physics ncert solutions class 11th

The approximate height from the surface of earth at which the weight of the body becomes $\frac{1}{3}$ of its weight on the surface of earth is :

[Radius of earth R = 6400 km and $\sqrt[]{3} = 1.732]$

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physics ncert solutions class 11th Circular Motion

Contributor-Level 10

$g = \frac{9}{{(1 + \frac{h}{R})}^{2}}$

$\Rightarrow \frac{g}{3} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$

$\Rightarrow {(1 + \frac{h}{R})}^{2} = 3$

1 + $\frac{h}{R} = \sqrt[]{3}$ $\frac{}{}$

$\Rightarrow \frac{h}{R} = \sqrt[]{3} - 1$

$\frac{h}{R} = 1.732 - 1$

$\frac{h}{R} = 0.732$

h = 0.732 * 6400

= 4684.8 km » 4685 km

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3 months ago

A boy ties a stone of mass 100 g to the end of 2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 8 N. If the maximum speed with which the stone can revolve is $\frac{K}{π}$ rev. /min. The value of K is :

(Assume the string is massless and unstretchable)

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Contributor-Level 10

By N L M : T sin θ = $^{}$ $m ω^{2} R$ - (1)

& R = $l$ sin θ - (2)

From (1) & (2)

$\Rightarrow T s i n θ = m ω^{2} l s i n θ$

$\Rightarrow T = m ω^{2} l$

$\Rightarrow 80 = (\frac{100}{1000}) ω^{2} * 2$

$\Rightarrow \frac{800}{2} = ω^{2}$

$\Rightarrow ω^{2} = 400$

$ω = 20 r a d / s e c$

And, $ω = \frac{2 π N}{2 π}$ $\frac{}{}$

$N = \frac{20 * 60}{2 π}$

$N = \frac{600}{π} r p m = \frac{k}{π}$

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Identify the pair of physical quantities which have different dimensions

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Contributor-Level 10

$Δ Q = m s Δ T \Rightarrow s = \frac{Δ Q}{m Δ T} \Rightarrow [s] = \frac{M L^{2} T^{- 2}}{M Q} = L^{2} T^{- 2} Q^{- 1}$

$Δ Q = m L \Rightarrow [L] = \frac{M L^{2} T^{- 2}}{M} = L^{2} T^{- 2}$

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A vessel contains 16g of hydrogen and 128g of oxygen at standard temperature and pressure. The volume of the vessel in cm³ is

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Contributor-Level 10

According to question, we can write

Total moles of gas = n = n_Oxygen+ n_Oxygen = $\frac{16}{2} + \frac{128}{32} = 12 m o l e s$

Volume of gas = 12 * 22.4 litre = 268.8 litre = 2.688 * 10⁵cm3 $\approx 27 * 1 0^{4} c m^{3}$

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Starting with the same initial conditions, an ideal gas expands form volume V₁ to V₂ in the three different ways. The work done by the gas in W₁ if the process is purely isothermal, W₂, if the process is purely adiabatic and W₃ if the process is purely isobaric. Then, choose the correct option

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Contributor-Level 10

Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-AD? Adiabatic

W₂ < W₁ < W₃

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What will be the effect on the root mean square velocity of oxygen molecules if the temperagture is doubled and oxygen molecule dissociates into atomic oxygen?

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alok kumar singh

Contributor-Level 10

V_rms = $\sqrt[]{\frac{3 R T}{M}}$

V_rms' = $\sqrt[]{\frac{3 R T_{f}}{m_{f}}} = \sqrt[]{\frac{3 R * 2 T * 2}{M}}$ $[\begin{matrix} G i v e n, T_{f} = 2 T & M_{f} = \frac{M}{2} \end{matrix}]$

