WBJEE 2020 Paper 2 (Physics & Chemistry) Question Paper with Solutions

We have, power = ${{work\,done} \over {time}}$

Given, time = 1 min = 60 s

Now, work done per cycle = (130 $-$ 30) = 100 cal = 100 $\times$ 4.2 J

As 2J of energy is consumed due to friction by the engine, hence work done per cycle

= 100 $\times$ 4.2 $-$ 2

= 420 $-$ = 418 J

Total work done for 90 cycles = 418 $\times$ 90 J

$\therefore$ Power = ${{418 \times 90} \over {60}} = 627$ W

Given, length of rectangular container, l = 5m

The number of moles of two gases are

$\mu$₁ = ${M \over {32}}$ and $\mu$₂ = ${M \over {18}}$

The given situation is shown in the figure given below.


where, x is the distance of movable partition P from the left wall A.

At equilibrium, p₁ = p₂ = p

and T₁ = T₂ = T

By ideal gas equation,

${{{p_1}{V_1}} \over {{\mu _1}}} = {{{p_2}{V_2}} \over {{\mu _2}}} = RT$

$\Rightarrow {{{V_1}} \over {{\mu _1}}} = {{{V_2}} \over {{\mu _2}}}$ ...................[$\because$ p₁ = p₂]

${{a.x} \over {{\mu _1}}} = {{a.(5 - x)} \over {{\mu _2}}}$ [$\because$ a = area of each wall A and B]

${x \over {{\mu _1}}} = {{5 - x} \over {{\mu _2}}}$

$\Rightarrow {x \over {M/32}} = {{5 - x} \over {M/18}}$

$\Rightarrow 32x = 18(5 - x)$

$\Rightarrow$ 32x = 90 $-$ 18x

$\Rightarrow$ 50x = 90

$\Rightarrow$ $x = {{90} \over {50}} = 1.8m$

According to given figure,


Given, a : r = 8 : $\pi$

${a \over r} = {8 \over \pi } \Rightarrow r = {{\pi a} \over 8}$ ...........(i)

Magnetic field due to square, ${B_1} = {{{\mu _0}I} \over {\pi {a \over 2}}} \times \sqrt 2$

outward

Magnetic field due to circular loop, ${B_2} = {{{\mu _0}I} \over {2r}}$ ,

inward

Strength of magnetic field at the common centre O,

$B = {{{\mu _0}I \times 8} \over {2\pi a}} - {{2\sqrt 2 {\mu _0}I} \over {\pi a}}$

Substituting value of r from Eq (i), we get

$B = {{{\mu _0}I \times 8} \over {2\pi a}} - {{2\sqrt 2 {\mu _0}I} \over {\pi a}}$

$= {{{\mu _0}I} \over {\pi a}}(4 - 2\sqrt 2 )$

$= {{{\mu _0}I} \over {\pi a}}2\sqrt 2 (\sqrt 2 - 1)$

When a DC voltage is applied at the two ends of a circuit kept in a closed box, it is observed that the current gradually increases from zero to a certain value and then remains constant. As capacitor blocks DC. So, the circuit must contains a resistor and an inductor in series as shown below.


From Graph

Growth of current, I = I₀ (1 $-$ e^{$-$ t/t L})

At t = 0, I = 0

At t = $\tau$_L

I = ${I_0}\left( {1 - {1 \over e}} \right) = 0.693{I_0}$


Where, ${\tau _L} = {L \over R}$ = time constant and

${I_0} = {E \over R}$ = maximum current

Let x be the distance from steel wire, where load is applied as shown in the figure below.


Given, Young's modulus of steel,

Y_steel = 2 $\times$ 10¹² dyne/cm²

Young's modulus of brass, Y_brass = 1 $\times$ 10¹² dyne/cm²

Cross-sectional area of wires, A = 0.005 cm²

Extension in wires, $\Delta$l = 0.1 cm

We know that,

Young's modulus, ${Y_{steel}} = {{stress} \over {strain}} = {{{F_s}/A} \over {{{\Delta l} \over l}}} = {{{F_s}.l} \over {\Delta lA}}$

$\Rightarrow$ ${F_s} = {Y_{steel}} \times {{\Delta l} \over l} \times A$

= $2 \times {10^{12}} \times {{0.1} \over {50}} \times 0.005$

$\Rightarrow$ F_s = 2 $\times$ 10⁷ dyne

Similarly,

${F_8} = {Y_{brass}} \times {{\Delta l} \over l} \times A$

= $1 \times {10^{12}} \times {{0.1} \over {50}} \times 0.005$

= 10⁷ dyne

From given figure,

F_s $\times$ x = F_B $\times$ (15 $-$ x)

