Tag: physics

Questions Related to physics

While is not basic transformation function 

  1. Translation

  2. Rotating

  3. Scaling

  4. Reflection


Correct Option: D
Explanation:

transformation is when you take a shape and you move it in some way. A basic rigid transformation is a movement of the shape that does not affect the size of the shape. The shape doesn't shrink or get larger. There are three basic rigid transformations: reflections, rotations, and translations. So, scaing is not a basic transformation function.

A ray of light is incident on a plane mirror along a vector $\hat i+\hat j-\hat k$. The noemal on incident point is along $\hat i+\hat j$. Then unit vector along the reflected ray is:

  1. $ \dfrac { 1 }{ \sqrt { 3 } } \left( \hat { i } +\hat { j } +\hat { k } \right)$

  2. $-\dfrac { 1 }{ \sqrt { 3 } } \left( \hat { i } +\hat { j } +\hat { k } \right)$

  3. $ \dfrac { 1 }{ \sqrt { 3 } } \left( -\hat { i } -\hat { j } +\hat { k } \right)$

  4. $ \dfrac { 1 }{ \sqrt { 3 } } \left( -\hat { i } +\hat { j } +\hat { k }\right)$


Correct Option: C

A circular coil of radius 6 cm and 20 turns rotates about its vertical diameter with an angular speed of 40 rad $s^{-1}$ in a uniform horizontal magnetic field of magnitude $2 \times 10^{-2}$ T. If the coil form a closed loop of resistance 8 $\Omega$, then the average power loss

  1. $2.07\times 10^{-3}$W

  2. $1.23\times 10^{-3}$W

  3. $3.14\times 10^{-3}$W

  4. $1.80\times 10^{-3}$W


Correct Option: A
Explanation:
Here, $r = 6 cm = 6 \times 10^{-2}m,$ 
$N = 20, \omega = 40 \, rad \, s^{-1}$
$B = 2 \times 10^{-2} T, R = 8 \Omega$
Maximum emf induced, $\epsilon \, = \, NAB \omega$
$ = \, N(\pi r^2)B \omega$
$ = 20 \times \pi \times (6 \times 10^{-2})^2 \times 2 \times 10^{-2} \times 40 \, = \, 0.18 V$

Average value of emf induced over a full cycle $\epsilon _{av}\, = \, 0$ 

Maximum value of current in the coil, $ I \, = \, \dfrac{\varepsilon }{R} = \dfrac{0.18}{8} = 0.023 \,A$

Average power dissipated, $ P \, = \, \dfrac{\varepsilon I}{2} \, = \, \dfrac{0.18 \times 0.023}{2} \, = \, 2.07 \times 10^{-3} W$

The axle of a circular wheel of radius R is held horizontally by two identical strings of equal lengths separated by a distance D. The tension in each string is $T _0$. The rim of the wheel carries a total charge $+$Q distributed uniformly on it. The wheel is vertical and is kept in a uniform vertical magnetic field $\vec{B}$. It is now rotated at an angular speed $\omega$. If the string break at a tension of $3T _0/2$, than the maximum possible value of $\omega$ at which the wheel can be rotated without breaking a string is $\dfrac{DT _0}{QBR^2}$.

  1. True

  2. False


Correct Option: A

A conductor of $3\ m$ length is moving perpendicular to its length as well as a magnetic field of ${10}^{-3}\ T$ with a speed of ${10}\ m/s$, then the force required to move it with this constant speed is

  1. $0.3\ N$

  2. $0.9\ N$

  3. Zero

  4. $3 \times {10}^{-3}\ N$


Correct Option: C
Explanation:

Force on rod moving in a magnetic fileld is given by:

$F=BIL$
Where $I$ is the current in the rod.

According to the question;
$I=0$
Therefore,
$F=0$

Thus,
No force is exerted on the conductor in the magnetic field.
Thus, no external field is required to move the conductor at constant speed.

A horizontal straight conductor (otherwise placed in a closed circuit) along east-west direction falls under gravity. There is:-

  1. No induced electromotive force along the length

  2. No induced current along the length

  3. An induced current from west to east

  4. None of the above.


Correct Option: C

In a uniform magnetic field of induction $B$ a wire in the form of semicircle of radius $r$ rotated about the diameter of the circle with angular frequency $\omega$. The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $R$ the mean power generated per period of rotation is:

  1. $\dfrac { B\pi { r }^{ 2 }\omega }{ 2R }$

  2. $\dfrac { { \left( B\pi { r }^{ 2 }\omega \right) }^{ 2 } }{ 2R }$

  3. $\dfrac { { \left( B\pi r\omega \right) }^{ 2 } }{ 2R }$

  4. $\dfrac { { \left( B\pi r{ \omega }^{ 2 } \right) }^{ 2 } }{ 8R }$


Correct Option: D

To produce a field of magnetic $\pi$ tesla at the center of circular loop of diameter 1 m, the current flowing through loop is :

  1. $5\times 10^6\;A$

  2. $10^7\;A$

  3. $2.5\times 10^6\;A$

  4. $2\times 10^6\;A$


Correct Option: C
Explanation:

Magnetic field due to circular loop at center O due to current $i$ is given as,

$B=\dfrac{\mu _0 I}{2r}=\dfrac{4\pi\times 10^{-7} \times I}{2\times 0.5}$
$\pi \times 1=4\pi \times 10^{-7} I$
$I=2.5\times 10^{6} A$

A metal ring of radius $r = 0.5\ m$ with its plane normal to a uniform magnetic field $B$ of induction $0.2\ T$ carries a current $I = 100\ A$. The tension in newtons developed in the ring is

  1. $100$

  2. $50$

  3. $25$

  4. $10$


Correct Option: D

A magnet is brought towards a coil (i) speedly (ii) slowly, then the induced electromotive force charge will be respectively

  1. More in first case / More in first case

  2. More in first case / Equal in both case

  3. Less in first case / More in second case

  4. Less in first case / Equal in both case


Correct Option: B