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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Class 11th

The velocity of liquid coming out of a small hole of a large vessel containing two different liquids shown in figure is

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics Wave Optics

Two pillars are at distance of 6.28 m from each other. What should be the maximum distance of an observer from pillar to see them separately?

(qR = 1 arc minute)

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Raj Pandey

Contributor-Level 9

qR =1/60º for human eye

⇒x = 21.6 km

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Class 12th

The work function of a metal is w0 and l is wavelength of incident radiation. There is no emission of photoelectron whenThe work function of a metal is w0 and l is wavelength of incident radiation. There is no emission of photoelectron when\

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Raj Pandey

Contributor-Level 9

For no emission of electron

λ > λ0

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Class 11th

An ideal fluid of density 800kgm^-³, flows smoothly through a bent pipe (as shown in figure) that tapers in cross-sectional area from a to $\frac{a}{2} .$ The pressure difference between the wide and narrow sections of pipe is 4100 Pa. At wider section, the velocity of fluid is $\frac{\sqrt[]{x}}{6} m s^{- 1}$ for x = ________. (Given g = 10 ms^-²)

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Raj Pandey

Contributor-Level 9

$ρ = 800 k g / m^{3}$

P₁ – P₂ = 4100 Pa

Bernoulli's question b/w (1) & (2)

$\frac{P_{1}}{ρ g} + \frac{v_{1}^{2}}{2 g} + z_{1} = \frac{P_{2}}{ρ g} + \frac{v_{2}^{2}}{2 g} + z_{2}$

⇒ $\frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{v_{2}^{2}}{2 g} + 0$ -(1)

Equation of continuity

A₁v₁ = A₂v₂

v₂ = 2v₁ -(2)

from (1) & (2)

$\Rightarrow \frac{P_{1} - P_{2}}{ρ g} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{{(2 v_{1})}^{2}}{2 g}$

$\frac{4100}{800 * 10} + \frac{v_{1}^{2}}{2 g} + 1 = \frac{4 v_{1}^{2}}{2 g}$

$\frac{12100}{8000} = \frac{3 v_{1}^{2}}{2 g}$

$\Rightarrow v_{1}^{2} = \frac{2 * 10}{3} * \frac{12100}{8000}$

$= \frac{2}{3} * \frac{121}{8}$

x = 363

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics Alternating Current

A 110 V, 50Hz, AC source is connected in the circuit (as shown in figure). The current through the resistance $55 Ω,$ at resonance in the circuit, will be__________A.

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Raj Pandey

Contributor-Level 9

At Resonance

X_L = X_C

then l_L = l_C

Now phasor diagram

for L & C

So, Net current = zero

$= (l_{L} + l_{C})$

Therefore current through R circuit at resonance will be zero

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Class 12th

As per the given circuit, the value of current through the battery will be__________A.

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Raj Pandey

Contributor-Level 9

$\begin{array}{l} D_{1} \\ D_{3} \end{array}}$ Forward biased offer zero Resistance

D₂} Reversed biased offers Infinite Resistance

$I = \frac{v}{R} = \frac{10}{10} = 1 A m p$

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Units and Measurement

In a vernier calipers, each cm on the main scale is divided into 20 equal parts. If tenth vernier scale division coincides with nineth main scale division. Then the value of vernier constant will be__________* 10^-²mm.

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Raj Pandey

Contributor-Level 9

Let 'x' he the value of one division of main scale

$x = \frac{1}{20} c m = 0.05 c m$

Let y be value of one division on venire scale given

10 y = 9 x

$y = \frac{9 x}{10}$

Least count = $x - \frac{9 x}{10} = \frac{x}{10} = \frac{0.05}{10}$

= 0.005 cm

= 5 * 10^-2 mm

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics Ray Optics and Optical Instruments

A light ray is incident, at an incident angle q₁, on the system of two plane mirrors M₁ and M₂ having an inclination angle 75° between them (as shown in figure). After reflecting from mirror M₁ it gets reflected back by the mirror M₂ with an angle of reflection 30°. The total deviation of the ray will be ___________ degree.

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Raj Pandey

Contributor-Level 9

$δ_{t o t a l} = 360 ° - 2 θ$

= 360 – 2 * 75° = 210°

Total deviation = $δ = 150 °$

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Electromagnetic Induction

A $10 Ω$ 20 mH coil carrying constant current is connected to a battery of 20 V through a switch. Now after switch is opened current becomes zero in 100 $μ s .$ The average e.m.f. induced in the coil is__________V.

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Raj Pandey

Contributor-Level 9

Decay of current in Inductor is given by,

$i = i_{0} e^{- t / τ} [W h e r e τ = \frac{L}{R}; i_{0} = \frac{v}{R} = \frac{20}{10} = 2]$

At t = 100 $μ s$

i = 0

i.e. i = i₀ $e^{- 100 / τ} = 0$ -(1)

e.m.f induced

$e = \frac{- L d i}{d t} = - L \frac{d}{d t} [i_{0} e^{- t / τ}]$

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Decay of current in Inductor is given by,

$i = i_{0} e^{- t / τ} [W h e r e τ = \frac{L}{R}; i_{0} = \frac{v}{R} = \frac{20}{10} = 2]$

At t = 100 $μ s$

i = 0

i.e. i = i₀ $e^{- 100 / τ} = 0$ -(1)

e.m.f induced

$e = \frac{- L d i}{d t} = - L \frac{d}{d t} [i_{0} e^{- t / τ}]$

= $= - L i_{0} \frac{- 1}{2} e^{- t / τ}$

$e = \frac{L i_{0}}{τ} e^{- t / τ}$

$e_{a v g} = \frac{\int_{0}^{200 μ s} e d t}{\int_{0}^{200 μ s} d t} = \frac{\int_{0}^{200 μ s} \frac{L i_{0}}{τ} e^{- t / τ} d t}{\int_{0}^{200 μ s} d t}$

$= \frac{L i_{0}}{200 * 1 0^{- 6} s e c} = \frac{20 * 1 0^{- 3} * 2}{200 * 1 0^{- 6}} = 400$

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics Mechanical Properties of Solids

The elongation of a wire on the surface of the earth is 10^-⁴m. The same wire of same dimensions is elongated by 6 * 10^-⁵ m on another planet. The acceleration due to gravity on the planet will be__________ms^-². (Take acceleration due to gravity on the surface of earth = 10 ms^-²)

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Raj Pandey

Contributor-Level 9

$Δ L$ (change in length) $= \frac{F l}{A E} = \frac{m g l}{A E}$

$Δ l \propto g$

$\frac{Δ l_{1}}{g_{1}} = \frac{Δ l_{2}}{g_{2}} \Rightarrow \frac{1 0^{- 4}}{10} = \frac{6 * 1 0^{- 5}}{g_{1}}$

$g_{1} = \frac{6 * 1 0^{- 4}}{1 0^{- 4}} =$ 6 m/sec²

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