$\Rightarrow$ 2 $\times$ 10⁷ $\times$ x = 10⁷ $\times$ (15 $-$ x)

$\Rightarrow$2x = 15 $-$ x

$\Rightarrow$ 3x = 15 $\Rightarrow$ x = 5 cm

Given, $\beta$ of transistor = 48

Base current, I_B = 200$\mu$A = 200 $\times$ 10^-6 A


We know that, current gain, $\beta$ = ${{{I_C}} \over {{I_B}}}$

$\Rightarrow$ $48 = {{{I_C}} \over {200 \times {{10}^{ - 6}}}}$

$\Rightarrow$ ${I_C} = 48 \times 200 \times {10^{ - 6}}$

= 96 $\times$ 10^-4 A

Now, V_CC = I_CR_C + V_CE

$\Rightarrow$ V_CE = V_CC $-$ I_CR_C

From figure, V_CC = 5 V and R_C = 500$\Omega$

V_CE = 5 $-$ 96 $\times$ 10^-4 $\times$ 500

= 5 $-$ 4.8 = 0.2 V

Given, q₁ = 1 $\times$ 10^-6 C, q₂ = 3 $\times$ 10^-6 C

$\therefore$ q₂ = 3q₁ ; q₂ > q₁

As we know that, electric field, $E\alpha {1 \over {{x^2}}}$

So, the graph showing variation of E_x in x-direction is as given below.

The process involved is as

${}_Z{X^A} \to z + {}_1{Y^A} + {}_{ - 1}{e^0} + v + Q$

$\therefore$ Mass defect, m_x = M_x $-$ ZM_e

and m_y = M_y $-$ (Z + 1)m_e

where, M_x is atomic mass of X, M_y is atomic mass of Y and m_e is mass of an electron.

Total mass defect, $\Delta m = {m_x} - {m_y} = ({M_x} - {M_y} - {m_e})$

$\therefore$ Binding energy, $E = \Delta m{c^2}$

So, maximum energy of $\beta$-particle emitted

= $({M_x} - {M_y} - {m_e}){c^2}$

Given, frequency, v = K$\delta$E

where, k is constant and $\delta$E is change in energy.

$\Rightarrow$ $k = {v \over {\delta E}} = {v \over {hv}}$ .....................[$\because$$\delta$E = hv]

= ${1 \over h}$

$\because$ We know that, energy, E = hv

$\Rightarrow$ $h = {E \over v} = {{M{L^2}{T^{ - 2}}} \over {{T^{ - 1}}}}$

h = [ML²T^-1]

$\therefore$ Dimension of $k = {1 \over h} = {1 \over {\left[ {M{L^2}{T^{ - 1}}} \right]}} = \left[ {{M^{ - 1}}{L^{ - 2}}T} \right]$

Given, a tennis ball hits the floor with a speed v making an angle $\theta$ with the normal as shown in the figure below.


The coefficient of restitution = $\varepsilon$

Now, from the above diagram,

In $\Delta AOB,\tan \phi = {{v\sin \theta } \over {\varepsilon v\cos \theta }} = {{\tan \theta } \over \varepsilon }$

$\therefore$ Angle of reflection, $\phi = {\tan ^{ - 1}}\left( {{{\tan \theta } \over \varepsilon }} \right)$

A conducting wire of length L bent in the form of circle of radius R and another conductor of length a is bent in the form of a square. The two loops are then placed in same plane such that the square loop is exactly at the centre of circular loop as shown in the figure

For circular loop,

2$\pi$R = L ............(i)

For square of side S,

4S = a .............(ii)

Flux linked, $\phi = {{{\mu _0}I{S^2}} \over {2R}}$

Mutual inductance, $M{\phi \over I} = {{{\mu _0}{S^2}} \over {2R}}$ Substituting R from Eq. (i) and S from Eq. (ii), we get

$M = {{{\mu _0}{a^2}} \over {2(16)}}{{2\pi } \over L} = {{{\mu _0}{a^2}\pi } \over {16L}}$

Given, speed of fighter plane, v = 360 km/h

= $360 \times {5 \over {18}}$ m/s = 100 m/s

Altitude, h = 500 m


Now, from equation of motion,

$h = ut + {1 \over 2}g{t^2}$

$\Rightarrow$ $500 = 0(t) + {1 \over 2} \times 10 \times {t^2}$ ........$\left[ {\matrix{ {u = 0and} \cr {g = 10m/{s^2}} \cr } } \right]$

$\Rightarrow$ t² = 100 $\Rightarrow$ t = 10 s

$\therefore$ The bomb should be dropped at the distance,

x = vt = 100 $\times$ 10 = 1000 m

A galvanometer can be converted to a voltmeter of full scale deflection V₀ by connecting a series resistance R₁, then

${R_{^1}} = {{{V_0}} \over {{I_g}}} - G$

$\Rightarrow {V_0} = {I_g}(G + {R_1})$ ................(i)

where, I_g = current through the galvanometer and G = resistance of galvanometer.

