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10 months ago

Exam Engineering Entrance Exam

My rank in AEEE 2025 is 18678. Can I get into Amrita campus at Coimbatore with this rank?

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10 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen physics ncert solutions class 11th

Show that the motion of a particle represented by y = sinω t – cos ω t is simple harmonic with a period of 2π/ω.

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Payal Gupta

Contributor-Level 10

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

As we know displacement y=sinwt-coswt

= $\sqrt[]{2} (\frac{1}{\sqrt[]{2}} s i n w t - \frac{1}{\sqrt[]{2}} c o s w t)$

= $\sqrt[]{2} (c o s \frac{π}{4} s i n w t - s i n \frac{π}{4} c o s w t)$

= $\sqrt[]{2} [s i n (w t - \frac{π}{4})]$

To comparing with standard equation

Y= asin (wt+ $\emptyset$ )

So T=2 $π$ /w

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10 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter Nine Coordination Compounds

Why are different colours observed in octahedral and tetrahedral complexes for the same metal and same ligands?

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Vishal Baghel

Contributor-Level 10

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

Ans: The octahedral and tetrahedral splitting up of d orbitals is different.

Δt= ( $\frac{4}{9}$ )Δ₀

Δt < Δ₀

Δt= CFSE in tetrahedral field

Δ₀= CFSE in octahedral field

Comparing CFSE with energy= $\frac{h c}{λ}$

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10 months ago

Ranchi University B.A

What is the total tuition fee for BA at Ranchi University?

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Shailja Rawat

Contributor-Level 10

The total tuition fee for Ranchi University BA courses ranges from INR 966 to INR 5.2 lakh. This fee is given in a range, as the fee varies from institute to institute.

Note: This fee has been sourced from official sources. However, it is subject to change. Hence, it is indicative.

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10 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter Nine Coordination Compounds

What is the relationship between observed colour of the complex and the wavelength of light absorbed by the complex?

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Vishal Baghel

Contributor-Level 10

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

Ans: When light wavelengths from a specific part of the spectrum are absorbed by a substance, the result is a complimentary colour. When a complex absorbs a wavelength of light, it reflects a complementary colour. If a violent colour is absorbed, for example, yellow is conveyed. The CFSE value, often known as the colour definer for any complex, comes next. To determine the wavelength value in order to determine which colour absorbs the most energy.

Δe= $\frac{h c}{λ}$

As λ has shorter wavelength.

Low spin complexes absorb shorter wavelengths, while high spin complexes

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

Δe= $\frac{h c}{λ}$

As λ has shorter wavelength.

Low spin complexes absorb shorter wavelengths, while high spin complexes absorb longer ones. The colour theory, complex, and wavelength are all determined using this method.

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10 months ago

MBA/PGDM MBA Ranking

What is the MBA ranking at Symbiosis Institute of Computer Studies and Research?

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Bhumika Yadav

Contributor-Level 10

The Symbiosis Institute of Computer Studies and Research has been ranked by recognised ranking bodies for the MBA category. The Business Today have ranked the SICSR Pune based on various parameters. In 2024, the institute claim 13 ranks to #103 in the 'MBA' category by the Business Today. Check out the following table for more information:

Publisher	2022	2023	2024
Business Today	121	116	103

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10 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen physics ncert solutions class 11th

Find the time period of mass M when displaced from its equilibrium positon and then released for the system shown in Fig.

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Payal Gupta

Contributor-Level 10

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

For calculation purpose, in this situation we will neglect gravity because it is constant throughout will not affect the net restoring force.

Let in the equilibrium position, the spring has extended by an amount x_o

Let displacement by spring is and string be x .

But string is extensible so only spring will contribute in extension x+x=2x

So net extension is 2x+x_o

So force is F= 2T

T=kx_o

F=2kx_o

But when mass is lowered down further by x

F’=2T’ but spring length is 2x+x_o

F’=2k (2x+x_o)

Restoring force on the system

F_restoring=- (F’-F)

So using above equati

...more

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

For calculation purpose, in this situation we will neglect gravity because it is constant throughout will not affect the net restoring force.

Let in the equilibrium position, the spring has extended by an amount x_o

Let displacement by spring is and string be x .

