Physics

A planar loop of wire rotates in a uniform magnetic field. Initially, at $t = 0$ , the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $10 s$ about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at:

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Physics Physics System of Particles and Rotational Motion

Contributor-Level 10

At any time $t$

$? = B A c o s ? \frac{π t}{5} (? ω = \frac{2 π}{T} = \frac{2 π}{10} = \frac{π}{5})$

$\frac{d ?}{d t} = - \frac{π}{5} B A s i n ? (\frac{π t}{5})$

$e = \frac{π}{5} B A s i n ? (\frac{π t}{5})$

e will be maximum when $\frac{π t}{5}$ is $\frac{π}{2}$ .

$t = 2.5 s e c$
e will be minimum when $\frac{π t}{5}$ is $π, 0$

$t = 0,5 s e c$

So, answer will be (1).

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5 months ago

Mass per unit are of a circular disc of radius a depends on the distance $r$ from its centre as $σ (r) = A + B r$ . The moment of inertia of the disc about the axis, perpendicular to the plane and passing through its centre is:

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Contributor-Level 10

Moment of inertia of ring

$d l = {d m r}^{2}$

I = \int_{0}^{a} ? (A + B r) 2 π r d r π r^{2}

= 2 π A \int_{0}^{a} ? r^{3} d r + 2 π B \int_{0}^{a} ? r^{4} d r

= 2 π  [A \frac{a^{4}}{4} + B \frac{a^{5}}{5}]

= 2 π a^{4}  [\frac{A}{4} + \frac{B a}{5}]

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5 months ago

In the circuit, the logical value of A = 1 or B = 1 when potential at A or B is 5 V and the logical value of A = 0 or B = 0 when potential at A or B is 0 V. The truth table of the given circuit will be:

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Contributor-Level 10

When both A and B have logical value 'l' both diode are reverse bias and current will flow in resistor hence output will be 5 volt i.e. logical value '1'.

In all other case conduction will take place hence output will be zero value i.e. logical value 'o'.

So truth table is

A B Y

0 0

0 1 0 (AND gate)

...more

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5 months ago

The magnetic moment of an electron (e) revolving in an orbit around nucleus with an orbital angular momentum is given by:

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Contributor-Level 10

Kindly consider the following answer

$\vec{μ} = - \frac{e}{2 m} \vec{L}$

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5 months ago

Physics Waves

A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much les than the speed of sound). The observers hears 2 beats/sec. The oscillation frequency of each tuning fork is $v_{0} = 1400 H z$ and the velocity of sound in air is $350 m / s$ . The speed of each tuning fork is close to:

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Contributor-Level 10

$f_{1} = f_{o} (\frac{v}{v - v_{s}}) f_{1} - f_{2} = 2$

$f_{2} = f_{o} (\frac{v}{v + v_{s}}) f_{0} v [\frac{v + v_{s} - (v - v_{s})}{v^{2} - v_{s}^{2}}] = 2; \frac{2 f_{0} v v_{s}}{v^{2} - v_{s}^{2}} = 2$

$? v_{s} ? v ∴ v_{s} = \frac{1}{4} m / s$

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Physics Atoms

The momentum of an electron revolving in nth orbit is given by:

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Contributor-Level 10

mvr = $\frac{n h}{2 π}$ $\frac{}{}$

So momentum (mv) = $\frac{n h}{2 π r}$

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5 months ago

A metal exposed to light of wavelength 800nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500nm is used. The work function of the metal is: (Take hc = 1230 eV –nm).

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Contributor-Level 10

$_{} \frac{}{}$ $K . E_{1} = \frac{1230}{800} - ?$ - (i)

$_{}_{} \frac{}{}$ $K . E_{2} = 2 K . E_{1} = \frac{1230}{500} - ?$ - (ii)

from (i) & (ii)

$? = 0.615$ ev

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5 months ago

A thin lens made of glass (refractive index $= 1.5$ ) of focal length $f = 16 c m$ is immersed in a liquid of refractive index 1.42. If its focal length in liquid is $f_{l}$ , then the ratio $f_{l}$ /f is closest to the integer:

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Contributor-Level 10

$\frac{1}{f} = (μ_{g} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$\frac{1}{f_{l}} = (\frac{μ_{g}}{μ_{l}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

(i) $\frac{f_{l}}{f} = \frac{μ_{g} - 1}{\frac{μ_{g}}{μ_{l}} - 1} = \frac{1.5 - 1}{\frac{1.5}{1.41} - 1} = 8.875 \approx 9$

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5 months ago

Time taken by light to travel in two different materials A and B refractive indices $μ_{A} a n d μ_{B}$ of same thickness is $t_{1} a n d t_{2}$ respectively. It $t_{2} - t_{1} = 5 * 1 0^{- 10} s$ and the ratio of $μ_{A} t o μ_{B}$ 1 : 2. Then, the thickness of material, in meter is: (Given $υ_{A} a n d υ_{B}$ are velocities of light in A and B materials respectively.)

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Contributor-Level 10

$\frac{μ_{A}}{μ_{B}} = \frac{C / v_{A}}{C / v_{B}} = \frac{v_{B}}{v_{A}} = \frac{1}{2}$

Let the thickness is d

$d = \frac{5 * 1 0^{- 10} * v_{A} v_{B}}{v_{A} - v_{B}}$

d = 5 * 10^-10 * v_A

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5 months ago

Which of the following statement is correct?

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