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Physics Physics Mechanical Properties of Solids

If the area of cross-section is halved and length of wire having young's modulus Y is doubled, then its young's modulus will become

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alok kumar singh

Contributor-Level 10

Young's modulus is property of material of wire and it is independent of geometrical factors.

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New answer posted

3 months ago

Physics Physics Moving Charges and Magnetism

A long wire carrying current $\sqrt[]{2}$ A is placed in uniform magnetic field of 3 * 10^–5 T. If magnetic field is perpendicular to wire, find the magnetic force on unit length of wire.

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alok kumar singh

Contributor-Level 10

F = iL B sinq

= $\sqrt[]{2}$ * 1 * 3 * 10^–5 sin90

= 3 $\sqrt[]{2}$ * 10^–5 N

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3 months ago

Physics Class 12th

Maximum wavelength of light source such that photoelectron can be ejected from material of work function 3 eV is

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

$λ = \frac{12400}{3} = 4133.3$ Å

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3 months ago

Physics Class 12th

Potential due to electric dipole on axial position at distance r from dipole is proportional to (assume r >> length of dipole)

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alok kumar singh

Contributor-Level 10

$| E | = \frac{2 k P}{r^{3}}$

$E = - \frac{d v}{d r}$ , $v \propto \frac{1}{r^{2}}$

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3 months ago

Physics Class 11th

Ball released from height 10 m strikes ground and rebounds height 5 m. Find impulse imparted by ground while collision, given mass of ball is 100 g. (Take g = 10 m/s²)

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alok kumar singh

Contributor-Level 10

$v_{1} = \sqrt[]{2 g 10}$

$v_{2} = \sqrt[]{2 g 5}$

$\vec{l} = Δ \vec{p}$

$l = 0.1 {\sqrt[]{2 g 10} + \sqrt[]{2 g 5}}$

$= 0.1 {10 \sqrt[]{2} + 10}$

$= (\sqrt[]{2} + 1) N s$

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3 months ago

Physics Physics Semiconductor Devices

Find current through zener diode if its breakdown voltage is 5V.

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alok kumar singh

Contributor-Level 10

$i_{b a t t e r y} = \frac{(20 - 5)}{200} = \frac{15}{200} A$

$i_{300 Ω} = \frac{5}{300} A$

$i_{z e n e r} = \frac{15}{200} - \frac{5}{300}$

= 58.33 mA

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3 months ago

Physics Class 11th

A ball is released from a height of 1 m on a smooth hemispherical surface as shown. Find its velocity when it is at a height of 0.5 m. (Take g = 10 m/s²)

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alok kumar singh

Contributor-Level 10

By conservation of mechanical energy

mg (1) = $\frac{1}{2}$ mv² + mg (0.5)

v² = 10

v = $\sqrt[]{10}$ m/s

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3 months ago

Physics Class 12th

In given circuit find potential difference across 700 W resistance (i.e. V₀).

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alok kumar singh

Contributor-Level 10

$i = \frac{7}{3.5 k + 0.9 k Ω} = \frac{7}{4.4 k}$

$V_{0} = i * 700 Ω = \frac{7}{4.4 k} * 7 k = \frac{9.9}{4.4} = 1.1 V$

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3 months ago

Physics Atoms

The ratio of kinetic energy & potential energy in 5^th excited state of Hydrogen atom is

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alok kumar singh

Contributor-Level 10

Kinetic energy: Potential energy = 1 : –2

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3 months ago

Physics Class 11th

A particle of mass m is projected with speed v at an angle of 30° with the horizontal, find its angular momentum about point of projection when it reaches its maximum height.

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alok kumar singh

Contributor-Level 10

Velocity at maximum height = vcoss30°

L = m (vcos30) H

$= m v (\frac{\sqrt[]{3}}{2}) * \frac{v^{2} s i n^{2} 30}{2 g}$

$= \sqrt[]{3} \frac{m v^{3}}{16 g}$

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