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Physics Work, Energy and Power

A body of mass 'm' dropped from a height 'h' reaches the ground with a speed of 0.8 $\sqrt[]{g h} .$ The value of work done by the air-friction is:

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alok kumar singh

Contributor-Level 10

Work done is equal to change in K.E.

$S o W_{1} + W_{2} = \frac{1}{2} M {(0.8 \sqrt[]{g h})}^{2} - 0$ W₁^-> work done by mg

$m g h + W_{2} = \frac{1}{2} m * 0.64 g h$ W₂ -> work done by air friction

$W_{2} = 0.32 m g h - m g h = - 0.68 m g h$

W₂ = -0.68 mgh

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5 months ago

Physics Oscillations

A bob of mass 'm' suspended by a thread of length I undergoes simple harmonic oscillations with time period T. If the bob is immersed in a liquid that has density $\frac{1}{4}$ times that of the bob and the length of the thread is increased by 1/3^rd of the original length, then the time period of the simple harmonic oscillations will be:-

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Vishal Baghel

Contributor-Level 10

Time period T = 2π $\sqrt[]{\frac{l}{g_{e f f}}}$

$\Rightarrow T_{i} = T = 2 π \sqrt[]{\frac{l}{g}}$

$\Rightarrow T_{f} = T' = 2 π \sqrt[]{\frac{4 l / 3}{g (1 - \frac{ρ}{σ})}} = 2 π \sqrt[]{\frac{16 l}{9 g}}$

$\Rightarrow T' = \frac{4}{3} T$

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5 months ago

Physics Class 12th

Due to cold weather a 1m water pipe of cross-sectional area 1 cm² is filled with ice at -10°C. Resistivity heating is used to melt the ice. Current of 0.5A is passed through 4 k $Ω$ resistance. Assuming that all the heat produced is used for melting, what is the minimum time required?

(Given latent heat of fusion for water/ice = 3.33 * 10⁵ J kg^-¹, specific heat of ice = 2 * 10³ J kg^-1 and density of ice = 10³ kg/m³)

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alok kumar singh

Contributor-Level 10

Energy required to melt

Q = $M S Δ T + M L$

$\Rightarrow 1 0^{- 1} * 2 * 1 0^{3} * 10 + 1 0^{- 1} * 3.33 * 1 0^{5}$

->3.53 * 10⁴ J

Heat produce in wire

H = l²RT

$Q = 3.53 * 1 0^{4} = {(\frac{1}{2})}^{2} * (4 * 1 0^{3}) * t$

$t = \frac{3.53 * 1 0^{4} * 4}{4 * 1 0^{3}} = 35.3 s e c$

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5 months ago

Physics Physics Laws of Motion

Statement-I: Two forces $(\vec{P} + \vec{Q}) a n d (\vec{P} - \vec{Q}) w h e r e \vec{P} ⊥ \vec{Q},$ when act at an angle θ₁ to each other the magnitude of their resultant is $\sqrt[]{3 (P^{2} + Q^{2})},$ when they act an angle θ₂, the magnitude of their resultant becomes $\sqrt[]{2 (P^{2} + Q^{2})} .$ This is possible only when θ₁ < ₂.

Statement-II: In the situation given above.

θ₁ = 60° and θ₂ = 90°

In the light of the above statements, choose the most appropriate answer from the options given below:-

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Vishal Baghel

Contributor-Level 10

$\vec{x} = | \vec{P} + \vec{Q} | = \sqrt[]{P^{2} + Q^{2}}$

=> $\vec{y} = | \vec{P} - \vec{Q} | = \sqrt[]{P^{2} + Q^{2}}$

$| \vec{x} + \vec{y} | = \sqrt[]{x^{2} + y^{2} + 2 x y}$

$\Rightarrow \sqrt[]{3 (P^{2} + Q^{2})} = \sqrt[]{{(\sqrt[]{P^{2} + Q^{2}})}^{2} + {(\sqrt[]{P^{2} + Q^{2}})}^{2} + 2 (P^{2} + Q^{2}) c o s θ_{1}}$

$\Rightarrow θ_{1} = 60 °$

=>Using same formula : θ₂ = 90°

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5 months ago

Physics Laws of Motion

A block of mass m slides on the wooden wedge, which in turn slides backward on the horizontal surface. The acceleration of the block with respect to the wedge is: Given m = 8 kg, M = 16kg. Assume all the surfaces shown in the figure to be frictionless.

