NEET-UG 2016

According to question,

L $\propto$ h^p c^q G^r

[M⁰LT⁰] = [ML²T^-1]^p [LT^-1]^q [M^-1L³T^-2]^r

Equating power both sides, we get

p - r = 0        ......(1)

2p + q + 3r = 1        ......(2)

- p - q - 2r = 0        ......(3)

Solving equation (1), (2), (3), we get

p = r = $\frac{1}{2}$, q = $-\frac{3}{2}$

$\therefore$ L = $\frac{\sqrt{hG}}{c^{3/2}}$

For car P,

x_P(t) = (at + bt²)

v_P(t) = $\frac{d{x}_p\left(t\right)}{dt}$ = a + 2bt

Similarly for car Q,

x_Q(t) = (ft - t²)

v_Q(t) = $\frac{d{x}_Q\left(t\right)}{dt}$ = f - 2t

When they have same velocity then, v_P(t) = v_Q(t)

$\therefore$ a + 2bt = f - 2t

$\Rightarrow$ 2t(b + 1) = f - a

$\Rightarrow$ t = $\frac{f-a}{2\left(1+b\right)}$

Here $a_c=a\cos 30^o$ = 15 $\times$ $\frac{\sqrt{3}}{2}$ m/s²

We know, $a_c=\frac{v^2}{R}$

$\Rightarrow$ $v=\sqrt{a_{c}R}$ = $\sqrt{15\times \frac{\sqrt{3}}{2}\times 2.5}$ = 5.7 m/s

Here $\overrightarrow{r_1}=\left(-2\hat{i}+5\hat{j}\right)m$
$\overrightarrow{r_2}=\left(-4\hat{j}+3\hat{k}\right)m$
$\overrightarrow{F}=\left(4\hat{i}+3\hat{j}\right)N$

Work done by force F in moving from $\overrightarrow{r_1}$ and $\overrightarrow{r_2}$

$W=\overrightarrow{F}\left(\overrightarrow{r_2}-\overrightarrow{r_1}\right)$

$\Rightarrow W=\left(4\hat{i}+3\hat{j}\right)\left(4\hat{j}+3\hat{k}+2\hat{i}-5\hat{j}\right)$

$\Rightarrow W=\left(4\hat{i}+3\hat{j}\right).\left(2\hat{i}-\hat{j}+3\hat{k}\right)$

$\Rightarrow W=8+\left(-3\right)=5J$

Mass of bullet, m = 10 g = 0.01 kg
Initial speed of bullet, u = 400 m s^–1
Mass of block, M = 2 kg
Length of string, l = 5 m
Speed of the block after collision = v1
Speed of the bullet on emerging from block, v = ?

Using energy conservation principle for the block,
(KE + PE)_Reference = (KE + PE)_h

$\Rightarrow \frac{1}{2}M{v}_1^2=Mghor,v_1=\sqrt{2gh}$

$v_1=\sqrt{2\times 10\times 0.1}=\sqrt{2}m{s}^{-1}$

Using momentum conservation principle for block and bullet system,
(M × 0 + mu)_{Before collision} = (M × v₁ + mv)_{After collision}

$\Rightarrow 0.01\times 400=2\sqrt{2}+0.01\times v$

$\Rightarrow v=\frac{4-2\sqrt{2}}{0.01}=117.15m{s}^{-1}\approx 120m{s}^{-1}$

Masses of the balls are same and collision is elastic, so their velocity will be interchanged after collision.

Given, p_i = p_f = mV
Change in momentum of the ball

$={\overrightarrow{p}}_f-{\overrightarrow{p}}_i$

$=\left(-p_{fx}\hat{i}-p_{fy}\hat{j}\right)-\left(-p_{ix}\hat{i}-p_{iy}\hat{j}\right)$

$=-\hat{i}\left(p_{fx}+p_{ix}\right)-\hat{j}\left(p_{fy}+p_{iy}\right)$

$=-2p_{ix}\hat{i}=-mV\hat{i}\left[\because p_{fy}-p_{iy}=0\right]$

Here, p_ix = p_fx = p_icos 60° = $\frac{mV}{2}$

$\because$ Impulse imparted by the wall = change in the momentum of the ball = mV.

