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New answer posted•a year ago

Exemplar Solution 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.

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Exemplar Solution 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.

6 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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•a year ago

Exemplar Solution 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.<!-- [if gte msEquation 12]> 1V

1 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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•a year ago

Exemplar Solution 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

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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 e

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New answer posted•a year ago

Exemplar Solution 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.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

So t= 300 $\frac{π}{1.414} = 66.6 s$

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

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

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

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New answer posted•a year ago

Exemplar Solution Physics NCERT Exemplar Solutions Class 11th Chapter Five

Figure shows (v_x ,t) and (v_y,t) diagrams for a body of unit mass. Find the force as a function of time.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Explanation- v_x=2t for 0

= 2 (2-t) for 1

=0 for t>2s

V_y= t for 0

= 1 for t>1s

F_x= ma_x= mdv_x/dt= 1 (2)

F_y= ma_y= mdv_y/dt

= 1 (1) for 0

F= F_x? +F_y?

= 2? +?

=-2?

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New answer posted•a year ago

Exemplar Solution Physics NCERT Exemplar Solutions Class 11th Chapter Five

When a body slides down from rest along a smooth inclined plane making an angle of 45° with the horizontal, it takes time T. When the same body slides down from rest along a rough inclined plane making the same angle and through the same distance, it is seen to take time pT, where p is some number greater than 1. Calculate the co-efficient of friction between the body and the rough plane.

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Explanation- as angle is 45

On smooth inclined plane acceleration will be a = gsin $θ$

So acceleration will become a= g/ $\sqrt[]{2}$

Using equation of motion s =ut +1/2at²

S= $\frac{1}{2} \frac{g}{\sqrt[]{2}} T^{2}$

On rough inclined plane a = g (sin $θ - μ c o s θ$ )

= g (sin $45 - μ c o s 45$ )= $\frac{g (1 - μ)}{\sqrt[]{2}}$

So s=ut +1/2at²

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New answer posted•a year ago

Exemplar Solution Chemistry NCERT Exemplar Solutions Class 11th Chapter Five

Match the following gas laws with the equation representing them.

(i) Boyle’s law

(a) V ∝ n at constant T and p

(ii) Charle’s law

(b) p_total = p₁ + p₂ + p₃……… at constant T, V

(iii) Dalton's law

(c) = constant

(iv) Avogadro law

(d) V ∝ T at constant n and p

(e) p ∝ at constant n and T

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

(i) Boyle’s law	(e) p ∝ at constant n and T
(ii) Charle’s law	(d) V ∝ T at constant n and p
(iii) Dalton's law	(b) p_total = p₁ + p₂ + p₃……… at constant T, V
(iv) Avogadro law	(a) V ∝ n at constant T and p

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New answer posted•a year ago

Exemplar Solution Chemistry NCERT Exemplar Solutions Class 11th Chapter Five

Match the graphs between the following variables with their names :

Graphs

Names

(i) Pressure vs temperature (a) graph at constant molar volume.

(a) Isotherms

(ii) Pressure vs volume graph at constant temperature.

(b) Constant temperature curve

(iii) Volume vs temperature graph

at constant pressure.

(c) Isochores

(d) Isobars

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

(i) Pressure vs temperature (a) graph at constant molar volume.

(ii) Pressure vs volume graph at constant temperature.

(a) Isotherms

(iii) Volume vs temperature graph

at constant pressure.

(d) Isobars

Upvote

New answer posted•a year ago

Exemplar Solution Chemistry NCERT Exemplar Solutions Class 11th Chapter Five

Pressure versus volume graph for a real gas and an ideal gas are shown in Fig. 5.4.

Answer the following questions on the basis of this graph.

1. Interpret the behaviour of real gas with respect to ideal gas at low pressure.

2. Interpret the behaviour of real gas with respect to ideal gas at high pressure.

3. Mark the pressure and volume by drawing a line at the point where real gas behaves as an ideal gas.

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

1. At low pressure, the curve of real gas coincides with that of ideal gas, this shows that the deviation of behaviour of real gas with respect to ideal gas is small or negligible.

2. At high pressure, the curve of r

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(i) Boyle’s law	(a) V ∝ n at constant T and p
(ii) Charle’s law	(b) p_total = p₁ + p₂ + p₃……… at constant T, V
(iii) Dalton's law	(c) = constant
(iv) Avogadro law	(d) V ∝ T at constant n and p
	(e) p ∝ at constant n and T

Graphical representation	x and y co-ordinates
(i)	(a) pV vs. V
(ii)	(b) p vs. V
(iii)	(c) p vs.<!-- [if gte msEquation 12]> 1V

Graphs	Names
(i) Pressure vs temperature (a) graph at constant molar volume.	(a) Isotherms
(ii) Pressure vs volume graph at constant temperature.	(b) Constant temperature curve
(iii) Volume vs temperature graph at constant pressure.	(c) Isochores
	(d) Isobars

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