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Cheran College of Physiotherapy is visited by top companies for placements, such as Biocon, BOSCH, ICICI Bank,  and others. The below-mentioned are the Cheran College of Physiotherapy placement statistics:

The four year BTech course is offered by the CARE. The fees against the course needs to be paid either annually or semester-wise. The total Tuition fee for the BTech course is INR 6 Lakh. The mentioned fee is taken from official website and could be subject to change.

Portal: eapcet-sche.aptonline.in (or cets.apsche.ap.gov.in EAPCET section)

Steps:

• Use your hall ticket and DOB
• Register & pay Upload certificates
• Fill web options
• Check seat allotment
• Self report
• Visit college

This is a True and False Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenequationofthecircleis{x}^2+y^2-2x+6y+1=0\\ Here,2g=-2\Rightarrow g=-1\\ 2f=6\Rightarrow f=3\\ \therefore Centre=\left(-g,-f\right)=\left(1,-3\right)\\ andradiusr=\sqrt[]{g^2+f^2-c}=\sqrt[]{1+9-1}=3\\ \therefore \text{\hspace{0.17em}Distance}betweenthe\text{\hspace{0.17em}point}\left(1,2\right)andthecentre\left(1,-3\right)\\ =\sqrt[]{{\left(1-1\right)}^2+{\left(2+3\right)}^2}=5\\ Here,5>3,sothe\text{\hspace{0.17em}point}liesoutsidethecircle.\\ Hence,thegivenstatementisFalse.\end{array}$

Yes, direct admissions are possible at the Cheran College of Physiotherapy. Students must meet the basic eligibility criteria and get admissions. Candidates must follow the application process which includes visiting the official website. Then, clicking on the Apply Now tab. Then, select the required course. Then, pay the application fee. Then, fill out and submit the application form with required information. 

This is a true and false Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenequationofthecircleis{x}^2+y^2=a^2\\ andthe\text{\hspace{0.17em} tangent}islx+my=1\\ Herecentreis\left(0,0\right)andradius=a\\ If\left(l,m\right)liesonthecircle\\ \therefore \sqrt[]{{\left(l-0\right)}^2+{\left(m-0\right)}^2}=a\\ \Rightarrow \sqrt[]{l^2+m^2}=a\Rightarrow l^2+m^2=a^2\left(whichisacircle\right)\\ So,the\left(l,m\right)liesonthecircle.\\ Hence,thegivenstatementisTrue.\end{array}$

Candidates looking for course admission at Cheran College of Physiotherapy must meet the eligibility set by the college. The college offers BPT and MPT course at the UG and PG levels. Cheran College of Physiotherapy is affiliated with the Dr. MGR Medical University - Chennai. The college offers UG and PG courses to students. These courses are provided across the Sports, Neurology, and various other streams.

Candidates must know their Cheran College of Physiotherapy eligibility criteria and apply accordingly. Follow the Cheran College of Physiotherapy application process mentioned below:

Step 1: Visit the official website of Cheran College of Physiotherapy.

Step 2: Select the required course.

Step 3: Pay the application fee.

Step 4: Fill out and submit the application form with the required information.

Here's a possible answer:

After considering my options, I would choose B.Tech from a private college like Medicaps or Acropolis if I'm interested in a specialised engineering programme with industry connections. However, if I'm looking for a more affordable option with a strong academic foundation, BSc from a government college like DAVV would be a better fit. Ultimately, the decision depends on my career goals, budget, and priorities.

There are about 60+ Nursing colleges in Bihar offering GNM courses. Some of them are mentioned below along with their tuition fees:

Source: Official site and may vary.

This is a true and false Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenequationofthecircleis{x}^2+y^2-14x-10y-151=0\\ Shortest\text{\hspace{0.17em}distance}=distancebetweenthe\left(2,-7\right)\\ andthecentre-radiusofthecircle\\ Centreofthegivencircleis\\ 2g=-14\Rightarrow g=-7\\ 2f=-10\Rightarrow f=-5\\ \therefore Centre=\left(-g,-f\right)=\left(7,5\right)\\ andr=\sqrt[]{{\left(-7\right)}^2+{\left(-5\right)}^2+151}=\sqrt[]{49+25+151}\\ =\sqrt[]{225}=15\\ \therefore Shortest=\sqrt[]{{\left(7-2\right)}^2+{\left(5+7\right)}^2}-15\\ =\sqrt[]{25+144}-15=13-15=\left|-2\right|=2\\ Hence,thegivenstatementisFalse.\end{array}$

The cost of BPT at Cheran College of Physiotherapy varies and depends on the level of competition and the available seats. The cost excludes the transportation charges, hostel charges, and various other charges. The cost is quite flexible and depends on the course and the choice of specialisation. 