Here, $l_1+l_2=l$

Centre of mass of the system,

$l_1=\frac{m_1\times 0+m_2\times l}{m_1+m_2}=\frac{m_{2}l}{m_1+m_2}$

$l_2=l-l_1=\frac{m_{1}l}{m_1+m_2}$

Required moment of inertia of the system,

$I=m_1{l}_1^2+m_2{l}_2^2$

$=\left(m_1{m}_2^2+m_2{m}_1^2\right)\frac{l^2}{{\left(m_1+m_2\right)}^2}$

$=\frac{m_1{m}_2\left(m_1+m_2\right)l^2}{{\left(m_1+m_2\right)}^2}=\frac{m_1{m}_2}{m_1+m_2}{l}^2$

$\frac{E_{Sphere}}{E_{Cylinder}}=\frac{\frac{1}{2}{I}_s{\omega }_s^2}{\frac{1}{2}{I}_c{\omega }_c^2}=\frac{I_s{\omega }_s^2}{I_c{\omega }_c^2}$

Here, $I_s=\frac{2}{5}m{R}^2$,   $I_c=\frac{1}{2}m{R}^2$

${\omega }_c=2{\omega }_s$

$\frac{E_{Sphere}}{E_{Cylinder}}=\frac{\frac{2}{5}m{R}^2\times {\omega }_s^2}{\frac{1}{2}m{R}^2\times {\left(2{\omega }_2\right)}^2}$

$\Rightarrow \frac{4}{5}\times \frac{1}{4}=\frac{1}{5}$

It is known that rotational kinetic energy is given as :

$E=\frac{I{\omega }^2}{2}$ = $\frac{L^2}{2I}$

Hence, L² $\propto$ I

Further, E_A = E_B

Also, $\frac{{L_A}^2}{2I_A}=\frac{{L_B}^2}{2I_B}$

$I_B>I_A$

$L_B>L_A$

Total mechanical energy of satellite is given by

E = – GMm/2r

Here, r = R + h

And, GM = g_oR²

So, E = – mg_oR²/2(R+h)

Since, acceleration due to gravity is given as

$g=\{\begin{array}{cc}\frac{4}{3}\pi \rho Gr& r\le r\\ \frac{4}{3}\pi \rho G\frac{R^3}{r^2}& r>R\end{array}$

Capillary rise, $h=\frac{2T\cos \theta }{r\rho g}$

For given value of T and $r,h\propto \frac{\cos \theta }{\rho }$

Also, $h_1=h_2=h_3$

or $\frac{\cos {\theta }_1}{{\rho }_1}=\frac{\cos {\theta }_2}{{\rho }_2}=\frac{\cos {\theta }_3}{{\rho }_3}$

Since, ${\rho }_1>{\rho }_2>{\rho }_3$, so $\cos {\theta }_1>\cos {\theta }_2>\cos {\theta }_3$

For $0\le \theta <\frac{\pi }{2},{\theta }_1<{\theta }_2<{\theta }_3$

Hence, $0\le {\theta }_1<{\theta }_2<{\theta }_3<\frac{\pi }{2}$

We have d = 2700 m, ρ = 10³ kg/m3, compressibility = 45.4 × 10⁻¹¹ /pascal

Now the pressure at bottom of ocean is

p = ρgd

= 10³ × 10 × 2700

= 27 × 10⁶ Pa

Hence, fractional compression

= 45.4 × 10⁻¹¹ × 27 × 10⁶

= 1.2 × 10⁻² = 0.125 N m^$-$1

According to Newton’s law of cooling,

$\frac{dT}{dt}=K\left(T-T_s\right)$

For two cases,

$\frac{d{T}_1}{dt}=K\left(T_1-T_s\right)$

and $\frac{d{T}_2}{dt}=K\left(T_2-T_s\right)$

Here, $T_s=T,T_1=\frac{3T+2T}{2}=2.5T$

and $\frac{d{T}_1}{dt}=\frac{3T-2T}{10}=\frac{T}{10}$

$T_2=\frac{2T+T^{\prime }}{2}$ and $\frac{d{T}_2}{dt}=\frac{2T-T^{\prime }}{10}$

So $\frac{T}{10}=K\left(2.5T-T\right)$   ...(i)

$\frac{2T-T^{\prime }}{10}=K\left(\frac{2T+T^{\prime }}{2}-T\right)$   ...(ii)

Dividing eqn. (i) by eqn. (ii), we get

$\frac{T}{2T-T^{\prime }}=\frac{\left(2.5T-T\right)}{\left(\frac{2T+T^{\prime }}{2}-T\right)}$

$\frac{2T+T^{\prime }}{2}-T=\left(2T-T^{\prime }\right)\times \frac{3}{2}$

$T^{\prime }=3\left(2T-T^{\prime }\right)\Rightarrow 4T^{\prime }=6T$

$\therefore T^{\prime }=\frac{3}{2}T$

If the final common temperature is T_c, C_c and C_h average heat capacities of cold and hot bodies then as per principle of calorimetry,

heat lost = heat gained

C_h (100°C – T_c) = C_c × T_c

Now T_c = C_h/(C_h + C_c) × 100°C

= 100/[1 + (C_c/C_h)],     where C_c/C_h < 1

It is seen that 1 + C_c/C_h < 2

Hence, T_c > (100/2)°C or T_c > 50°C

As PV = nRT

$\Rightarrow n=\frac{PV}{RT}=\frac{mass}{molarmass}$   ...(i)

Density, $\rho$ = $\frac{mass}{volume}=\frac{\left(molarmass\right)P}{RT}=\frac{\left(m{N}_A\right)P}{RT}$
   [From eqn. (i)]

$\rho =\frac{mP}{kT}$    $\left(\because R=N_{A}k\right)$

Temperature inside refrigerator = t₂ °C

Room temperature = t₁ °C

For refrigerator,

$\frac{\mathrm{Heat}\mathrm{given}\mathrm{to}\mathrm{high}\mathrm{temperature}\left(Q_1\right)}{\mathrm{Heat}\mathrm{taken}\mathrm{from}\mathrm{lower}\mathrm{temperature}\left(Q_2\right)}=\frac{T_1}{T_2}$

$\frac{Q_1}{Q_2}=\frac{t_1+273}{t_2+273}$

$\Rightarrow \frac{Q_1}{Q_1-W}=\frac{t_1+273}{t_2+273}$

$\Rightarrow 1-\frac{W}{Q_1}=\frac{t_2+273}{t_1+273}$

$\Rightarrow \frac{W}{Q_1}=\frac{t_1-t_2}{t_1+273}$

The amount of heat delivered to the room for each joule of electrical energy (W = 1 J)

$Q_1=\frac{t_1+273}{t_1-t_2}$

Process described by the equation,

PV³ = constant

For a polytropic process, PV^$$\alpha$$ = constant

$C=C_V+\frac{R}{1-\alpha }=\frac{3}{2}R+\frac{R}{1-3}=R$

The time period of oscillation is given by

$T=2\pi \sqrt{m/k}$

$T_1=3=2\pi \sqrt{m/k}$

$T_2=5=2\pi \sqrt{\left(\frac{m+1}{k}\right)}$

On dividing :

$3/5=\sqrt{m/\left(m+1\right)}$

$9/25=\sqrt{m/\left(m+1\right)}$

$9m+9=25m$

$16m=9\Rightarrow m=9/16$

(n – 1) and (n + 1) suppose to form frequency n

n and n will be at resonance

n – 1 and n produce 1 beat

n + 1 and n produce 1 beat

$\therefore$ Number of beats formed are ‘2’

Second overtone of an open organ pipe

(Third harmonic) $=3\times v{}_0^{\prime }=3\times \frac{v}{2L^{\prime }}$

First overtone of a closed organ pipe

(Third harmonic) $=3\times v_0=3\times \frac{v}{4L}$

According to question,
$3v{}_0^{\prime }=3v_0\Rightarrow 3\times \frac{v}{2L^{\prime }}=3\times \frac{v}{4L}$

$\therefore$ L' = 2L

Torque is given by τ = pE sin θ

τ = pE sin θ = q$l$E sinθ

$\Rightarrow$ q = τ /$l$E sinθ

= 4/(2 × 10^-2 × 0.5 × 2 × 10⁵) = 2 mC

On applying KVL in the circuit, the potential difference between the two points can be obtained.

V_A – V_B = (2W × 2A) + 3V + (2A × 1W) = 4V + 3V + 2V = 9 V

Resistance of bulb,
$R_B=\frac{V^2}{P}=\frac{{\left(100\right)}^2}{500}=20\Omega$

Power of the bulb in the circuit,
P = VI

$I=\frac{P}{V_B}$

$=\frac{500}{100}=5A$

V_R = IR $\Rightarrow$ (230 – 100) = 5 × R

$\therefore$ R = 26 $\Omega$

Here, $C_1=\frac{2{\epsilon }_0{k}_{1}A}{3d},C_2=\frac{2{\epsilon }_0{k}_{2}A}{3d}$

$C_3=\frac{2{\epsilon }_0{k}_{3}A}{3d},C_4=\frac{2{\epsilon }_0{k}_{4}A}{d}$

Given system of C₁, C₂, C₃ and C₄ can be simplified as


$\therefore$ $\frac{1}{C_{AB}}=\frac{1}{C_1+C_2+C_3}+\frac{1}{C_4}$

Suppose, $C_{AB}=\frac{k{\epsilon }_{0}A}{d}$

$C_{AB}=\frac{1}{k\left(\frac{{\epsilon }_{0}A}{d}\right)}=\frac{1}{\frac{2}{3}\frac{{\epsilon }_{0}A}{d}\left(k_1+k_2+k_3\right)}+\frac{1}{\frac{2{\epsilon }_{0}A}{d}{k}_4}$

$\Rightarrow \frac{1}{k}=\frac{1}{2\left(k_1+k_2+k_3\right)}+\frac{1}{2k_4}$

$\therefore \frac{2}{k}=\frac{3}{k_1+k_2+k_3}+\frac{1}{k_4}$

Frequency of revolution of charge in magnetic field is given as

F = $\frac{qB}{2\pi m}$

= $\frac{\left(1.76\times {10}^{11}\right)\times 3.57\times {10}^{-2}}{2\times 3.14}$

= 10⁹ Hz

= 1 GHz

For One turn loop :

B = $\frac{{\mu }_{0}i}{2r}$

For n turn loop :

r = $\frac{R}{n}$

B' = $\frac{{\mu }_{0}ni}{2\frac{R}{n}}$

= $\left(\frac{{\mu }_{0}i}{2R}\right)n^2$

= n²B

W = MB (cos 0° – cos 60°)

$\Rightarrow$ W = MB$\left(1-\frac{1}{2}\right)=\frac{MB}{2}$ ....(1)

Required torque for this position

$\tau$ = MB sin $\theta$

= MB sin 60°

= $\frac{\sqrt{3}}{2}MB$ = $\sqrt{3}W$ [From (1)]

Emf generated in loop 1,

${\epsilon }_1=-\frac{d\varphi }{dt}=-\frac{d}{dt}\left(BA\right)=-A\times \frac{dB}{dt}$

$\Rightarrow$ ${\epsilon }_1=-\pi r^2\times \frac{dB}{dt}$

Emf generated in loop 2,

${\epsilon }_2=-\frac{d\varphi }{dt}=-\frac{d}{dt}\left(BA\right)=-\frac{d}{dt}\left(0\times A\right)$ = 0

Net impedance, Z = $\sqrt{R^2+X_L^2}$ = $100\sqrt{2}\Omega$

Peak value of displacement current = Maximum conduction current in the circuit

= $\frac{{\epsilon }_0}{Z}$ = $\frac{220\sqrt{2}}{100\sqrt{2}}$ = 2.2 A

Power factor, cos $\varphi$ = $\frac{R}{Z}$

= $\frac{V_R}{V}$

= $\frac{V_R}{\sqrt{V_R^2+{\left(V_L-V_C\right)}^2}}$

= $\frac{80}{\sqrt{{80}^2+{\left(100-40\right)}^2}}$

= 0.8

Quality factor of an L-C-R circuit is given by,

Q = $\frac{1}{R}\sqrt{\frac{L}{C}}$

So, Q₁ = $\frac{1}{20}\sqrt{\frac{1.5}{35\times {10}^{-6}}}$ = 10.35

Q₂ = $\frac{1}{25}\sqrt{\frac{2.5}{45\times {10}^{-6}}}$ = 9.43

Q₃ = $\frac{1}{15}\sqrt{\frac{3.5}{30\times {10}^{-6}}}$ = 22.77

Q₄ = $\frac{1}{25}\sqrt{\frac{1.5}{45\times {10}^{-6}}}$ = 7.30

As Q₃ is maximum of Q₁, Q₂, Q₃, and Q₄. Hence, option (c) should be selected for better tuning of an L-C-R circuit.

Seeing from one end, h₁ = $\mu$ × (h – b) = 3/2 × 5 = 15/2 cm

From other end of the slab, h₂ = $\mu$ × h = 3/2 × 3 = 9/2 cm

Now total height, (15/2 + 9/2) = 24/2 = 12 cm

Given u = 400 cm = 4 m

v = $\infty$ and f = ?

