Physics NCERT Exemplar Solutions Class 11th Chapter Five: Overview, Questions, Preparation

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Answer- c

Explanation- in a uniform translatory motion, all parts of the ball have the same velocity in magnitude and direction and this velocity is constant . in this situation body A will be in translatory motion.

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Answer-b

Explanation- to solve this question we have to apply newton's law of motion, in terms of force and change in momentum.

as we know F= dp/dt

As body moving uniform velocity so dp= 0

So F=0

As all part of scale is moving with uniform velocity and total force is zero . hence torque will also be zero.

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Answer- c

Explanation- given u= (3? +4? )m/s

And v= - (3? +4? )m/s, and mass of ball is 150g=0.15kg

Change in momentum = final momentum – initial momentum

 = mv-mu

= m (v-u) = 0.15 [- (3? +4? )- (3? +4? )]

= - [0.15 $\times 6i+0.15\times 8j$ ]

= - [0.9? +1.2? ]

$?$ P =- [0.9? +1.2? ]

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Answer- c

Explanation- change in momentum $?p$ =- [0.9? +1.2? ]

Magnitude = $\sqrt[]{{0.9}^2+{1.2}^2}$

$=\sqrt[]{0.81+1.44}=$ 1.5 kg m s^-1       

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Answer- d

Explanation-we know that for a system F_ext=dp/dt

If F_ext=0, dp =0,

P = constant

Hence, momentum of a system will remain conserve if external force on the system is zero.

In case of collision between particles, equal and opposite forces will act on individual particles by newton's third law.

according to newton's 2^nd law F=dp/dt

If F =0 then p = constant so momentum remains constant if external force is zero.

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Answer- c

Explanation –consider the adjacent diagram

 OA= p₁= initial momentum of player northward

AB= p₂= final momentum of player towards west

OB= OA+AB

Change in momentum = P₂-P₁= AB-OA so we can say that AR will be along south west.

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Answer- a

Explanation- x (t)= pt+qt²+rt³

V= dx/dt=p+2qt+3rt²

So a= o+2q+6rt

At t=2s a= 2q+6 $*$ 2 (r)

= 2q+12r

= 2 (4)+12 (5)=68m/s

So F =ma= 2 (68)= 136N

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Answer- b

Explanation-  we know that m= 5 kg

F= (–3? + 4? ) N and

initial velocity v= (6? -12? ) m/s

So retardation a= $\frac{F}{m}$  = ( $-\frac{3?}{5}+\frac{4?}{5}$ ) m/s²

We know v=u+at, for X-component only ,0 = 6? - $\frac{3?}{5}t$

So t= $\frac{5\times 6}{3}$ = 10s

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Answer- b

Explanation – mass of car m =0

Initial velocity =0,

velocity at east direction = v?

Time =2s

So v=u+at,

v? =0+a (2)

so a =v/2?

F=ma = $\frac{mv}{2}?$  so force = mv/2 towards east.

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Answer- a, b, d

Explanation – x=0 for t<0s

X (t)=Asin4 $\pi t$  for 0

X=0, for t>1/4s

For,0 $\pi t$

Acceleration will be, a= dv/dt=-16 ${\pi }^2$ Asin4 $\pi t$

At t= 1/8 s, a (t)=-16 ${\pi }^2$ Asin4 $\pi \left(\frac{1}{8}\right)$ =-16 ${\pi }^2$ A

So force F =ma =-16 ${\pi }^2$ Am

Impulse = change in linear momentum = f (t)= -16 ${\pi }^2$ Am (1/4)

Impulse = change in linear momentum

=F $\times t$ =-16 ${\pi }^{2}Am\times \frac{1}{4}=-4{\pi }^{2}Am$

= -4 ${\pi }^2$ Am so clearly force depends upon a so force is also not constant .

The impulse (change in linear momentum)

At t=0 is same as t=1/4s

Clearly, force depends upon A which is not constant. Hence, force is also not constant.

