Best viewed in Google Chrome

Advertisement

Popular Posts

Services

Showing posts with label JEE Advanced. Show all posts
Showing posts with label JEE Advanced. Show all posts

NEET 2019 and JEE Mains 2019 complete schedules released by NTA - National Testing Agency

- No comments
NTA (National Testing Agency) has released the Exam Dates, Registration, Modes of Exam and other details for NEET 2019, JEE Main I and II 2019.
The revised schedule for NTA NEET 2019 and NTA JEE Mains were released on 21st August, 2018 by MHRD. Dates of registration, releasing admit card, exam date, modes of exam (format), results etc. of NEET 2019 and JEE Main 2019 are provided in the table below.
MHRD has decided against conducting the NEET 2019 examination twice next year. As per the earlier announcement by MHRD, NEET 2019 and JEE Main 2019 were to be conducted by NTA twice a year which has been changed now. Following a recommendation from Health Ministry, MHRD has scrapped the plan to conduct the examination in computer based format and twice a year. Now, NEET 2019 would be conducted on May 5, 2019 and the online application forms would begin from November 30, 2018. The results would be announced on June 5, 2019.
The forms for NTA JEE Main 2019 I would be released and begin from 1st September, 2019. There would be no change in the proposal of conducting NTA JEE Main 2019 in online format only. The examination would be conducted in multiple sessions from January 6, 2019 to January 20, 2019 and students would have the option of appearing on any of the days by slot booking and NTA JEE Main II 2019 would be conducted in April’19. 

NTA NEET 2019: New Rules and Changes
1. The only change for NEET 2019 is that the conducting authority has been changed. Till 2018, NEET for undergraduate medical and dental admissions i.e. NEET-UG was conducted by CBSE.
2. From 2019 onwards, the examination would be conducted by NTA. Candidates would be required to fill the online application forms on nta.ac.in. Details of how to apply for NEET NTA 2019 would be released when the forms are released on November 30.

Everything else for NEET NTA 2019 remains unchanged. For example:
1. NTA NEET 2019 examination would be a pen and paper test.
2. NTA NEET 2019 syllabus would remain the same as was the case for 2018.
3. NTA NEET 2019 eligibility is expected to remain the same.
4. As earlier, it will be conducted only once a year, which would be on May 5, 2019. 
5. NTA NEET 2019 would be conducted in all the 8 regional languages as well as in English, Hindi and Urdu.
Representative Image - NTA NEET 2019, NTA JEE Main 2019 details
For more details and registrations please visit official website of NTA (National Testing Agency) https://www.nta.ac.in

 Related Articles



Circular and Rotational Motions - MCQ Test Series solved questions for NEET-UG, IIT JEE, COMEDK, MBBS and BDS admission tests, Engineering Entrance Exams

- No comments

Physics: Circular Motion, Rotational Motion

Physics MCQ Test Series solved questions for NEET-ug, IIT JEE, COMEDK, MBBS and BDS admission tests, Engineering Entrance Exams and other such competitive exams.
Syllabus for NEET and IIT JEE:
Kinematics of Circular Motion; Dynamics of Uniform Circular Motion; Centripetal Force; Examples: vehicle on level circular road, vehicle on baked track; Angular Velocity and is relation with Linear Velocity; Torque or Moment of Force and Angular Momentum; Laws of Conservation of Angular Momentum; Theorems of perpendicular axes and parallel axes; Kinematics of Rotational Motion about a fixed axis; Rolling Motion; General relation among Position-Velocity-Acceleration for Motion in a plane and Uniform Circular Motion; Centrifugal Force; Motion in a vertical circle; Rotational Motion of a rigid body; Moment of Inertia; Angular Impulse.

