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Physics NCERT Exemplar Solutions Class 11th Chapter Six

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Physics NCERT Exemplar Solutions Class 11th Chapter Six Physics Nuclei

From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire:

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Raj Pandey

Contributor-Level 9

Kindly consider the following Image

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Physics NCERT Exemplar Solutions Class 11th Chapter Six Physics Nuclei

A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64 . The ratio of radius of two daughter nuclei respectively is :

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Raj Pandey

Contributor-Level 9

$\begin{array}{r} R = R_{0} (A)^{1 / 3} \\ \frac{R_{1}}{R_{2}} = {(\frac{125}{64})}^{1 / 3} = \frac{5}{4} \end{array}$

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2 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Six Class 12th

Match List I with List II

List I Pollutant		List II Disease / sickness
A.	Benzenesulphonyl chloride	I.	Test for primary aminies
B.	Hoffmann bromamide reaction	II.	Anti saytzeff
C.	Carbylamine reaction	III.	Hinsberg reagent
D.	Hoffman orientation	IV.	Known reaction of isocyanates

Choose the correct answer from the options given below ;

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alok kumar singh

Contributor-Level 10

Intermediate : R – N = C = O (Isocyanate)

(C) Carbylamine reaction ->Test for primary amine.

$R - N H_{2} o r A r - N H_{2} + C H C l_{3} + 3 K O H \to R - N C o r A r - N C + 3 K C l + 3 H_{2} O$

(D) Hoffmann orientation -> Anti saytzeff (Formation of less substituted alkene as major product)

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Physics NCERT Exemplar Solutions Class 11th Chapter Six Class 12th

In carius method of estimation of halogen, 0.45g of an organic compound gave 0.36g of AgBr. Find out the percentage of bromine in the compound.

(Molar masses: AgBr = 188 g mol^-¹, Br = 80 g mol^-)

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alok kumar singh

Contributor-Level 10

Mass of organic compound = 0.45 gm

Mass of AgBr obtained = 0.36 gm

$∴ M o l e s o f A g B r = \frac{0.36}{188}$

$∴ M a s s o f B r o m i n e = \frac{0.36}{188} * 80 = 0.1532 g m$

$∴ % B r i n c o m p o u n d = \frac{0.1532}{0.45} * 100 = 34.04 %$

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Physics NCERT Exemplar Solutions Class 11th Chapter Six Class 12th

Match List I with List II

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alok kumar singh

Contributor-Level 10

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Physics NCERT Exemplar Solutions Class 11th Chapter Six Biomolecules

A sugar 'X' dehydrates very slowly under acidic condition to give furfural which on further reaction with resorcinol given the coloured product after sometime.

Sugar 'X' is

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alok kumar singh

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Physics NCERT Exemplar Solutions Class 11th Chapter Six Class 12th

Match List I with List II

List I Polymers		List II Commercial names
A.	Phenol-formaldehyde resin	I.	Glyptal
B.	Copolymer of 1,3-butadiene and styrene	II.	Novolac
C.	Polyester of glycol and phtalic acid	III.	Buna-S
D.	Polyester of glycol and terephthalic acid	IV.	Dacron

Choose the correct answer from the options given below;

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alok kumar singh

Contributor-Level 10

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Physics NCERT Exemplar Solutions Class 11th Chapter Six Chemistry Hydrocarbon

Given below are two statements. One is labeled as Assertion A and, the other is labeled as Reason R.

Assertion A: [6] Annulene, [8] Annulene and cis-[10] Annulene, as respectively aromatic, not-aromatic and aromatics.

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alok kumar singh

Contributor-Level 10

Assertion -> Not correct

But Reason R -> correct

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3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Six Class 11th

Two blocks M₁ and M₂ having equal mass are free to move on a horizontal frictionless surface. M₂ is attached to a massless spring as shown in Fig. 6.10. Initially M₂ is at rest and M₁ is moving toward M₂ with speed v and collides head-on with M₂.

(a) While spring is fully compressed all the KE of M₁ is stored as PE of spring.

(b) While spring is fully compressed the system momentum is not conserved, though final momentum is equal to initial momentum.

(c) If spring is massless, the final state of the M₁ is state of rest.

(d) If the surface on which blocks are moving has friction, then collision cannot be elastic.

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force.

a) So spring will continue compress until the blocks moves with the same velocity.

b) As the surfaces are frictionless momentum of the system will conserved.

c) If the spring is massless whole energy of M1 will be imparted and M1 will be rest .

d) collision is elastic even if friction is not involved.

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3 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Six Class 11th

A bullet of mass m fired at 30° to the horizontal leaves the barrel of the gun with a velocity v. The bullet hits a soft target at a height h above the ground while it is moving downward and emerges out with half the kinetic energy it had before hitting the target. Which of the following statements are correct in respect of bullet after it emerges out of the target?

(a) The velocity of the bullet will be reduced to half its initial value.

(b) The velocity of the bullet will be more than half of its earlier velocity.

(c) The bullet will continue to move along the same parabolic path.

(d) The bullet will move in a different parabolic p

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Payal Gupta

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

0+1/2mv²= mgh + 1/2mv'

$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v')²=v²-2gh = v^'= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv'²]

1/2m(v₁)²=1/4m(v')²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

From eqn 1 and 2

$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

So v₁ = v'/ $\sqrt[]{2}$ =v²(v'/2)

v₁>v'/2

hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v'

(d) as the velocity of the bullet changes to v' which is less than v' hence , path, followed will change and the bullet reaches at point B instead of A

(f) as the bullet is passing through the target

...more

This is a multiple choice answer as classified in NCERT Exemplar

(b) conserving energy between “O” ans ”A”

U_i+ K_i= U_f+ K_f

0+1/2mv²= mgh + 1/2mv'

$\frac{{(v')}^{2}}{2} = \frac{v^{2}}{2} = - g h$

(v')²=v²-2gh = v^'= $\sqrt[]{v^{2} - 2 g h}$ ……….1

Let speed after emerging be v₁then

=1/2mv₁²=1/2[1/2mv'²]

1/2m(v₁)²=1/4m(v')²=1/4m[v²-2gh]

V₁= $\sqrt[]{\frac{v^{2}}{2} - g h}$ ………….2

From eqn 1 and 2

$\frac{v'}{v_{1}} = \frac{\sqrt[]{v^{2} - 2 g h}}{\sqrt[]{v^{2} - 2 g h / \sqrt[]{2}}} = \sqrt[]{2}$

So v₁ = v'/ $\sqrt[]{2}$ =v²(v'/2)

v₁>v'/2

hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v'

(d) as the velocity of the bullet changes to v' which is less than v' hence , path, followed will change and the bullet reaches at point B instead of A

(f) as the bullet is passing through the target the loss in energy of the bullet is transferred to particles of the target . therefore their internal energy increases.

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