This is a true and false Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenequationofthecircleis\\ x^2+y^2+6x+2y=0\\ Centreis\left(-3,-1\right)\\ Ifx+3y=0istheequationofdiameter,thenthecentre\left(-3,-1\right)willlieonx+3y=0\\ -3+3\left(-1\right)=0\\ \Rightarrow -6\ne 0\\ So,x+3y=0isnotthediameterofthecircle.\\ Hence,thegivenstatementisFalse.\end{array}$

Candidates wishing to secure a seat in any programme offered by Indian School of Business Management and Administration, Chennai are required to follow the steps given below:

Step 1: Fill out the application form on the official website of the institute. 

Step 2: Appear for the entrance exam if required. 

Step 3: Candidates with the final admission offer can pay the course fee and secure their admissions. 

In Andhra Pradesh EAMCET (AP EAPCET) counselling, the number of rounds usually depends on the availability of seats after each phase. Based on the past few years, candidates can generally expect three main rounds of counselling—Round 1, Round 2, and a Final Phase—along with a special/spot round for any leftover seats.

For Andhra Pradesh EAMCET 2026, while the exact schedule will be announced by APSCHE after results, it is likely to follow the same pattern as 2025, where three rounds of counselling were conducted, and a spot admission round was held for vacant seats.

No, candidates do not require appearing for JEE Main for admission at CMJ University. The university offers admission to BTech course based on merit of the aspirants in the last qualifying exam. Candidates who have passed Class 12 with Mathematics and Physics as compulsory subjects are offered admission in the BTech programme.

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

Sol:

$\begin{array}{l}\left(i\right)Giventhatvertices\left(\pm 5,0\right),foci\left(\pm 7,0\right)\\ Vertexofhyperbola=\left(\pm a,0\right)andfoci\left(\pm ae,0\right)\\ \therefore a=5andae=7\Rightarrow 5\times e=7\Rightarrow e=\frac{7}{5}\\ Now{b}^2=a^2\left(e^2-1\right)\\ \Rightarrow b^2=25\left(\frac{49}{25}-1\right)\Rightarrow b^2=25\times \frac{24}{25}\Rightarrow b^2=24\\ Theequationofthehyperbolais\\ \frac{x^2}{25}-\frac{y^2}{24}=1\\ \left(ii\right)Giventhatvertices\left(0,\pm 7\right),e=\frac{4}{3}\\ Clearly,thehyperbolaisvertical.\\ \therefore a=5ande=\frac{4}{3}\\ Weknowthat{b}^2=a^2\left(e^2-1\right)\\ \Rightarrow b^2=49\left(\frac{16}{9}-1\right)\Rightarrow b^2=49\times \frac{7}{9}\Rightarrow b^2=\frac{343}{9}\\ Theequationofthehyperbolais\\ \frac{y^2}{49}-\frac{9x^2}{343}=1\Rightarrow 9x^2-7y^2+343=0\\ \left(iii\right)Giventhat:foci\left(0,\pm \sqrt[]{10}\right)\\ \therefore ae=\sqrt[]{10}\Rightarrow a^2{e}^2=10\\ Weknowthat{b}^2=a^2\left(e^2-1\right)\\ \Rightarrow b^2=a^2{e}^2-a^2\Rightarrow b^2=10-a^2\\ Equationofthehyperbolais\\ \frac{y^2}{a^2}-\frac{x^2}{b^2}=1\Rightarrow \frac{y^2}{a^2}-\frac{x^2}{10-a^2}=1\\ Ifitpassesthroughthe\left(2,3\right)then\\ \Rightarrow \frac{9}{a^2}-\frac{4}{10-a^2}=1\Rightarrow \frac{90-9a^2-4a^2}{a^2\left(10-a^2\right)}=1\\ \Rightarrow 90-13a^2=10a^2-a^4\Rightarrow a^4-23a^2+90=0\\ \Rightarrow a^4-18a^2-5a^2+90=0\\ \Rightarrow a^2\left(a^2-18\right)-5\left(a^2-18\right)=0\\ \Rightarrow \left(a^2-18\right)\left(a^2-5\right)=0\\ \Rightarrow a^2=18,a^2=5\\ \therefore b^2=10-18=-8and{b}^2=10-5=5\\ \therefore b\ne -8and{b}^2=5\\ Hence,therequiredequationis\frac{y^2}{5}-\frac{x^2}{5}=1or{y}^2-x^2=5.\end{array}$