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Physics Physics Thermodynamics

A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52K, its efficiency is doubled. The temperature is Kelvin of the source will be…………

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Payal Gupta

Contributor-Level 10

Efficiency in the Carnot engine is defined as :

$η = \frac{w o r k}{h e a t} = \frac{1}{4}$

$η = 1 - \frac{T_{c}}{T_{h}}, 2 η = 1 - \frac{T_{c} - 52}{T_{h}} \Rightarrow T_{h} = 208 K$

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Physics Class 11th

Two particles having masses 4g and 16g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is n : 2. The value of n will be……….

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Payal Gupta

Contributor-Level 10

$P = \sqrt[]{2 K m}$

$\frac{n}{2} = \sqrt[]{\frac{m_{1}}{m_{2}}} = \frac{1}{2}$

n = 1

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Physics Physics Electrostatic Potential and Capacitance

Two small spheres each of mass 10mg are suspended from a point by threads 0.5m long. They are equally charged and repel each other to a distance of 0.20m. The charge on each of the sphere is $\frac{a}{21} * 1 0^{- 8} C .$ The value of 'a' will be……. [Given g = 10 ms^-2]

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Payal Gupta

Contributor-Level 10

$T s i n θ = \frac{k q^{2}}{r^{2}}$

$T c o s θ = m g$

$t a n θ = \frac{k q^{2}}{m g r^{2}}$

Using the above equation find the value of 'q'

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Physics Physics Oscillations

Two identical springs of spring constant '2k' are attached to a block of mass m and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is

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Payal Gupta

Contributor-Level 10

Since both spring applied force in same direction.

Equivalent spring constant = 4k

$T = 2 π \sqrt[]{\frac{m}{4 k}}$

$T = π \sqrt[]{\frac{m}{k}}$

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Physics Class 12th

Match List I with List II.

List I List II

(a) Rectifier (i) Used either for stepping up or stepping down the a.c. voltage

(b) Stabilizer (ii) Used to convert a.c. voltage into d.c. voltage

(c) Transformer (iii) Used to remove any ripple in the rectified output volt

...more

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5 months ago

Physics Class 11th

Y = A sin $(ω t + ?_{0})$ is the time-displacement equation of a SHM. At t = 0 the displacement of the particle is y = $\frac{A}{2}$ and it moving along negative x-direction. Then the initial phase angle $?_{0}$ will be

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Payal Gupta

Contributor-Level 10

Using general equation of SHM :

$y = A s i n (ω t + ?_{0}) = \frac{A}{2}$

$ω t + ?_{0} = \frac{π}{6}, \frac{5 π}{6}$

At t = 0

$?_{0} = \frac{π}{6}, \frac{5 π}{6}$

Since it is moving in – x direction

$? = \frac{5 π}{6}$

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5 months ago

Physics Physics Electric Charge and Field

An electron with kinetic energy K₁ enters between parallel plates of a capacitor at an angle 'α' with the plates. It leaves the plates at angle 'β' with kinetic energy K₂. Then the ratio of kinetic energies K₁ : K₂will be

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Payal Gupta

Contributor-Level 10

Since force will act in horizontal direction so vertical component of speed will be same

v₁ cos α = v₂ cos β

$\Rightarrow \frac{K_{1}}{K_{2}} = {(\frac{v_{1}}{v_{2}})}^{2} = \frac{c o s^{2} β}{c o s^{2} α}$

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Physics Physics Magnetism and Matter

In a ferromagnetic region with zero magnetization.

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Payal Gupta

Contributor-Level 10

In ferromagnetic material, below Curie's temperature, a domain is defined as macroscopic region with zero magentisation.

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Physics Physics Atoms

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n = 2 to n = 1 state is

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Payal Gupta

Contributor-Level 10

Using Bohr's theory :

$λ = \frac{h c}{E_{2} - E_{1}} = \frac{1242 e V - n m}{(- 3.4) - (- 13.6)} = 121.8 n m$

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Physics Work, Energy and Power

A sphere of radius 'a' and mass 'm' rolls along a horizontal plane with constant speed v₀. It encounters and inclined plane at angle θ and climbs upward. Assuming that it rolls without slipping, how far up the sphere will travel?

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Payal Gupta

Contributor-Level 10

Using conservation of energy between initial and final position shown :

$[\frac{1}{2} (\frac{2}{5} m a^{2}) {(\frac{V_{0}}{a})}^{2} + \frac{1}{2} m V_{0}^{2}] + 0 = m g l s i n θ$

$\Rightarrow \frac{7}{10} m V_{0}^{2} = m g l s i n θ$

$\Rightarrow l = \frac{7 v_{0}^{2}}{10 g s i n θ}$

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