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Physics Class 11th

A body is projected from the ground at an angle of $45 °$ with horizontal. Its velocity after 2s is 20 ms^-1. The maximum height reached by the body during its motion is ______m. (use g = 10ms^-2)

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Payal Gupta

Contributor-Level 10

after 2 sec, v = 20 m/sec

$\begin{array}{l} \vec{v} = u c o s 45 ° \hat{i} + (u s i n 45 ° - g t) \hat{j} \\ v = 20 = \sqrt[]{{(u c o s 45 °)}^{2} + {(u s i n 45 ° - 20)}^{2}} \end{array}$

$\Rightarrow 400 = u^{2} + 400 - 40 u s i n 45 °$

u = 40 sin 45°

$H_{m a x} = \frac{u^{2} s i n^{2} 45}{2 g} = \frac{4 0^{2} s i n^{2} 45 ° * s i n^{2} 45}{2 g}$

$= \frac{40 * 40}{2 * 10} * \frac{1}{2} * \frac{1}{2} = 20 m$

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6 months ago

Physics Physics Atoms

In Bohr's atomic model of hydrogen, let K, P and E are the kinetic energy, potential energy and total energy of the electron respectively. Choose the correct option when the electron undergoes transitions to a higher level

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Payal Gupta

Contributor-Level 10

Based on theory

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6 months ago

Physics Physics Wave Optics

Two light beams of intensities in the ratio of 9 : 4 are allowed to interfere. The ratio of the intensity of maxima and minima will be

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Payal Gupta

Contributor-Level 10

$\frac{l_{1}}{I_{2}} = \frac{9}{4}$ $\frac{l_{m a x}}{l_{m a x}} = \frac{l_{1} + l_{2} + 2 \sqrt[]{l_{1} l_{2}}}{l_{1} + l_{2} - 2 \sqrt[]{l_{1} l_{2}}}$

$= \frac{\frac{9}{4} + 1 + 2 \sqrt[]{\frac{9}{4}}}{\frac{9}{4} + 1 - 2 \sqrt[]{\frac{9}{4}}}$

$= \frac{\frac{13}{4} + \frac{2 * 3}{2}}{\frac{13}{4} - 2 * \frac{3}{2}} = \frac{25}{1}$

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6 months ago

Physics Dual Nature of Radiation and Matter

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speed of emitted electrons for the two frequencies respectively will be

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Payal Gupta

Contributor-Level 10

$h υ_{1} = h υ_{0} + K . E_{1}$

$\Rightarrow 3.8 e V = 0.6 e V + K . E_{1}$

K. E₁ = 3.2 eV (i)

$\Rightarrow h υ_{2} = h υ_{01} + K . E_{2}$

1.4 = 0.6 + K.E₂

KE₂ = 0.8 eV (ii)

$\frac{K . E_{1}}{K . E_{2}} = \frac{3.2}{0.8} = 4$

$\frac{v_{1}}{v_{2}} = 2 : 1$

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6 months ago

Physics Class 12th

An electric bulb is rated as 200W. What will be the peak magnetic field at 4m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.

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Payal Gupta

Contributor-Level 10

I_R = I_B (due to magnetic field)

(due to Radiation)

$\frac{P}{A} * E f f i c i e n c y = \frac{B_{0}^{2}}{2 μ_{0}} c$

$\Rightarrow \frac{200}{4 π * 4^{2}} * \frac{3.5}{100} = \frac{B_{0}^{2}}{2 * 4 π * 1 0^{- 7}} * 3 * 1 0^{8}$

$\Rightarrow B_{0} = 1.71 * 1 0^{- 8} T$

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6 months ago

Physics Physics Electrostatic Potential and Capacitance

A long cylindrical volume contains a uniformly distributed charge of density $ρ$ . The radius of cylindrical volume is R. A charge particle (q) revolves around the cylinder in a circular path. The kinetic energy of the particle is

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Payal Gupta

Contributor-Level 10

$F_{C} = F_{q}$

$\Rightarrow \frac{m v^{2}}{R} = \frac{ρ R}{2 \in_{0}} q$

$m v^{2} = \frac{ρ R^{2} q}{2 ε_{0}} \Rightarrow \frac{1}{2} m v^{2} = k . E . = \frac{1}{4} \frac{ρ R^{2} q}{\in_{0}}$

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6 months ago

Physics Physics Electrostatic Potential and Capacitance

If the charge on a capacitor is increased by 2C, the energy stored in it increases by 44%. The original charge on the capacitor is (in C)

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Payal Gupta

Contributor-Level 10

$U_{i} = \frac{Q^{2}}{2 C}$

Uf = 1.44 Ui

Qf = Q + 2

$1.44 U_{i} = \frac{{(Q + 2)}^{2}}{2 C}$

$\Rightarrow 1.44 \frac{Q^{2}}{2 C} = \frac{{(Q + 2)}^{2}}{2 C}$

$\Rightarrow 1.2 Q = Q + 2$

0.2 Q = 2

Q = 10 C

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6 months ago

Physics Physics Thermodynamics

A 100g of iron nail is hit by a 1.5 kg hammer striking at a velocity of 60ms^-1. What will be the rise in the temperature of the nail if one fourth of energy of the hammer goes into heating the nail?

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Payal Gupta

Contributor-Level 10

Heating energy $= \frac{1}{4} t h$ energy of the hammer

$= \frac{1}{4} * \frac{1}{2} m v^{2}$

$= \frac{1}{4} * \frac{1}{2} * 1.5 * 60 * 60$

$= \frac{15 * 6 * 60}{4 * 2}$

= 675 J

$Δ T = \frac{675}{42} = 16.07 ° C$

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6 months ago

Physics Physics System of Particles and Rotational Motion

A fly wheel is accelerated uniformly from rest and rotates through 5 rad in the first second. The angle rotated by the fly wheel in the next second, will be

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Payal Gupta

Contributor-Level 10

$θ = ω_{0} t + \frac{1}{2} α t^{2}$

$5 = 0 * 1 + \frac{1}{2} * α * 1^{2}$

= 10 rad/sec²

(ln 2 seconds) = $ω_{0} t + \frac{1}{2} α t^{2}$

$θ_{2} = 20 r a d$

Angle rotated by wheel in 2^nd second

= $θ_{2} - θ$

= 20 – 5

= 15 rad

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6 months ago

Physics Physics Motion in Straight Line

An object of mass 5kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use g = 10ms^-2].

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Payal Gupta

Contributor-Level 10

$a = \frac{m g + 10}{g} = \frac{5 * 10 + 10}{10}$

$\Rightarrow a = 6 m / s e c^{2}$

v = u + at

0 = u – 6t

$t_{A} = \frac{u}{6}$ (time of Ascent)

$= u * \frac{u}{6} - \frac{u}{2} * 6 * \frac{u}{6} * \frac{u}{6}$

= $\frac{u^{2}}{6} - \frac{u^{2}}{12}$

$h = \frac{u^{2}}{12}$

for time of descent

$a = \frac{50 - 10}{10}$

a = 4 m /sec²

$\Rightarrow \frac{u^{2}}{12} = 0 * t + \frac{1}{2} * 4 t^{2}$

$t_{B} = \frac{u}{\sqrt[]{24}}$

$\frac{t_{A}}{t_{B}} = \frac{u * 24}{6 * u} =$ $\sqrt[]{2}$

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