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Answer- a, b, d, e

Explanation- suppose A and B are moving together a_common= $\frac{F-f_1}{m_A+m_B}=\frac{2\left(F-t_1\right)}{3m}$

Pseudo force = m_A (a_common)= $\frac{m}{2}*$ $\frac{2\left(F-t_1\right)}{3m}$ = $\frac{F-t_1}{3}$

Force will be maximum when pseudo force and frictional force are equal to one another

$\frac{F_{max}-f_1}{3}$ = $?m_{A}g$

= 0.2 $*\frac{m}{2}*g=0.1mg$

F_max= 0.3mg+f₁

= 0.3mg+0.1 (3/2)mg= 0.45mg

(a) for F=0.25mgmax bodies will move together

(b) for F=0.5mg>F_max body A will slip with respect to B

(c) for F=0.5mg>F_max bodies will slip

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Answer- b, d

Explanation – as we know f = $\mu N=\mu m_{1}gcos\theta$

For system m₁+m₂ to move up

So m₂g- (m₁gsin $\theta +f$ )>0

So m₂g- (m₁gsin $\theta +\mu m_{1}gcos\theta$ )>0

So m₂>m₁ (sin $\theta +\mu cos\theta$ )

But if bodies moves downward

m₁gsin $\theta -f>m$ ₂g

m₁gsin> $\theta +\mu m_{1}gcos\theta$ m₂g

m₂1 (sinθ - uCosθ)

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Answer- b, c

Explanation- when A starts moving up

Mgsin ${\theta }_1+f=mgsin{\theta }_2$

mgsin ${\theta }_1$ + $\mu mgcos{\theta }_1$ = $mgsin{\theta }_2$

$\mu =\frac{\mathrm{s}\mathrm{i}\mathrm{n}{\mathrm{\theta }}_2-\mathrm{s}\mathrm{i}\mathrm{n}{\mathrm{\theta }}_1}{\mathrm{c}\mathrm{o}\mathrm{s}{\mathrm{\theta }}_1}$

When A moves downward f= $mgsin{\theta }_2-$ mgsin ${\theta }_1$ so clearly ${\theta }_2>{\theta }_1$

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Answer- c, d

Explanation – m₁=m₂ = 50g= 1/20kg

Initial velocity u₁=u₂= 5m/s

Final velocity v₁=v₂= -5m/s

Time duration = 10^-3s

Change in linear ₁momentum = m (v-u)= 1/20 (-5-5)=-0.5N-s

Force = impulse /time=0.5/10^-3= 500N

So impulse and force are opposite in directions.

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Answer- a, c

Explanation – mass m= 10kg

F₁ =6N, F₂= 8N

Resultant force F= $\sqrt[]{36+64}=10N$

a=f/m= 10/10= 1m/s²

let $\theta$ ₁ be the angle between R and F₁

tan $\theta$ ₁=8/6=4/3

_{$\theta$ 1}= tan $\theta$ ₁ (4/3) w.r.t F₁=6N

let $\theta$ ₂ be the angle between F and F₂

tan $\theta$ ₁=8/6=4/3

_{$\theta$ 2}= tan $\theta$ ₂ (4/3) w.r.t F₂=8N

Power gain = (i_c² R_c) / (i_b² R_B) = (i_c/i_b)² (R_c/R_B) = (10²)² (10? /10³) = 10?
i_c/i_b = 100 ⇒ β = i_c/i_b = 100

1/C = 5/ (ε? × 100) + 5/ (10ε? × 100)
⇒ C = (ε? × 1000) / 55 = (8.85 × 10? ¹² × 1000) / 55 = 1.61 × 10? ¹? F = 161 pF

R_eq = 10 + (50 × 20) / (50 + 20) = 170/7 Ω
⇒ I = 170 / (170/7) = 7A
⇒ x = 10 × 7 = 70V ⇒ Voltage across 10Ω resistor

Deceleration, a = u² / 2S = 10² / (2 × 0.5) = 100 m/s².
Retarding force, F = MA = 0.1 × 100 = 10N

Apply conservation of momentum along y-axis, we can write
10v? - 10v? sin 30° = 0
⇒ v? = 20m/s

Angle from direction of flow = 90° + θ = 120°
⇒ θ = 30°
sin 30° = u/v ⇒ u/10 = 1/2 ⇒ u = 5 m/s

For Wire-A
F / (πr_A²) = Y (l_A / 2) . (1)
For Wire-B
F / (πr_B²) = Y (l_B / 4) . (2)
From equations (1) and (2), we can write
(l_A / (2r_A²) = (l_B / (4r_B²) ⇒ l_B/l_A = 2 (r_B/r_A)² ⇒ x = l_B = 32

Potential difference across resistor at time t = V = 30/3 = 10V
Current, I = 10 / (5 * 10? ) = 2? A

Using Conservation of Mechanical Energy at point-A and at point-B, we can write
K_B = U_A - U_B [Since K_A = 0]
⇒ (1/2)mv_B² = mg (h_A - h_B)
⇒ v_B = √ (2 × 10 × (10 - 5) = 10m/s