But string is extensible so only spring will contribute in extension x+x=2x

So net extension is 2x+x_o

So force is F= 2T

T=kx_o

F=2kx_o

But when mass is lowered down further by x

F’=2T’ but spring length is 2x+x_o

F’=2k (2x+x_o)

Restoring force on the system

F_restoring=- (F’-F)

So using above equations

F_restoring= [2k (2x+x₀)-2Kx_o]=-4kx

Ma=-4kx

So a is directly proportional to negative displacement so it follows SHM.

So time period is T= $\frac{2 π}{w}$

w= $\sqrt[]{\frac{4 k}{m}}$ so T= 2 $π \sqrt[]{\frac{M}{4 K}}$

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10 months ago

PTET Process PTET Counselling

When does the PTET counselling process start?

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Aayushi Harsha

Contributor-Level 7

The PTET counselling starts after the results and merit list are declared. The PTET counselling is conducted across multiple rounds to accommodate candidates with different ranks for the BEd course from participating colleges in Rajasthan. Candidates should regularly check the official website for updates regarding the exact schedule and important dates to avoid missing any deadlines. We will also be sharing the direct link for candidates to view the counselling schedule.

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10 months ago

IIITDM Jabalpur - Indian Institute of Information Technology Design and Manufacturing B.Des

How much cut-off have increased/decreased for B.Des at IIITDM Jabalpur in 2025?

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Vishakha

Contributor-Level 10

The IIITDM Jabalpur BDes cutoff for 2025, in Round 3, has been released. The closing rank for the general AI category is 202. Moreover, after comparing the round 3 closing ranks of the same course, they were 170 and 165 in 2024 and 2023, respectively. Therefore, in the year 2025, the rank increases, which reflects that the competition slightly decreases.

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10 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter Nine Coordination Compounds

CoSO₄Cl.5NH₃ exists in two isomeric forms ‘A’ and ‘B’. Isomer ‘A’ reacts with AgNO₃ to give white precipitate but does not react with BaCl₂ . Isomer ‘B’ gives white precipitate with BaCl₂ but does not react with AgNO₃.

Answer the following questions.

(i) Identify ‘A’ and ‘B’ and write their structural formulas.

(ii) Name the type of isomerism involved.

(iii) Give the IUPAC name of ‘A’ and ‘B’.

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Vishal Baghel

Contributor-Level 10

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

Ans: (i) 'A' is [Co (NH₃)₅SO₄]Cl

'B' is [Co (NH₃)₅Cl]SO₄

_(ii)_{Type of isomerism is ionisation isomerism}

_(iii)IUPAC name of isomer 'A' is pentaaminesulphatocobalt (III)chloride

IUPAC name of isomer 'B' is pentaaminechlorocobalt (III)sulphate

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10 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen physics ncert solutions class 11th

A simple pendulum of time period 1s and length l is hung from a fixed support at O, such that the bob is at a distance H vertically above A on the ground (Fig) The amplitude is θ o . The string snaps at θ = θ₀ /2. Find the time taken by the bob to hit the ground. Also find distance from A where bob hits the ground. Assume θ₀ to be small so that sin _o ₀ and cos ₀

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Payal Gupta

Contributor-Level 10

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

Let us assume t=0 when $θ = θ$ _o then $θ$ = $θ$ ₀coswt

Given a seconds pendulum w=2 $π$

$θ$ = $θ_{o} c o s 2 π t$

At time t, let $θ = θ_{o} / 2$

Cos2 $π t_{1} = \frac{1}{2}$ , t₁=1/6

$\frac{d θ}{d t}$ =-( $θ_{o} 2 π$ )sin2 $π t$

t=t₁=1/6

d $\frac{θ}{d t}$ =- $θ_{o} 2 π$ sin $\frac{2 π}{6} = - \sqrt[]{3} π θ_{o}$

the linear velocity is u=- $\sqrt[]{3} π θ_{o} l$

the vertical component is U_y= $- \sqrt[]{3} π l s i n (\frac{θ_{o}}{2})$

the horizontal component U_x=- $- \sqrt[]{3} π l c o s (\frac{θ_{o}}{2})$

at the time it snaps the vertical height is

H’=H+l(1-cos $(\frac{θ_{o}}{2})$ )

Let the time require for fall be t , then

H’= H+l (1 $- c o s (\frac{θ_{o}}{2})$ )

