Tags Class 11th

Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question

Questions

Discussions

Active Users

Followers

New answer posted

a year ago

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

A cricket fielder can throw the cricket ball with a speed v_o . If he throws the ball while running with speed u at an angle θ to the horizontal, find

(a) The effective angle to the horizontal at which the ball is projected in air as seen by a spectator.

(b) What will be time of flight?

(c) What is the distance (horizontal range) from the point of projection at which the ball will land ?

(d) Find q at which he should throw the ball that would maximise the horizontal range as found in (iii).

(e) How does q for maximum range change if u >v_o , u = v_o , u < v_o?

(f) How does q in (v) compare with that for u = 0 (i.e.45⁰) ?

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- a) for x direction u_x= u+v_ocos $θ$

u_y=velocity in y direction= v₀sin $θ$

now tan $θ = \frac{u_{y}}{u_{x}} = \frac{u_{o} s i n θ}{u + u_{o} c o s θ}$

$θ = {t a n}^{- 1} \frac{u_{o} s i n θ}{u + u_{o} c o s θ}$

b) let t be the time flight y =0 u_y=v_osin $θ, a_{y} = - g, t = T$

y= u_yt+1/2 a_yt²

0= v_osin $θ T$ + $\frac{1}{2} (- g T^{2})$

So T = $\frac{2 u_{o} s i n θ}{g}$

c) horizontal range R, = (u+v_ocos $θ$ T= (u+v_ocos $θ$ ) $\frac{2 u_{o} s i n θ}{g}$

d) for range to be maximum dR/d $θ = 0$

$\frac{v_{o}}{g} [2 u c o s θ + v_{o} c o s 2 θ * 2] = 0$

$2 u c o s θ + 2 v_{o} [2 {c o s}^{2} θ - 1] = 0$

4v_ocos^{2
$θ + 2 u c o s θ - 2 v_{o} = 0$}

So cos $θ$ = $\frac{- u ? \sqrt[]{u^{2} + 8 v_{o}^{2}}}{4 v_{o}}$

$θ = {c o s}^{- 1} [\frac{- u \pm \sqrt[]{u^{2} + 8 v_{o}^{2}}}{4 v_{o}}]$

e) cos $θ$ = $\frac{- v_{o} ? \sqrt[]{u^{2} + 8 v_{o}^{2}}}{4 v_{o}} = \frac{- 1 + 3}{4} = \frac{1}{2}$

so $θ = 60$

f) if u=0 $θ_{m a x} = {c o s}^{- 1} [\frac{- 0 \pm \sqrt[]{u^{2} + 8 v_{o}^{2}}}{4 v_{o}}] = {c o s}^{- 1} (\frac{1}{\sqrt[]{2}}) = 45$ ⁰

0 0

New answer posted

a year ago

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

A river is flowing due east with a speed 3m/s. A swimmer can swim in still water at a speed of 4 m/s .

(a) If swimmer starts swimming due north, what will be his resultant velocity (magnitude and direction)?

(b) If he wants to start from point A on south bank and reach opposite point B on north bank,

(a) Which direction should he swim?

(b) What will be his resultant speed?

(c) From two different cases as mentioned in (a) and (b) above, in which case will he reach opposite bank in shorter time?

0 Follower 8 Views

Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – speed of river V_r= 3m/s

Speed of swimmer V_s= 4m/s

(a) when swimmer starts swimming due north then its resultant velocity

V= $\sqrt[]{v_{r}^{2} + v_{s}^{2}} = \sqrt[]{3^{2} + 4^{2}} = 5 m / s$

tan so 'N

(b) to reach at point B resultant velocity will be

V= $\sqrt[]{v_{s}^{2} - v_{r}^{2}} = \sqrt[]{4^{2} - 3^{2}} = \sqrt[]{7} m / s$

tan $θ = \frac{v_{r}}{v} = \frac{3}{\sqrt[]{7}}$

in case b time taken by swimmer to cross the river

t₁=d/v=d/ $\sqrt[]{7}$

so t1

0 0

New answer posted

a year ago

Class 11th Thermodynamics

Assertion (A): A liquid crystallises into a solid and is accompanied by decrease in entropy.

Reason (R): In crystals, molecules organise in an ordered manner.

(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 2 Views

alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

option (i)

Explanation: The entropy of a liquid reduces as it crystallises. Because the molecules are more organised in crystalline form.

0 0

New answer posted

a year ago

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

A girl riding a bicycle with a speed of 5 m/s towards north direction, observes rain falling vertically down. If she increases her speed to 10 m/s, rain appears to meet her at 45° to the vertical. What is the speed of the rain? In what direction does rain fall as observed by a ground based observer?

0 Follower 4 Views

Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – V_r= a? +b?

Velocity v_g= 5m/s

Velocity of rain w.r.t girl = V_r-V_g= a? +b? -5?

= (a-5)? +b?

a-5=0, a=5

case II

v_g = 10m/s?

V_rg= V_r- V_g

= a? +b? -10? = (a-10)? +b?

Rain appear to be fall at 45 degree so = b/a-10 =1

So b =-5

Velocity of rain = a? +b?

V_r = 5? -5?