While the particle moves from mean position to displacement, half of its amplitude, its phase changes by π/6 rad. So,
Time taken, t = (π/6)/ω = T/12 = (2/12)s = (1/6)s
a = 6

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explanation-total mass of bicycle and stone =m₁=50+0.5= 50.5kg

Velocity of bicycle u₁= 5m/s mass of stone m₂= 0.5kg

Velocity of stone u₂= 15m/s mass of girl and bicycle m= 50kg

 Also the speed of bicycle change when stone is thrown

According to conservation of linear momentum

m₁u₁=m₂u₂+mv

50.5 (5)= 0.5 (15)+50 (v)

V= 4.9m/s

Change in speed 5-4.9= 0.1m/s

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Explanation – w^'=R=m (g-a)

Mass of person =50kg

Descending acceleration= 9m/s²

Acceleration due to gravity g= 10m/s²

Apparent weight of person R= m (g-a)

= 50 (10-9)=50N

Reading in weighing scale = 50/10 =5kg

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Explanation- mass of body =2kg

From the above graph the body at x=0 when t=0 so body at rest.

So impulse is also zero.

From t=0s to t=4s graph becomes straight line which shows body is at uniform velocity

Beyond 4s graph is parallel to time axis, body is at rest

So tan $\theta$ =3/4

Impulse at t=4s= change in momentum= mv-mu

 =m (v-u)

 = 2 (0-3/4)= -3/2kgm/s

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Explanation- when a person driving car suddenly applies brakes the lower comes to rest but the upper part of the body will remain in motion. so if the driver not wearing seat belt will falls forward and his head hit against the steering wheel.

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Explantion -  given mass of the body = 2kg

Velocity of the body = 2t? +t²?

Velocity of body at t=2s

V= 2 $\times 2$ ? +2²? = 4? +4?

Momentum of body =mv = 2 (4? +4? )= 8? +8? kgm/s

Acceleration a= dv/dt

= d/dt (2t? +t²? )= 2? +2t?

At t=2s, a= (2i+2 (2)j)= 2? +4?

Force acting on the body F=ma

=2 (2? +4? )

= 4? +8? N

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Explanation -the frictional force shown on vertical axis and the applied force shown on horizontal axis. The portion OA of graph represents static friction which is a self-adjusting .in this f=F.

Whereas the point B represents force of limiting friction . Cd|OX represents kinetic friction. The force of kinetic friction does not change by applying force.

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Explanation- porcelain objects are wrapped in paper while transporting because they are very fragile can easily break. while we wrapped them in paper to increase the time so the damage to them is as less as possible.

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Explanation-when she falls on cemented floor the time is very less so force is very large which increase the pain but F $\times ?t$ =change in momentum =constant . while in muddy ground the time is very large as compared to cemented floor so less will be applied . so she feel less pain.

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Explanation- mass of the objects =0.5kg

Speed of the objects =25m/s

a) impulse imparted = change in momentum = mv-mu

= m (v-u)= 0.5 (25-0)=12.5N-s

b) velocity of the object after rebounding= 25/2=-12.5m/s

change in momentum = m (v-u)= 0.5 (-12.5-25)= -18.75N-s

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Explanation- while going up a mountain the force of friction acting on a vehicle of mass m is f= $\mu R$

= $\mu mgcos\theta .$ To avoid skidding force of friction should be large and therefore cos $\theta$ should be large and $\theta$ should be small. That is why mountain roads generally made winding upwards rather than going straight up.

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Explanation- The thread AB will break earlier than the thread CD. This is because force acting on thread CD= applied force and force acting on thread AB = applied force +weight of 2kg mass. Hence force acting on thread Ab is larger than the force acting on thread CD

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Explanation – when the lower thread CD is pulled with a * the thread Cd itself break because the thread CD is not transmitted to the thread Ab instantly.

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Explanation- m₁= 5kg, m₂= 3kg

And g= 9.8m/s² and a= 2m/s²

For the upper block T₁-T₂-5g=5a

T₁-T₂= 5 (g+a)

For the lower block T₂-3g =3a

T₂=3 (g+a)=3 (9.8+2)=35.4N

T₁=T₂+5 (g+a)=94.4N

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Explanation- in equilibrium, The force mgsin $\theta$ acting on the block A is parallel to the plane should be balanced by the tension in the string

mgsin $\theta =T=F$

and for block B, w=T=F

 from these two above equation 

w=mgsin $\theta$ = 100sin30= 100 (1/2)=50N

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Explanation-given mass of the block =M

