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Class 11th The P Block Elements

Draw the structure of boric acid showing hydrogen bonding. Which species is present in water? What is the hybridisation of boron in this species?

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alok kumar singh

Contributor-Level 10

This is a Short Answers Type Questions as classified in NCERT Exemplar

H₃BO₃

Boric acid forms a hexagon and rings through hydrogen bonding and has a layer-like structure.

Boric acid is present in water as [B (OH)₄]−.H₃BO₃ electron from the OH of water and forms the complex BOH for negative for sp³and is present in sp³ hybridisation.

Reaction:

B (OH)₃+2H₂O→ [B (OH)₄]⁻+ H₃O⁺

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Class 11th System of Particles and Rotational Motion

A disc of radius R is rotating with an angular speed ωo about a horizontal axis. It is placed on a horizontal table. The coefficient of kinetic friction is µk.

(a) What was the velocity of its centre of mass before being brought in contact with the table?

(b) What happens to the linear velocity of a point on its rim when placed in contact with the table? (c) What happens to the linear speed of the centre of mass when disc is placed in contact with the table?

(d) Which force is responsible for the effects in (b) and (c).

(e) What condition should be satisfied for rolling to begin?

(f) Calculate the time taken for the rolling to begin

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) Before being bought in contact with the table the disc was in pure rotational motion. Hence V_cm=0

(b) when the disc is placed in contact with table due to friction centre of mass acquires some linear velocity.

(c) when the rotating disc is placed in contact with the table due to friction centre of mass acquires some linear velocity.

(d) friction is responsible for the effects ib b and c

(e) when rolling starts V_cm=wR

(f) time period for rolling to begin is t= $\frac{R w_{o}}{μ g (1 + \frac{m R^{2}}{I})}$

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10 months ago

Class 11th System of Particles and Rotational Motion

Two discs of moments of inertia I₁ and I₂ about their respective axes (normal to the disc and passing through the centre), and rotating with angular speed ω₁ and w₂ are brought into contact face to face with their axes of rotation coincident.

(a) Does the law of conservation of angular momentum apply to the situation? why?

(b) Find the angular speed of the two-disc system.

(c) Calculate the loss in kinetic energy of the system in the process.

(d) Account for this loss.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

(a) yes the law of conservation of angular momentum can be applied because there is no net external torque on the system of the two discs.

External forces gravitation and normal reaction act through the axis of rotation, hence produce no torque.

(b) by conservation of angular momentum

L_f= L_i

Iw=I₁w₁+I₂w₂

So w= $\frac{I_{1} w_{1} + I_{2} w_{2}}{I_{1} + I_{2}}$

(c) K_f= $\frac{1}{2} (I_{1} + I_{2}) \frac{{(I_{1} w_{1} + I_{2} w_{2)}}^{2}}{I_{1} + I_{2}}$

K_f= ½ (I₁w₁²+I₂w₂²)

$? k =$ K_f-K_i =- $\frac{- I_{1} I_{2}}{2 (I_{1} + I_{2})} (w_{1} - w_{2})$ ²<0

(d) hence there is loss of KE of the system. The loss of kinetic energy is mainly due to work against the friction.

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Class 11th System of Particles and Rotational Motion

Find the centre of mass of a uniform (a) half-disc, (b) quarter-disc.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

let M and R be mass and radius of half disc , mass per unit area of half disc

So m = $\frac{M}{\frac{1}{2} π R^{2}}$

(a) the half disc be supposed to be consist of a large number of semicircular ring of mass dm and thickness dr and radii ranging from r=0 to r=R.

surface area of semicircular ring of radius r and thickness dr = $\frac{1}{2} 2 π r d r = π r d r$

so mass of the elementary ring dm = $π r d r * \frac{2 M}{π R^{2}}$

dm= $\frac{2 M}{R^{2}} r d r$

if x,y are coordinates of centre of mass of this element,

then (x,y)=(0,2r/ $π$ )

so x=0 and y =2r/ $π$

let x_cm and y_cm be the coordinates of the centre of mass of the semicircular disc

so x_cm= $\frac{1}{M} \int_{o}^{R} x d m = \frac{1}{M} \int_{o}^{R} 0 d m = 0$

Y_cm= $\frac{1}{M} \int_{o}^{R} y d m = \frac{1}{M} \int_{0}^{R} \frac{2 r}{π} * (\frac{2 M}{R^{2}} r d r)$

= $\frac{4}{π R^{2}} \int_{0}^{R} r^{2} d r = \frac{4}{π R^{2}} [\frac{r^{3}}{3}]$ ₀^R

= 4R/3 $π$

Centre of mass of a uniform quar

...more

This is a long answer type question as classified in NCERT Exemplar

let M and R be mass and radius of half disc , mass per unit area of half disc

So m = $\frac{M}{\frac{1}{2} π R^{2}}$

(a) the half disc be supposed to be consist of a large number of semicircular ring of mass dm and thickness dr and radii ranging from r=0 to r=R.

surface area of semicircular ring of radius r and thickness dr = $\frac{1}{2} 2 π r d r = π r d r$

so mass of the elementary ring dm = $π r d r * \frac{2 M}{π R^{2}}$

dm= $\frac{2 M}{R^{2}} r d r$

if x,y are coordinates of centre of mass of this element,

then (x,y)=(0,2r/ $π$ )

so x=0 and y =2r/ $π$

let x_cm and y_cm be the coordinates of the centre of mass of the semicircular disc

so x_cm= $\frac{1}{M} \int_{o}^{R} x d m = \frac{1}{M} \int_{o}^{R} 0 d m = 0$

Y_cm= $\frac{1}{M} \int_{o}^{R} y d m = \frac{1}{M} \int_{0}^{R} \frac{2 r}{π} * (\frac{2 M}{R^{2}} r d r)$

