Apply For National Eligibility cum Entrance Test (UG) – 2019 [NEET(UG)-2019]

NATIONAL ELIGIBILITY CUM ENTRANCE TEST (UG) 2019 [NEET (UG) – 2019] will be conducted on Sunday, 5th May, 2019.

As per regulations framed under the Indian Medical Council Act -1956 and the Dentists Act-1948 and as amended from time to time, (NEET (UG) - 2019) will be conducted by National Testing Agency (NTA) for admission to MBBS/BDS Courses in Indian Medical/Dental Colleges run with the approval of Medical Council of India/Dental Council of India under the Union Ministry of Health and Family Welfare, Government of India except for the institutions established through an Act of Parliament i.e. AIIMS and JIPMER Puducherry.

The responsibility of the NTA is limited to the conduct of the entrance examination, declaration of result and for providing an “All India Rank merit list” to the Directorate General Health Service, Government of India for the conduct of counselling for 15% All India Quota Seats and for supplying the result to States/other Counselling Authorities.

Candidates can apply for NEET (UG) - 2019 “Online” only. The online application process has already started on 1^st November 2018 and will end on 30^th November 2018. You may visit the website www.ntaneet.nic.in for all details

NEET 2018 Questions on Electronics

“There is only one corner of the universe that you can be certain of improving, and that’s your own self.”

– Aldous Huxley

 

The following questions were included in the NEET 2018 question paper. These are of the usual type and are simple.

(1) In the circuit shown in the figure, the input voltage Vi is 20 V, V_BE = 0 and V_CE = 0. The values of I_B, I_C and β are given by

(1) I_B = 20 μA, I_C = 5 mA, β = 250

(2) I_B = 25 μA, I_C = 5 mA, β = 200

(3) I_B = 40 μA, I_C = 10 mA, β = 250

(4) I_B = 40 μA, I_C = 5 mA, β = 125

Since the emitter to base voltage of the transistor is zero, the entire input voltage 20 volt appears across the 500 kΩ base resistor. The base current is therefore given by

            I_B = 20 V/500 kΩ = (20/500) mA =0.04 mA = 40 μA

Since the emitter to collector voltage is zero, the entire 20 volt supply appears across the 4 kΩ collector resistor. The collector current is therefore given by

            I_C = 20 V/4 kΩ = 5 mA   

The current gain is given by

            β = I_C/ I_B = (5 mA)/(40 μA) = (5×10^–3)/(40×10^–6) = 125

So option (4) is the answer/

(2) In the combination of the following gates the output Y can be written in terms of inputs A and B as

The AND gate at the top has inputs A and B̅. Therefore its output is A • B̅. The bottom AND gate has inputs A̅ and B so that its output is A̅ • B.

Since the outputs of the NAND gates are applied to the two inputs of the OR gate, the final output (from the OR gate) is A • B̅ + A̅ • B as given in option (2).

 

Apply For National Eligibility cum Entrance Test (UG) – 2018 [NEET(UG)-2018]

National Eligibility cum Entrance Test (UG) – 2018 [NEET(UG)-2018] will be conducted on Sunday, the 6^th May, 2018, by the Central Board of Secondary Education (CBSE). This test is conducted for admission to MBBS/BDS Courses in India in Medical/Dental Colleges run with the approval of Medical Council of India/Dental Council of India under the Union Ministry of Health and Family Welfare, Government of India, except for the institutions established through an Act of Parliament e.g. AIIMS and JIPMER Pondicherry.

Online application for NEET 2018 has already started on 8^th February 2018 and will be allowed upto 9^th March 2018 Friday (23.50 Hrs)

Complete details of NEET 2018 (UG) can be obtained from the website http://cbseneet.nic.in

You will find many questions (with solution) of the type useful for NEET 2018 on this site. Click on the label NEET below this post to access all useful questions discussed on this site. You will need to click on the tab ‘Older Posts’ to access all posts in this connection.