Speed of rain V_r= $\sqrt[]{5^{2} + {(- 5)}^{2}} = 5 \sqrt[]{2} m / s$

0 0

New answer posted

a year ago

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

A particle falling vertically from a height hits a plane surface inclined to horizontal at an angle θ with speed Vo and rebounds elastically . Find the distance along the plane where if will hit second time.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation- y=O, u_y= V_ocos $θ$

a_y=-gcos $θ$ , t =T

applying equation of kinematics

y=u_yt+ $\frac{1}{2} a_{y}$ t²

0 = V_ocos $θ T$ +T^{2
$\frac{1}{2} (- g c o s θ)$}

T= $\frac{2 V_{o} c o s θ}{g c o s θ}$

T= 2V₀/g

X= L, u_x=V_osin $θ$ , a_x= gsin $θ$ , t=T= $\frac{2 V_{o}}{g}$

X=u_xt+ $\frac{1}{2} a_{x} t^{2}$

L= V_osin $θ T + \frac{1}{2} g s i n θ T^{2}$

L= $\frac{4 v_{o}^{2}}{g}$ sin $θ$

0 0

New question posted

a year ago

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

0 Follower 8 Views

New answer posted

a year ago

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

A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal (Fig).

(a) Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).

(b) Time of flight.

(c) β at which range will be maximum.

0 Follower 5 Views

Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – particle is projected from the point O.

Let time taken in reaching from point O to point P is T.

for journey O to P

y=0,u_y= V_osin $β$ ,a_y= -gcos $α, t = T$

y=u_yt + $\frac{1}{2} a_{y} t^{2}$

0= V_osin $β T + \frac{1}{2} (- g c o s α) T^{2}$

T[V_osin $β - \frac{g c o s α}{2}$ T]=0

T = time of flight = $\frac{2 V_{o} s i n β}{g c o s α}$

Motion along OX

x= L ,u_x= V_ocos $β$ , a_x= -gsin $α$

t =T = $\frac{2 V_{o} s i n β}{g c o s α}$

x= u_xt+ $\frac{1}{2} a_{x} t^{2}$

L= V₀cos $β T$ + $\frac{1}{2} (- g s i n α) T^{2}$

L= T[V₀cos $β - \frac{1}{2} g s i n α T$ ]

L= $\frac{2 V_{o} s i n β}{g c o s α}$ [V_ocos $β - \frac{V_{o} s i n α s i n β}{c o s α}$ ]

L= $\frac{2 v_{o}^{2} s i n β}{g {c o s}^{2} α} c o s ? (α + β)$

Z= sin $β c o s ? (α + β)$

= sin $β [c o s α c o s β - s i n α s i n β]$

= $\frac{1}{2} (c o s α + s i n 2 β - 2 s i n α . {s i n}^{2} β)$

= ½ [sin2] $β c o s α - s i n α (1 - c o s 2 β)$

= $\frac{1}{2} [s i n 2 β c o s α + c o s 2 β . s i n α - s i n α]$

= $\frac{1}{2}$ [sin(2 $β + α$ )-sin $α$ ]

For z maximum

2 $β + α = \frac{π}{2}$ , $β = \frac{π}{4} - \frac{α}{2}$

0 0

New question posted

a year ago

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

A particle is projected in air at an angle β to a surface which itself is inclined at an angle α to the horizontal (Fig).

(a) Find an expression of range on the plane surface (distance on the plane from the point of projection at which particle will hit the surface).

(b) Time of flight.

(c) β at which range will be maximum.

0 Follower 6 Views

New answer posted

a year ago

Class 11th Thermodynamics

Assertion (A): Spontaneous process is an irreversible process and may be reversed by some external agency.

Reason (R): Decrease in enthalpy is a contributory factor for spontaneity.

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

(ii) Both A and R is 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 6 Views

alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: (ii)

Explanation: The energy factor for a spontaneous process should be favourable (i.e., -ve) and the randomness should be positive.

0 0

New answer posted

a year ago

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

A gun can fire shells with maximum speed vo and the maximum horizontal range that can be achieved is R= $\frac{v_{o}^{2}}{g}$ .If a target farther away by distance ∆x (beyond R) has to be hit with the same gun (Fig), show that it could be achieved by raising the gun to a height at least h = $∆ x [1 + \frac{∆ x}{R}] .$

0 Follower 6 Views

Vishal Baghel

Contributor-Level 10

This is a Long Answer Type Question as classified in NCERT Exemplar

Explanation – target T is at horizontal distance x= R+ $? x$ and between point of projection y= -h

Maximum horizontal range R= $\frac{v_{o}^{2}}{g} θ = 45$ …………1

Horizontal component of initial velocity = V_ocos $θ$

Vertical component of initial velocity = -V_osin $θ$

So h = (-V_osin $θ$ )t + $\frac{1}{2} g t$ ²………….2

R+ $? x$ = V_ocos $θ * t$

So t= $\frac{R + ? x}{v_{o} c o s θ}$

Substituting value of t in 2 we get

So h = (-V₀sin $θ$ ) $(\frac{R + ? x}{v_{o} c o s θ}) + \frac{1}{2} g ({\frac{R + ? x}{v_{o} c o s θ})}^{2}$

H = -(R+ $? x$ )tan $θ$ + $\frac{1}{2} g (\frac{({R + ? x)}^{2}}{v_{o}^{2} {c o s}^{2} θ})$

$θ = 45$ , h = -(R+ $? x$ )tan $45$ + $\frac{1}{2} g (\frac{({R + ? x)}^{2}}{v_{o}^{2} {c o s}^{2} 45})$

So h = -(R+ $? x$ ) $1$ + $\frac{1}{2} g (\frac{({R + ? x)}^{2}}{v_{o}^{2} (\frac{1}{2})})$

So h = -(R+ $? x$ )+ $\frac{{(R + ? x)}^{2}}{R}$

So h = -R- $? x$ +(R+ $\frac{{? x}^{2}}{R} + 2 ? x$ )

h= $? x + \frac{{? x}^{2}}{R}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
703k Reviews
1850k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 11th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience