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#### The value of    is : Option: 1   Option: 2   Option: 3   Option: 4

Trigonometric Identities -

Trigonometric Identities-

These identities are the equations that hold true regardless of the angle being chosen.

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Allied Angles (Part 1) -

Allied Angles (Part 1)

Two angles or numbers are called allied iff their sum or difference is a multiple of π/2

• sin (900 - θ) = cos (θ)

• cos (900 - θ) = sin (θ)

• tan (900 - θ) = cot (θ)

• csc (900 - θ) = sec (θ)

• sec (900 - θ) = csc (θ)

• cot (900 - θ) = tan (θ)

• sin (900 + θ) = cos (θ)

• cos (900 + θ) = - sin (θ)

• tan (900 + θ) = - cot (θ)

• csc (900 + θ) = sec (θ)

• sec (900 + θ) = - csc (θ)

• cot (900 + θ) = - tan (θ)

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#### The strength of an aqueous NaOH solution is most accurately determined by titrating: (Note: consider that an appropriate indicator is used) Option: 1 Aq.NaOH in a pipette and aqueous oxalic acid in a burette Option: 2 Aq.NaOH in a volumetric flask and concentrated in a conical flask Option: 3 Aq.NaOH in a burette and concentrated in a conical flask Option: 4 Aq.NaOH in a burette and aqueous oxalic acid in a conical flask

As we have learnt,

To calculate the strength of NaOH, it is titrated against oxalic acid. For this purpose, NaOH is kept in a burette and oxalic acid in a conical flask.

Therefore, Option(4) is correct.

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#### If  and   then  is equal to ________. Option: 1   Option: 2  Option: 3  Option: 4

Trigonometric Identities -

Trigonometric Identities-

These identities are the equations that hold true regardless of the angle being chosen.

-

Double Angle Formula and Reduction Formula -

Double Angle Formula and Reduction Formula

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Correct Option (2)

#### Reaction of an inorganic sulphite X with dilute $H_{2}SO_{4}$ generates compound Y.Reaction of Y with gives X. Further, the reaction of X with Y and water affords compound Z. Y and Z, respectively, are :Option: 1Option: 2Option: 3Option: 4

The reaction will be-

So, Y and Z, respectively, are $SO_{2}\; \; and \; \; NaHSO_{3}$

Therefore, the correct option is (3).

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#### If you spill a chemical toilet cleaning liquid on your hand, your first aid would be:Option: 1 aqueous NH3Option: 2  gaseous NaHCO3Option: 3 aqueous NaOHOption: 4 Vinegar

Chemical toilet cleaning liquid contains acid and hence the first aid would be a weak base like NaHCO3.

Therefore, Option(2) is correct.

#### Two poles, of length a metres and  of length metres are erected at the same horizontal level with bases at . If and , then : Option: 1 Option: 2 Option: 3 Option: 4

$\angle ACB= \angle ACD-\angle BCD\\$

$\Rightarrow \tan\angle ACB=\tan\left ( \angle ACD-\angle BCD \right )\\$

$\tan\angle ACD= \frac{x}{b}, \tan\angle BCD= \frac{x}{a+b}\\$

$\Rightarrow \frac{1}{2}= \frac{\frac{x}{b}-\frac{x}{a+b}}{1+\frac{x}{b}\cdot \frac{x}{a+b}}\\$

$\Rightarrow \frac{1}{2}= \frac{\left ( a+b \right )x-bx}{b\left ( a+b \right )+x^{2}}\\$

$\Rightarrow x^{2}-2ax+b\left ( a+b \right )= 0$

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#### The addition of dilute salt solution will give :Option: 1 Precipitate of Option: 2 Precipitate of Option: 3 Precipitate of Option: 4 A solution of

Addition of dilute $\mathrm{NaOH}$ to a salt solution of $\mathrm{Cr^{3+}}$ will lead to the formation of a precipitate of hydrated chromium (III) oxide.

$\mathrm{Cr}^{3+}+\underset{\mathrm{dil}}{\mathrm{NaOH}} \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{3} \cdot \mathrm{nH}_{2} \mathrm{O}$

Hence, the correct answer is option (1)

#### Acidic ferric chloride solution on treatment with excess of potassium ferrocyanide gives a Prussian blue coloured colloidal species. It is :Option: 1Option: 2Option: 3Option: 4

$\mathrm{Fe^{3+}}$ ions react with excess of potassium ferrocyanide to form $\mathrm{KFe\left [ Fe\left ( CN \right )_{6} \right ]}$.

The reaction is given below:

$\mathrm{\underset{excess}{K_{4}\left [ Fe\left ( CN \right ) _{6}\right ]}+Fe^{3+}\rightarrow KFe\left [ Fe\left ( CN \right )_{6} \right ]}$

Hence,the correct answer is Option (4)

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#### Match List - I with List - II : List-I (Metal Ion) List-II (Group in Qualitative analysis) (i) Group - III (ii) Group - IIA (iii) Group - IV (iv) Group - IIB Choose the most appropriate answer from the options given below :Option: 1Option: 2Option: 3Option: 4

The correct match of basic radicalsand their groups in Qualitative analysis is given below

$\mathrm{(a)Mn^{2+}:Group\; I V (iii)}$

$\mathrm{(b)As^{3+}:Group\; IIB(iv)}$

$\mathrm{(c)Cu^{2+}:Group\; II A(ii)}$

$\mathrm{(d)Al^{3+}: Group (III)(i )}$

Hence the correct answer is option (1)

#### Consider the sulphides and . Number of these sulphides soluble in is_________.

The sulphide which are soluble in $\mathrm{50% \ HNO_{3}}$ are $\mathrm{PbS,\ CuS,\ As_{2}S_{3},\ CdS}$

Sulphides insoluble in $\mathrm{50%\ HNO_{3}}$ are $\mathrm{HgS}$ and $\mathrm{Sb_{2}S_{3}}$

Hence, the correct answer is 4