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6 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Five

Assertion (A): At critical temperature liquid passes into gaseous state imperceptibly and continuously.

Reason (R) : The density of liquid and gaseous phase is equal to critical temperature.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) A is false but R is true.

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Payal Gupta

Contributor-Level 10

This is a assertion and reason answer type question as classified in NCERT Exemplar

(i) Both A and R are true and R is the correct explanation of A.

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New answer posted

6 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Five

Assertion (A): Gases do not liquefy above their critical temperature, even on applying high pressure.

Reason (R) : Above critical temperature, the molecular speed is high and intermolecular attractions cannot hold the molecules together because they escape because of high speed.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) A is false but R is true.

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Payal Gupta

Contributor-Level 10

This is a assertion and reason answer type question as classified in NCERT Exemplar

(i) Both A and R are true and R is the correct explanation of A.

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New answer posted

6 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Five

Assertion (A): The temperature at which vapour pressure of a liquid is equal to the external pressure is called boiling temperature.

Reason (R) : At high altitude atmospheric pressure is high.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) A is false but R is true.

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Payal Gupta

Contributor-Level 10

This is a assertion and reason answer type question as classified in NCERT Exemplar

option (iii). A is true but R is false.

At high altitude, the atmospheric pressure is low.

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New answer posted

6 months ago

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

There are four forces acting at a point P produced by strings as shown in Fig. , which is at rest. Find the forces F₁ and F₂

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alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- on resolving forces into rectangular components in equilibrium forces (F₁+1/ $\sqrt[]{2}$ )N are equal to $\sqrt[]{2} N$ and F₂ is equal to ( $\sqrt[]{2} + 1 / \sqrt[]{2}$ )N

F₁+1/ $\sqrt[]{2}$ = $\sqrt[]{2}$

F₁= $\sqrt[]{2} - 1 / \sqrt[]{2}$ = $\frac{2 - 1}{\sqrt[]{2}} = \frac{1}{\sqrt[]{2}} = 0.707 N$

F₂= $\sqrt[]{2} + \frac{1}{\sqrt[]{\sqrt[]{2}}} = \frac{2 + 1}{\sqrt[]{2}} = \frac{3}{\sqrt[]{2}} = 2.121 N$

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New answer posted

6 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Five

Assertion (A): At constant temperature, pV vs V plot for real gases is not a straight line.

Reason (R) : At high pressure all gases have Z > 1 but at intermediate pressure most gases have Z < 1.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) A is false but R is true.

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Payal Gupta

Contributor-Level 10

This is a assertion and reason answer type question as classified in NCERT Exemplar

At constant temperature, the pV vs V plot for real gases is not a straight line because there is intermolecular attraction present in real gases which is absent in ideal gases; hence ideal gases form a straight line in the pV vs V plot at constant temperature.

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6 months ago

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

A cricket bowler releases the ball in two different ways

(a) Giving it only horizontal velocity, and

(b) Giving it horizontal velocity and a small downward velocity.

The speed vs at the time of release is the same. Both are released at a height H from the ground. Which one will have greater speed when the ball hits the ground? Neglect air resistance.

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alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation – horizontal velocity u_x= v_s

During projectile motion horizontal velocity remains unchanged

V_x=u_x=v_s

In vertical direction v_y²= u_y²+2gH

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6 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Five

In the following questions a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each question.

Assertion (A): Three states of matter are the result of balance between intermolecular forces and thermal energy of the molecules.

Reason (R): Intermolecular forces tend to keep the molecules together but thermal energy of molecules tends to keep them apart.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) A is false but R is true.

0 Follower 3 Views

Payal Gupta

Contributor-Level 10

This is a assertion and reason answer type question as classified in NCERT Exemplar

option (i). Both A and R are true and R is the correct explanation of A.

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New answer posted

6 months ago

Class 11th Chemistry NCERT Exemplar Solutions Class 11th Chapter Five

Match the following graphs of ideal gas with their co-ordinates:

Graphical representation	x and y co-ordinates
(i)	(a) pV vs. V
(ii)	(b) p vs. V
(iii)	(c) p vs.

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Payal Gupta

Contributor-Level 10

This is a matching answer type question as classified in NCERT Exemplar

(i) - B

(ii) - C

(iii) - A

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New answer posted

6 months ago

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

The displacement vector of a particle of mass m is given by r(t)=ȋ A cos $w t +$ ĵ B sinwt.

(a) Show that the trajectory is an ellipse.

(b) Show that F= -mw²/r

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alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation- displacement vector of particle is r (t)=? Acos $w t +$ ? Bsinwt

X=Acoswt

x/A= coswt………1

displacement along y axis is

y=Bsinwt

y/B= sinwt……….2

squaring and then adding eqn1 and 2 we get

x²/A²+y²/B²=cos²wt+sin²wt =1

this is an equation of ellipse. Therefore trajectory of particle is an ellipse.

b)v= dr/dt= id/dt (Acoswt)+jd/dt (Bsinwt)

? [A (-sinwt)w]+? [B (coswt).w]

= -? Awsinwt+? Bwcoswt

Acceleration a= dv/dt

So a= -? Awd/dt (sinwt)+? Bw [-sinwt]w

=? Aw²coswt-? Bsinwt

= -w²r

So force acting on the particle f=ma=-mrw²

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6 months ago

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

A racing car travels on a track (without banking) ABCDEFA (Fig). ABC is a circular arc of radius 2 R. CD and FA are straight paths of length R and DEF is a circular arc of radius R = 100 m. The coefficient of friction on the road is µ = 0.1. The maximum speed of the car is 50 m s–1. Find the minimum time for completing one round.

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alok kumar singh

Contributor-Level 10

This is a Long Answer type Questions as classified in NCERT Exemplar

Explanation – centripetal force, F= mv²/r= f= $μ N = μ m g$

V= $\sqrt[]{μ r g}$

For path ABC path length =3/4 (2 $π 2 R$ )= $3 π R = 3 π (100) = 300 π$

V₁= $\sqrt[]{μ 2 R g} = \sqrt[]{0.1 * 100 * 10} = 1.414 m / s$

So t= 300 $\frac{π}{1.414} = 66.6 s$

For path DEF path length = $\frac{1}{4} (2 π R) = 50 π$

V₂= $\sqrt[]{μ R g} = \sqrt[]{0.1 * 100 * 10} = \frac{10 m}{s}$

So t₂= 50 $\frac{π}{10}$ = 15.7s

For path CD and FA

Path length =R+R= 200m

T= 200/50= 4s

Total time = 66.6+15.7+4= 86.3s

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