Physics

Physics Physics Moving Charges and Magnetism

Contributor-Level 9

Steam point (T₁) = 100°C = 273 + 100 = 373 K

Ice point (T₂) = 0°C = 273 K

$η = (1 - \frac{T_{2}}{T_{1}}) * 100 = (1 - \frac{273}{373}) * 100 = 26.81 %$

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2 months ago

Two long current carrying conductors are placed parallel to each other at a distance of 8 cm between them. The magnitude of magnetic field produced at mid-point between the two conductors due to current flowing in them is 300 $μ T$ . the equal current flowing in the two conductors is:

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Contributor-Level 10

If currents are flowing in same direction, magnetic field will cancel each other, so the currents must flowing in opposite direction

$B_{P} = \frac{μ_{0} I}{2 π r} * 2$

$\Rightarrow 300 * 1 0^{- 6} = \frac{4 π * 1 0^{- 7}}{2 π * 4 * 1 0^{- 2}} * 2$

=> I = 30 A

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2 months ago

The variation of acceleration due to gravity (g) with distance (r) from the centre of the earth is correctly represented by: (Given R = radius of earth)

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Contributor-Level 9

$g \propto r 0 \leq r \leq R$

$g \propto \frac{1}{r^{2}} r > R$

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Why is there a negative sign in the total energy formula of a satellite?

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Aadit Singh Uppal

Contributor-Level 10

The minus sign in the formula indicates that:

A gravitational force exists.
The satellite cannot leave the orbit by itself.
A minimum threshold of energy will be required to move the satellite from its existing orbit.

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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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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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2 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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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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2 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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