= $\frac{4}{π R^{2}} \int_{0}^{R} r^{2} d r = \frac{4}{π R^{2}} [\frac{r^{3}}{3}]$ ₀^R

= 4R/3 $π$

Centre of mass of a uniform quarter disc

Mass per unit area of the quarter disc = $\frac{M}{\frac{π R^{2}}{4}} = \frac{4 M}{π R^{2}}$

For half disc along yaxis and xaxis =4R/3 $π$

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10 months ago

Class 11th The P Block Elements

Explain the nature of boric acid as a Lewis acid in water

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alok kumar singh

Contributor-Level 10

This is a Short Answers Type Questions as classified in NCERT Exemplar

H₃BO₃ or B (OH)₃ is an electron deficient compound or Lewis acid which easily accepts electron from OH of water and releases its proton hence it is a Monobasic acid.

Reaction:

B (OH)₃+ 2H₂O → [B (OH)₄]⁻+ H₃O⁺

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10 months ago

Class 11th The P Block Elements

Draw the structures of BCl₃.NH₃ and AlCl₃ (dimer).

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alok kumar singh

Contributor-Level 10

This is a Short Answers Type Questions as classified in NCERT Exemplar

The central B atom in BCl₃ has six electrons in its valence shell. As a result, it is an electron-deficient molecule in need of two more electrons to complete its octet. To put it another way, BCl₃ acts as a Lewis acid. NH₃ on the other hand, has a lone pair of electrons that it can easily donate. As a result, NH₃ serves as a Lewis base. As shown below, the Lewis acid (BCl₃) and Lewis base (NH₃) combine to form an adduct:

The valence shell of AlCl₃ contains six electrons. As a result, it is an electron-deficient molecule that requires two additional electrons to compl

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Class 11th The P Block Elements

A tetravalent element forms monoxide and dioxide with oxygen. When air is passed over a heated element (1273 K), producer gas is obtained. Monoxide of the element is a powerful reducing agent and reduces ferric oxide to iron. Identify the element and write formulas of its monoxide and dioxide. Write chemical equations for the formation of producer gas and reduction of ferric oxide with the monoxide

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alok kumar singh

Contributor-Level 10

This is a Long Answers Type Questions as classified in NCERT Exemplar

2C (s) + O₂+ 4N₂ (g) 2CO (g) + 4N₂ ( g)

Fe₂O₃ (S)+3CO (g) 2Fe (S)+3CO₂

C= tetravalent carbon

CO= carbon monoxide

Fe₂O₃= ferric oxide

CO₂= carbon dioxide

Tetravalent elements i.e. carbon combines with oxygen to produce carbon monoxide. The reaction occurs at high temperatures and in the presence of nitrogen gas, which acts as a producer gas only. It is not consumed in the reaction. The carbon monoxide formed acts as a reducing agent for ferric oxide, reduces the oxidation state of iron from +3 to zero, and oxidiz

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Class 11th The P Block Elements

A non-metallic element of group 13, used in making bullet proof vests is an extremely hard solid of black color, It can exist in many allotropic forms and has unusually high melting points. Its trifluoride acts as Lewis acid towards ammonia. The element exhibits a maximum covalency of four. Identify the element and write the reaction of its trifluoride with ammonia. Explain why the trifluoride acts as a Lewis acid

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alok kumar singh

Contributor-Level 10

This is a Long Answers Type Questions as classified in NCERT Exemplar

Boron is the only non-metallic and extremely hard element in group 13, and it is also used to make bulletproof vests. Boron exists in a variety of allotropic forms. It usually has a high melting point and no d orbital. Using 2s and 2p orbitals, it can achieve a maximum covalency of 4 . Because the octet of boron is not completed in trivalent halides of boron, it acts as Lewis acid. It forms an adduct when it reacts with Lewis base.

BF₃+NH₃→H₃N−B−F₃

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Class 11th The P Block Elements

A compound (A) of boron reacts with NMe₃ to give an adduct (B) which on hydrolysis gives a compound (C) and hydrogen gas. Compound (C) is an acid. Identify the compounds A, B and C. Give the reactions involved.

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alok kumar singh

Contributor-Level 10

This is a Long Answers Type Questions as classified in NCERT Exemplar

B₂H₆+ 2NMe₃→ 2BH₃NMe₃BH₃NMe₃+H₂O → H₃BO₃+ NMe₃+ 6H₂

(A) (B) (C)

(A): B₂H₆

(B): 2BH₃NMe₃

(C): H₃BO₃

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10 months ago

Class 11th The P Block Elements

(i) What are silicones? State the uses of silicones. (ii) What are boranes? Give chemical equation for the preparation of diborane.

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alok kumar singh

Contributor-Level 10

This is a Long Answers Type Questions as classified in NCERT Exemplar

(i) Organosilicon polymers having (R₂SiO₂) as monomer units are called silicones. Silicones contain organic side groups which surround it giving alkane like nature and makes it hydrophobic. Silicones are applicative in electrical insulators, water proofing, sealant. They have been utilized in the biological field in cosmetic implants and other surgeries.

(ii) Boranes correspond to alkane-like compounds of boron. They consist of boron and hydrogen. Most common borane existing is dibecane.

4BF₃+ 3LiAlH₄?2B₂H₆+ 3LiF + 3AIF₃

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