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11 months ago

Physics Waves

A wave given by equation y = 1mm sin $(\frac{n x}{30} - 5 n t)$ is produced in a string 100m long of mass 1 kg. What is the tension (in N) in string? x is in m & t in sec.

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Vishal Baghel

Contributor-Level 10

$v = \frac{ω}{k} = \frac{5 π * 30}{π} = 150 = \sqrt[]{\frac{\frac{T}{1}}{100}}$

$\Rightarrow T = \frac{2250000}{100} = 225 N$

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New answer posted

11 months ago

Physics Physics Wave Optics

When light of wavelength 540 nm passes through a single slit of unknown width, the diffraction pattern displays a second maximum where the first minimum of light of an unknown wavelength had been observed to fall. What is the unknown wavelength?

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Vishal Baghel

Contributor-Level 10

$\frac{λ_{1}}{w} = \frac{3 * 540}{2 w}$

$λ_{1} = 810 n m$

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New answer posted

11 months ago

Physics Class 12th

Drawing power form a 12V car battery, a 9V stabilized DC voltage is required to power a car stereo system, attached to the terminals A and B, as shown in the figure. If a Zener diode with ratings, V_z = 9V and P_max = 0.27 W, is connected as shown in the figure, for the above purpose, the minimum series resistance R_s must be:-

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Vishal Baghel

Contributor-Level 10

$i_{m a x} = \frac{P_{m a x}}{V_{z}} = \frac{3}{100}$

$∴ R_{s_{m i n}} = \frac{V_{R_{S}}}{i_{m a x}}$

$= \frac{3}{3 / 100} = 100 Ω$

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New answer posted

11 months ago

Physics Class 12th

Mass of Bromine, Selenium and Krypton are given as:-

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Vishal Baghel

Contributor-Level 10

$Q_{(i)} = (M_{B r} - M_{K r}) c^{2} > 0$

$\begin{array}{l} Q_{(i i)} = (M_{B r} - M_{S e} - 2 M_{e}) c^{2} > 0 \\ Q_{(i i)} = (M_{B r} - M_{s e}) c^{2} > 0 \end{array}$

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New answer posted

11 months ago

Physics Dual Nature of Radiation and Matter

The sketch shown below illustrates the apparatus and results for a famous experiment. The graph on the right is a polar plot of the number of electrons received in the detector. From the result, the experimenters were able to conclude that:-

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Vishal Baghel

Contributor-Level 10

This experiment is the famous Davission-Germer experiment which establishes wave nature of electrons.

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New answer posted

11 months ago

Physics Communication Systems

A signal containing multiple frequencies is shown in figure A. It is passed through 2 different filters and then outputs are shown in figure B & C.

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Vishal Baghel

Contributor-Level 10

Low pass filter will allow low frequency signal to pass while high pass filter allow high frequency to pass through

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New answer posted

11 months ago

Physics Electromagnetic Induction

A metal sheet is being pulled out of a region of magnetic field. IF the sheet is not magnetic:-

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Vishal Baghel

Contributor-Level 10

Heat will develop due to eddy current which will oppose the motion and hence a force is required.

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11 months ago

Physics Physics Moving Charges and Magnetism

An infinite cylindrical current carrying wire carries a current that is uniformly distributed over its cross section. The magnetic field as a function of distance form it's axis is-

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Vishal Baghel

Contributor-Level 10

Apply Ampere's Law

For inside

$B 2 π r = μ_{0} J π r^{2}$

$\Rightarrow B \propto r$

For outside

$B 2 π r = μ_{0} J π R^{2}$

$\Rightarrow B \propto \frac{1}{r}$

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New answer posted

11 months ago

Physics Class 12th

Two air filled capacitors are connected as shown. The left capacitor is suddenly filled with a dielectric of constant K. The work done by battery after that is $24 μ J .$ Dielectric constant K is:-

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Vishal Baghel

Contributor-Level 10

$W D = Δ Q V$

$\Rightarrow Δ Q = 4 μ C$

$Q_{i} = \frac{2 * 4}{2 + 4} * 6 = 8 μ C$

$Q_{f} = 12 μ C = C_{e q} * 6$

$C_{e q} = 2 μ F \Rightarrow C_{1} = C_{2} = 4 μ F$

$k = \frac{C_{1}}{C_{0}} = 2$

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New answer posted

11 months ago

Physics Class 12th

In half deflection method, the resistance of resistance box in series with the galvanometer is 10,000 $Ω$ . When the galvanometer is shunted with 10 $Ω$ resistance, its reading drops to half the initial value. The resistance of galvanometer is:-

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Vishal Baghel

Contributor-Level 10

$i = \frac{v}{R + G}$

$\frac{i}{2} = \frac{v}{\frac{S G}{S + G} + R} * \frac{S}{S + G}$

$\Rightarrow \frac{2 S}{S G + R G + R S} = \frac{1}{R + G}$

$\frac{S R}{R - S} = G$

R > > S

$\Rightarrow G \approx S$

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