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Option : (d) 7+

A series of f-block elements having electronic configuration (n – 2)f^1-14(n – I) d^0-1 ns² placed separately in the periodic table represents inner transition series. The f-orbitals lie much inside the e-orbitals.

Since the last electron enters pre-penultimate shell, these elements are inner transition elements.

There are two inner transition series as follows :

4f-series ₅₈Ce → ₇₁Lu 

5f-series ₉₀Th → ₁₀₃Lr

Option : (b) d → d transition

Option : (d) [Ar] 3d⁵ 4s¹

(d)  (n-1)d subshell

Transition elements is another name for d block elements. In d block elements electrons enter in subshell.

Option : (c) (n – 1) d subshell

(A)

• Elements in which differentiating electron enters into the pre-penultimate shell the (n – 2) f-orbital are known as f.block elements.
• They include 28 elements with atomic numbers ranging from 58-71 and atomic numbers 90 to 103 collectively.
• There are two f-series or two f-block elements, namely 4f and 5f series.
• The f-block includes two inner transition series namely the lanthanoid series. Cerium (58) to LuteUum (71) or the 4 f-block elements and the actinoid series. Thorium (90) to I.awrencium (103) or the 5f block elements.

(B) f-block elements are called inner transition elements since f-orbital lies much inside the forbital in relation to the transition metals, These elements have 1 to 14 electrons in their f-orbital.

m₁ = 2 kg,

T₁ = 80°C,

s = 4.2 × 10³ J kg^-1 °C^-1,

m₂ = 3kg,

T₂ = 20°C

Let the final temperature be T₃

Heat lost by 3 kg water = Heat gained by 2 kg water.

⇒ m₁s(T₁ – T₃) = m₂s(T₃ – T₂)

⇒ 2 × (80 – T₃) = 3 × (T₃ – 20)

⇒ T₃ = 44° C

∴ Final temperature = 44° C.

The compounds of transition metals exhibit magnetic properties due to the presence of unpaired electrons in their atoms or ions.

The common chemical properties of the dblock elements are : 

• All d-block elements are electropositive metals. 
• They exhibit variable oxidation states and form coloured salts and complexes. 
• They are good reducing agents. 
• They form insoluble oxides and hydroxides. 
• Iron, cobalt, copper, molybdenum and zinc are biologically important metals.
• They catalyse biological reactions. 

(b) A will go higher than B.

Explanation:

The force of resistance of air will add to retardation of both the body but its magnitude will depend upon mass of the objects as per "Newton's Second Law of Motion" and it will be equal to Force/mass. Since the force is equal on both the objects, the object having greater mass (A) will be retarded less, so it will go higher than B.  

Option : (d) 3rd group 7th period.

Option : (c) inner transition elements

Properties of actinoids :

• Actinoids are silvery white (similar to lanthanoids). 
• They are highly reactive radioactive elements. 
• Most of these elements are not found in nature. They are radioactive and man made. 
• They experience decrease in the atomic and ionic radii from Ac to Lw, known as actinoid contraction. 
• The common oxidation state is +3. Elements of the first half of the series exhibit higher oxidation states. 

• Thorium oxide (ThO₂) with 1% CeO₂ is used as a major source of indoor lighting, as well as for outdoor camping.
• Uranium is used in the nuclear reactors.
• The isotopes of Thorium and Uranium have very long half-life, so that we get very negligible radiation from them: Hence they can be used safely. 

Option : (b) f – f 

Actinoid contraction : The gradual decrease in atomic and ionic radii of actinoids with the increase in atomic number is called actinoid contraction.

• The electronic configuration of Ce³⁺ is, [Xe] 4f⁷
• Even though there is one unpaired electron in 4f sub-shell, the f → f transition involves very low energy. Hence, Ce³⁺ ion does not absorb radiation in the visible region. 

Therefore Ce³⁺ ion is colourless.

• The common oxidation state of the Lanthanoids is 3 + due to the loss of 2 electrons from outermost 6s orbital and one electron from the penultimate 5d sub-shell.
• Gd³⁺ and Lu³⁺ show extra stability due to their half-filled and completely filled f-orbitals, Gd³⁺ = [Xe]4f⁷, Lu³⁺ = [Xe]4f¹⁴
• Ce and Tb attain the 4f° and 4f⁷ configurations in the 4 + oxidation states. Eu and Yb attain the 4f⁷ and 4f¹⁴ configurations in the 2 + oxidation states. Sm and Tm also show the 2+ oxidation state although their stability can be explained based on thermodynamic factors.
• Some lanthanoids show 2 + and 4 + oxidation states even though they do not have stable electronic configuration of 4f°, 4f⁷ or 4f¹⁴. E.g. Pr⁴⁺ (4f¹), Nd²⁺ (4f⁴), Sm²⁺(4f⁶), Dy⁴⁺(4f⁸) etc 

Gd and Lu show only one oxidation state 3 +, since they acquire electronic configurations with extra stability namely 4f⁷ and 4f¹⁴ respectively.

