Tags Physics

Physics

Get insights from 5.6k questions on Physics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics

Follow Ask Question

5.6k

Questions

Discussions

Active Users

Followers

New answer posted

8 months ago

Physics Physics Oscillations

If two similar spring each of spring constant K₁ are joined in series, the new spring constant and time period would be changed by a factor

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

T $1 = 2 π \sqrt[]{\frac{m}{k}}$

$k_{e q} = \frac{k * k}{k + k} = \frac{k}{2}$

$T_{2} = 2 π \sqrt[]{\frac{m}{(k / 2)}} = 2 π \sqrt[]{\frac{2 m}{k}}$

$\Rightarrow \frac{T_{2}}{T_{1}} = \sqrt[]{2} a n d \frac{k_{2}}{k_{1}} = \frac{1}{2}$

0 0

New answer posted

8 months ago

Physics Class 12th

The temperature q at the junction of two insulating sheets,having thermal resistances R₁ and R₂ as well top and bottom temperatures q₁ and q₂ (as shown in figure) is given by

0 Follower 10 Views

Payal Gupta

Contributor-Level 10

taking θ₁ > θ₂

Heat current = $\frac{θ_{1} - θ}{R_{1}} = \frac{θ - θ_{2}}{R_{2}} \Rightarrow \frac{θ_{1}}{R_{1}} + \frac{θ_{2}}{R_{2}} = θ (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$\Rightarrow θ = \frac{θ_{1} R_{2} + θ_{2} R_{1}}{R_{1} + R_{2}}$

0 0

New answer posted

8 months ago

Physics Class 12th

LED is constructed from Ga – As – P semiconducting material. The energy gap of this LED is 1.9 eV. Calculate the wavelength of light emitted and its colour.

[h = 6.63 * 10^-34 Js and c = 3 * 10⁸ ms^-1]

0 Follower 11 Views

Payal Gupta

Contributor-Level 10

$λ = \frac{h c}{E} = \frac{6.63 * 1 0^{- 34} * 3 * 1 0^{8}}{1.9 * 1.6 * 1 0^{- 19}}$ = 654 nm (red)

0 0

New answer posted

8 months ago

Physics Class 11th

Four identical solid spheres each of mass 'm' and radius 'a' are placed with their centres on the four corners of a square of side 'b'. The moment of inertia of the system about side of square where the axis of rotation is parallel to the plane of the square is

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$I_{A B} = \frac{2}{5} m a^{2} * 2 + (\frac{2}{5} m a^{2} + m b^{2}) * 2 = \frac{8 m a^{2}}{5} + 2 m b^{2}$

0 0

New answer posted

8 months ago

Physics Physics Electrostatic Potential and Capacitance

Consider the combination of 2 capacitors C₁ and C₂, with C₂ > C₁, when connected in parallel, the equivalent capacitance is $\frac{15}{4}$ times the equivalent capacitance of the same connected in series, Calculate the ratio of capacitors, $\frac{C_{2}}{C_{1}} .$

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

According question, we can write

$\frac{C_{1} * C_{2}}{C_{1} + C_{2}} = \frac{15}{4} (C_{1} + C_{2}) \Rightarrow 4 C_{1} C_{2} = 15 (C_{1}^{2} + C_{2}^{2} + 2 C_{1} C_{2})$

$\Rightarrow 4 (\frac{C_{2}}{C_{1}}) = 15 {(\frac{C_{2}}{C_{1}})}^{2} + 30 (\frac{C_{2}}{C_{1}}) + 15 \Rightarrow 15 x^{2} + 26 x + 15 = 0,$ where x = $\frac{C_{2}}{C_{1}}$

$\Rightarrow x = \frac{- 26 \pm \sqrt[]{676 - 900}}{30}$

Since x cannot be real, so this Question has been dropped by NTA

0 0

New answer posted

8 months ago

Physics Class 11th

A large number of water drops, each of radius r, combine to have a drop of radius R. If the surface tension is T and mechanical equivalent of heat is J, the rise in heat energy per unit volume will be

