Physics

A thin circular ring of mass M and radius R is rotating with a constant angular velocity 2 rads^-1 in a horizontal plane about an axis vertical to its plane and passing through the centre of the ring. If two objects each of mass m be attached gently to the opposite ends of a diameter of ring, the will then rotate with an angular velocity (in rads^-¹).

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Contributor-Level 9

By conservation of Angular momentum

Li = Lf

$M R^{2} ω = (m R^{2} + 2 m R^{2}) ω'$

$\frac{2 M}{M + 2 m} = ω'$

$\Rightarrow$ $ω' = \frac{2 M}{M + 2 m}$

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8 months ago

Physics Thermodynamics

Starting with the same initial conditions, an ideal gas expands form volume V₁ to V₂ in the three different ways. The work done by the gas in W₁ if the process is purely isothermal, W₂, if the process is purely adiabatic and W₃ if the process is purely isobaric. Then, choose the correct option

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Contributor-Level 10

Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃

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8 months ago

A ball is released from rest point P of a smooth semi-spherical vessel as shown is figure. The ratio of the centripetal force and normal reaction on the ball at point Q is A while angular position of point Q is a with respect to point P. Which of the following graphs represent the correct relation between A and a when ball goes from Q to R?

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Contributor-Level 9

NLM at point Q

N – mg sin a = $\frac{m v^{2}}{R}$

$N = \frac{m v^{2}}{R} + m g s i n α$ - (1)

COE : between P & Q

$\frac{m v^{2}}{R} = 2 m g s i n α$ - (2)

Centripetal force = 2mg sin a

N (Normal Reaction) = 2mg sin a + mg sin a

= 3mg sin a

$\frac{C e n t r i p e t a l f o r c e}{N o r m a l R e a c t i o n} = \frac{2 m g s i n α}{3 m g s i n α} = \frac{2}{3}$

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8 months ago

Physics Class 12th

A capacitor is discharging through a resistor R. Consider in time t₁ the energy stored in the capacitor reduces to half of its initial value and in time t₂, the charge stored in the capacitor reduces to half of its initial value and in time t₂ the charge stored reduces to one eight of its initial value. The ratio t₁/t₂ will be

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Contributor-Level 10

According to Law of discharging of capacitor, we can write

$Q = Q_{0} e^{- \frac{t}{τ}} \Rightarrow \frac{Q_{2}}{8} = Q_{0} e^{- \frac{t_{2}}{τ}} \Rightarrow t_{2} 3 τ l n (2), a n d$

^{$U = \frac{Q^{2}}{2 C} = \frac{Q_{0}^{2}}{2 C} e^{- \frac{2 t}{τ}} \Rightarrow \frac{1}{2} (\frac{Q_{0}^{2}}{2 C}) = \frac{Q_{0}^{2}}{2 C} e^{- \frac{2 t}{τ}} \Rightarrow t_{1} = \frac{τ}{2} l n (2)$}

^{$\Rightarrow \frac{t_{1}}{t_{2}} = \frac{1}{6}$}

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8 months ago

Physics Class 12th

The electric field at a point associated with a light wave is given by

E = 200 $[s i n (6 * 1 0^{15}) t + s i n (9 * 1 0^{15}) t] V m^{- 1}$

Given : h = 4.14 * 10-¹⁵ eVs

If the light fall on a metal surface having a work function of 2.50 eV, the maximum kinetic energy of the photoelectrons will be

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Physics Physics Motion in Straight Line

Contributor-Level 10

The light wave contains two lights of different frequencies, so

$E_{1} = h υ_{1} = \frac{4.14 * 1 0^{- 15} * 6 * 1 0^{15}}{2 π} = 3.96 e V, a n d$

$E_{2} = h υ_{2} = \frac{4.14 * 1 0^{- 15} * 9 * 1 0^{15}}{2 π} = 5.92 e V$

Maximum kinetic energy of the photoelectron = 5.9-2.5 = 3.42 eV

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8 months ago

An object is thrown vertically upwards. At its maximum height, which of the following quantity becomes zero?

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Physics Physics Ray Optics and Optical Instruments

Contributor-Level 9

At maximum height velocity (v) = 0

So, momentum = mv = m * 0 = 0

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8 months ago

The speed of light in media 'A' and 'B' are 2.0 * 10¹⁰ cm/s and 1.5 * 10¹⁰ cm/s respectively. A ray of light enters from the medium B to an incident angle 'q'. If the ray suffers total internal reflection, then

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Contributor-Level 10

$\frac{s i n θ_{C}}{V_{B}} = \frac{s i n 90 °}{V_{A}} \Rightarrow s i n θ_{C} = \frac{V_{B}}{V_{A}} = \frac{1.5 * 1 0^{10}}{2.0 * 1 0^{10}} = \frac{3}{4}$

According to question, we can write

$θ > θ = s i n^{- 1} (\frac{3}{4})$

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8 months ago

A person is standing in an elevator. In which situation, he experiences weight loss?

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Contributor-Level 9

.B.D of person w.r.t. elevator

mg – N = ma

N = mg – ma

N decreases so, person experiences weight loss

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8 months ago

Physics Class 12th

Two point charge Q each are placed at a distance d apart. A third point charge q is placed at a distance x from mid-point on the perpendicular bisector. The value of x at which charge q will experience the maximum Coulomb's force is:

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Contributor-Level 10

According to question, we can write

$F_{C A} = F_{C B} = \frac{K q Q}{x^{2} + {(\frac{d}{2})}^{2}} \Rightarrow F = 2 F_{C A} c o s θ = \frac{2 K q Q x}{{[x^{2} + {(\frac{d}{2})}^{2}]}^{\frac{3}{2}}}$

For maxima of force $\frac{d F}{d x} = 0, s o$

$x = \frac{d}{2 \sqrt[]{2}}$

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8 months ago

An expression for a dimensionless quantity P is given by $P = \frac{α}{β} l o g_{e} (\frac{k t}{β t});$ where a and b are constants, x is distance; k is Boltzmann constant and t is the temperature. Then the dimensions of a will be:

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