(1) Consider Pr³⁺ and Tm³⁺ ions. Tm³⁺ (4f¹²) has nf electron 12 electrons. Pr²⁺ (4f²) has (14 – n) = (14 – 2) = 12 electrons. Both, Tm³⁺ and Pr³⁺ are green.

(2) Consider Nd³⁺ and Er³⁺ ions. Er³⁺ (4f¹¹) has nf electrons 11. 

Nd³⁺ (4f³) has (14 – n) is (14 – 3) = 11 electrons. These both ions Er³⁺, Na³⁺ are pink in colour.

Due to dispersion of light and different wavelength of colours in medium.

1. A white light from a slit ‘S’ is made to pass through prism P which forms spectrum on a white screen AB. 

2. A narrow slit H is made on the screen AB, parallel to slit S to allow the light of particular colour to pass through it. 

3. The light of a particular colour is made to fall on the second prism Q placed with its base in opposite direction to that of the prism P. 

4. The light after passing through the second prism Q is received on another white screen M. 

5. It is observed that the colour of light obtained on the screen M is same as that of the light incident on the second prism Q through the slit H.

Surface area of a shephe = 4πr² 

Ratio of surface areas of spheres = square of ratio of their radius 

Given, 

Diameter of the moon is approximately one fourth of the diameter of the earth 

∴ r_m = 1/4 × r_e 

⇒ r_m : r_e = 1 : 4 

Ratio of their surface area = 1 : 16

Volume of a sphere = (4/3)πr³ 

(i) radius is 2 cm 

⇒ Volume of the sphere = (4/3) × (22/7) × 2³ 

= 33.52 cm³ 

(ii) radius is 3.5 cm 

⇒ Volume of the sphere = (4/3) × (22/7) × 3.5³ 

= 179.67 cm³ 

(iii) radius is 10.5 cm 

⇒ Volume of the sphere = (4/3) × (22/7) × (10.5)³ 

= 4851 cm³

Volume of a hemisphere = (2/3)πr³ 

Given, 

hemispherical tank has inner radius of 2.8 m 

⇒ Volume of the tank = (2/3) × (22/7) × 2.8³ 

⇒ Volume of the tank = 45.976 m³ 

= 45976 litres

Volume of a sphere = \(\frac{4}{3}\)πr³ 

Given, 

diameter of the moon is approximately one fourth of the diameter of the earth 

Radius of moon = 1/4 × radius of the earth.

Ratio of their volume = \(\frac{\frac{4}{3}\pi(\frac{r}{4})^3}{\frac{4}{3}\pi r^3}\)= 1 : 64

Volume of the moon is \(\frac{1}{64^{th}}\) times the volume of the earth.

Total surface area of a hemisphere = 3πr² 

Total surface area of a hemisphere of radius 10 cm = 3 × π × 10² 

= 942 cm²

We have, 

Radius = 10 cm 

Total surface area of a hemisphere = 2πr² 

⇒ Total surface area of a hemisphere = 2 × 3.14 × 10 × 10 

= 628 cm² 

Total surface area of a solid hemisphere = 3πr² 

⇒ Total surface area of a solid hemisphere = 3 × 3.14 × 10 × 10 

= 942 cm²

Option : (D)

Total surface area of a sphere = 4πr² 

Total surface area of a hemisphere = 2πr² + πr² = 3πr²

Ratio of the total surface area of a sphere and a hemisphere of same radius = 4 : 3

Volume of a hemisphere = \(\frac{4}{3}\)πr³ 

Given, 

capsule of medicine is in the shape of a sphere of diameter 3.5 mm.

Radius = \(\frac{3.5}{2}\) = 1.75 mm

Volume of medicine filled inside = \(\frac{4}{3}\) × π × 1.75³ 

= 22.458 mm³

Height of cone (h) = 21 cm 

Slant height of cone (l) = 28 cm

We know, 

l² = r² + h² 

28² = r² + 21² 

or r = 7√7 cm 

Now, 

We know, 

Volume of cone = \(\frac{1}{3}\)πr²h 

= \(\frac{1}{3}\) x π x (7√7)² x 21

= 2401 π 

Therefore, Volume of cone is 2401 π cm³.