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

From volume conservation

$n * \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3} \Rightarrow R^{3} = n r^{3} . . . . . . . . . (1)$

Decrease in surface area = $n * 4 π r^{2} - 4 π R^{2}$

$(Δ A) = 4 π [n r^{2} - R^{2}] = 4 π [\frac{n * r^{3}}{r} - R^{2}] = 4 π [\frac{R^{3}}{r} - R^{2}] = 4 π R^{3} [\frac{1}{r} - \frac{1}{R}]$

Energy released (W) = $T * Δ A = 4 π R^{3} T [\frac{1}{r} - \frac{1}{R}]$

Heat produced (Q) = $\frac{W}{J} = \frac{4 π T R^{3}}{J} [\frac{1}{r} - \frac{1}{R}]$

$N o w, Q = m s Δ θ$

$\Rightarrow \frac{4 π R^{3} T}{J} [\frac{1}{r} - \frac{1}{R}] = (\frac{4}{3} π R^{3}) * | x | * Δ θ = \frac{3 T}{J} [\frac{1}{r} - \frac{1}{R}]$

0 0

New answer posted

8 months ago

Physics Class 11th

Assume that a tunnel is dug along a chord of the earth, at a perpendicular distance (R/2) from the earth's centre, where 'R' is the radius of the Earth. The wall of the tunnel is frictionless. If a particle is released in this tunnel, it will execute a simple harmonic motion with a time period

0 Follower 10 Views

Payal Gupta

Contributor-Level 10

F = m E cos θ

$= m (\frac{G M}{R^{3}} * r) * \frac{x}{r}$ $m a = \frac{m g}{R} * \Rightarrow a = \frac{g}{R} x$

$\Rightarrow T = 2 π \sqrt[]{\frac{R}{g}}$

0 0

New answer posted

8 months ago

Physics Class 11th

Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Body 'P' having mass M moving with speed 'u' has head – on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, Body 'Q' will have a maximum speed equal to '2u' after collision.

Reason R : During elastic collision, the momentum and kinetic energy are both conserved. In the light of the above statements, choose the most appropriate answer from the options given below

0 Follower 6 Views

Payal Gupta

Contributor-Level 10

Mass m will acquire velocity 2u. Total momentum of system will be conserved but total kinetic energy is not conserved during collision.

0 0

New answer posted

8 months ago

Physics Physics Motion in Plane

A particle is moving with uniform speed along the circumference of a circle of radius of radius R under the action of a central fictitious force F which is inversely proportional to R³. Its time period of revolution will be given by

0 Follower 5 Views

Payal Gupta

Contributor-Level 10

$F = \frac{k}{R^{3}} = \frac{m v^{2}}{R} \Rightarrow v = \sqrt[]{\frac{k}{m R^{2}}} = \sqrt[]{\frac{k}{m}} * \frac{1}{R}$

$T = \frac{2 π R}{v} = \frac{2 π R}{\sqrt[]{\frac{k}{m} * (\frac{1}{R})}} \Rightarrow T α R^{2}$

0 0

New answer posted

8 months ago

Physics Units and Measurement

In a typical combustion engine the work done by a gas molecule is given by W = $α^{2} β e^{- \frac{β x^{2}}{k T}},$ where x is the displacement, k is the Boltzmann constant and T is the temperature. If and are constants, dimensions of will be

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$W = α^{2} β e^{- \frac{- β x^{2}}{k T}}$

$[β x^{2}] = [k T] = [M L^{2} T^{- 2}]$

$β = M T^{- 2}$

$[W] = [α^{2} β] \Rightarrow [M L^{2} T^{- 2}] = [α^{2}] [M T^{- 2}] \Rightarrow [α^{2}] = [L^{2}] \Rightarrow [α] = [L]$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
688k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Physics

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience