Moving Charges and Magnetism

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New answer posted•a year ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th 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?

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 Short Answer Type Questions 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 acceleratio

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New answer posted•a year ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th 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?

5 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 – 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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New answer posted•a year ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th 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 ṝ =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 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.

7 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 – 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

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New answer posted•a year ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th 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)?

7 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- 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 rang

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New answer posted•a year ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th 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?

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

(c) time taken by sw

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New answer posted•a year ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th 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?

27 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_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/

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New answer posted•a year ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th 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.

5 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- 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 answer posted•a year ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th 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.

6 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 – 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

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New answer posted•a year ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th 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 = $∆ x [1 + \frac{∆ x}{R}] .$

6 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 – 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

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New answer posted•a year ago

Moving Charges and Magnetism Physics NCERT Exemplar Solutions Class 12th Chapter Four

A hill is 500 m high. Supplies are to be sent across the hill using a canon that can hurl packets at a speed of 125 m/s over the hill. The canon is located at a distance of 800m from the foot of hill and can be moved on the ground at a speed of 2 m/s; so that its distance from the hill can be adjusted. What is the shortest time in which a packet can reach on the ground across the hill ? Take g =10 m/s².

6 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- speed of jackets = 125m/s

Height of hill = 500m

To cross the hill vertical component of velocity should be grater than this value u_y= $\sqrt[]{2 g h}$

$= \sqrt[]{2 * 10 * 500} = 100 m / s$

So u²= u_x²+u_y²

Horizontal component of initial velocity u_x= $\sqrt[]{u^{2} - u_{y}^{2}} = \sqrt[]{125^{2} - 100^{2}} = 75 m / s$

Time taken to reach the

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