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10 months ago

Physics Physics Motion in Plane

A planet revolving in elliptical orbit has:

(A) a constant velocity of revolution.

(B) has the least velocity when it is nearest to the sun.

(C) its areal velocity is directly proportional to its velocity.

(D) areal velocity in inversely proportional to its velocity.

(E) To follow a trajectory such that the areal velocity is constant.

Choose the correct answer from the options given below

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Payal Gupta

Contributor-Level 10

In elliptical orbit, areal velocity is constant

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Physics Class 11th

The trajectory of a projectile in an vertical plane is y = αx - βx², where α and β are constants and x & y are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection θ and the maximum height attained H are respectively given by:

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Vishal Baghel

Contributor-Level 10

$y = α x - β x^{2}$

$\Rightarrow \frac{d y}{d x} = α - 2 β x = 0$

$\Rightarrow x = \frac{α}{2 β}$

$y_{m a x} = α * \frac{α}{2 β} - β * {(\frac{α}{2 β})}^{2} = \frac{α^{2}}{4 β}$

Range = $2 x = \frac{α}{β} = \frac{2 u^{2} s i n θ . c o s θ}{g}$

On comparing with

y = x tan θ - $\frac{g x^{2}}{2 u^{2} c o s^{2} θ}$

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10 months ago

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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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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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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10 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]

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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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10 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

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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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10 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, $\frac{C_{2}}{C_{1}} .$

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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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10 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

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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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10 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

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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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