Solved MCQ Test Series – Set 3 (Q. No. 21-30)
Question 21: The angular speed of a fly wheel making 120 rev/min is:
a. 2π rad/s
b. 4π2 rad/s
c. π rad/s
d. 4π rad/s

Question 22: A child is swinging a swing, minimum and maximum heights of swing from earth’s surface are 0.75 m and 2 m respectively. The maximum velocity of this swing is –
a. 5 m/s
b. 10 m/s
c. 15 m/s
d. 20 m/s

Question 23: An aeroplane is flying with a uniform speed of 100 m/s along a circular path of radius 100 m. The angular speed of the aeroplane will be:
a. 1 rad/s
b. 2 rad/s
c. 3 rad/s
d. 4 rad/s

Question 24: A block of mass m slides down along the surface of the bowl from the rim to the bottom as shown in fig. The velocity of the block at the bottom will be –
https://www.indiastudysolution.com















Question 25: When a body moves with a constant speed along a circle:
a. Its velocity remains constant
b. No force acts on it
c. No work is done on it
d. No acceleration is produced on it

Question 26: A 1 kg ball is rotated in a vertical circle by using a string of length 0.1 m. If the tension in the string at the lowest point is 29.4 N, its angular velocity at that position is –
a. 7 rads-1
b. 14 rads-1
c. 3.5 rads-1
d. 25.6 rads-1

Question 27: A cyclist moves in a circular track of radius 100 m. If the coefficient of friction is 0.2. Then the maximum speed with which the cyclist take a turn without leaning inwards is:
a. 14.0 m/s
b. 140 m/s
c. 1.4 m/s
d. 9.8 m/s

Question 28: If all objects on the equator of earth feel weightless then the duration of the day will nearly become –
a. 6.2 hr
b. 4.4 hr
c. 2.2 hr
d. 1.41 hr

Question 29: Two racing cars of masses m1 and m2 are moving in circles of radii r1 and r2 respectively. Their speeds are such that each makes a complete circle in the same time t. The ratio of the angular speeds of the first to the second car is:
a. 1: 1
b. m1: m2
c. r1: r2
d. m1m2: r1r2

Question 30: A 2 kg stone is swung in a vertical circle by attaching it at the end of a string of length 2m. If the string can withstand a tension 140.6 N, the maximum speed with which the stone can be rotated is –
a. 22 ms-1
b. 44 ms-1
c. 33 ms-1
 Circular and Rotational Motions More Questions Answers  

<<Prev (Q.No 11 - 20)                                (Q.No 31-40) Next>>


 Recent Posts
BIOLOGY
CHEMISTRY
PHYSICS

EduNews: IITJEE Main 2018 cut-off score is lowest ever

- No comments

The cut-off score used for this year's IIT JEE Main to determine general category candidate's eligibility for JEE Advanced is the lowest ever and caps a four-year downward trend.
Professors at the IIT, which admit students on the basis of their JEE Advanced scores, suggested two possible reasons. One: an increase in available IIT seats; Two: top students' growing preference for non-engineering courses such as medicine. All those who scored more than 74 in the IIT JEE Main, out of a total marks of 360,will be allowed to take the JEE Advanced 2018.
Year
Cut-off score (out of 360) for JEE Advanced
2018
74
2017
81
2016
100
2015
105
2014
115
2013
113
When JEE Main and JEE Advanced were started in 2013, the cut-off used was 113. This increased to 115 the following year but has been falling every year since then.
Dr.Panigrahi of IIT Kharagpur and former JEE Advanced chairperson, said that in 2013, the IITs wanted the 1.5 lakh best IIT JEE Main candidates to take the JEE Advanced. In 2018, they want 2.2 lakh to do so. This could be because the institutes along with their seats have increased from about 9500 in 2013 to almost 12000 now.
Also, the number of students taking IITJEE Main has been declining from 13 lakh in 2013 to about 11 lakh in 2018. "Demand for engineering courses is falling. Large numbers of seats remain vacant in engineering colleges," Rajeev Kumar former professor of IIT Kharagpur now in JNU, said.
This fact is also supported by the data since number of students taking the NEET (National Eligibility-cum-Entrance Test), which determines admission to undergraduate medical and dental courses, has grown from 8 lakh in 2016 to 13 lakh in 2018.
(Source: Telegraph, edited)

