From 2020 AIIMS exam has been merged with NEET.

# AIIMS MBBS 2018 Higher Questions and Answers with Solutions Part 1

Get unlimited access to the best preparation resource for competitive exams : get questions, notes, tests, video lectures and more- for all subjects of your exam.

## Part-A (Physics)

Q: 1. In the diagram 100 kg block is moving up with constant velocity, then find out the tension at point P:

Title: Q-1-The Diagram 100 kg Block is Moving up with Constant Velocity

Description: Q-1- The Diagram 100 kg Block is Moving up with Constant Velocity

Ans. 1470

Solution:

Title: Q-1-The Diagram 100 kg Block is Moving up with Constant Velocity

Description: Q-1- The Diagram 100 kg Block is Moving up with Constant Velocity

Q: 2. In a simple microscope of focus length 5 cm final image is formed at D, then its magnification will be:

(1)

(2)

(3)

(4)

Ans. (1)

Solution:

Q:3. Centre of mass of a ring will be at a position.

Ans.

Q: 4. In a full wave rectifier in which input voltage is represented by , then peak inversion voltage of non-conducting diode will be:

(1)

(2)

(3)

(4)

Ans. (3)

Solution:

Q: 5. A long cylindrical wire carrying current of amp. has radius of , then find its magnetic field induction at a point from the centre of the wire

(1)

(2)

(3)

(4)

Ans. (1)

Solution:

Q: 6. A parallel plate capacitor of capacity is discharging through a resistor. If its energy reduces to half in one second. The value of resistance will be

(1)

(2)

(3)

(4)

Ans. (1)

Solution:

When energy is

Then

Q: 7. Water is flowing in a non-viscous tube as shown in the diagram. The diameter at point A and point B are and respectively. The pressure difference between points A & B are , then find out the rate of flow:

Title: Q-7-Image of Non-Viscous Tube

Description: Q-7-Image of Image of Non-Viscous Tube

Ans.

Solution:

Q: 8. ‘Biota Savart’ law of magnetism is analogous to:

Ans. Columbus Law’s

Solution:

Bio Savart’ law is analogous to coulomb’s law but if it was not in option then Gauss’s law is correct

Q: 9. In an electromagnetic wave the expression for electric field is given by the permeability is given permittivity, then find the average intensity delivered:

(1)

(2)

(3)

(4)

Answer: (1)

Solution: