WBJEE 2021 Paper 2 (Physics & Chemistry) Question Paper with Solutions

The net voltage in the circuit is (10 $-$ 9) V = 1 V

This leads to reverse bias of diode in the circuit. Hence, current through the diode will equal to 0 mA.

Charge per unit length of the thin rod = $\lambda$

If l be the length of the rod, the charge on the rod,

q = l$\lambda$ ..... (i)

According to given situation,

l = 2$\pi$R

$\therefore$ From Eq. (i), we get

q = 2$\pi$R$\lambda$

Current associated,

$I = {q \over T}$

$I = {{2\pi R\lambda } \over T}$ ..... (ii)

Magnetic field on the axis at a distance d away from the centre,

$B = {{{\mu _0}I{R^2}} \over {2{{({R^2} + {d^2})}^{3/2}}}}$

For d >> R,

$B = {{{\mu _0}I{R^2}} \over {2{{({d^2})}^{3/2}}}}$

$= {{{\mu _0}I{R^2}} \over {2{d^3}}}$

$= {{{\mu _0}\left( {{{2\pi R\lambda } \over T}} \right){R^2}} \over {2{d^3}}}$

$= {{{\mu _0}\pi \lambda } \over T}.{{{R^3}} \over {{d^3}}}$

$\Rightarrow B \propto {{{R^3}} \over {{d^3}}}$ [given, $B \propto {{{R^m}} \over {{d^n}}}$]

$\therefore$ m = 3, n = 3

In projectile motion, horizontal component at velocity remains constant with time. Hence, correct representation of graph between t and u_x is shown in option (b).

During first half of the motion (i.e. upto x = 20 m) acceleration in increasing linearly is given by

$a = {5 \over {20}}x + 5 \Rightarrow a = {x \over 4} + 5$

$\Rightarrow {{dv} \over {dt}} = {x \over 4} + 5$ $\because$ $\left( {a = {{dv} \over {dt}}} \right)$

$\Rightarrow {{v\,.\,dt} \over {dx}} = {x \over 4} + 5$ $\because$ $\left( {{{dx} \over {dt}} = v \Rightarrow dt = {{dx} \over v}} \right)$

$\Rightarrow \int\limits_5^v {v\,.\,dv = \int\limits_0^{20} {\left( {{x \over 4} + 5} \right)\,.\,dx} }$

$\Rightarrow \left[ {{{{v^2}} \over 2}} \right]_5^v = \left[ {{{{x^2}} \over 8} + 5x} \right]_0^2$

$\Rightarrow {{{v^2}} \over 5} - {{25} \over 2} = {{400} \over 8} + 100$

$\Rightarrow {{{v^2}} \over 2} = 150 + {{25} \over 2} \Rightarrow {v^2} = 325$

Now, in second half of motion, acceleration is constant, i.e. a = 10 m/s²

$\Rightarrow$ v² $-$ u² = 2as

$\Rightarrow$ v'² $-$ v² = 2as (take, v = v', u = v)

$\Rightarrow$ v'² $-$ 325 = 2 $\times$ 10 $\times$ (35 $-$ 20) ($\because$ v² = 325)

$\Rightarrow$ v' = $\sqrt {625}$ = 25 m/s

The given impulse acts as both linear and an angular impulse.

Linear impulse = P = mv_CM (where, v_CM = velocity of centre at mass of rod)

Angular impulse = $P \times {L \over 2} = {I_{CM}}\omega$

where, ${I_{CM}} = {{m{L^2}} \over {12}}$

and $\omega$ is angular velocity at rod about centre of mass,

i.e. $\omega = {{PL} \over {2{I_{CM}}}}$

Kinetic energy

$= \left[ {{1 \over 2}mv_{CM}^2 + {1 \over 2}{I_{CM}}{\omega ^2}} \right]$

$= \left[ {{1 \over 2} \times m \times {{\left( {{P \over m}} \right)}^2} + {1 \over 2} \times {{m{L^2}} \over {12}} \times {{{P^2}{L^2} \times 144} \over {4 \times {m^2} \times {L^2}}}} \right]$

$= {{{P^2}} \over {2m}}[1 + 3]$

$= {{2{P^2}} \over m}$

As the gravitational field is uniform, therefore centre of gravity and the centre of mass are at same location.

$\therefore$ The location of centre of mass is

$h = {{\int_0^\infty {hdm} } \over {\int_0^\infty {dm} }}$

$h = {{\int_0^\infty {h\rho dh} } \over {\int_0^\infty {\rho dh} }}$ ..... (i)

But from barometric formula and gas law,

$\rho = {\rho _0}{e^{ - Mgh/RT}}$

$\therefore$ Putting the value of $\rho$ in Eq. (i), we have

$h = {{\int_0^\infty {h{\rho _0}{e^{( - Mgh/RT)}}dh} } \over {\int_0^\infty {{\rho _0}{e^{ - (Mgh/RT)}}dh} }} = {{RT} \over {Mg}}$

Given, D = 1.2 m, d = 0.02 cm = 0.02 $\times$ 10^$-$2 m

According to question,

${{3D\lambda } \over d} - {{5D\lambda } \over {2d}} = 0.18 \times {10^{ - 2}}$

$\lambda \left[ {{{3 \times 1.2} \over {0.02 \times {{10}^{ - 2}}}} - {{5 \times 1.2} \over {2 \times 0.02 \times {{10}^{ - 2}}}}} \right] = 0.18 \times {10^{ - 2}}$

$\Rightarrow \lambda \left[ {3 - {5 \over 2}} \right] \times 1.2 = 0.18 \times {10^{ - 2}} \times 0.02 \times {10^{ - 2}}$

$\Rightarrow \lambda = {{0.18 \times {{10}^{ - 4}} \times 0.02 \times 2} \over {1.2}}$ m

$\Rightarrow \lambda = 0.006 \times {10^{ - 4}}$ m

$\Rightarrow \lambda = 600 \times {10^{ - 9}}$ m

= 600 nm

The situation is as shown in the figure below


Given,

P = 5 D

$\therefore$ $f = {1 \over P} = {1 \over 5}$ m = 20 cm

Let u = $-$ $\infty$ (object distance)

Then, v = f (by sign convention)

Using formula for refraction at spherical surface,

${{{\mu _2}} \over v} - {{{\mu _1}} \over u} = {{{\mu _2} - {\mu _1}} \over R}$

$\Rightarrow {4 \over {3 \times (20)}} - {1 \over {( - \infty )}} = {{{4 \over 3} - 1} \over R}$ [$\because$ v = 20 cm]

$\Rightarrow {1 \over {15}} = {1 \over {3R}}$

$\Rightarrow$ R = 5 cm

Magnetic field at the centre O of the circular coil due to bar magnet (axial position)

$B = {{{\mu _0}} \over {4\pi }}.{{2M} \over {{x^3}}}$

Hence, option (a) is not true.

Induced emf, $e = - {{d\phi } \over {dt}} = - {d \over {dt}}.\,BA$

$= - {d \over {dt}}.\,{{{\mu _0}} \over {4\pi }}\,.\,{{2M} \over {{x^3}}}\pi {a^2}$

$= - {{{\mu _0}M{a^2}} \over 2}{d \over {dt}}.\,{1 \over {{x^3}}}$

$= - {\mu _0}M{a^2}\,.\,\left( {{{ - 3} \over {{x^4}}}} \right){{dx} \over {dt}}$

$e = {{3{\mu _0}M{a^2}} \over {{x^4}}}.\,{{dx} \over {dt}}$

$\Rightarrow e \propto {1 \over {{x^4}}}$

Hence, option (b) is true.

Induced current, $I = {e \over R} = {{3{\mu _0}M{a^2}} \over {{x^4}R}}{{dx} \over {dt}}$ ..... (i)

$\therefore$ Magnetic moment, $\mu = IA = {{3{\mu _0}M{a^2}} \over {{x^4}R}}{{dx} \over {dt}}\pi {a^2}$

$\Rightarrow \mu = {{3\pi {\mu _0}M} \over {{x^4}R}}{{dx} \over {dt}}({a^4})$

$\mu \propto {a^4}$

$\Rightarrow$ Hence, option (c) is true.

Net heat produced,

H $\propto$ I²

$\Rightarrow$ $H \propto {\left( {{1 \over {{x^4}}}} \right)^2}$ [from Eq. (i)]

$\Rightarrow H \propto {1 \over {{x^8}}}$

Hence, option (d) is not true.

The given situation is shown below


Since, charge particle enters into magnetic field B perpendicular to it, hence it performs a circular path in magnetic field. Radius of circular path,

$r = {{mv} \over {Bq}}$

Charge particle will enter in region c, when L < 2r

$\Rightarrow L < 2{{mv} \over {Bq}}$

$\Rightarrow {{LBq} \over {2m}} < v \Rightarrow v > {{qLB} \over {2m}}$

$\Rightarrow v > \left( {{1 \over 2}} \right){{qLB} \over m}$

i.e. When $v > {{qLB} \over m}$, then particle surely will inter into region c, because ${{qLB} \over m} > {{qLB} \over {2m}}$

Hence, option (a) is correct.