Similarly, a galvanometer can be converted to an ammeter by connecting a shunt resistance R₂, then

${R_2} = \left( {{{{I_g}} \over {{I_0} - {I_g}}}} \right)G$

$\Rightarrow {I_g} = {I_0}\left( {{{{R_2}} \over {{G_{}} + {R_2}}}} \right)$ ..........(ii)

where, I₀ = total current.

Now, from Eq. (i), we get

${{{V_0}} \over {{I_g}}} - {R_1} = G$ ...................(iii)

From Eq. (ii), we get

$G + {R_2} = {{{I_0}} \over {{I_g}}} \times {R_2}$

$\Rightarrow$ $G = {{{I_0}} \over {{I_g}}} \times {R_2} - {R_2}$ .............(iv)

From Eqs. (iii) and (iv), we get

${{{V_0}} \over {{I_g}}} - {R_1} = {{{I_0}} \over {{I_g}}} \times {R_2} - {R_2}$

$\Rightarrow$ ${{{V_0}} \over {{I_g}}} - {{{I_0}} \over {{I_g}}} \times {R_2} = {R_1} - {R_2}$

$\Rightarrow$ ${I \over {{I_g}}}({V_0} - {I_0}{R_2}) = {R_1} - {R_2}$

$\Rightarrow$ ${I_g} = {{{V_0} - {I_0}{R_2}} \over {{R_1} - {R_2}}}$

Given, slit width, a = 0.5 mm = 0.5 $\times$ 10^-3 m

Wavelength, $\lambda$ = 600 nm = 600 $\times$ 10^-9 m

Distance of screen, D = 50 cm = 50 $\times$ 10^-2 m

$\therefore$ Linear separation = ${{2D\lambda } \over a}$

= ${{2 \times 50 \times {{10}^{ - 2}} \times 600 \times {{10}^{ - 9}}} \over {0.5 \times {{10}^{ - 3}}}}$

= 12 $\times$ 10^-4 m

= 1.2 $\times$ 10^-3 m = 1.2 mm

Time period of second pendulum, T = 2s

Displacement equation of simple pendulum,

Given, $x = A\sin \omega t$ ...................(i)

t = 2.25 s = 2 + 0.25

= $2 + {1 \over 4}$

$\therefore$ t = $T + {T \over 8}$

Velocity of second pendulum,

$v = {{dx} \over {dt}}$

= ${d \over {dt}}.(A\sin \omega t)$ .............[from Eq. (i)]

v = A$\omega$ cos $\omega$ t

Velocity at t = T,

$v = A\omega \cos \left( {{{2\pi } \over T}.T} \right) = A\omega = {v_0}$ .............(ii)

Velocity at $t = T + {T \over {8'}}$

${v_1} = A\omega \cos \omega \left( {T + {T \over 8}} \right)$

= $A\omega \cos {{2\pi } \over T}\left( {T + {T \over 8}} \right)$

= $A\omega \cos 2\pi .{9 \over 8}$

= $A\omega \cos \left( {2\pi + {\pi \over 4}} \right)$

= $A\omega \cos {\pi \over 4}$

= ${{A\omega } \over {\sqrt 2 }}$

${v_1} = {{{v_0}} \over {\sqrt 2 }}$ ...........[from Eq. (ii)]

By equation of motion,

$v_1^2 = v_0^2 - 2gh$

$\Rightarrow$ ${\left( {{{{v_0}} \over {\sqrt 2 }}} \right)^2} = v_0^2 - 2gh \Rightarrow h = {{v_0^2} \over {4g}}$

The given situation is as shown in figure below.


