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Quizzes (598) Questions (13441)

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Multiple choice motional emf physics

A square coil of side 0.5$\mathrm { m }$ has movable side It is placed such that its plane is perpendicularuniform magnetic field of induction 0.2$\mathrm { T }$ . If all sides are allowed to move with a speed of 0.1for 4 sec outwards, average indaced emf is

  1. Zero
  2. 0.01$\mathrm { V }$
  3. 0.028$\mathrm { V }$
  4. 0.072$\mathrm { V }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The change in flux is Delta(phi) = B * Delta(Area). The side increases from 0.5m to 0.5 + (0.1 * 4) = 0.9m. The area changes from 0.5^2 = 0.25 to 0.9^2 = 0.81. Delta(phi) = 0.2 * (0.81 - 0.25) = 0.2 * 0.56 = 0.112. Emf = Delta(phi) / Delta(t) = 0.112 / 4 = 0.028V.

Multiple choice motional emf physics

The amplitude of a magnetic field, which is part of a harmonic electromagnetic wave in vacuum, is  $\mathrm { B } _ { 0 } =510\mathrm { nT } .$  What is the amplitude of the electric field of the wave? 

  1. $140 \mathrm { NC } ^ { - 1 }$
  2. $153 \mathrm { NC } ^ { - 1 }$
  3. $163 \mathrm { NC } ^ { - 1 }$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In an electromagnetic wave, the relationship between electric field amplitude E_0 and magnetic field amplitude B_0 is E_0 = c * B_0, where c is the speed of light (3 * 10^8 m/s). E_0 = (3 * 10^8) * (510 * 10^-9) = 153 N/C.

Multiple choice motional emf physics

A  $1\mathrm { m }$  long conducting rod rotates with an angular frequency of  $400\mathrm { rad } \mathrm { s } ^ { - 1 }$  about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant magnetic field of  $0.5\mathrm { T }$  parallel to the axis exists everywhere.Calculate the e.m.f developed between the center and the ring?

  1. $100 \mathrm { V }$
  2. $101 \mathrm { V }$
  3. $102 \mathrm { V }$
  4. $103 \mathrm { V }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The emf induced in a rotating rod is E = (1/2) * B * omega * L^2. Substituting B = 0.5 T, omega = 400 rad/s, and L = 1 m, we get E = 0.5 * 0.5 * 400 * 1^2 = 100 V.

Multiple choice motional emf physics

A charged particle enters in a uniform magnetic field with velocity at an angle of ${ 60 }^{ o }$ with the magnetic field. The pitch of helical path is x, the radius of helix is 

  1. $\dfrac { x }{ 2\sqrt { 3 } x } $
  2. $\dfrac { 2x }{ \sqrt { 3 } \pi } $
  3. $\dfrac { \sqrt { 3 } x }{ \pi } $
  4. $\dfrac { \sqrt { 3 } x }{ 2\pi } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Pitch x = v * cos(theta) * T = v * cos(theta) * (2 * pi * m) / (q * B). Radius R = (m * v * sin(theta)) / (q * B). Dividing R by x gives R / x = (sin(theta)) / (2 * pi * cos(theta)) = tan(60) / (2 * pi) = sqrt(3) / (2 * pi). Thus R = (sqrt(3) * x) / (2 * pi).

Multiple choice motional emf physics

When a charged particle is projected perpendicular to a magnetic field then the

  1. Velocity of the particle remain constant

  2. Momentum of the particle remain constant

  3. Kinetic energy of the particle remain constant

  4. Path of particle is straight line
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When a charged particle enters a magnetic field perpendicularly, the magnetic force acts perpendicular to the velocity. This force changes the direction of velocity but not its magnitude, meaning kinetic energy remains constant.

Multiple choice motional emf physics

A $0.1m$ long conductor carrying a current of $50A$ is perpendicular to a magnetic field of $1.21mT$. The mechanical power to move the conductor with a speed of $1m{s}^{-1}$ is

  1. $0.25mW$
  2. $6.25mW$
  3. $0.625W$
  4. $1W$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Mechanical power required $=$ electrical power expended due to the induced emf 

                                               $= (Blv) \times I \ =( 0.00121 \times 0.1 \times 1 ) \times 50\ =6.25 m W $

Multiple choice motional emf physics

A non conducting ring having charge q uniformly distributed over its circumference is placed on a rough horizontal surface. A vertical time varying magnetic field $B=4t^2$ is switched on a time $t=0$. Mass of the ring is m and radius is R. The ring starts rotating after $2$ seconds. The coefficient of friction between the ring and the tablets.

  1. $\displaystyle\frac{4qmR}{g}$
  2. $\displaystyle\frac{2qmR}{g}$
  3. $\displaystyle\frac{8qR}{mg}$
  4. $\displaystyle\frac{qR}{2mg}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics range of hearing sound: production of sound audible, infrasonic and ultrasonic sounds different sounds

Which waves are used in sonography?

  1. Microwaves

  2. Infrared rays

  3. Radio waves

  4. Ultrasonic waves
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Ultrasonic waves are used in sonography for the mapping of body organs. The Ultrasonic waves that are sent inside the body are reflected from the different parts by creating the echo. They are more useful and less harmful than radiation.

Multiple choice physics range of hearing sound: production of sound audible, infrasonic and ultrasonic sounds different sounds


A microwave and an ultrasonic sound wave have the same wavelength. Their frequencies are in the ratio (approximately) 

  1. $10^2$
  2. $10^4$
  3. $10^6$
  4. $10^8$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Frequency of microwaves, $\upsilon _m \approx 10^{11}Hz$
Frequency of ultrasonic sound waves,$\upsilon _m \approx 10^5Hz$
$\therefore \, \dfrac{\upsilon _m}{\upsilon _u} = \dfrac{10^{11}}{10^{5}} = 10^6$

Multiple choice physics range of hearing sound: production of sound audible, infrasonic and ultrasonic sounds different sounds

Which one of the following animals can hear infrasound?

  1. rhinoceros

  2. bat

  3. dog

  4. honey bee
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

Rhinos have extremely good ears, picking up infra-sound far deeper than the range of human hearing.
Rhinos, can hear down to a frequency of four hertz, whereas even a human baby, with entirely undamaged ears, can normally only pick up sounds as low as 20 hertz. Giraffes and elephants can also hear in this infra-sound range.