$= 2 \sqrt[]{\frac{3 R T}{M}}$

V_rms' = 2V_rms

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4.22 $\hat{i}$ and $\hat{j}$ are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors $\hat{i}$ + $\hat{j}$ , and $\hat{i}$ - $\hat{j}$ ? What are the components of a vector A= 2 $\hat{i}$ + 3 $\hat{j}$ along the directions of $\hat{i}$ + $\hat{j}$ , and $\hat{i}$ - $\hat{j}$ ? [You may use graphical method]

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physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

Contributor-Level 10

4.22

Let us consider a vector $\overset{?}{P}$ . The equation can be written as

Px = Py = 1 $|\overset{?}{P}|$ = $\sqrt (P_{x}^{2} + P_{y}^{2})$ $|\overset{?}{P}|$ = $\sqrt (1^{2} + 1^{2}) |\overset{?}{P}|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\overset{?}{i}$ + $\overset{?}{j}$ = $\sqrt{2}$

Let $θ$ be the angle made by vector $\overset{?}{P}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(P x / P y) θ$ = $\tan^{- 1} ? (1 / 1)$ , $θ$ = 45 $°$ with the x axis

Let $\overset{?}{Q}$ = $\overset{?}{i}$ - $\overset{?}{j}$

$\overset{?}{Q_{x}} \overset{?}{i}$ – $\overset{?}{Q_{y}}$ $\overset{?}{j}$ = ( $\overset{?}{i}$ – $\overset{?}{j})$

$\overset{?}{Q_{x}}$ = $\overset{?}{Q_{y}}$ = 1

$|\overset{?}{Q}|$ = $\sqrt{{Q_{x}}^{2} + {Q_{y}}^{2}}$ = $\sqrt 2$

Hence $|\overset{?}{Q}|$ = $\sqrt 2$ . Therefore the magnitude of ( $\overset{?}{i}$ + $\overset{?}{j})$ = $\sqrt 2$

Let $θ$ be the angle made

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4.22

Let us consider a vector $\overset{?}{P}$ . The equation can be written as

Px = Py = 1 $|\overset{?}{P}|$ = $\sqrt (P_{x}^{2} + P_{y}^{2})$ $|\overset{?}{P}|$ = $\sqrt (1^{2} + 1^{2}) |\overset{?}{P}|$ = $\sqrt{2}$ …….(i)

So the magnitude of vector $\overset{?}{i}$ + $\overset{?}{j}$ = $\sqrt{2}$

Let $θ$ be the angle made by vector $\overset{?}{P}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(P x / P y) θ$ = $\tan^{- 1} ? (1 / 1)$ , $θ$ = 45 $°$ with the x axis

Let $\overset{?}{Q}$ = $\overset{?}{i}$ - $\overset{?}{j}$

$\overset{?}{Q_{x}} \overset{?}{i}$ – $\overset{?}{Q_{y}}$ $\overset{?}{j}$ = ( $\overset{?}{i}$ – $\overset{?}{j})$

$\overset{?}{Q_{x}}$ = $\overset{?}{Q_{y}}$ = 1

$|\overset{?}{Q}|$ = $\sqrt{{Q_{x}}^{2} + {Q_{y}}^{2}}$ = $\sqrt 2$

Hence $|\overset{?}{Q}|$ = $\sqrt 2$ . Therefore the magnitude of ( $\overset{?}{i}$ + $\overset{?}{j})$ = $\sqrt 2$

Let $θ$ be the angle made by vector $\overset{?}{Q}$ , with the x axis as given in the above figure

$\tan ? θ$ = $(Q x / Q y) θ$ = ${- \tan}^{- 1} ? (1 / 1)$ , $θ$ = - 45 $°$ with the x axis

It is given that $\overset{?}{A}$ = 2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ . We can write Ax $\overset{?}{i}$ + Ay $\overset{?}{j}$ = 2 $\overset{?}{i}$ + 3 $\overset{?}{j}$

