A = 0.05 m², 

B = 0.01 Wb/m², 

N = 10, C 

= 5 × 10^-9 N∙m per degree, 

I = 30 μA = 30 × 10 A,

I = \(\left(\cfrac{C}{NAB}\right)\) θ

∴ The deflection of the coil,

θ = \(\cfrac{NIAB}C\) = \(\cfrac{10\times30\times10^{-6}\times0.05\times0.01}{5\times10^{-9}}\) = 30°

Data ; A = 10^-3 m² , N = 100, C = 10^-8 

N∙m/degree, B = 0.05 Wb/m² , 7 = 1.6 × 10^-5 A

I = \(\left(\cfrac{C}{NAB}\right)\) θ

∴ The deflection of the coil,

θ = \(\cfrac{NIAB}C\) = \(\cfrac{(100)(1.6\times10^{-5})(10^{-3})(0.05)}{10^{-8}}\) = 8

Principle : A current-carrying coil suspended in a magnetic field experiences a torque which rotates the plane of the coil and tends to maximize the magnetic flux through the coil.

The deflection of the coil in a moving-coil galvanometer is linearly related to the current through it and, therefore, can be used to measure current in terms of the deflection.

The angle of rotation, θ ∝ IB (in the usual notation)

As I₂B₂ = (2I₁) \(\left(\cfrac{B_1}2 \right)\)

= I₁B₁, 

the angle of rotation will be the same, i.e., 30°.

(D) the torsion constant of the suspension fibre.

1. Advantage of radial magnetic field in a moving- coil galvanometer:

• As the coil rotates, its plane is always parallel to the field. That way, the deflecting torque is always a ximum depending only on the current in the coil, but not on the position of the coil.
• The restoring torque is proportional to the deflection so that a radial field makes the deflection proportional to the current. The instrument then has a linear scale, i.e., the divisions of the scale are evenly spaced. This makes it particularly straight forward to calibrate and to read.

2. Producing radial magnetic field :

• The pole pieces of the permanent magnet are made cylindrically concave, concentric with the axis of the coil.
• A soft iron cylinder is centred between the pole pieces so that it forms a narrow cylindrical gap in which the sides of the coil can move. Together, they produce a radial magnetic, field; that is, the magnetic lines of force in the gap are along radii to the central axis.

(A) the elastic torsion of the suspension fibre

The given statement is false. Consider a coil of N turns, each of area A. If the current through the coil is I, the magnetic dipole moment of the coil is, in magnitude, μ = NIA. That is, μ ∝ I, for given N and A. Thus, for a coil of given geometry, its magnetic dipole moment varies with the current through it.

For an ideal solenoid, the magnetic field induction outside is negligible, nearly zero. Inside the solenoid, the field lines are parallel to the axis of the solenoid and the magnitude of the magnetic induction, B = μ₀nI, where μ₀ is the permeability of free space, n is the number of turns of wire per unit length and I is the steady current in the solenoid.

(C) not experience any torque

Data : l = 30 cm = 0.3 m, 

I = 5 A, B = 0.2 T, θ = 60°

The magnitude of the force on the conductor,

F = I\(\vec l\) x\(\vec B\)| = IlB sin θ 

= (5 A) (0.3 m) (0.2 T) sin 60° 

= 0.3 × 0.866 = 0.2598 N 

The direction of the force is given by the cross product rule.

Correct option is (D) (0.08 N/m)

Correct option is (B) 4R

Solution :

Dark or black

(iii) Fraction from of ratio 9 : 19 is 9/19

In figure, 

Number of triangles = 4 

Number of circles = 2 

Number of squares = 2 

Total figures = 4 + 2 + 2 = 8 

(i) Number of triangles to the number of circle, 

Required ratio = 4/2 = 2/1 = 2 : 1 

(ii) Number of squares to all the figures, 

Required ratio = 2/8 = 1/4 = 1 : 4 

(iii) Number of triangles to all the figures. 

Required ratio = 4/8 = 1/2 = 1 : 2

(i) Required ratio = 25/150 = 1/6 = 1 : 6 

(ii) Required ratio = 72/36 = 2/1 = 2 : 1 

(iii) Required ratio = 55/121 = 5/11 = 5 : 11

(iv) Required ratio = 35/55 = 7/11 = 7 : 11

(i) 3 : 5 = 1 : 15 = 3 × 15 = 5 × 1 

45 ≠ 5 

∴ 3 : 5 and 1 : 15 are not in proportion 

(ii) 4 : 12 and 9 : 27 

⇒ 4 : 12 = 9 : 27 

⇒ 4 × 27 = 12 × 9 

⇒ 108 = 108 

∴ 4 : 12 and 9 : 27 are in proportion 

(iii) 10 : 15 = 4 : 6 

⇒ 10 × 6 = 15 × 4

 ⇒ 60 = 60 

∴ Given numbers are in proportion.

(iii) Equivalent ratio of 2 : 3 is 6 : 9