Coefficient of friction between block and the wall = $\mu$

In equilibrium  condition vertical and horizontal component balance each other

So f=mg

And F=N but the force of friction f= $\mu N=\mu F$

  $\mu F=Mg$

F= Mg/ μ $\mu$

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Explanation- mass of gun =100kg

Mass of ball =1kg and height of cliff = 500m

So horizontal distance travelled by the ball is x = 400m

So h =1/2gt²

500= $\frac{1}{2}10\left(t\right)$ ² so t= 10s

X=ut then u= 400/10=40m/s

 According to principle of conservation pf momentum

Mv=mu

V= 40/100= 0.4m/s

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Explanation- from the diagram x= t=dx/dt=1m/s

So a_x= 0

From the diagram y =t²

dy/dt=2t  or a_y=d²y/dt²=2m/s²

I_y= ma_y=500 (10^-3) (2) =1N

F_x=ma_x=0

F= $\sqrt[]{F_x^2+F_y^2}$ = 1N

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Explanation -initial speed of coin u= 20m/s

Acceleration of elevator =2m/s² acceleration due to gravity = 10m/s²

Effective acceleration =g+a=12m/s²

We know that v=u +at

0= 20+ (-12)t

So t= 20/12=5/3 s

Time of ascent = time of descent

Total time coin fall back into hand = (5/3+5/3)=10/3s=3.33s

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Explanation – as body moving with uniform acceleration a=0

The sum of forces is zero F₁+F₂+F₃=0

a)let F₁, F₂, F₃ be the three forces passing through the point . let F₁and F₂ be in the plane A

so F₃ =- (F₁+F₂) so F₃ is also in plane A.

b)consider the torque of the forces about P . since all the forces pass through P the torque is zero

torque = OP (F₁+F₂+F₃)

since F₁+F₂+F₃=0 so torque =0

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Explanation- as angle is 45

On smooth inclined plane acceleration will be a = gsin $\theta$

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 $\theta -\mu cos\theta$ )

= g (sin $45-\mu cos45$ )= $\frac{g(1-\mu )}{\sqrt[]{2}}$

So s=ut +1/2at²

S= 0+ $\frac{1}{2}\frac{g\left(1-\mu \right)}{\sqrt[]{2}}\left(pT\right)$ ²

Comparing two above distance

$\frac{\mathrm{}}{}$$\frac{1}{2}\frac{g\left(1-\mu \right)}{\text{exception 25:}}\left(pT\right)$²= $\frac{1}{2}\frac{g}{\text{exception 25:}}{T}^2$

So after solving we get $$$\mu =(1-1/p^2)$

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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?

=0

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Explanation – centripetal force, F= mv²/r= f= $\mu N=\mu mg$

V= $\sqrt[]{\mu rg}$

For path ABC path length =3/4 (2 $\pi 2R$ )= $3\pi R=3\pi \left(100\right)=300\pi$

V₁= $\sqrt[]{\mu 2Rg}=\sqrt[]{0.1\times 100\times 10}=1.414m/s$

So t= 300 $\frac{\pi }{1.414}=66.6s$

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

V₂= $\sqrt[]{\mu Rg}=\sqrt[]{0.1\times 100\times 10}=\frac{10m}{s}$

So t₂= 50 $\frac{\pi }{10}$ = 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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Explanation- displacement vector of particle is r (t)=? Acos $wt+$ ? 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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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

V_y= $\sqrt[]{2gH}$

Resultant speed of the ball at the bottom

V= $\sqrt[]{v_x^2+v_y^2}$

V= $\sqrt[]{v_s^2+2gH}$

b) when ball is given horizontal velocity and a small downward velocity

in horizontal direction V'_x=u_x=v_s

in vertical direction v_y'²= u_y²+2gH

resultant velocity of the ball at the bottom

v'= $\sqrt[]{v_s^2+u^2+2gH}$

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

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

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Explanation -for the box to just starts sliding down mg

sin $\theta =t=\mu N=\mu mgcos\theta$

tan $\theta$ = $\mu$

$\theta ={tan}^{-1}\left(\mu \right)$

b) when angle of inclination is increased to $\alpha >\theta$

F₁= mgsin $\alpha -f=mgsin\alpha -\mu N$

= mg ( $sin\alpha -\mu cos\alpha$ )

c) to keep the box stationary, upward force needed F₂= m $gsin\alpha$ +f= mg ( $sin\alpha +\mu cos\alpha$ )

d) if the box is to move with an upward acceleration a then upward force needed 

F₃=mg ( $sin\alpha +\mu cos\alpha$ )+ma

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Explanation- mass of helicopter m =2000kg