Let the time required for fall be at t then

H’=u_yt+1/2 gt²

1/2gt²+ $\sqrt[]{3} π θ_{o} l s i n \frac{θ_{o}}{2} t - H^{'} = 0$

t= $\frac{- \sqrt[]{3} π θ_{0} s i n \frac{θ_{o}}{2} \pm \sqrt[]{3 π^{2} θ_{o}^{2} {s i n}^{2} \frac{θ_{o}}{2} + 2 g H'}}{g}$

given that $θ_{o}$ is small , hence neglecting terms of order $θ_{o}^{2}$ and higher

t= $\sqrt[]{\frac{2 H^{'}}{g}}$

H’ $\approx H + l (1 - 1)$

t= $\sqrt[]{\frac{2 H}{g}}$

the distance travelled in the x direction is

...more

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

Let us assume t=0 when $θ = θ$ _o then $θ$ = $θ$ ₀coswt

Given a seconds pendulum w=2 $π$

$θ$ = $θ_{o} c o s 2 π t$

At time t, let $θ = θ_{o} / 2$

Cos2 $π t_{1} = \frac{1}{2}$ , t₁=1/6

$\frac{d θ}{d t}$ =-( $θ_{o} 2 π$ )sin2 $π t$

t=t₁=1/6

d $\frac{θ}{d t}$ =- $θ_{o} 2 π$ sin $\frac{2 π}{6} = - \sqrt[]{3} π θ_{o}$

the linear velocity is u=- $\sqrt[]{3} π θ_{o} l$

the vertical component is U_y= $- \sqrt[]{3} π l s i n (\frac{θ_{o}}{2})$

the horizontal component U_x=- $- \sqrt[]{3} π l c o s (\frac{θ_{o}}{2})$

at the time it snaps the vertical height is

H’=H+l(1-cos $(\frac{θ_{o}}{2})$ )

Let the time require for fall be t , then

H’= H+l (1 $- c o s (\frac{θ_{o}}{2})$ )

Let the time required for fall be at t then

H’=u_yt+1/2 gt²

1/2gt²+ $\sqrt[]{3} π θ_{o} l s i n \frac{θ_{o}}{2} t - H^{'} = 0$

t= $\frac{- \sqrt[]{3} π θ_{0} s i n \frac{θ_{o}}{2} \pm \sqrt[]{3 π^{2} θ_{o}^{2} {s i n}^{2} \frac{θ_{o}}{2} + 2 g H'}}{g}$

given that $θ_{o}$ is small , hence neglecting terms of order $θ_{o}^{2}$ and higher

t= $\sqrt[]{\frac{2 H^{'}}{g}}$

H’ $\approx H + l (1 - 1)$

t= $\sqrt[]{\frac{2 H}{g}}$

the distance travelled in the x direction is u_xt to the left of where the bob is snapped

X= Uxt= $\sqrt[]{3} π θ_{o} l c o s (\frac{θ_{o}}{2}) \sqrt[]{\frac{2 H}{g}} s$

cos $\frac{θ_{o}}{2} \approx 1$

X= $\sqrt[]{3} π θ_{o} l \sqrt[]{\frac{2 h}{g}} = \sqrt[]{\frac{6 H}{g}} θ_{o} l π$

At the time of snapping , the bob was at a horizontal distance of $l s i n \frac{θ_{o}}{2} \approx l \frac{θ_{0}}{2}$ from A

Thus, the distance of bob from A where it meets the ground is

= $\frac{l θ_{o}}{2} - X = \frac{l θ_{o}}{2} - \sqrt[]{\frac{6 H}{g}} θ_{o} l π$

= $θ_{o} l (\frac{1}{2} - π \sqrt[]{\frac{6 H}{g}} θ_{o} l π)$

= $θ_{o} l (\frac{1}{2} - π \sqrt[]{\frac{6 H}{g}})$

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10 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter Nine Coordination Compounds

Using valence bond theory, explain the following in relation to the complexes given below:

[Mn(CN)₆]^3- ,[Co(NH₃)₆]³⁺ ,[Cr(H₂O)₆]³⁺ ,[FeCl]^4-

(i) Type of hybridisation.

(ii) Inner or outer orbital complex.

(iii) Magnetic behaviour.

(iv) Spin only magnetic moment value

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Vishal Baghel

Contributor-Level 10

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

Ans:

[Mn (CN)₆]³⁻

Electronic configuration is Mn³⁺= [Ar]3 d⁴hence box electronic structure

(i) Type of hybridisation d²sp³

(ii) Inner orbital complex

(iii) paramagnetic, due to presence of three unpaired electrons.

