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D) Reflexive angle

D) all of the above

1. Straight angle

2. Complete angle

3. Protractor

B) Measurement

1. Aryabhatta and Brahmagupta

2.Earth’s measurement

3. Earth’s measurement

Correct option is  D) Greek

Correct option is  C) arms

1. Angle

2.°

3. Right angle

Correct option is  B) Vertex

A) Line segment

Correct option is  A) Line

Correct option is  D) Ray

i) A line has two end points. (False) 

ii) Ray is a part of line. (True) 

iii) A line segment has two end points. (True) 

iv) We can draw many lines through two points. (False)

Acute angles : 10°, 30°, 45°, 60°, 89° (< 90°) 

Obtuse angles : 110°, 150°, 160°, 172°, 178° (90° < obtuse < 180°) 

Reflex angles : 210°, 270°, 300°, 345°, 359° (reflex > 180°)

Correct Answer is :

Correct option is  B) l // m

C) always same

The given equations are : 4x + 7y – 10 = 0 and 10x + ky – 25 = 0 

Here, a₁ = 4, b₁ = 7, c₁ = –10 

a₂ = 10, b₂ = k, c₂ = – 25 

For the given equation to represent coincident lines, 

\(\frac{a_1}{a_2}\)=\(\frac{b_1}{b_2}\)=\(\frac{c_1}{c_2}\) i.e., \(\frac{4}{10}\) = \(\frac{7}{k}\) = \(\frac{-10}{-25}\)

\(\Rightarrow\) \(\frac{4}{10}\) = \(\frac{7}{k}\) \(\Rightarrow\) \(\frac{70}{4}\) = \(\frac{35}{2}\)

i) Intersecting lines : (l, m); (l, n); (n, o); (m, o); (l, o); (m, n) 

ii) Intersecting lines: (p, q); (p, r); (p, s); (q, r); (q, s) 

Concurrent lines : (p, q, s)

Answer is : (c) 17

Let the four prime numbers in the ascending order of their magnitudes be a, b, c and d. 

Given, a × b × c = 715 and b × c × d = 2431 

HCF of (abc and bcd) = bc

 ∴ Now, 715 = 11 × 13 × 5 and 2431 = 11 × 13 × 17

⇒ HCF of 715 and 2431 = 11 × 13 = 143 

∴ Largest number (d) = \(\frac{bcd}{bc} = \frac{2431}{143}\) = 17

The given pair of equation is 

2/x + 3/y = 9/xy ………(i) 

4/x + 9/y = 21/xy ………(ii) 

Multiplying (i) and (ii) by xy, we have 

3x + 2y = 9 ……….(iii) 

9x + 4y = 21 ………(iv) 

Now, multiplying (iii) by 2 and subtracting from (iv), we get 

9x – 6x = 21 – 18 ⇒ x = 3/3 = 1 

Putting x = 1 in (iii), we have 

3 × 1 + 2y = 9 ⇒ y = 9−3/2 = 3 

Hence, x = 1 and y = 3.

Nitrogen is found in a diatomic form with a triple bond between two atoms. N₂ molecules are held together by weak van der Waal's force of attraction which can be easily broken by the collision of the molecules. So, N₂ is a gas at room temperature.

[NO– ] Nitrate

Cu + 4HNO₃ → Cu(NO₃ )₂ + 2NO₂ + 2H₂O

1. C + 4HNOs → CO₂ + 2H₂O + 4NO₂ 

2. 3Cu + 8HNO₃ → 3Cu (NO₃ )2 + 4H₂O + 2NO

Nitric acid works as an oxidising agent. 

The oxidising property of nitric acid is based on the fact that when nitric acid undergoes decomposition, it yields nascent oxygen, which is very reactive.

2HNO₃ (cone.) → H₂O + 2NO₂ + [O] 

2HNO₃ (dil.) → H₂O + 2NO + 3[O]

This nascent oxygen oxidises metals, non-metals, organic and inorganic compounds. During the process, nitric acid itself gets reduced to various products (NO, NO₂ , N₂O, NH₃ , etc.) depending upon the concentration of the acid, reaction temperature and activity of the metal with which it is reacting.

