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Physics System of Particles and Rotational Motion

A cord is wound round the circumference of wheel of radius r. The axis of the wheel is horizontal and the moment of inertial about it is I. A weight mg is attached to the cord at the end. The weight falls from rest. After falling through a distance 'h', the square of angular velocity of wheel will be:

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Vishal Baghel

Contributor-Level 10

mg – T = ma. (1)

T * R = l . (2)

a = R. (3)

With the help of equations (1), (2) and (3), we get

$a = \frac{m g}{m + \frac{l}{R^{2}}}$

$v = \sqrt[]{2 a h} = ω R$

$\Rightarrow ω^{2} = \frac{2 m g h}{l + m R^{2}}$

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a year 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

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Payal Gupta

Contributor-Level 10

T $1 = 2 π \sqrt[]{\frac{m}{k}}$

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

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

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

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a year 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

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Payal Gupta

Contributor-Level 10

taking θ₁ > θ₂

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

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

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New answer posted

a year 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]

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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)

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a year 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

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Payal Gupta

Contributor-Level 10

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

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New answer posted

a year 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, $\frac{C_{2}}{C_{1}} .$

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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

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New answer posted

a year 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

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

$N o w, Q = m s Δ θ$

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

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New answer posted

a year 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

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Payal Gupta

Contributor-Level 10

F = m E cos θ

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

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

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New answer posted

a year 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

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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.

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New answer posted

a year 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

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Payal Gupta

Contributor-Level 10

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

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

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