(iv) Spin only magnetic moment is calculated using the formula : n=2 in this case, we get spin only magnetic moment in BMas = $\sqrt[]{2 (2 + 2}$ = $\sqrt[]{8}$ = 2.87BM

[Co (NH₃)₆]³⁺

Electronic configuration of Co³⁺= [Ar]3 d⁶

(i) Hyb As shown in the above box electronic structure the type of hybridisation is . d²sp³

(ii) Inner orbital complex

(iii) Diama

...more

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

Ans:

[Mn (CN)₆]³⁻

Electronic configuration is Mn³⁺= [Ar]3 d⁴hence box electronic structure

(i) Type of hybridisation d²sp³

(ii) Inner orbital complex

(iii) paramagnetic, due to presence of three unpaired electrons.

(iv) Spin only magnetic moment is calculated using the formula : n=2 in this case, we get spin only magnetic moment in BMas = $\sqrt[]{2 (2 + 2}$ = $\sqrt[]{8}$ = 2.87BM

[Co (NH₃)₆]³⁺

Electronic configuration of Co³⁺= [Ar]3 d⁶

(i) Hyb As shown in the above box electronic structure the type of hybridisation is . d²sp³

(ii) Inner orbital complex

(iii) Diamagnetic

(iv) Zero ( Since no unpaired electrons are present, as depicted above)

[Cr (H₂O)₆]³⁺

(i) type of hybridisation. d²sp³

(ii) Inner orbital complex

(iii) paramagnetic

(iv) 3.87BM

[Fe (Cl)₆]⁴⁻

electronic configuration is of Fe²⁺= [Ar]3 d⁶

(i) type of hybridisation will be sp³d²

(ii) Outer orbital complex

(iii) Paramagnetic

(iv) 4.9BM

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10 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen physics ncert solutions class 11th

A tunnel is dug through the centre of the Earth. Show that a body of mass ‘m’ when dropped from rest from one end of the tunnel will execute simple harmonic motion.

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Payal Gupta

Contributor-Level 10

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

The gravitational force on the particle at a distance r from the centre of the earth arises entirely from that portion of matter of the earth in shells internal to the positiin of the particles . the external shells exert no force on the particle.

Let g’ be the acceleration at P

So g’ =g (1-d/R)=g (R-d/R)

R-d=y

g’=gy/R’

F=-mg’= -mgy/R

F $\propto - y$

Ma=-Mgy/R, a = -gy/R

Comparing a=-w²y

w²=g/R

T=2 $π \sqrt[]{\frac{R}{g}}$

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10 months ago

IIM Jammu - Indian Institute of Management IIM Jammu - Indian Institute of Management Eligibility

Which entrance exams are accepted by IIM Jammu?

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Abhay Arora

Contributor-Level 10

Indian Institute of Management Jammu offers undergraduate, postgradute, and Diploma courses. The institute accepts admission based on merit and entrance exams. IIM Jammu accepts national entrance exams such as JEE Main, CAT, MAT, GATE, and others. Aspiring candidates must qualify the eligibility criteria set by the institute for the desired course.

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10 months ago

Chemistry NCERT Exemplar Solutions Class 12th Chapter Nine Coordination Compounds

Using crystal field theory, draw energy level diagram, write electronic configuration of the central metal atom/ion and determine the magnetic moment value in the following:

(i) [CoF₆]^3- ,[Co(H₂O)₆]²⁺ ,[Co(CN)₆]^3-

^{(ii) [FeF₆]3- ,[Fe(H₂O)₆]2 + ,[Fe(CN)₆]4 -}

0 Follower 11 Views

Vishal Baghel

Contributor-Level 10

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

Ans: (i) Electronic cnfiguration: Co³⁺ =[Ar]3d⁶

Energy level diagram:

Magnetic moment:

Number of unpaired electrons (n)=4

Magnetic moment = μ= $\sqrt[]{n (n + 2}$ = $\sqrt[]{4 (4 + 2}$

= $\sqrt[]{24}$ = 4.9 BM

[Co(H₂O)₆]²⁺

Electronic cnfiguration: Co²⁺=[Ar]3 d⁷

Energy level diagram:

Magnetic moment: Since ,number of unpaired electrons (n)=3, therefore magnetic moment = $\sqrt[]{3 (3 + 2}$ = $\sqrt[]{15}$ = 3.87BM

