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6 months ago

Class 12th System of Particles and Rotational Motion

A rolling wheel of 12kg is on an inclined plane at position P and connected to a mass of 3kg through a string of fixed length and pulley as shown in figure.

Consider PR as friction free surface. The velocity of centre of mass of the wheel when it reaches at the bottom Q of the inclined plane PQ will be $\frac{1}{2} \sqrt[]{x g h} m / s .$ The value of x is…………

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Payal Gupta

Contributor-Level 10

By conservation of Energy

$m g h = \frac{1}{2} l_{P} ω^{2}$

3gh $= \frac{1}{2} (M R^{2} + M R^{2}) ω^{2}$

$3 g h = m v^{2}$

$v^{2} = \frac{3 g h}{12}$

$v = \frac{\sqrt[]{g h}}{2} = \frac{\sqrt[]{x g h}}{2}$

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Class 12th Indian School of Technology and Management

Does Indian School of Technology and Management, Mumbai accept Class 12th marks?

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Parul Shukla

Contributor-Level 10

Candidates seeking admission to UG, PG, PhD, Diploma programme can enrol for admission with Class 12/graduation/PG marks as per course requirement. The college offers various UG courses in various streams such as Management Studies, Computer Science, etc. ISTM, Mumbai admissions are based on merit.

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6 months ago

Class 12th Laws of Motion

A mass of 10kg is suspended vertically by a rope of length 5m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is q = tan^-¹ (x * 10^-¹). The value of x is…………….

(Given, g = 10 m/s²)

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Payal Gupta

Contributor-Level 10

F.B.D. of point 'P'

$T s i n θ = 30$ - (1)

$T c o s θ = 100$ - (2)

$(1) \div (2)$

$t a n θ = \frac{3}{10} \Rightarrow θ = t a n^{- 1} (3 * 1 0^{- 1})$

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Class 12th Chemistry in Everyday Life

Match List I with List II

List I		List II
(Anion)		(gas evolved on reaction with dil H₂SO₄)
A.	CO₃^2-	I.	Colourless gas which turns lead acetate paper black
B.	S²^-	II.	Colourless gas which turns acidified potassium dichromate solution green
C.	SO₃²-	III.	Brown fumes which turns acidified Kl solution containing starch blue.
D.	NO₂^-	IV.	Colourless gas evolved with brisk effervescence, which turns lime water milky

Choose the correct answer from the option given below:

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Payal Gupta

Contributor-Level 10

A. $\underset{S a l t}{N a_{2} C O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a S O_{4} + H_{2} O + C O_{2} ↑$

CO₂ is colourless gas with brisk effervescence and turns lime water milky

$C O_{2} + C a {(O H)}_{2} \to C a C O_{3} ↓ + H_{2} O$

B. $\underset{S a l t}{N a_{2} S} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} S ↑$

H₂S + (CH₃COO)₂Pb → $\underset{B l a c k}{P b S} + 2 C H_{3} C O O H$

C. $\underset{S a l t}{N a_{2} S O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} O + S O_{3} ↑$

SO₃ is colourless gas which turns acidified potassium dichromate solution green as

$S O_{3} + K M n O_{4} + H^{+} \to \underset{G r e e n}{M n S O_{4}} + S O_{4}^{2 -} + H_{2} O$

D. $\underset{S a l t}{2 N a N O_{2}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + 2 H N O_{2}$ &nbs

...more

A. $\underset{S a l t}{N a_{2} C O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a S O_{4} + H_{2} O + C O_{2} ↑$

CO₂ is colourless gas with brisk effervescence and turns lime water milky

$C O_{2} + C a {(O H)}_{2} \to C a C O_{3} ↓ + H_{2} O$

B. $\underset{S a l t}{N a_{2} S} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} S ↑$

H₂S + (CH₃COO)₂Pb → $\underset{B l a c k}{P b S} + 2 C H_{3} C O O H$

C. $\underset{S a l t}{N a_{2} S O_{3}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + H_{2} O + S O_{3} ↑$

SO₃ is colourless gas which turns acidified potassium dichromate solution green as

$S O_{3} + K M n O_{4} + H^{+} \to \underset{G r e e n}{M n S O_{4}} + S O_{4}^{2 -} + H_{2} O$

D. $\underset{S a l t}{2 N a N O_{2}} + \underset{d i l}{H_{2} S O_{4}} \to N a_{2} S O_{4} + 2 H N O_{2}$

$3 H N O_{2} \to N H O_{3} + 2 N O + H_{2} O$

$2 N O + O_{2} \to \underset{B r o w n g a s}{2 N O_{2}}$

$2 H N O_{2} + 2 K l + 2 C H_{3} C O O H \to 2 C H_{3} C O O K + 2 H_{2} O + 2 N O + I_{2}$

I₂ + starch → Blue colour complex.

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Class 12th Chemistry in Everyday Life

Match List I with List II

List I List II

A. Antipyretic I. Reduces pain

B. Analgesic II. Reduces stress

C. Tranquilizer &nbs

...more

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Payal Gupta

Contributor-Level 10

A. Antipyretic reduces fever.

B. Analgesic reduces pain.

C. Tranquilizer reduces stress.

D. Antacid reduces acidity of stomach.

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Class 12th Biomolecules

Given below are two statements.

Statements I: Maltose has two a-D-glucose units linked at C₁ and C₄ and is a reducing sugar:

Statements II: Maltose has two manosaccharides: a-D-glucose and b-D-glucose linked at C₁ and C₆ and its is a non-reducing sugar.

In the light of the above statements, choose the correct answer from the options given below.

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Payal Gupta

Contributor-Level 10

$C_{12} H_{22} O_{11} + H_{2} O \to α - D - g l u c o s e + α - D - g l u c o s e$

In maltose, two units of a-D-glucose are linked by glycosidic linkage at C₁ and C₄ as shown.

a - D – glucose a - D - glucose

Maltose has hemiacetal link so it is reducing sugar.

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Class 12th Polymers

Which is true about Buna-N

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Payal Gupta

Contributor-Level 10

Buna-N is obtained by copolymerization of 1, 3-butadiene and acrylonitrile,

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Class 12th Amines

Decarboxylation of all six possible forms of diaminobenzoic acid $C_{6} H_{3} {(N H_{2})}_{2} C O O H$ yields three product A, B and C. Three acids give a products 'A' two acids gives a products 'B' and one acids give a products 'C'. The melting point of product 'C' is

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Payal Gupta

Contributor-Level 10

Above are the six possible forms of diaminobenzoic acid.

On decarboxylation we get;

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Class 12th Biomolecules

Match List I with List II

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Payal Gupta

Contributor-Level 10

...more

A. Phenol is converted to salicylaldehyde by Riemer-Tiemann reaction

B. Phenol is converted to benzene using Zn powder

C. Phenol is converted to benzoinone using oxidizing agent which is $N a_{2} C r_{2} O_{7} / H^{+}$

D. Phenol can be converted to p-bromophenol using Br₂/CS₂

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6 months ago

Class 12th Environmental chemistry

On the surface of polar stratospheric clouds, hydrolysis of chlorine nitrate given A and B while its reaction with HCl produces B and C.A, B and C are, respectively.

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Payal Gupta

Contributor-Level 10

Chlorine nitrate as hydrolysis gives hypohalous acid and nitric acid as,

$C l O N O_{2} + H_{2} O \to \underset{A}{H O C l} + \underset{B}{H N O_{3}}$

Chlorine nitrate on reaction with HCl produces Cl₂ and HNO₃ as,

$C l O N O_{2} + H C l \to \underset{C}{C l_{2}} + \underset{B}{H N O_{3}}$

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