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10 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)& (iii)

Aluminium is the element which is a metal and good conductor of electricity.

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10 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)& (iii)

Ionic radii decreases with the increase of the effective nuclear charge and increases with the increase of shielding effect or e- - e - repulsion as it outweighs the effective nuclear charge effect.

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10 months ago

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)& (iv)

Both electronegativity and metallic character do not possess units as they both are qualitative properties not a quantitative property.

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10 months ago

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A
Anya Singh

Contributor-Level 9

Ahmedabad Business School campus is located in Gujarat and offers a safe environment to students. The college offers all the essential facilities for students, including academic blocks, a library, collaboration spaces, a café, a convenience store, etc. ABS Gujarat campus offers a modern infrastructure and caters to all the needs of students. Further, it has state-of-the-art infrastructure which includes designed classrooms, 24/7 computer labs, a modern library, a huge auditorium, etc.

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10 months ago

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H
Himanshu Singh

Contributor-Level 10

The final step of BTech admission at Neelkanth involves paying the tuition fees, which range between INR 2.1 Lacs to INR 2.4 lakh, and submitting original documents for verification at the campus. Once completed, the admission is finalized, allowing students to start their academic journey.

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10 months ago

0 Follower 14 Views

A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (ii)& (iii)

The ionization enthalpy of N is higher than that of F because it possess half filled orbital which provide it extra stability due to symmetry.

 As the atomic size of F is much smaller thus its electron gain enthalpy is lower than that of the Cl

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10 months ago

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H
Himanshu Singh

Contributor-Level 10

BTech selection is based on the candidate's entrance exam score in UPTAC or JEE Main. After the Neelkanth application submission, candidates participate in counselling and seat allotment rounds conducted by the authorities. Document verification and seat confirmation are also part of the selection process.

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10 months ago

0 Follower 23 Views

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (ii)& (iii)

Isoelectronic species/ions are those species that possess the same number of electrons.

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10 months ago

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H
Himanshu Singh

Contributor-Level 10

Students can apply online by visiting the official Neelkanth Group of Institutions website. They must register by providing personal details, verify their email, complete the application form, upload necessary documents, pay the application fee, and submit the form to complete the process for BTech admission.

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10 months ago

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i), (iii)& (iv)

He with 1s2 electronic configuration has the highest electron gain enthalpy in the periodic table. In any period the alkali metals have the lowest effective nuclear charge in that particular period for which its atomic radius is the highest in that period.

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10 months ago

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H
Himanshu Singh

Contributor-Level 10

Admission is primarily entrance exam-based. Candidates are selected based on their scores in exams like UPTAC or JEE Main. Merit in these exams determines eligibility for counselling and seat allotment, making entrance exam performance crucial for BTech admission at Neelkanth Group of Institutions.

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10 months ago

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Vishal Baghel

Contributor-Level 10

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

Explanation- When we are considering a point source of sound wave. The disturbance due to the source propagates in spherical symmetry that is in all directions. The formation of

wavefront is in accordance with Huygen's principle.

So, Huygen's principle is valid for longitudinal sound waves also.

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10 months ago

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Pragati Taneja

Contributor-Level 10

Yes, admission to Oriental College of Management is completely based on entrance exams's scores. Candidates must complete graduation with a 50% aggregate (45% for reserve category). The selection criteria for this couse is based on CMAT scores.

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10 months ago

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alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (i)& (iv)

Both S and Cl have the higher tendency to gain electrons in order to attain stable the nearest noble gas configuration i.e. of Argon.

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10 months ago

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Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar
Explanation-
refractive index = 1.38 refractive index = 1.5

λ = 5500 A 0

Consider a ray incident at an angle i. A part of this ray is reflected from the air-film interface And apart refracted inside.

This is partly reflected at the film-glass interface and a part transmitted. A part of the

reflected ray is reflected at the film-air interface and a part transmitted as r2 parallel to r 1. Of course successive reflections and transmissions will keep on decreasing the amplitude of the wave. Hence, rays r 1 and r2 shall dominate the behaviour. If incident light is to be transmitted thro

...more

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10 months ago

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A
alok kumar singh

Contributor-Level 10

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option   (i)& (iv)

The s-block elements and p-block elements together are called representative elements.

New Question

10 months ago

The optical properties of a medium are governed by the relative permittivity  (εr) and relative permeability (μr). The refractive index is defined as μ r ε r = n.For ordinary material εr > 0 and μr > 0 and the positive sign is taken for the square root. In 1964, a Russian scientist V. Veselago postulated the existence of material with εr < 0 and μr < 0. Since then such ‘metamaterials’ have been produced in the laboratories and their optical properties studied. For such materials n = – μ r ε r   . As light enters a medium of such refractive index the phases travel away from the direction of propagation.  (i) According to the description above show that if rays of light enter such a medium from air (refractive index =1) at an angle θ in 2nd

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Vishal Baghel

Contributor-Level 10

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

Explanation-All points with the same optical path length must have the same phase.

So – μ r ε r A E   =BC-–  μ r ε r C D

BC= μ r ε r (CD-AE)

BC>0, si must be greater than AD

But in other figure

–  μ r ε r A E = B C     μ r ε r C D

So BC= –  μ r ε r C D - A E

But clearly here BE is less than zero

To proving snells law we know that

BC=ACsin θ and CD-AE=ACsin θ

So n= sini/sinr

New Question

10 months ago

0 Follower 7 Views

V
Vishal Baghel

Contributor-Level 10

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

Explanation-consider the disturbance at the receiver R1 which is at a distance d from B

YA= acos(wt)  and path difference is λ 2  hence phase difference is π .

Thus the wave R1 because of B

YB= acos(wt- π )= - acoswt here path difference is λ  and hence phase difference is 2 π

Thus R1 because of C

Yc= acos(wt-2 π )= acoswt

(i)let the signal picked up at R2 from B be YB= a1cos(wt)

The path difference between signal at D and that B is λ 2

YD= -a1cos(wt)

The path difference between signal at A and that atB is

d 2 + ( λ 2 ) 2 -d = d( 1 + λ 2 4 D 2 ) 1 / 2 -d = λ 2 8 d 2

a s d ? λ  therefore path difference os 0

p h a s e d i f f e r e n c e = 2 π γ λ 2 8 d 2

Y A=a1co

...more

New Question

10 months ago

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New Question

10 months ago

0 Follower 91 Views

V
Vishal Baghel

Contributor-Level 10

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

Explanation- as the refractive index of the class μ , the path difference will be calculated as ? x =2dsin θ +( μ - 1 )L

For principal maxima ,(path difference is zero)

2dsin θ 0+( μ - 1 )L=0

Sin θ 0= - L ( μ - 1 ) 2 d = - L ( 0.5 ) 2 d

Sin θ 0=-1/16

OP=Dtan θ 0= Dsin θ 0=-D/16

For pat ? h difference ? λ 2

2dsin θ 1+0.5L= ? λ 2

Sin θ 1= ? λ 2 - 0.5 L 2 d = ? λ 2 - d 8 2 d

= λ 2 - λ 8 2 λ = ? 1/4 -1/16

So two possible values 1 4 - 1 16 = 3 16  and- 1 4 - 1 16 = - 5 16

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