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a. ggana – Ganga
b. uhokmg – Gomukh
c. srdiahwa – Haridwar
d. etirvin – Triveni
e. masgna – Sangam

Correct option is (iii) three rivers

Answer is (a) Nuclear charge.

a. I have a book.
b. The table has four legs.
c. The tiger has black stripes.
d. Butterflies have coloured wings.

Correct option is (b) acculturation.

I, Br, F, and Cl.

Correct option is (d) None of the above

(Be²⁺, He); (F^-, N^3-); (Fe³⁺, CO³⁺); (S^2-, Ar)

Correct option is (b) Denmark, New Zealand, Sweden and the U.S.

(d)  G.T.Seaborg & Mendelinium 

(c)  It is the strongest oxidising agent because it has maximum electron negativity 

It is tendency of an atom in a molecule to attract the shared pair of electron to itself.

The molecular formula of compound formed between B & C is  C₂B₃

(c)   [Xe] 4f¹⁴ 5d⁶ 6s²

The element  exhibits highest oxidation number is Os

An element X belongs to Gp16 & 5^th period. Its atomic number is 52.

(d)  Statement 1 is false, statement 2 is true

The element with highest electronic affinity belongs to  Period 3 gp 17

(b) 1–s, 2–r, 3–p, 4–q

Correct option  (b) TFFF

Correct option  (c)  TTTF

(b)  Statement 1 is true, statement 2 is true but is not correct explanation for statement 1 

(a)  N = 6 , l = 1 , m = ⁺ -1 , m_s = ⁺-1/2

The element with atomic number 44 belongs d-Block

(b) Statement 1 is true, statement 2 is true but is not correct explanation for statement 1 

The electronic configuration of an element of chalcogen family is [Ar] 3d¹⁰ 4s² 4p⁴

The order of ionization energy of K, Ca, & Ba are  Ca > Ba > K

(a)  1–t, 2–s, 3–p, 4–q

The element having valence shell electronic configuration 3s² 3p¹ must be placed in the p - block of the periodic table as its last electron enters in p-subshell (3p).

The ionisation potential of N > O because N posses stable half filled p–orbital.

(b)  B < C < N < O  (Increasing first ionisation enthalpy)

(c)  1-C, 2-D, 3-B

(b) 1–s, 2–r, 3–p, 4–q

(b)  Li < Be < B < C ΔiHi 

(i) General outer electronic configuration of s-block elements is ns^1-2 i.e., either ns¹ or ns²

(ii) General outer electronic configuration of p-block elements is ns² np^1-6

(iii) General outer electronic configuration of d-block elements is (n - 1) d^1-10 ns^1-2

(iv) General outer electronic configuration of f-block elements is (n - 2) f^1-14 (n - 1) d^0-1 ns^-2

Correct option   (b) TTTF

Correct option  (a)  FTFTF

Let's go for a little walk. Only you must tell me about things. I shall be so glad when today is over. The other bad days can't be half as bad as this. Having a leg tied up and hopping about on a crutch is almost fun, I guess. I don't think I'll mind being deaf for a day - at least not much. But being blind is so frightening.

Because Mendeleev gave priority to similarities in properties. 

Germanium and gallium

• In Mendeleev ′s Periodic Table there are instances where elements with higher atomic mass is placed before the element with lower atomic mass. 
• This was done to ensure that elements with similar properties were included in the same group. 
• Hence Cobalt was placed before Nickel despite of higher atomic number of Cobalt than Nickel.

According to.Mendeleev, the properties of the elements are a periodic function of their atomic weights, while according to modem periodic law, the properties of the elements are periodic functions of their atomic numbers.

(b) Power cord

Correct answer is (c) Speaker

(b) Personal computer

1. True

2. True

(D) Operating System

Data: ∆ABC and ∆ABD are two triangles on the same base AB. 

Line segment CD bisected by AB at O’. 

To Prove: ar.(∆ABC) = ar.(∆ABD) 

Prove: AB bisects line segment CD at O’. 

∴ OC = OD In ∆ADC, AO is the median on CD. 

∴ ar.(∆AOD) = ar.(∆AOC) …………… (i) 

In ∆BDC, BO is the median on CD. 

∴ ar.(∆BOD) = ar.(∆BOC) ……………. (ii) 

Adding (i) and (ii), 

ar.(∆AOD) + ar.(∆BOD) = ar.(∆AOC)+ ar.(∆BOC) 

∴ ar.(∆ABD) = ar.(∆ABC)

We know that ABCD is a rhombus with P, Q, R and S as the midpoints of AB, BC, CD and DA.

AC and BD diagonals are joined.

Using the midpoint theorem

We know that

PQ = ½ AC

By substituting the values we get

PQ = ½ (16)

By division

PQ = 8cm

In △ DAC we know that S and R are the midpoints of AD and DC

Using the midpoint theorem

SR = ½ AC

By substituting the values

SR = ½ (12)

By division

SR = 6cm

Consider the rectangle PQRS

Area of the rectangle PQRS = length × breadth

So we get

Area of the rectangle PQRS = 6 × 8

By multiplication

Area of the rectangle PQRS = 48 cm²

Therefore, area of the figure is 48 cm².

From the figure we know that D and E are the midpoints of AB and AC

So we get

DE || BC || PQ

Consider △ ACP

We know that AP || DE and E is the midpoint of AC

Using the midpoint theorem we know that D is the midpoint of PC

So we get

DE = ½ AP

It can be written as

AP = 2DE …… (1)

Consider △ ABQ

We know that AQ || DE and D is the midpoint of AB

Using the midpoint theorem we know that E is the midpoint of BQ

So we get

DE = ½ AQ

It can be written as

AQ = 2DE ……. (2)

Using equations (1) and (2)

We get

AP = AQ

We know that △ ACP and △ ABQ lie on the bases AP and AQ between the same parallels BC and PQ

So we get

Area of △ ACP = Area of △ ABQ

Therefore, it is proved that ar (△ ABQ) = ar (△ ACP).

Correct option is  D) 3