Since, charge particle enters into region c, hence path of the particle is a circle not in region b.

Hence, option (c) is not correct.

Time spent in region b is given as $T = {{2\pi m} \over {Bq}}$, which is independent of v. Hence, option (d) is correct.

We know that, latent heat of fusion at ice is 79.7 cal per gram.

Let final temperature be T.

Then,

m₁s$\Delta$T = m₂L

$\Rightarrow$ 300 $\times$ 1 $\times$ (25 $-$ T) = 100 $\times$ 75

$\Rightarrow$ 25 $-$ T = 25

$\Rightarrow$ T = 0$^\circ$C

After that total energy left, Q = 4.7 $\times$ 100 = 470 cal

Total mass of water = 400 g

Amount of water again converted into ice,

$m = {Q \over {{L_{ice}}}} = {{470} \over {79.7}}$

$\Rightarrow$ m = 5.9 g

Thus, whole mass is converted into water at 0$^\circ$C and about 5.9 g water is again converted into ice whose temperature is also 0$^\circ$C.

After achieving the temperature of 0$^\circ$C, latent heat of fusion is required firstly for conversion of water into ice, then further lowering of temperature as possible. So, the final temperature will be 0$^\circ$C.

The given situation is shown in the figure


We have to calculate electric field at point P (0, 2a, 0).

Net electric field at point P,

$E = {{ - \sigma } \over {2{\varepsilon _0}}}\widehat j - {{2\sigma } \over {2{\varepsilon _0}}}\widehat j - {{4\sigma } \over {2{\varepsilon _0}}}\widehat j$

$E = {{ - 7\sigma } \over {2{\varepsilon _0}}}\widehat j$

Given,


Since, the table is frictionless, hence acceleration of the whole system is given as

$a = {F \over {m + 2m + 3m}} = {F \over {6m}}$

Free body diagram at 1st block (left most)


From Newton's equation of motion,

$F - {N_1} = ma = m \times {F \over {6m}}$ $\because$ $\left( {a = {F \over {6m}}} \right)$

$\Rightarrow {N_1} = {{5F} \over 6}$

Free body diagram of 3rd block (right most)


From Newton's equation of motion,

${N_2} = 3ma = 3m \times {F \over {6m}}$ $\because$ $\left( {a = {F \over {6m}}} \right)$

$\Rightarrow {N_2} = {F \over 2}$

Hence, F > N₁ > N₂.

In pure inductive circuit, alternating voltage leads ${\pi \over 2}$ angle from alternating current.

i.e. Phase difference, $\phi = {\pi \over 2}$

Average power consumed,

$P = {V_{rms}} \times {I_{rms}} \times \cos {\pi \over 2}$

$\Rightarrow P = 0$ $\because$ $\left[ {\cos {\pi \over 2} = 0} \right]$

Given densities of both spheres are same. Let it be $\rho$ unit and density of liquid be $\sigma$ and $\eta$ be its viscosity.

Let radius of sphere S₁ be r₁ and that of sphere S₂ be r₂ and volumes are V₁ and V₂, respectively.

Now,

${{{m_1}} \over {{V_1}}} = {{{m_2}} \over {{V_2}}}$ ($\because$ density is same)

$\Rightarrow {{{m_1}} \over {{4 \over 3}\pi r_1^3}} = {{{m_2}} \over {{4 \over 3}\pi r_2^3}}$

$\Rightarrow {\left( {{{{r_1}} \over {{r_2}}}} \right)^3} = \left( {{{{m_1}} \over {{m_2}}}} \right) = {8 \over 1}$

$\Rightarrow {{{r_1}} \over {{r_2}}} = {2 \over 1}$

Now, terminal velocity is given by

$v = {{2{r^2}(\rho - \sigma )g} \over {9\eta }}$

$\Rightarrow {{{v_1}} \over {{v_2}}} = {{{{2r_1^2(\rho - \sigma )g} \over {9\eta }}} \over {{{2r_2^2(\rho - \sigma )g} \over {9\eta }}}}$

$\Rightarrow {{{v_1}} \over {{v_2}}} = {\left( {{{{r_1}} \over {{r_2}}}} \right)^2} = {2^2} = 4$

The electric field along Z-axis due to uniformly charged circular ring of radius a in XY-plane is given by

$E = {{kqz} \over {{{({z^2} + {a^2})}^{3/2}}}}$

where, q is the net charge on the ring and z is axial distance on Z-axis from the centre of the ring. Now, at point M in the graph shown, electric field is maximum.