[Co(CN)₆]³⁻

Electronic configuration: [Ar]Co³⁺=3 d⁶

Energy level diagram:

(ii)

Ans: [FeF₆]³⁻

Electronic configuration: Fe³⁺=[Ar]3 d⁵

Ener

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

Ans: (i) Electronic cnfiguration: Co³⁺ =[Ar]3d⁶

Energy level diagram:

Magnetic moment:

Number of unpaired electrons (n)=4

Magnetic moment = μ= $\sqrt[]{n (n + 2}$ = $\sqrt[]{4 (4 + 2}$

= $\sqrt[]{24}$ = 4.9 BM

[Co(H₂O)₆]²⁺

Electronic cnfiguration: Co²⁺=[Ar]3 d⁷

Energy level diagram:

Magnetic moment: Since ,number of unpaired electrons (n)=3, therefore magnetic moment = $\sqrt[]{3 (3 + 2}$ = $\sqrt[]{15}$ = 3.87BM

[Co(CN)₆]³⁻

Electronic configuration: [Ar]Co³⁺=3 d⁶

Energy level diagram:

(ii)

Ans: [FeF₆]³⁻

Electronic configuration: Fe³⁺=[Ar]3 d⁵

Energy level diagram:

Number of unpaired electrons = μ= $\sqrt[]{n (n + 2} = \sqrt[]{5 (5 + 2}$ = μ= $\sqrt[]{35}$ = 5.95BM [Fe(H₂O)₆]²⁺

Fe²⁺=[Ar]3 d⁶

Energy level diagram:

Magnetic moment

Since, number of unpaired electrons (n)=4, therefore magnetic moment is-

$\sqrt[]{4 (4 + 2}$ = $\sqrt[]{24}$ = 4.9BM

[Fe(CN)₆]⁴⁻

Electronic configuration: Fe²⁺=[Ar]3 d⁶

Energy level diagram

Since CN⁻is strong field ligand, it forces all the electrons to get paired up Magnetic moment.

In the absence of unpaired electrons it behaves as diamagnetic ( ie. no magnetic behaviour ).

So magnetic moment = zero

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10 months ago

AIIMS Rishikesh Radiology

What is the starting salary after B.Sc radiology from AIIMS Rishikesh? Does it have a good scope?

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10 months ago

Physics NCERT Exemplar Solutions Class 11th Chapter Fourteen physics ncert solutions class 11th

One end of a V-tube containing mercury is connected to a suction pump and the other end to atmosphere. The two arms of the tube are inclined to horizontal at an angle of 45° each. A small pressure difference is created between two columns when the suction pump is removed. Will the column of mercury in V-tube execute simple harmonic motion? Neglect capillary and viscous forces. Find the time period of oscillation.

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Payal Gupta

Contributor-Level 10

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

Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $ρ =$ density of the liquid

PE= dmgx=A $ρ d x g x$

So total PE of the column

= $\int_{0}^{h_{1}} A d x g ρ x$ = $A g ρ \int_{o}^{h_{1}} x d x = A ρ g \frac{h_{1}^{2}}{2}$

But h₁=lsin45

PE=A $ρ$ gl²sin²45/2

Similarly PE of right column = A $ρ$ gl²sin²45/2

Total PE = A $ρ$ gl²sin²45/2+ A $ρ$ gl²sin²45/2

= A $ρ$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A ρ g}{2} [{(l - y)}^{2} + {(l + y)}^{2} - l^{2}$ ]

= $\frac{A ρ g}{2} [2 {(l}^{2} + y^{2})]$ =A $ρ g {(l}^{2} + y^{2})$

Change in KE = ½ mv²

m=A $ρ 2 l$

change in KE= 1/2A $ρ 2 l v^{2}$ =A $ρ l v^{2}$

...more

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

Let us consider an infinitesimal liquid column of length dx at a height x from horizontal line.