Comparing the components on both sides, we get

Ax =2 and Ay = 3

$|\overset{?}{A}|$ = $\sqrt{{(2}^{2}} + 3^{2}$ ) = $\sqrt 13$

Let $\overset{?}{A_{x}}$ make an angle $θ$ with the x axis, then

$\tan ? θ$ = (Ax / Ay), $θ$ = $\tan^{- 1} (? 3 / 2)$ = 56.31 $°$

So the angle between (2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ ) and ( $\overset{?}{i}$ + $\overset{?}{j}$ )

$θ'$ = 56.31 – 45 = 11.31 $°$

Component of vector $\overset{?}{A}$ , along the direction of $\overset{?}{P}$ , making an angle $θ$

= ( A cos $θ'$ ) $\overset{?}{P}$

= ( A cos 11.31) $\frac{(\overset{?}{i} + \overset{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ $* \frac{0.9806}{\sqrt{2}}$ ( $\overset{?}{i}$ + $\overset{?}{j}$ )

= 2.5 $*$ $\sqrt{2}$

= $\frac{5}{\sqrt{2}}$

Let $θ'$ be the angle between the vectors (2 $\overset{?}{i}$ + 3 $\overset{?}{j}$ ) and ( $\overset{?}{i}$ - $\overset{?}{j})$

$θ " = 45 + 56.31 = 101.31 °$

Component of vector $\overset{?}{A}$ , along the direction of $\overset{?}{Q}$ , making an angle $θ$

= (A cos $θ "$ ) $\frac{(\overset{?}{i} - \overset{?}{j})}{\sqrt{2}}$

= $\sqrt{13}$ cos ( $101.31 °) \frac{(\overset{?}{i} - \overset{?}{j})}{\sqrt{2}}$

= -0.5( $\overset{?}{i}$ - $\overset{?}{j})$

= $- \frac{1}{\sqrt{2}}$

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Which of the following statements are true for a stationary wave?

(a) Every particle has a fixed amplitude which is different from the amplitude of its nearest particle.

(b) All the particles cross their mean position at the same time.

(c) All the particles are oscillating with same amplitude.

(d) There is no net transfer of energy across any plane.

(e) There are some particles which are always at rest.

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physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, b, d, e) a) clearly every particle at x will have amplitude =asinkx=fixed

b) for mean position =0

coswt=0

wt= (2n-1) $\frac{π}{2}$

hence for a fixed of n all particles are having same value of time t= (2n-1) $\frac{π}{2 w}$

c) amplitude of all the particles are asinkx which is different for different particles at different values of x

d) the energy is a stationary wave is confined between two nodes.

e) particles at different nodes are always at rest.

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A train, standing in a station yard, blows a whistle of frequency 400 Hz in still air. The wind starts blowing in the direction from the yard to the station with a speed of 10m/s. Given that the speed of sound in still air is 340m/s,

(a) The frequency of sound as heard by an observer standing on the platform is 400Hz.

(b) The speed of sound for the observer standing on the platform is 350m/s.

(c) The frequency of sound as heard by the observer standing on the platform will increase.

(d) The frequency of sound as heard by the observer standing on the platform will decrease.

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physics ncert solutions class 11th Physics NCERT Exemplar Solutions Class 11th Chapter Fifteen

Contributor-Level 10

This is a multiple choice answer as classified in NCERT Exemplar

(a, b) v_o=400Hz, v=340m/s

V_m=10m/s

(a) as both source and observer are stationary, hence frequency observed will be same as natural frequency v_o=400Hz

(b) the speed of sound v=v+v_w= 340+10=350m/s

(c) there will be on effect on frequency because there is no relative motion between source and observer hence c, d are incorrect.

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4 months ago

The transverse displacement of a string (clamped at its both ends) is given by y (x,t) = 0.06 sin (2πx/3) cos (120πt). All the points on the string between two consecutive nodes vibrate with

(a) Same frequency

(b) Same phase

(c) Same energy

(d) Different amplitude.

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