So,

${{dE} \over {dZ}} = {{kq{{({z^2} + {a^2})}^{3/2}} - {3 \over 2}{{({z^2} + {a^2})}^{3/2}}.\,2{z^2}} \over {({z^2} + {a^2})}}$

For maxima, ${{dE} \over {dZ}} = 0$

$\Rightarrow {({z^2} + {a^2})^{3/2}} = {3 \over 2} \times 2{z^2}{({z^2} + {a^2})^{1/2}}$

$\Rightarrow {z^2} + {a^2} = 3{z^2} \Rightarrow 2{z^2} = {a^2}$

$\Rightarrow {z \over a} = {1 \over {\sqrt 2 }}$

Centre of mass of three particles (i.e. 1 kg, 2kg and 3kg)

${{1{r_1} + 2{r_2} + 3{r_3}} \over 6} = [1\widehat i + 2\widehat j + 3\widehat k]$

$\because$ $\left[ {CM = {{{m_1}{r_1} + {m_2}{r_2} + {m_3}{r_3}} \over {{m_1} + {m_2} + {m_3}}}} \right]$

where, r₁, r₂ and r₃ are position vector of mass 1 kg, 2kg and 3kg, respectively.

Now, centre of mass of two particles (i.e. 3kg and 2 kg)

${{3{r_4} + 2{r_5}} \over 5} = [ - \widehat i + 3\widehat j - 2\widehat k]$

where, r₄ and r₅n are position vector of mass 1 kg and 2 kg.

Now, according to question, if 5 kg mass is added to whole particle system, then entire mass lies on CM of first system,

${{1{r_1} + 2{r_2} + 3{r_3} + 3{r_4} + 2{r_5} + 5{r_6}} \over {16}} = [1\widehat i + 2\widehat j + 3\widehat k]$

where, r₆ is position vector of 5 kg mass.

$\Rightarrow 6\widehat i + 12\widehat j + 18\widehat k - 5\widehat i + 15\widehat j - 10\widehat k + 5{r_6} = 16\widehat i + 32\widehat j + 48\widehat k$

$\Rightarrow 5{r_6} = 15\widehat i + 5\widehat j + 40\widehat k$

$\Rightarrow {r_6} = 3\widehat i + \widehat j + 8\widehat k$

Hence, position of 5 kg mass is (3, 1, 8).

RMS value of potential difference is given as

${V_{rms}} = \sqrt {{{V_0^2 + {0^2}} \over 2}}$

$= {{{V_0}} \over {\sqrt 2 }}$

The given situation is shown below


A radial electric field E exists in the region between the two shells due to charge on inner shell only.

Electric field at point P is calculated according to Gauss's law,

i.e. ${\phi _E} = E\,.\,4\pi {r^2} = {q \over {{\varepsilon _0}}}$

$\Rightarrow E = {q \over {4\pi {\varepsilon _0}{r^2}}}$ .... (i)

The potential at the centre of sphere,

$V = - \int_a^b {E\,.\,dr = \int_a^b {E\,.\,dr = \int_a^b {{q \over {4\pi {\varepsilon _0}{r^2}}}dr} } }$ [from Eq. (i)]

$\Rightarrow V = {q \over {4\pi {\varepsilon _0}}}\left( {{1 \over a} - {1 \over b}} \right)$

Since, both wires are identical, so they have same cross-section A and same length l. Also, in series combination of wires, effective resistance,

R = R₁ + R₂

$\Rightarrow {{(l + l)} \over {{\sigma _{eff}}A}} = {1 \over {{\sigma _1}A}} + {1 \over {{\sigma _2}A}}$

$\Rightarrow {2 \over {{\sigma _{eff}}}} = {1 \over {{\sigma _1}}} + {1 \over {{\sigma _2}}}$

$\Rightarrow {\sigma _{eff}} = {{2{\sigma _1}{\sigma _2}} \over {{\sigma _1} + {\sigma _2}}}$

The colours of the four bands of carbon resistors are red, yellow, orange and no fourth band.


$\therefore$ R = 24 $\times$ 10³$\Omega$ $\pm$ 20%

= 24k$\Omega$ $\pm$ 20%

Under isothermal conditions,

${p_1}{V_1} + {p_2}{V_2} = pV$

$\Rightarrow \left( {p - {{4T} \over a}} \right)\,.\,{4 \over 3}\pi {a^3} + \left( {p - {{4T} \over b}} \right)\,.\,{4 \over 3}\pi {b^3} = \left( {p - {{4T} \over c}} \right)\,.\,{4 \over 3}\pi {c^3}$

where, T is surface tension of soap bubble.

$\Rightarrow p[{c^3} - ({a^3} + {b^3})] - 4T({a^2} + {b^2} - {c^2}) = 0$

$\Rightarrow T = {{p({c^3} - {a^3} - {b^3})} \over {4({a^2} + {b^2} - {c^2})}}$

The van der Waals' equation for real gas is as follows

$p + {{a{n^2}} \over {{V^2}}}(V - nb) = nRT$

where, p is the pressure, a and b are constants, V is the volume, T is the temperature and n is the number of moles.

van der Waals' equation can be rearranged as

$p = {{nRT} \over {V - nb}} - {{a{n^2}} \over {{V^2}}}$

Dividing first part with n and the 2nd part with n², we get

$p = {{{{nRT} \over n}} \over {{V \over n} - {{nb} \over n}}} - {{{{a{n^2}} \over {{n^2}}}} \over {{{{V^2}} \over {{n^2}}}}}$

$\Rightarrow p = {{RT} \over {{V \over n} - b}} - {a \over {{{{V^2}} \over {{n^2}}}}}$ ..... (i)

The equation given in question is as follows

$p = {{RT} \over {2V - b}} - {a \over {4{b^2}}}$ ..... (ii)

By comparing Eqs. (i) and (ii), we get

${V \over n} = 2v$

$\Rightarrow n = {V \over {2V}} = {1 \over 2}$ (i.e. $n = {1 \over 2}$ mole)

By using mole formula to determine the amount of O₂ as follows

$Mole = {{Mass} \over {Molar\,mass}}$ ...... (iii)

The molar mass of O₂ is 32 g/mol. Substituting in Eq. (iii), we get

$\Rightarrow {1 \over 2} = {{Mass} \over {32g/mol}}$

$\Rightarrow$ Mass of O₂ = 16 g

Initial length of rod = L

Initial angular velocity, $\omega$_i = $\omega$

When rod pivoted at one end is freely rotated in horizontal plane, then its angular momentum,

L_i = I₁$\omega$₁

${L_i} = \left( {{{M{L^2}} \over 3}} \right).\,\omega$


When temperature of the system is increased with temperature $\Delta$T, then change in length of rod $\Delta$L is given as

$\Delta$L = $\alpha$L$\Delta$T

Now, new length of the rod,

L' = $\Delta$L + L = $\alpha$L$\Delta$T + L = L($\alpha$$\Delta$T + 1)

$\Rightarrow$ L' = L(1 + $\alpha$$\Delta$T) ....... (i)

Now, new angular momentum,

L_f = I₂$\omega$₂ = ${{{ML{'^2}} \over 3}.{\omega \over 2}}$

By the law of conservation of angular momentum,

L_f = L_i

I₂$\omega$₂ = ${{{M{L^2}} \over 3}.\,\omega }$

${{{ML{'^2}} \over 3}.{\omega \over 2}}$ = ${{{M{L^2}} \over 3}.\,\omega }$

$\Rightarrow {{L{'^2}} \over 2} = {L^2}$

$\Rightarrow {{{L^2}{{(1 + \alpha \Delta T)}^2}} \over 2} = {L^2}$ [from Eq. (i)]

$\Rightarrow {(1 + \alpha \Delta T)^2} = 2$

$\Rightarrow 1 + {\alpha ^2}\Delta {T^2} + 2\alpha \Delta T = 2$

Since, $\alpha$ is very small, hence the term $\alpha$²$\Delta$T² may be neglected.

$\therefore$ $1 + 2\alpha \Delta T = 2 \Rightarrow 2\alpha \Delta T = 1$

$\Rightarrow \Delta T = {1 \over {2\alpha }}$

Since, susceptibility $\chi$ of diamagnetic material is independent of temperature.


Hence, ${1 \over \chi }$ is also independent of temperature.

Therefore, material B is diamagnetic. For paramagnetic substance, susceptibility varies inversely as temperature.

i.e. $\chi$ $\propto$ ${1 \over T}$ or ${1 \over \chi}$ $\propto$ T

Therefore, material A is paramagnetic.

