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7 months ago

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

A cyclist starts from centre O of a circular park of radius 1km and moves along the path OPRQO as shown Fig. 4.3. If he maintains constant speed of 10ms–1, what is his acceleration at point R in magnitude and direction?

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Vishal Baghel

Contributor-Level 10

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

Explanation- the cyclist covers OPRQO path.

As we know whenever an object performing circular motion, acceleration is called centripetal acceleration and is always directed towards the centre.

so there will be centripetal acceleration a= v²/r

So a= 100/1km= 100/1000=0.1m/s² along RO.

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7 months ago

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

A man wants to reach from A to the opposite corner of the square C (Fig. 4.10). The sides of the square are 100 m. A central square of 50m * 50m is filled with sand. Outside this square, he can walk at a speed 1 m/s. In the central square, he can walk only at a speed of v m/s (v < 1). What is smallest value of v for which he can reach faster via a straight path through the sand than any path in the square outside the sand?

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Vishal Baghel

Contributor-Level 10

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

Explanation – T_sand= $\frac{A P + Q C}{1} + \frac{P Q}{v} = \frac{25 \sqrt[]{2} + 25 \sqrt[]{2}}{1} + \frac{50 \sqrt[]{2}}{v}$

= 50 $\sqrt[]{2} + \frac{50 \sqrt[]{2}}{v} = 50 \sqrt[]{2} (\frac{1}{v} + 1)$

Time taken T_outside= $\frac{A R + R C}{1} s$

AR= $\sqrt[]{75^{2} + 25^{2}} = 25 \sqrt[]{10} m$

RC= AR= $25 \sqrt[]{10} m$

T_outside= 2AR= 50 $\sqrt[]{10} s$

T_sandoutside . so After solving we get velocity is greater v= 0.81m/s

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7 months ago

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

Motion in two dimensions, in a plane can be studied by expressing position, velocity and acceleration as vectors in Cartesian co-ordinates where are unit vector along x and y directions, respectively and Ax and Ay are corresponding components of (Fig). Motion can also be studied by expressing vectors in circular polar co-ordinates as A=A_rr + $A_{θ} θ$ where ? =r/r=cos $θ i + s i n θ j$ and $θ = - s i n θ i + c o s θ j$ are unit vectors along direction in which 'r' and 'q ' are increasing.

(a) Express in terms of q.

(b) Show that both q are unit vectors and are perpendicular to each other.

(c) Show that d(r)/dt , where w =dq/dt q and dq/dt = -wr

(d) For a particle

...more

(a) Express in terms of q.

(b) Show that both q are unit vectors and are perpendicular to each other.

(c) Show that d(r)/dt , where w =dq/dt q and dq/dt = -wr

(d) For a particle moving along a spiral given by r=aqr , where a = 1 (unit), find dimensions of 'a'.

(e) Find velocity and acceleration in polar vector representation for particle moving along spiral described in (d) above.

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Vishal Baghel

Contributor-Level 10

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

Explanation – r=cos $θ ? + s i n θ$ ?…….1

$θ = - s i n θ ? + c o s θ ?$ …….2

Multiplying eq1 by sin $θ$ and 2 with cos $θ$ and adding

Rsin $θ + θ c o s θ = s i n θ c o s θ ? + {s i n}^{2} θ j + {c o s}^{2} θ ? - s i n θ c o s θ ?$

= ?( ${c o s}^{2} θ + {s i n}^{2} θ$ )=j

= rsin $θ + θ c o s θ = j$

n(rcos $θ - θ s i n θ$ )=i

b)r $θ = c o s θ ? + s i n θ ? (- s i n θ ? + c o s θ ?)$

= -cos $θ s i n θ + s i n θ . c o s θ = 0$ $θ = 90$

c)r=cos $θ ? + s i n θ ?$

dr/dt=d/dt(cos $θ ? + s i n θ ?$ )=w[-cos $θ ? + s i n θ ?$ ]

d)L= M^oLT⁰

e)a=1unit , r= $θ r = θ [c o s θ ? + s i n θ ?]$

v= dr/dt= $\frac{d θ}{d t} r + θ \frac{d}{d t} [c o s θ ? + s i n θ ?]$

v= $\frac{d θ}{d r} r + θ [- c o s θ ? + s i n θ ?] \frac{d θ}{d t}$

= $\frac{d θ}{d t} r + θ w θ = w r +$ w $θ θ$

a= $\frac{d}{d t} [w r + w θ θ] = \frac{d}{d t} [\frac{d θ}{d t} r + \frac{d θ}{d t} (θ θ)]$

a= $\frac{d^{2} θ}{d t^{2}} r + \frac{d θ}{d t}$ $\frac{d r}{d t} + \frac{d^{2} θ}{d t^{2}} θ θ + \frac{d θ}{d t} \frac{d}{d t} (θ θ)$

= $\frac{d^{2} θ}{d t^{2}} r + w^{2} θ + \frac{d^{2} θ}{d t^{2}} θ θ + w^{2} θ + w^{2} θ (- r)$

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7 months 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) ?

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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

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New answer posted

7 months 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?

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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

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New answer posted

7 months 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.

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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.

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New answer posted

7 months 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?

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

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7 months 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.

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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 $θ$

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New question posted

7 months ago

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

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New answer posted

7 months 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.

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

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