If $ρ =$ density of the liquid

PE= dmgx=A $ρ d x g x$

So total PE of the column

= $\int_{0}^{h_{1}} A d x g ρ x$ = $A g ρ \int_{o}^{h_{1}} x d x = A ρ g \frac{h_{1}^{2}}{2}$

But h₁=lsin45

PE=A $ρ$ gl²sin²45/2

Similarly PE of right column = A $ρ$ gl²sin²45/2

Total PE = A $ρ$ gl²sin²45/2+ A $ρ$ gl²sin²45/2

= A $ρ$ gl²/2

If due to pressure difference is created y element of left side moves on the right side then liquid present in the left arm =l-y

But liquid present in the right arm =l+y

Total PE = PE_final-PE_initial

Change in PE = $\frac{A ρ g}{2} [{(l - y)}^{2} + {(l + y)}^{2} - l^{2}$ ]

= $\frac{A ρ g}{2} [2 {(l}^{2} + y^{2})]$ =A $ρ g {(l}^{2} + y^{2})$

Change in KE = ½ mv²

m=A $ρ 2 l$

change in KE= 1/2A $ρ 2 l v^{2}$ =A $ρ l v^{2}$

so from above eqn

change in PE +change in KE = A $ρ g {(l}^{2} + y^{2})$ +A $ρ l v^{2}$

system being conservative

so change in total energy =0

A $ρ g {(l}^{2} + y^{2}) + A ρ l v^{2}$ =0

Differentiating both with respect to time t

A $ρ g (2 y v) + A ρ l (2 v) a = 0$ =0

Dy/dt =v and dv/dt=a

(gy+la)2A $ρ v$ =0

2A $ρ v = c o n s t a n t$ and 2A $ρ v \neq 0$

La+gy=0

A+(gy/l)=0

d²y/dt²+gy/l=0

d²y/dt²+w²y=0

w= $\sqrt[]{\frac{g}{l}}$

T=2 $π \sqrt[]{\frac{l}{g}}$

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10 months ago

IGNOU - Indira Gandhi National Open University B.Ed

How can I register for IGNOU B.Ed counselling?

1 Follower 25 Views

Mayank Kumari

Contributor-Level 7

To register for the counselling process candidates will have to follow the steps mentioned below:

Visit the official website of IGNOU and check the dates.
Visit the regional centre in offline mode.
Pay the counselling registration fee.
Fill your preferences for the centre.
Submit and lock your choices before the deadline.
Ensure that all details are accurate and the choices are finalized before submission.

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10 months ago

Humanities & Social Sciences Paramedical

I have to do MA in clinical psychology and 3rd round is going to happen. My score is 63%. Will I get admission in GU Navrangpura?

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JITESH SINGH

Contributor-Level 6

It is not confirmed that you will get admission into the MA Clinical Psychology programme at Gujarat University's Navrangpura campus with a 63% score. The eligibility criteria for most postgraduate courses, including psychology, usually require a minimum of 55% to 60%, and some programs, like Amity's, specify 65%. Since the third round is happening, it's possible that the seats are limited and the competition is high.

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10 months ago

Maths NCERT Exemplar Solutions Class 12th Chapter Seven Maths

Kindly consider the following

$\int e^{- 3 x} c o s^{3} x d x$

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alok kumar singh

Contributor-Level 10

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

Sol:

$\begin{array}{l} L e t ? ? ? ? I = ? \underset{I I}{e^{? 3 x}} \underset{I}{c o s^{3} x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . ? e^{? 3 x} ? d x ? ? (D (c o s^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 c o s^{2} x (? s i n x) . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? c o s^{2} x s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? (1 ? s i n^{2} x) s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + ? \underset{I}{s i n^{3} x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x ? e^{? 3 x} d x ? ? (D (s i n^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 s i n^{2} x . c o s x . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? s i n^{2} x c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? (1 ? c o s^{2} x) c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? ? c o s^{3} x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x + s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? I \\ ? ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] ? [s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{3} ? c o s x . e^{? 3 x} d x + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} + \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ N o w, ? I_{1} = \frac{2}{3} ? \underset{I}{c o s x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . ? e^{? 3 x} d x ? ? (D (c o s x) . ? e^{? 3 x} ? d x) d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? ? ? s i n x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? \frac{1}{3} ? s i n x . e^{? 3 x} d x] \end{array}$

$\begin{array}{l} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} ? s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{2}{27} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . I_{1} \\ I_{1} + \frac{1}{9} I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? \frac{10 I_{1}}{9} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ S o, ? ? ? ? ? 2 I = ? \frac{1}{3} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{6} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} + \frac{1}{30} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{5} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{24} [s i n 3 x ? c o s 3 x] + \frac{3 e^{? 3 x}}{40} [s i n x \end{array}$

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