Given figure is


Process 1 is isobaric expansion (p = constant). Hence, temperature of gas will increase, therefore $\Delta$U₁ will be positive.

Process 2 is an isothermal process, $\Delta$U₂ = 0

Process 3 is an adiabatic expansion, hence temperature of gas will fall.

$\therefore$ $\Delta$U₃ = negative

Therefore, $\Delta$₁ > $\Delta$U₂ > $\Delta$U₃

The given situation is shown below


Case 1 If charge q₃ is taken as positives.

In this situation, force on charge q₃ due to charges q₁ and q₂ will be repulsive and in opposite direction. Hence, at a certain location between +q₁ and +q₂, charge q₃ will be in equilibrium, when both repulsive forces will be equal in magnitude.

Case 2 If charge q₃ is taken as negative.

In this situation, force on charge q₃ due to charges q₁ and q₂ will be attractive and in opposite direction. Hence, at a certain location between +q₁ and +q₂, charge q₃ will be in equilibrium, when both attractive forces will be equal in magnitude.

The given logic circuit is


Output is written as

$Y = A\overline B + \overline A B$

When A = 0, B = 1,

then Y = 0 . $\overline 1$ + $\overline 0$ . 1

= 0 . 0 + 1 . 1

= 0 + 1

= 1

When A = 0, B = 0,

then Y = 0 . $\overline 0$ + $\overline 0$ . 0

= 0 . 1 + 1 . 0

= 0 + 0 = 0

At maximum compression of spring,

total kinetic energy of first block = potential energy of spring

$\Rightarrow {1 \over 2}$ mv² = ${1 \over 2}$ kx²

$\Rightarrow x = \sqrt {{m \over k}} \,.\,v$

Radius (r) of nth orbit in H-atom $= {{{n^2}} \over Z} \times 0.529\mathop A\limits^o$

Speed (v) of electron in nth orbit in H-atom $= 2.19 \times {10^6}\left( {{Z \over n}} \right)$ m/s

Energy (E) of electron in nth orbit in H-atom $= {{ - 13.6} \over {{n^2}}}$ eV

So, the quantity which is proportional to quantum number n is rv.

Mass, m = 0.02 kg

Potential energy,

$U = Ax(x - 4)J$

$\Rightarrow U = A{x^2} - 4Ax$

Force, $F = - {{dV} \over {dx}} = - {d \over {dx}}(A{x^2} - 4Ax)$

$\Rightarrow F = - 2Ax + 4A$ .... (i)

This force is dependent on x, hence particle is not acted upon by a constant force.

From Eq. (i), it is clear that f $\propto$ $-$ x

Hence, particle executes simple hormonic motion. Speed of particle is maximum at equilibrium position, i.e. when F = 0

i.e. $-$2Ax + 4A = 0

$\Rightarrow$ x = 2m

Period of oscillation,

$T = 2\pi \sqrt {{m \over k}}$ .... (i)

From Eq. (i), F = $-$2Ax + 4A

Since, value of k will be obtained in terms of A.

Therefore, the value of time period will be also obtained in terms of A, hence option (d) is not correct.

The given situation is shown below


Here, i = r

Since, i + r = 90$^\circ$

$\Rightarrow$ i + i = 90$^\circ$

$\Rightarrow$ i = 45$^\circ$

i = r = 45$^\circ$

Coordinate at point P is

(R cos$\theta$, R sin$\theta$) = (R cos135$^\circ$, R sin135$^\circ$)

$= \left[ {R\left( {{{ - 1} \over {\sqrt 2 }}} \right),R\left( {{1 \over {\sqrt 2 }}} \right)} \right]$

$= \left( {{{ - R} \over {\sqrt 2 }},{R \over {\sqrt 2 }}} \right)$

Let initially the electron in hydrogen atom be in ground state, i.e. n₁ = 1 and after bombarding, it jumps to nth energy level i.e. n₂ = n, then

12.5 eV $= \left[ {{{13.6} \over {{{(1)}^2}}} - {{13.6} \over {{n^2}}}} \right]$ eV

$\Rightarrow 1 - {1 \over {{n^2}}} = {{12.5} \over {13.6}}$

$\Rightarrow 1 - {{12.5} \over {13.6}} = {1 \over {{n^2}}}$

$\Rightarrow {1 \over {{n^2}}} = 0.080$

$\Rightarrow n = 3.53 \approx 4$

So, n₂ = n = 4th excited state.