Object distance from convex mirror, u = $-$ 60 cm

Focal length of convex mirror, f = 30 cm

Using mirror formula,

${1 \over v} + {1 \over u} = {1 \over f} \Rightarrow {1 \over v} - {1 \over {60}} = {1 \over {30}}$

$\Rightarrow {1 \over v} = {1 \over {30}} + {1 \over {60}} = {{2 + 1} \over {60}} = {3 \over {60}}$

$v = 20$ cm

Distance between the object and image = 60 + 20 = 80 cm

As there is no parallex between the image formed by the two mirrors, i.e. the images are formed at the same place.

So, the plane mirror is at a distance of 40 cm from the object.

The given circuit is as shown below.


$\because$ In steady state, the combination of resistance and capacitance in loop CEFD will not work. So, we can eliminate this combination and hence the equivalent circuit is as shown below.


$\because$ In the above circuit, the current i will flow from B to C and same current i will flow from C to B. Hence, total current through the 2$\Omega$ resistor is zero.

Let force acting due to electrostatic attraction of charges be F_e as shown in the figure below.


Now,
Tsin$\theta$ = F_e ................(i)

Tcos$\theta$ = mg ................(ii)

On dividing Eq. (i) by Eq. (ii), we get

${{T\sin \theta } \over {T\cos \theta }} = {{{F_e}} \over {mg}} = k{{{q_1} \times {q_2}} \over {{r^2}}}$

Here, q₁ = q₂ = q

Distance, r = 2L sin$\theta$

$\Rightarrow$ tan$\theta$ = ${{k{q^2}} \over {{{(2L\sin \theta )}^2}.mg}}$ ...........(iii)

Now, in the second case, when the charges are tripled, new charge q' = 3q

As the angle between the strings are doubled, $\theta$ becomes 2$\theta$, hence from Eq. (iii), we get

$\tan 2\theta = {{k(3q)(3q)} \over {{{(2L\sin 2\theta )}^2}.mg}}$

$\tan 2\theta = {{9k{q^2}} \over {{{(2L\sin 2\theta )}^2}.mg}}$ ..........(iv)

On dividing Eq. (iv) by Eq. (iii), we get

${{\tan 2\theta } \over {\tan \theta }} = {{9k{q^2}} \over {4{L^2}{{\sin }^2}2\theta .mg}} \times {{4{L^2}{{\sin }^2}\theta .mg} \over {k{q^2}}}$

${{\tan 2\theta } \over {\tan \theta }} = {{9{{\sin }^2}\theta } \over {{{\sin }^2}2\theta }}$

As, $\tan 2\theta = {{2\tan \theta } \over {1 + {{\tan }^2}\theta }}$

and $\sin 2\theta$ = 2 $\sin \theta$ $\cos \theta$

$\Rightarrow$ ${2 \over {1 - {{\tan }^2}\theta }} = {{9{{\sin }^2}\theta } \over {4{{\sin }^2}\theta {{\cos }^2}\theta }} = {9 \over {4{{\cos }^2}\theta }}$

$\Rightarrow$ ${2 \over {1 - {{\tan }^2}\theta }} = {9 \over 4}{\sec ^2}\theta$

$\Rightarrow$ ${2 \over {1 - {{\tan }^2}\theta }} = {9 \over 4}(1 + {\tan ^2}\theta )$

............[$\because$ 1 + tan²$\theta$ = sec²$\theta$]

Let tan²$\theta$ = x

${2 \over {1 - x}} = {9 \over 4}(1 + x)$

$\Rightarrow$ 8 = 9 $-$ 9x²

$\Rightarrow$ 9x² = 1

$\Rightarrow$ x² = ${1 \over 9}$ $\Rightarrow$ x = ${1 \over 3}$

$\Rightarrow$ tan²$\theta$ = ${1 \over 3}$

$\Rightarrow$ tan$\theta$ = ${1 \over {\sqrt 3 }} = \tan 30^\circ$

$\theta$ = 30$^\circ$

Hence, the angle between the strings

= 2$\theta$ = 2 $\times$ 30$^\circ$ = 60$^\circ$

The block diagram is as shown below,



According to above diagram,

mg = N + Fsin$\theta$

$\Rightarrow$ N = mg $-$ Fsin$\theta$ ......(i)

and Fcos$\theta$ = $\mu$N

$\Rightarrow$ Fcos$\theta$ = $\mu$(mg $-$ Fsin$\theta$).............[using Eq. (i)]

For F_min'

${d \over {d\theta }}(\cos \theta + \mu \sin \theta ) = 0$

$\Rightarrow - \sin \theta + \mu \cos \theta = 0$

$\tan \theta$ = $\mu$

$\sin \theta = {\mu \over {\sqrt {1 + {\mu ^2}} }}$ and $\cos \theta = {1 \over {\sqrt {1 + {\mu ^2}} }}$

$\therefore$ ${F_{\min }} = {{\mu mg} \over {{1 \over {\sqrt {1 + {\mu ^2}} }} + {{\mu .\mu } \over {\sqrt {1 + {\mu ^2}} }}}}$

$= {{\mu mg} \over {{{1 + {\mu ^2}} \over {\sqrt {1 + {\mu ^2}} }}}} = {{\mu mg} \over {\sqrt {1 + {\mu ^2}} }}$

$\Rightarrow$ ${F_{\min }} = {{\mu .mg} \over {\sqrt {1 + {\mu ^2}} }}$

Given, resistance of the circular loop = 20$\Omega$

Electric charge flowing due to the induced current,

$q = I.dt = {{\Delta \phi } \over R} = {{{\phi _2} - {\phi _1}} \over R}$ ,

where ${{\phi _1}}$ , ${{\phi _2}}$ are magnetic flux at t₁ = 0 and t₂ = 20 ms.

As the loop is perpendicular to the magnetic field B, the flux passing through the loop is,

${\phi _1} = B.A = BA\cos 90^\circ = 0$

Hence, $q = {{\Delta \phi } \over R} = {{{\phi _2} - {\phi _1}} \over R} = 0$

$\therefore$ q = 0 C

The given system of charges is as shown below,


$\because$ Potential energy, $PE = {1 \over {4\pi {\varepsilon _0}}}{{{q_1}{q_2}} \over r} = {{k{q_1}{q_2}} \over r}$

Here, q₁ = q₂ = q

and r = a

$\Rightarrow$ $PE = {{kqq} \over a} = {{k{q^2}} \over a}$

Due to four identical point masses,

$P{E_i} = {{4k{q^2}} \over a} + {{2k{q^2}} \over {\sqrt 2 a}}$

According to law of conservation of energy,

KE_i + PE_i = KE_f + PE_f

Initially, KE_i = 0

and at infinity, PE_f = 0

$\therefore$ $K{E_f} = P{E_i} = {{4k{q^2}} \over a} + {{2k{q^2}} \over {\sqrt 2 a}}$

= ${{k{q^2}} \over a}(4 + \sqrt 2 )$

$\because$In CGS unit, k = 1

$\therefore$ KE = ${{{q^2}} \over a}(4 + \sqrt 2 )$

A wire is bent to form a D-shaped loop carrying current I, where the curved part is semi-circle of radius R. The loop is placed in a uniform magnetic field B which is directed into the plane of the paper. As single current by flowing in the loop, so net magnetic force on a closed current loop in a uniform magnetic field B is zero.

Given, radius of solid cylinder = a

Charge density of cylinder = p

Dielectric constant of material of cylinder = k

The electric field at a radial distance x(x < a) as shown in calculated as


From Gauss's law,

$\int {E.dA = {{{Q_{in}}} \over {k{\varepsilon _0}}}}$ , where k is dielectric constant of material.

$\Rightarrow$ E(2$\pi$xl) = ${{p(\pi {x^2}l)} \over {k{\varepsilon _0}}}$ .................[$\because$ charge, Q_in = p($\pi$x²l)]

$\Rightarrow$ E = ${{px} \over {2k{\varepsilon _0}}}$

We know that,

terminal velocity, ${v_T} = {{2{r^2}} \over {9\eta }}(\rho s - \rho L)g$

$\Rightarrow$${v_T} \propto {1 \over \eta }$ , where $\eta$ is viscosity of the liquid.

Hence, graph between terminal velocity v_T and viscosity of liquid $\eta$ is as shown below.

Given, mass of boiling water = 100 g

Mass of cold water = 300 g

Fall in temperature = 100 $-$ 20 = 80$^\circ$ C

Let heat capacity of calorimeter = ms

Specific heat of water = 1 cal g^-1 $^\circ$ C^-1

$\because$ Heat lost by boiling water = $\Delta$ T = 100 $\times$ 1 $\times$ 80

and heat gained by cold water

= 300 $\times$ 10 $\times$ 10 + ms $\times$ 10

Now, according to the principle of calorimetry,

heat lost = heat gained

$100 \times 1 \times 80 = 300 \times 1 \times 10 + ms \times 10$

$\Rightarrow$ ms = 500 cal/$^\circ$ C

Let specific heat of metal is s_b

Mass of block = 1 kg = 1000 g

Rise in temperature of metallic block,

$\Delta$ t = 19 $-$ 10 = 9$^\circ$ C

Then, again from principle of calorimetry,

m_mixture $\times$ s$\Delta$ T + ms$\Delta$ T = m_block $\times$ s_b $\times$ $\Delta$ t

$(100 + 300) \times 1 \times (20 - 19) + 500 \times (20 - 19)$

= 1000 $\times$ s_b $\times$ 9

$\Rightarrow$ s_b = 0.1 cal/g$^\circ$ C

A thin convex lens is placed just above an empty vessel of depth 80 cm. The image (I) of a coin kept at the bottom of the vessel is thus formed 20 cm above the lens. If now water is poured in the vessel upto a height of 64 cm as shown in the figure below.