HNO₃ (Nitric acid)

Hydrogen |H⁺ | ions and Nitrate ions. 

 It is because it does not liberate hydrogen gas when treated with metals. Instead it liberates oxides of nitrogen, such as nitric oxide, nitrogen dioxide, etc., as it is very powerful oxidising agent

(4) Metallurgy

Tincture of iodine- It is a dilute solution of iodine prepaired in ethanol (2 to 3%). This chemical prevent the growth of micro-organisms and may also kill them.

Reaction of Iodine with Cu₂I₂- When iodine reacts with Cu₂I₂ it gives cupric iodine.

Cu₂I₂ + I₂ → 2CuI₂

Reaction of Iodine with NH₃- It reacts with liquid ammonia to form nitrogen tri-iodide

2NH₃ + 3I₂ → 2NI₃NH₃ + 3HI

Nitrogen tri-iodide behaves as explosive in the dry state. Vapours of iodine are liberated when it explodes.

8NH₃NI₃ → 6NH₄I + 9I₄ + 5N₂

Correct Answer is: (B) Natural polymers

Natural Polymers since all of these polymers are found in nature.

Reaction of Iodine with NaOH- In cold stat when iodine reacts with dilute solution of NaOH it gives sodium iodine and hypo iodide.

I₂ + 2NaOH → NaI + NaOI + H₂O

when it reacts with conc. solution of NaOH it gives sodium iodide and iodate.

3I₂ + 6NaOH → 5NaI + NaIO₃ + 3H₂O

Reactions of Iodine with Starch- It turns a freshly prepared starch solution or moist starch paper blue.

The correct option is (B) Tertiary Alcohol.

The reducing power of an element is measured in terms of standard electrode potential (E⁰) corresponding to the following transformation i.e, tendency to lose electron.

M_(s) \(\longrightarrow\) M⁺_(aq) + e^-

i. Reducing nature of group 1 elements:

All the alkali metals have high negative values of E⁰ which indicates that they have strong reducing nature and hence, they can be used as strong reducing agents. 

Lithium is the most powerful and sodium is the least powerful reducing agent in the group.

ii. Reducing nature of group 2 elements:

• All the alkaline earth metals have high negative values of stanard reduction potential (E⁰) and are strong reducing agents.
• However, reducing power of alkaline earth metals is less than that of alkali metals.

Ethylamine to ethyl alcohol:

C₂H₅NH₂ + {HNO₂}/{N₂,H₂O} → C₂H₅OH (Ethyl alcohol)

Correct option is a. H₂O because of hydrogen bonding

H₂S < H₂Se < H₂Fe + H₂O

(C) The monovalent ions of alkali metals are larger in size than the parent atoms.

The elements of group 2 are collectively called alkaline earth metals because they occur as minerals in rocks.

Nitrogen can not extend its valence from 3 to 5 due to absence of d-orbital while phosphorous shown penta valence as d-orbital are present in its valence cell.

HF < HCl < HBr < HI

Correct option is (C) Francium

i. Group 1 (alkali metals):

• Two elements of group 1 i.e., sodium and potassium are the sixth and seventh most abundant elements present in the earth’s crust.
• However, francium does not occur appreciably in nature because it is radioactive and has short half-life period.

ii. Group 2 (alkaline earth metals):

• The elements magnesium and calcium are found abundantly in earth’s crust. 
• Radium is radioactive and is not easy to find.

HI < HBr < HCl < HF

• In the free state hydrogen exists as dihydrogen gas.
• Hydrogen is most abundant element in the universe and constitutes 70% of the total mass of the universe.
• Hydrogen is also the principal element in the solar system.
• On the earth, hydrogen is the tenth most abundant element on mass basis and the third most abundant element on atom basis.