Ask & Answer: India's Largest Education Community

Velocity of block in equilibrium, in first case,

$v=A\omega =A.\sqrt[]{\frac{k}{M}}$

Velocity of block in equilibrium, is second case,

$v\text{'}=A\text{'}\omega \text{'}=A\text{'}\sqrt[]{\frac{k}{M+m}}$

From conservation of momentum,

Mv = (M + m) v’

$\Rightarrow MA\sqrt[]{\frac{k}{M}}=\left(M+m\right)A\text{'}\sqrt[]{\frac{k}{M+m}}\Rightarrow A\text{'}=A\sqrt[]{\frac{M}{M+m}}$

${\overrightarrow{a}}_1=x\widehat{i}-\widehat{j}+\widehat{k}\&{\overrightarrow{a}}_2=\widehat{i}+y\widehat{j}+z\widehat{k}$           

given  ${\overrightarrow{a}}_1\&{\overrightarrow{a}}_2$ are collinear then ${\overrightarrow{a}}_1=\lambda {\overrightarrow{a}}_2$  

$\Rightarrow \left(x\widehat{i}-\widehat{j}+\widehat{k}\right)=\lambda \left(\widehat{i}+y\widehat{j}+z\widehat{k}\right)$         

Since $\widehat{i},\widehat{j}\&\widehat{k}$ are not collinear so

  $Sox\widehat{i}+y\widehat{j}+z\widehat{k}=\lambda \widehat{i}-\frac{1}{\lambda }\widehat{j}+\frac{1}{\lambda }\widehat{k}$         

Hence possible unit vector parallel to it be  $\frac{1}{\sqrt[]{3}}\left(\widehat{i}-\widehat{j}+\widehat{k}\right)$ for $\lambda$ =

Since, $\frac{x^2}{\alpha kT}$ $\frac{{}^{}}{}$ should be dimensionless.

So, dimension of $\alpha ,\left[\alpha \right]=\frac{L^2}{M{L}^2{T}^{-2}}=M^{-1}{T}^2$ $\frac{{}^{}}{{}^{}{}^{}}{}^{}{}^{}$

Dimension of $\alpha {\beta }^2$ ${}^{}$ should be that of W.

So, $\left[\alpha {\beta }^2\right]=M{L}^2{T}^{-2}$ ${}^{}{}^{}{}^{}$

$\Rightarrow \left[{\beta }^2\right]=\frac{M{L}^2{T}^{-2}}{M^{-1}{T}^2}=M^2{L}^2{T}^{-4}\Rightarrow \left[\beta \right]=ML{T}^{-2}$

$\frac{dq}{dt}=t=20t+8t^2$

$\Rightarrow {\displaystyle \underset{0}{\overset{q}{\int }}dq={\displaystyle \underset{0}{\overset{15}{\int }}\left(20t+8t^2\right)dt}}$

$\Rightarrow q=20\times \frac{15^2}{2}+8.\frac{15^3}{3}=11250C$

Given f(x) = ${\displaystyle {\int }_e^x{e}^{t}f\left(t\right)dt+e^x..........\left(i\right)}$  

using Leibniz rule then

f’(x) = e^xf(x) + e^x

  _{$\Rightarrow \frac{dy}{dx}=e^{x}y+e^{x}wherey=f\left(x\right)then\frac{dy}{dx}=f\text{'}\left(x\right)$}                

P = -e^x, Q = e^x

Solution be y. (I.F.) =   ${\displaystyle \int Q\left(I.F.\right)dx+c}$

I. f. =   $e^{{\displaystyle \int -e^{x}dx}=e^{-e^x}}$

$\Rightarrow y.\left(e^{-e^x}\right)={\displaystyle \int e^x.e^{-e^x}dx+c}$          

  $y.e^{-e^x}=-{\displaystyle \int dt+c=-t+c=-e^{-e^x}+c..........\left(ii\right)}$          

Put x = 0 , in (i) f (0) = 1

$From\left(ii\right),\frac{1}{e}=-\frac{1}{e}+cgivenc=\frac{2}{e}$        

Hence f(x) = 2. $e^{\left(e^x-1\right)}-1$  

Difference is the level of classes they teach. Trained Graduate teaches from class 6 to 10, whereas PGT teaches class 11 and 12.  A master's degree is required in PGT, unlike TGT.

9 methods of teaching are Lecture, Game, Experiential, Flipped Classrooms, Inquiry, Project, Collaborative, Student-centered approach, and technology-based 

The duration of the MBA course at Centre for Distance and Online education, JIIT is of two years. Admissions to the course are both merit and entrance-based since both graduation marks and entrance scores are accepted. Candidates willing to take admission in the course must apply online on the official website of the institute. 

In case of errors in hall ticket report to exam body by sharing screenshot of error, along with proof to justify. If there is no response, carry admit card along with valid proof.

Four types depending on the classes they teach.  Early Childhood Teachers teach nursery students, Primary are for grades 1 through 5, Trained Graduates from 6 through 8, and Post-graduate for 11 and 12.

The VSAT admit card consists of the following details - 

• Candidates name and roll number
• Candidates application number
• Exam date and timings
• Exam day guidlines
• Candidates scanned photography and signature

NCHMCT JEE is conducted once a year. The National Testing Agency (NTA) annually conducts NCHMCT JEE on behalf of the National Council for Hotel Management & Catering Technology (NCHMCT).

NCHMCT JEE full form is National Council for Hotel Management and Catering Technology Joint Entrance Examination.

NTA released NCHMCT JEE 2026 application form in December 26, 2025. Registration details and application form link is available on the official website.

NCHMCT JEE is a national-level hospitality entrance exam in India. The exam is conducted by the National Testing Agency (NTA) on behalf of the National Council for Hotel Management & Catering Technology (NCHMCT). 

The VSAT registration last for around a month, however, the exam body also extends it to a certain period for candidates who miss on applying during the deadline.

The VSAT examination is conducted to shortlist candidates for B.Tech courses at the university campus.

Courses like BElEd, NTT, MEd, and BEd are available

The VSAT exam is conducted by the Vignan's Foundation for Science, Technology & Research. It is a deemed university located in Andhra Pradesh. The VSAT entrance examination is conducted to shortlist candidates for BEtech and other UG courses at the university.

JJ Thomson, through the cathode ray tube experiment, discovered electrons. He passes the electricity from the tube and sees rays coming from the negative end. These are tiny negatively charged particles, he named them electrons.