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alok kumar singh

Contributor-Level 10

Suppose acceleration of wedge is a and acceleration of block w. r.t. wedge is a₁ then N cos60° = Ma = 16 a -> N = 32 a

For block w.r.t. wedge

N + 8a sin 30° = 8g cos 30°

N = 8g cos 30° - 8a sin 30°

->32a = 8g cos 30° - 8a sin 30°

->a = $\frac{\sqrt[]{3}}{9} g$

Now for 8 kg,

$8 g s i n 30 ° + 8 a c o s 30 ° = m a_{1}$

$a_{1} = g * \frac{1}{2} + \frac{\sqrt[]{3}}{9} g * \frac{\sqrt[]{3}}{2}$

$= \frac{2}{3} g$

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5 months ago

Physics Work, Energy and Power

A block moving horizontally on a smooth surface with a speed of 40 m/s splits into parts with masses in the ratio of 1:2. If the smaller part moves at 60 m/s in the same direction, then the fractional change in kinetic energy is:-

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Vishal Baghel

Contributor-Level 10

Using conservation of linear momentum:

=> 40 * 3m = 60 * m + v * 2m

=> v = 30 m/s

$K E_{i} = \frac{1}{2} * 3 m * {(40)}^{2}$

$K E_{f} = \frac{1}{2} * m * {(60)}^{2} + \frac{1}{2} * 2 m * {(30)}^{2}$

$\Rightarrow \frac{K E_{f}}{K E_{i}} = \frac{54}{48}$

Fractional change in kinetic energy = 1 $- \frac{K E_{f}}{K E_{i}} = \frac{1}{8}$

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5 months ago

Physics Atoms

A free electron of 2.6 eV energy collides with a H⁺ ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. (h = 6.6 * 10^-34 Js)

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Vishal Baghel

Contributor-Level 10

Energy of electron in first existed state will be -3.4 eV.

So total energy difference will be (2.6 + 3.4) eV.

Wavelength $(λ) = \frac{1242 e V - n m}{6 e V} = 207 n m$

$F r e q u e n c y = \frac{c}{λ} = \frac{3 * 1 0^{8}}{207 * 1 0^{- 9}} = 1.45 * 1 0^{9} M H z$

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5 months ago

Physics Dual Nature of Radiation and Matter

The temperature of an ideal gas in 3-dimensions is 300K. The corresponding de-Broglie wavelength of the electron approximately at 300 K, is:

[m_e = mass of electron = 9 * 10^-³¹kg

h = Planck constant = 6.6 * 10^-³⁴ Js

k_B = Boltzmann constant = 1.38 * 10^-²³ JK^-¹]

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alok kumar singh

Contributor-Level 10

De Broglie wavelength

$λ = \frac{h}{m V} = \frac{h}{\sqrt[]{2 m E}} E \to K . E .$

$E = \frac{3}{2}$ KT for gas

$S o λ = \frac{h}{\sqrt[]{3 m K T}} = \frac{6.6 * 1 0^{- 34}}{\sqrt[]{3 * 9 * 1 0^{- 31} * 1.38 * 1 0^{- 23} * 300}}$

$\begin{array}{l} λ = 6.26 * 1 0^{- 9} m \\ λ = 6.26 n m \end{array}$

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5 months ago

Physics Class 11th

Statement-I: If three forces ${\vec{F}}_{1}, {\vec{F}}_{2} a n d {\vec{F}}_{3}$ are represented by three sides of triangle and, ${\vec{F}}_{1} + {\vec{F}}_{2} = - {\vec{F}}_{3},$ then these three forces are concurrent forces and satisfy the condition for equilibrium.

Statement-II: A triangle made up of three forces ${\vec{F}}_{1}, {\vec{F}}_{2} a n d {\vec{F}}_{3}$ as its sides taken in the same order, satisfy the condition for translator equilibrium.

In the light of the above statements, choose the most appropriate answer from the options given below:

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Vishal Baghel

Contributor-Level 10

For equilibrium net force acting on the system should be zero.

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5 months ago

Physics Thermodynamics

A mixture of hydrogen and oxygen has volume 500 cm³, temperature 300K, pressure 400 kPa and mass 0.76g. The ratio of masses of oxygen to hydrogen will be:-

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Vishal Baghel

Contributor-Level 10

Using Ideal gas Equation :

PV = nRT

$\Rightarrow n = \frac{P V}{R T} = \frac{400 * 1 0^{3} * 500 * 1 0^{- 6}}{8.3 * 100} = 0.008$

=> $m_{1} = 0.12, m_{2} = 0.64$

=> $\frac{m_{2}}{m_{1}} = \frac{16}{3}$

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