Mass of the crew and passengers m= 500kg

Acceleration a =15m/s² and g = 10m/s²

a) force on the floor of the helicopter by the crew and passengers

m (g+a)= 500 (10+15)N= 12500N

b) action of rotor of the helicopter on the surroundings air= (m₁+m₂) (g+a)

= (2000+500) (10+15)= 2500 (25)= 62500N

c) force on the helicopter due to surroundings air

= reaction of force applied by helicopter

= 62500N

y = x5 (1 – x) = x tan θ$\left(1-\frac{x}{R}\right)$

tan = 5, R = 1

$sin\theta =\frac{5}{\sqrt[]{26}},cos\theta =\frac{1}{\sqrt[]{26}}$

$R=\frac{u^{2}sin2\theta }{g}=1$

$\Rightarrow u^2=26\Rightarrow u=\sqrt[]{26}m/s$

y – component of initial velocity

= u sin θ

= $=\sqrt[]{26}\times \frac{5}{\sqrt[]{26}}$

= 5 m/s

Loss in P.E. = Gain in k.E

2 mg R = $\frac{1}{2}\left(\frac{1}{2}m{R}^2+m{R}^2\right){\omega }^2$

$\omega =\sqrt[]{\frac{8g}{3R}}=4\sqrt[]{\frac{g}{2\times 3R}}$

$x=\frac{g}{2}=5$

L = 1 m

$\Delta L=0.4\times 10^{-3}m$

$m=1kg$

d = 0.4 × 103 m

$\frac{F}{A}=y\frac{\Delta L}{L}$

$y=\frac{FL}{A\Delta L}=\frac{\left(mg\right)\times 1}{\left(\frac{\pi d^2}{4}\right)\times 0.4\times 10^{-3}}$

$\Delta y=0.1\times 0.199\times 10^{12}=1.99\times 10^{10}$

= 1.99

$f_1=100=fo\left(\frac{C}{C-V_S}\right)$

C = Speed of sound

Vs = Speed of source

f2 = 50 = fo = $\left(\frac{C}{C+V_s}\right)$

$\frac{f_1}{f_2}=2=\frac{C+V_S}{C-V_S}$

$fo=\frac{200}{3}=\frac{x}{3}$

x = 200

Capacitance of each capacitor

$C_1=\frac{A3{\epsilon }_0}{\frac{1}{2}}=6A{\epsilon }_0$

$C_2=A4{\epsilon }_0=4A{\epsilon }_0$

Equivalent capacitance

$\Delta v_2=\frac{240A{\epsilon }_0}{4A{\epsilon }_0}=60v$

$v_{foil}=60v$

Initially,  $\frac{P}{Q}=\frac{40}{60}=\frac{2}{3}$ - (1)

Finally,  $\frac{P+x}{Q}=\frac{80}{20}=\frac{4}{1}$

(2) (1)

$\frac{P+x}{P}=\frac{4\times 3}{2}=6$

$\Rightarrow 1+\frac{x}{P}=6$

$\frac{x}{P}=5$

x = 5 P = 5 × 4 = 20 $\Omega$

At very high frequencies

$X_C=\frac{1}{\omega C}\to 0$

$X_L=\omega L\to \infty$

Thus equivalent circuit

$Z=1+2+2=5\Omega$

$l=\frac{220}{5}=44A$

for point B,  $\frac{1}{u}=-0.10c{m}^{-1},\frac{1}{v}=0$

Thus u = 10 cm, v = $\infty$

i.e., f = 10 cm

$\frac{1}{10}=\left(1.5-1\right)\left(\frac{2}{R}\right)\Rightarrow \frac{1}{R}=\frac{1}{10}$

R = 10 cm

For first line of Lyman

$\frac{1}{\lambda }=R\left(1-\frac{1}{4}\right)=R\frac{3}{4}$

$\lambda =\frac{4}{3R}$

3rd line (Paschen)

$\frac{1}{{\lambda }_3}=R\left(\frac{1}{3^2}-\frac{1}{6^2}\right)=\frac{R}{9}\times \frac{3}{4}$

2nd line (B almer)

${\frac{1}{\lambda }}_2=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{R}{4}\times \frac{3}{4}$

$a\left(\frac{4}{3R}\right)=\frac{20}{3R}\Rightarrow a=5$

$l=\frac{120-60}{4000}=0.015A$

Thus l₂ = l - L_L

= 0.015 – 0.006

= 0.009 = 9mA