Related Articles

CBSE to declare IIT-JEE Main 2018 results on April 30 afternoon

- No comments

The results for the Indian Institutes of Technology-Joint Entrance Exam (IIT-JEE) Main 2018 will be announced today on April 30 by the Central Board of Secondary Education (CBSE). Around fourteen lakh students who appeared for the IIT-JEE Main 2018 online and offline examination can check their results for both JEE Main which are expected to be declared late afternoon today on the official websites - jeemain.nic.in, cbseresults.nic.in or results.nic.in
The future of these engineering aspirants who have appeared for the exam on April 8 (offline) and April 15 will be decided after CBSE releases the result.
The scores of candidates -- All India and category ranks -- obtained will be displayed. The rank card can also be downloaded by candidates for admission purposes.


www.indiastudysolution.com graphics

How and where to check your IIT JEE Main result

  • CBSE announces the result at jeemain.nic.in and cbseresults.nic.in.
  • To check the result, you have to log in and enter your JEE Main roll number and date of birth.
  • The result which is displayed on the screen can be downloaded as the rank card.
  • Use the option to print the rank card. It is advised to take a copy.
  • It is also better to thoroughly check all details mentioned on the rank card.

How to confirm if you qualified for JEE Advanced or not? (IIT JEE Advanced 2018)
The JEE main is conducted every year for admission to various engineering and architecture courses in NITs, IITs and other CFTIs across the country. The exam is also an entrance examination for JEE Advanced.
The candidates who clear the JEE Main 2018 will be considered qualified for appearing in JEE Advanced 2018 examination. The JEE Advanced examination is conducted for admission to IITs.
The CBSE Board will also release the JEE main 2018 result along with the mark on - cbseresults.nic.in and results.nic.in.
Since JEE Main is the screening test for JEE Advanced, the top 2,24,000 candidates who successfully clear it must seriously look into preparing for JEE Advanced 2018.
CBSE will announce the JEE Main cutoff 2018 which will specify the scores required to be eligible to apply and appear for JEE Advanced 2018 which will be held on May 20 .
You can fill the JEE Advanced application form 2018 between May 2 and May 7.
Eligibility for NITs, IIITs and CFTIs Admissions
The Joint Seat Allocation Authority (JoSAA) will conduct common counseling for the IITs, NITs, IIITs and GFTIs from June 19 tentatively.
Allotment will be done according to the ranks and preferences selected during JoSAA registration.
Candidates who have a valid rank in the JEE Main or JEE Advanced are eligible to apply for the admissions in these institutes.

NEET and JEE Main to be conducted twice a year by NTA instead of CBSE

- No comments
The Center is considering to conduct the Entrance tests (NEET, JEE Main) for admissions to undergraduate courses in medical and engineering twice a year from 2019.

The government is in the process of setting up a National Testing Agency (NTA) which may start its operations from 2019. The agency will take over all the entrance tests now conducted by CBSE (Central Board of Secondary Education). These entrance tests are NEET (National Eligibility cum Entrance Test) for MBBS admission, the JEE Main (Joint Entrance Examination Main) for B.Tech courses and the NET (National Eligibility Test) for selecting teachers eligible to be appointed in colleges and universities.

"The NTA (National Testing Agency) will initially conduct those entrance examinations which are currently being conducted by the CBSE ... Other examinations, tests will be taken up gradually after the NTA is fully geared up ... The entrance examinations will be conducted in online mode at least twice a year, giving adequate opportunity to candidates to bring out their best," HRD Minister of State Upendra Kushwaha said in a written answer in the Lok Sabha.

At present all these entrance tests NEET, JEE Main, NET etc. are held once in a year. If the proposal is passed, then from 2019 National Eligibility cum Entrance Test (NEET) and Joint entrance Examination Main (JEE Main) will be conducted twice a year by NTA instead of Central Board of Secondary Education (CBSE). NTA will also locate the examination centers at sub-district and district levels.

   Related Articles

CBSE announced tentative dates for IITJEE Main and JEE Advanced 2018

- No comments
The offline version of IIT JEE Main 2018 will be held on April 8 next year, the Central Board of Secondary Education (CBSE) has announced.
It has not declared the dates for the computer-based version that only a few among the 12 lakh candidates take.
The IIT JEE Advanced 2018, to be conducted by IIT Kanpur, will be held on May 20. It will be an entirely computer-based test. 

The CBSE has also asked its affiliated schools not to pass off students of unaffiliated schools as their own for the board exams.