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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven

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2 weeks ago

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

The elastic behavior of material for linear stress and linear strain, is shown in the figure. The energy density for a linear strain of 5 * 10^-⁴ is __________KJ/m³. Assume that material is elastic upto the linear strain of 5 * 10^-⁴.

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Raj Pandey

Contributor-Level 9

$Y = \frac{σ}{\in} = \frac{20}{1 0^{- 10}},$ Y → young's modules of elasticity.

Stress $(σ)$ at strain $(\in)$ 5 * 10^-4

$σ = \in Y$

$= 5 * 1 0^{- 4} * 20 * 1 0^{10}$

= $1 0^{2} * 1 0^{6}$

= 10⁸

$U_{d} (E n e r g y d e n s i t y) = \frac{1}{2} σ \in$

$= \frac{1}{2} * 1 0^{8} * 5 * 1 0^{- 4}$

$\begin{array}{l} = 2.5 * 1 0^{4} \\ = 25 * 1 0^{3} = 25 k J / m^{3} \end{array}$

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Physics Motion in Straight Line

A ball of mass 0.5 kg is dropped from the height of 10m. The height, at which the magnitude of velocity becomes equal to the magnitude of acceleration due to gravity, is __________m.

[Use g = 10m/s²]

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Raj Pandey

Contributor-Level 9

Let 'h' be the height at which velocity becomes equal to magnitude of Acceleration

v = g = 10

v = u + at

10 = 0 + 10t

t = 1 sec

$h = u t + \frac{1}{2} a t^{2}$

$= 0 * 1 + \frac{1}{2} * 10 * 1 * 1$

= 5m

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

A fighter jet is flying horizontally at a certain with a speed of 200ms^-1. When it passes directly overhead ad anti-aircraft gun, a bullet is fired from the gun, at an angle (θ)with the horizontal, to hit the jet. If the bullet speed is 400m/s, the value of (θ)will be _____________.

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Raj Pandey

Contributor-Level 9

B → fighter jet

A → anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos q = 200

&nbs

...more

B → fighter jet

A → anti-Air craft gun

Draw velocity diagram of A w.r.t. B

If A hits B

Then Relative velocity perpendicular to the line joining A to B will be zero.

That means 400 cos q = 200

$\begin{array}{l} c o s θ = \frac{1}{2} \\ θ = 60 ° \end{array}$

less

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

In experiment of find acceleration due to gravity (g) using simple pendulum, time period of 0.5s is measured from time of 100 oscillation with a watch of 1 s resolution. If measured value of length is 10 cm known to 1 mm accuracy, The accuracy in the determination of g is found to be x%. The value of x is_______.

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Vishal Baghel

Contributor-Level 10

T = 0.5 sec

No. of oscillation = 100

Resolution = 1 sec

$l = 10 c m \pm . 1 c m$

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Kinetic Theory

According to kinetic theory of gases,

A. The motion of the gas molecules freezes at 0°C.

B. The mean free path of gas molecules decreases if the density of molecules is increased.

C. The mean free path of gas molecules increases if temperature is increased keeping pressure constant.

D. Average kinetic energy per molecules per degree of freedom is $\frac{3}{2} k_{B} T$ (for monoatomic gases).

Choose the most appropriate answer form the options given below

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Payal Gupta

Contributor-Level 10

$\bar{x}$ (mean free path) = $\frac{1}{\sqrt[]{2} π d^{2} n_{v}}$

$n_{v} = \frac{n N_{A}}{v},$ n → No. of moles in volume v N_A ® Avogadro's Number

$\frac{n}{v} = \frac{P}{R_{T}}$

$\bar{x} \propto v \propto \frac{1}{ρ (d e n s i t y} & \bar{x} α T a t c o n s t a n t P)$

^{$ρ ↑ \Leftrightarrow \bar{x} ↓ | \bar{x} ↑ \Leftrightarrow T ↑$}

The motion of the gas molecules freezes at 0K not 0° C

Average kinetic Energy per molecule per degree of freedom is $= \frac{1}{2} k_{B} T$ (for Mono atomic gases)

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

For a perfect gas, two pressures P1 and P2 are shown in figure. The graph shows:

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Payal Gupta

Contributor-Level 10

At constant Temp

$P α \frac{1}{v}$

v₂ > v₁

⇒ P₁ > P₂

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Physics NCERT Exemplar Solutions Class 11th Chapter Eleven Gravitation

Four spheres each of mass m form a square of side d (as shown in figure). A fifth sphere of mass M is situated at the centre of square. The total gravitational potential energy of the system is