Given, object distance, u = $-$ 80 cm

Image distance, v = + 20 cm

Using lens formula, we get

${1 \over f} = {1 \over v} - {1 \over u}$

${1 \over f} = {1 \over {20}} + {1 \over {80}} = {5 \over {80}}$

$\Rightarrow$ f = 16 cm

Now, water is poured in the vessel upto height of 64 cm, then object distance,

$u = 16 + {{64} \over \mu } = 16 + {{64 \times 3} \over 4} = 64$ cm

(given, refractive index, $\mu$ = 4/3)

Again, using lens formula,

${1 \over f} = {1 \over v} - {1 \over u}$

$\Rightarrow$ ${1 \over {16}} = {1 \over v} + {1 \over {64}}$ .......($\because$ u = $-$ 64 cm)

$\Rightarrow {1 \over v} = {3 \over {64}}$

$\therefore$ v = 21.33 cm, above the lens.

Let intensity from one slit = I₁
Intensity from other slit = I₂
According to question,
I₁ = 1.5I₂
$\Rightarrow$ ${{{I_1}} \over {{I_2}}} = 1.5 = {3 \over 2}$
$\therefore$ Ratio of intensity, ${{{I_{\max }}} \over {{I_{\min }}}} = {{{{\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)}^2}} \over {{{\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)}^2}}}$ = ${\left( {{{\sqrt 3 + \sqrt 2 } \over {\sqrt 3 - \sqrt 2 }}} \right)^2}$
= ${\left( {{{1.732 + 1.414} \over {1.732 - 1.414}}} \right)^2}$
= ${\left( {{{3.146} \over {0.318}}} \right)^2}$ = ${\left( {9.89} \right)^2}$ $\simeq$ 98

Wavelength of emitted radiation is given by

${1 \over \lambda } = R\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

For Lyman series,

${1 \over \lambda } = R\left( {{1 \over {{1^2}}} - {1 \over {n_{}^2}}} \right)$, where n = 2, 3, 4, ....

The range of wavelength in this series is given by

${1 \over {{\lambda _{\max }}}} = R\left( {{1 \over {{1^2}}} - {1 \over {2_{}^2}}} \right) = {{3R} \over 4}c{m^{ - 1}}$

and ${1 \over {{\lambda _{\min }}}} = R\left( {{1 \over {{1^2}}} - {1 \over {\infty _{}^2}}} \right) = Rc{m^{ - 1}}$

Thus for Lyman : $\left[ {{1 \over R}to{4 \over {3R}}cm} \right]$

For Balmer series,

${1 \over {{\lambda _{}}}} = R\left( {{1 \over {{2^2}}} - {1 \over {n_{}^2}}} \right)$ , where n = 3, 4, 5, ....

Given, during ab, heat rejected, Q_ab = $-$ 50J

During ca, heat absorbed, Q_ca = 80J

During bc, heat transferred, Q_bc = 0J

Work done by the gas, W = 40J

$\therefore$Area of closed curve abca = Q_ab + Q_bc + Q_ca

Area = $-$ 50 + 0 + 80

Area = 30J

Distance between the parallel plates = d

The axial distance of beam from the centre of parallel plates = L

As, time, t = ${{dis\tan ce} \over {velocity}} = {L \over {{v_0}}}$


y-component of velocity, ${v_y} = \left( {{{qE} \over m}} \right)t = {e \over m}.{V \over d} \times {L \over {{v_0}}}$

$\Rightarrow$ ${v_y} = {{eVL} \over {md{v_0}}}$

$\therefore$ $\tan \theta = {{{v_y}} \over {{v_z}}} = {{eLV} \over {md{v_0}.{v_0}}} = {{eVL} \over {mdv_0^2}}$

$\Rightarrow$ $\theta$ = ${\tan ^{ - 1}}\left( {{{eVL} \over {mdv_0^2}}} \right)$