Consider △ BCD

Using the Pythagoras theorem

We can write it as

DB² + BC² = DC²

By substituting the values

DB² + 8² = 17²

By subtraction

DB² = 17² – 8²

DB² = 289 – 64

By subtraction

DB² = 225

By taking the square root

DB = √ 225

So we get

DB = 15cm

We can find

Area of △ BCD = ½ × b × h

By substituting the values

Area of △ BCD = ½ × 8 × 15

On further calculation

Area of △ BCD = 60 cm²

Consider △ BAD

Using the Pythagoras theorem

We can write it as

DA² + AB² = DB²

By substituting the values

AB² + 9² = 15²

By subtraction

AB² = 15² – 9²

AB² = 225 – 81

By subtraction

AB² = 144

By taking the square root

AB = √ 144

So we get

AB = 12cm

We can find

Area of △ DAB = ½ × b × h

By substituting the values

Area of △ DAB = ½ × 9 × 12

On further calculation

Area of △ DAB = 54 cm²

So we get

Area of quadrilateral ABCD = area of △ DAB + area of △ BCD

By substituting the values

Area of quadrilateral ABCD = 54 + 60

By addition

Area of quadrilateral ABCD = 114 cm²

Therefore, the area of quadrilateral ABCD is 114 cm².

Consider △ ALD

Based on the Pythagoras theorem

AL² + DL² = AD²

By substituting the values

4² + DL² = 5²

So we get

DL² = 5² – 4²

DL² = 25 – 16

By subtraction

DL² = 9

By taking square root

DL = √ 9

So we get

DL = 3 cm

Consider △ BMC

Based on the Pythagoras theorem

MC² + MB² = CB²

By substituting the values

MC² + 4² = 5²

So we get

MC² = 5² – 4²

MC² = 25 – 16

By subtraction

MC² = 9

By taking square root

MC = √ 9

So we get

MC = 3 cm

From the figure we know that LM = AB = 7cm

So we know that CD = DL + LM + MC

By substituting the values

CD = 3 + 7 + 3

By addition

CD = 13cm

Area of Trapezium ABCD = ½ (sum of parallel sides × distance between them)

So we get

Area of Trapezium ABCD = ½ × (CD + AB) × AL

By substituting the values

Area of Trapezium ABCD = ½ × (13 + 7) × 4

On further calculation

Area of Trapezium ABCD = 20 × 2

By multiplication

Area of Trapezium ABCD = 40 cm²

Therefore, length of DC = 13cm and area of trapezium ABCD = 40 cm².

From the figure

Using the Pythagoras theorem in △ RTQ

We get

RT² + TQ² = RQ²

By substituting the values

RT² + 8² = 17²

On further calculation

RT² = 17² – 8²

So we get

RT² = 289 – 64

By subtraction

RT² = 225

By taking square root

RT = √ 225

RT = 15cm

We can find the area of trapezium

Area of trapezium PQRS = ½ (sum of parallel sides × distance between them)

So we get

Area of trapezium PQRS = ½ ((8 + 16) × 15)

On further calculation we get

Area of trapezium PQRS = 180 cm²

Therefore, the area of trapezium PQRS is 180 cm².

Correct option is (C) 25, 16

Let the numbers are a and b.

Sum of numbers is 41.

\(\therefore\) a+b = 41        __________(1)

And product of the number is 400.

\(\therefore\) ab = 400        __________(2)

Now, \((a-b)^2=(a+b)^2-4ab\)

\(=41^2-4\times400\)

= 1681 - 1600

= 81 \(=9^2\)

\(\Rightarrow\) a - b = 9        __________(3)

By adding equations (1) & (3), we get

(a+b) + (a - b) = 41+9

\(\Rightarrow\) 2a = 50

\(\Rightarrow\) a = \(\frac{50}2\) = 25

\(\therefore\) b = 41 - a         (From (1))

= 41 - 25 = 16

Hence, required numbers are 25 and 16.

Correct option is C) 25, 16

Correct option is (B) 30 and 15

Let A's present age be x years and B's present age be y years.

Since, A is twice as old as B.

\(\therefore\) x = 2y              __________(1)

10 years ages, A's age = (x - 10) years

& B's age = (y - 10) years

Given that 10 years ago, A was four times as old as B.

\(\therefore\) x - 10 = 4 (y - 10)

\(\Rightarrow\) 2y - 10 = 4y - 40    (From (1))

\(\Rightarrow\) 4y - 2y = 40 - 10

\(\Rightarrow\) 2y = 30

\(\Rightarrow\) y = \(\frac{30}2\) = 15

\(\therefore\) x = 2 \(\times\) 15 = 30     (From (1))

Hence, their present ages are 30 and 15 years.

Correct option is B) 30 and 15

Correct option is (C) 23, 33

Let the numbers are a and b.

\(\therefore\) Their sum = a+b

and their difference = a - b

Then according to given conditions, we have

\(\frac14(a+b)=14\)

\(\Rightarrow\) a+b = 56       ____________(1)

And \(\frac12(a-b)=5\)

\(\Rightarrow\) a - b = 10      ____________(2)

By adding equations (1) & (2), we get

(a + b) + (a - b) = 56+10

\(\Rightarrow\) 2a = 66

\(\Rightarrow\) a = 33

\(\therefore\) b = 56 - a    (From (1))

= 56 - 33 = 23

Hence, the numbers are 33 and 23.

