TS Polycet (Polytechnic) 2018 Previous Question Paper with Answers And Model Papers With Complete Analysis

Q). Ohm’s law is valid for
A) Gaseous conductors
B) Metal conductors
C) Semiconductors
D) Insulators

View Answer
B) Metal conductors

Q). Which of the following is used to prevent damages due to overloading to the household circuit?
A) Switch
B) Fuse
C) Resistance
D) Semiconductor

View Answer
B) Fuse

Q). Induction stove’ works on the principle of
A) Electro magnetic induction
B) Generator
C) Fuse
D) Switch

View Answer
A) Electro magnetic induction

Q). The number of field lines passing through the plane of area ‘A’ perpendicular to the magnetic field is called
A) Magnetic induction
B) Magnetic flux
C) Magnetic flux density
D) Solenoid

View Answer
B) Magnetic flux

Q). Electric power generated in a circuit placed in uniform magnetic field in electro magnetism is
A) P = \frac{I\;\bigtriangleup t}{\bigtriangleup\varnothing}
B) P = \frac{\bigtriangleup\varnothing}{I\;\bigtriangleup t}
C) P =I\frac{\bigtriangleup\varnothing}{\bigtriangleup t}
D) P =I

View Answer
C) P =I\frac{\bigtriangleup\varnothing}{\bigtriangleup t}

Q). The induced current will appear in such a direction that it opposes the changes in the flux in the coil is
A) Ampere law
B) Lenz’s law
C) Ohm’s law
D) Faraday’s law

View Answer
B) Lenz’s law

Q). S.I. unit of magnetic flux is
A) Tesla
B) Weber
C) Weber/metre
D) Newton

View Answer
B) Weber

Q). Which of the following converts mechanical energy into electrical energy?
A) Motor
B) Battery
C) Generator
D) Switch

View Answer
C) Generator

Q). The value of magnetic flux density which is uniform is 2T, what is the flux passing through a surface of area 0.5 m2 perpendicular to it?
A) 1 Weber
B) 10 Weber
C) 1.25 Weber
D) 0.25 Weber

View Answer
A) 1 Weber

Q). A 10 N force acts on a rectangular conductor of 5 cm long placed perpendicular to a magnetic field. If the current in conductor is 50 A, then B
A) 0.4 tesla
B) 2.5 tesla
C) 25 tesla
D) 4 tesla

View Answer
D) 4 tesla
Explanation: Force = 10 N
Length of the conductor = 5cm=5/100m
Current = I = 50 A
∴ F = BIL => B = F/IL
= \frac{10x100}{50x5} = 4 telsa
∴ B = 4 Tesla

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