Class 12th

An electric dipole is placed at an angle of $30^{?}$ with an electric field of intensity $2 * 10^{5} {N C}^{- 1}$ . It experiences a torque equal to 4Nm. Calculate the magnitude of charge on the dipole, if the dipole length is 2 cm.

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Class 12th Physics Ncert Solutions Class 12th

Contributor-Level 9

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The ratio of frequencies of fundamental harmonic produced by an open pipe to that of a closed pipe having the same length is

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Contributor-Level 9

Fundamental harmonic frequency open pipe

$= \frac{V}{2 L} = v_{1}$

(say)

Fundamental harmonic frequency of closed pipe $= \frac{V}{4 L}$

$= v_{2}$ (say)

$\Rightarrow \frac{v_{1}}{v_{2}} = \frac{\frac{V}{2 L}}{\frac{V}{4 L}} 2 : 1$

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Class 12th Sai Tirupati University

An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance ' R' are connected in series to an ac source of potential difference ' V volts as shown in figure. Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10√2 A. The impedance of the circuit is:

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Vishal Baghel

Contributor-Level 10

Diameter = main scale reading + (circular scale reading * least count)
Diameter = 0 + (52 * 0.01 mm) = 0.52 mm = 0.052 cm.
In the RLC circuit:
Given I? = 10√2 A, so I? = I? /√2 = 10 A.
V? = √ [V? ² + (V? - V? )²] = √ [40² + (40 - 10)²] = √ [1600 + 30²] = √ [1600 + 900] = √2500 = 50 V.
Impedance Z = V? / I? = 50 V / 10 A = 5 Ω.
For Hindi: I? = 10√2 A, V? = 50V, Z = V? /I? = 50/ (10√2) = 5/√2 Ω.

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Can I get admission in Sai Tirupati University for BA with 60% in Class 12th?

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Sejal Baveja

Contributor-Level 10

Yes, candidates can get admission in Sai Tirupati University for BA with 60% in Class 12. The minimum required aggregate for almost all the UG courses is 45%-55% including relaxations for the reserved category. Candidates must ensure that they submit the marksheets and certificate of Class 12 to apply for and confirm admission at Sai Tirupati University.

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Consider the following statements (A) and (B) and identify the correct answer.

(A) A zener diode is connected in reverse bias, when used as a voltage regulator.
(B) The potential barrier of p-n junction lies between 0.1 V to 0.3 V.

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Contributor-Level 10

A reverse-biased Zener diode is used as a voltage regulator.
The potential barrier for Germanium (Ge) is approximately 0.3 V.
The potential barrier for Silicon (Si) is approximately 0.7 V.

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A dipole is placed in an electric field ns shown. In which direction will it move?

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Column -I gives certain physical terms associated with flow of current through a metallic conductor. Column - II gives some mathematical relations involving electrical quantities. Match Column - I and Column - II with appropriate relations.

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Class 12th Physics Nuclei

From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire:

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Contributor-Level 9

Kindly consider the following Image

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Class 12th Physics Nuclei

A nucleus of mass number 189 splits into two nuclei having mass number 125 and 64 . The ratio of radius of two daughter nuclei respectively is :

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Contributor-Level 9

$\begin{array}{r} R = R_{0} (A)^{1 / 3} \\ \frac{R_{1}}{R_{2}} = {(\frac{125}{64})}^{1 / 3} = \frac{5}{4} \end{array}$

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Two charged spherical conductors of radius R₁ and R₂ are connected by a wire. Then the ratio of surface charge densities of the spheres (σ₁/σ₂) is:

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