Correct option is C) 23, 33

The given equations are: 

7(y + 3) – 2(x + 2) = 14 

⇒ 7y + 21 – 2x – 4 = 14 

⇒-2x + 7y = -3 ……..(i) 

and 4(y – 2) + 3(x – 3) = 2 

⇒4y – 8 + 3x – 9 = 2

⇒3x + 4y = 19 ……….(ii) 

On multiplying (i) by 4 and (ii) by 7, we get: 

-8x + 28y = -12 ……(iii) 

21x + 28y = 133 ……(iv) 

On subtracting (iii) from (iv), we get: 

29x = 145 

⇒x = 5 

On substituting x = 5 in (i), we get: 

-10 + 7y = -3 

⇒7y = (-3 + 10) = 7 

⇒y = 1 

Hence, the solution is x = 5 and y = 1.

The given system of equations is 

0.3x + 0.5y = 0.5 …….(i) 

0.5x + 0.7y = 0.74 …….(ii) 

Multiplying (i) by 5 and (ii) by 3 and subtracting (ii) from (i), we get 

2.5y - 2.1y = 2.5 - 2.2 

⇒0.4y = 0.28 

⇒y = 0.28/0.4 = 0.7 

Now, substituting y = 0.7 in (i), we have 

0.3x + 0.5 × 0.7 = 0.5 

⇒0.3x = 0.50 – 0.35 = 0.15 

⇒x = 0.15/0.3 = 0.5 

Hence, x = 0.5 and y = 0.7

The given system of equations is 

0.4x + 0.3y = 1.7 …….(i) 

0.7x – 0.2y = 0.8 …….(ii)

Multiplying (i) by 0.2 and (ii) by 0.3 and adding them, we get 

0.8x + 2.1x = 3.4 + 2.4

 ⇒ 2.9x = 5.8

⇒x = \(\frac{5.8}{2.9}\) = 2

Now, substituting x = 2 in (i), we have 

0.4 × 2 + 0.3y = 1.7 

⇒ 0.3y = 1.7 – 0.8

⇒ y = \(\frac{0.9}{0.3}\) = 3

Hence, x = 2 and y = 3.

The given equations are: 

7(y + 3) – 2(x + 2) = 14 

⇒ 7y + 21 – 2x – 4 = 14 

⇒ -2x + 7y = -3 ……..(i) 

and 4(y – 2) + 3(x – 3) = 2 

⇒ 4y – 8 + 3x – 9 = 2

⇒ 3x + 4y = 19 ……….(ii) 

On multiplying (i) by 4 and (ii) by 7, we get: 

- 8x + 28y = -12 ……(iii) 

21x + 28y = 133 ……(iv) 

On subtracting (iii) from (iv), we get: 

29x = 145 

⇒ x = 5 

On substituting x = 5 in (i), we get: 

-10 + 7y = -3 

⇒ 7y = (-3 + 10) = 7 

⇒ y = 1 

Hence, the solution is x = 5 and y = 1.

y/7 = 3

y = 21 

y – 21 = 0 

a = 0 

b = 1 

c = -21

y – 2 = 0 

a = 0 

b = 1 

c = – 2

0.4x + 0.3y = 1.7

0.7x – 0.2y = 0.8

Multiply both the equations by 10, we get

4x + 3y = 17 …………(1)

7x – 2y = 8 ..………..(2)

Multiply (1) by 2 and (2) by 3,

8x + 6y = 34

21x – 6y = 24

Adding both the equations

29x = 58

x = 2

From (1); 4 x 2 + 3y = 17

8 + 3y = 17

3y = 17 – 8 = 9

y = 3

Answer: x = 2, y = 3

x = - 14/13

⇒ 13x = - 14 

⇒ 13x + 14 = 0 

a = 13 

b = 0 

c = 14

Correct option is (D) x + y = 12

Let both numbers are x & y.

Then their sum is \(x+y=12.\)

D) x + y = 12

Correct option is (C) x + y – 180 = 0

Sum of supplementary angles is \(180^\circ,\) Let both supplementary angles are x & y.

\(\therefore\) x+y = \(180^\circ\)

\(\Rightarrow\) \(x+y-180^\circ=0\) is required linear form.

C) x + y – 180 = 0

Correct option is (D) x – 2y – 14 = 0

Let first number is x and second number is y.

\(\therefore\) x - 2y = 14     (According to given condition)

\(\Rightarrow\) x – 2y – 14 = 0 is required linear form.

D) x – 2y – 14 = 0

Correct option is (D) (-3, 9)

Given equation is 3x – y = 0

\(\Rightarrow\) y = 3x

Hence, (- 3, 9) is not a solution of 3x - y = 0.

Correct option is  D) (- 3, 9)