Since, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

= $\frac{1}{\infty }-\frac{1}{4}$

so, f = – 4 m

Hence, lens will be concave.

Now P = $\frac{1}{f}$ = $\frac{1}{-4}$ = - 0.25 D

Focal length (f) of glass convex lens is given by

$\frac{1}{f}=\left({\mu }_g-1\right)\left(\frac{2}{R}\right)$

$\Rightarrow$ $\frac{1}{f}=\left(\frac{3}{2}-1\right)\left(\frac{2}{R}\right)$

$\Rightarrow$ f = R

Focal length (f₁) of water filled concave lens is given by

$\frac{1}{f_1}=\left({\mu }_w-1\right)\left(-\frac{2}{R}\right)$

$\Rightarrow$ $\frac{1}{f_1}=\left(\frac{4}{3}-1\right)\left(-\frac{2}{R}\right)$ = $-\frac{2}{3R}$ = $-\frac{2}{3f}$

Equivalent focal length (f_eq) of lens system

$\frac{1}{f_{eq}}=\frac{1}{f}-\frac{2}{3f}+\frac{1}{f}$ = $\frac{4}{3f}$

$\therefore$ f_eq = $\frac{3f}{4}$

Here, a = 0.02 cm = 2 × 10^–4 m

$\lambda$ = 5 × 10^–5 cm = 5 × 10^–7 m

D = 60 cm = 0.6 m

Position of first minima on the diffraction pattern,

y₁ = $\frac{D\lambda }{a}$

= $\frac{0.6\times 5\times {10}^{-7}}{2\times {10}^{-4}}$ = 15 $\times$ 10^-4 m = 0.15 cm

Maximum Intensity is given as I_max = ${\left(\sqrt{I_1}+\sqrt{I_2}\right)}^2$

Minimum intensity is given as I_min = ${\left(\sqrt{I_1}-\sqrt{I_2}\right)}^2$

Given $\frac{I_1}{I_2}$ = n

$\frac{I_{\max }}{I_{\min }}={\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)}^2$

= ${\left(\frac{\sqrt{\frac{I_1}{I_2}}+1}{\sqrt{\frac{I_1}{I_2}}-1}\right)}^2$ = ${\left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)}^2$

$\frac{I_{\max }-I_{\min }}{I_{\max }+I_{\min }}=$ $\frac{\frac{I_{\max }}{I_{\min }}-1}{\frac{I_{\max }}{I_{\min }}+1}$

= $\frac{{\left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)}^2-1}{{\left(\frac{\sqrt{n}+1}{\sqrt{n}-1}\right)}^2+1}$

= $\frac{{\left(\sqrt{n}+1\right)}^2-{\left(\sqrt{n}-1\right)}^2}{{\left(\sqrt{n}+1\right)}^2+{\left(\sqrt{n}-1\right)}^2}$

= $\frac{4\sqrt{n}}{2\left(n+1\right)}$ = $\frac{2\sqrt{n}}{n+1}$

N₀ = Nuclei at time t = 0

N₁ = Remaining nuclei after 40% decay

= (1 – 0.4) N₀ = 0.6 N₀

N₂ = Remaining nuclei after 85% decay

= (1 – 0.85) N₀ = 0.15 N₀

$\therefore$ $\frac{N_2}{N_1}$ = $\frac{0.15N_0}{0.6N_0}$ = ${\left(\frac{1}{2}\right)}^2$

Hence number of half-lives is 2, so the time needed is :

t = 2 × 30 = 60 mints

When electron jumps from higher orbit to lower orbit then, wavelength of emitted photon is given by,

$\frac{1}{\lambda }=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$

On jumping from 3^rd orbit to 2^nd orbit,

$\frac{1}{\lambda }=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ = $\frac{5R}{36}$

On jumping from 4^th orbit to 3^rd orbit,

$\frac{1}{{\lambda }^{\prime }}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$ = $\frac{7R}{144}$

$\therefore$ $\lambda$' = $\frac{144}{7}\times \frac{5\lambda }{36}$ = $\frac{20\lambda }{7}$

Kinetic energy of electrons

K = $\frac{p^2}{2m}$ = $\frac{{\left(\frac{h}{\lambda }\right)}^2}{2m}$ = $\frac{h^2}{2m{\lambda }^2}$

For certain frequency, maximum wavelength that can be emitted is $\lambda$₀ which is cut off wavelength obtained at cut off frequency,

E₀ = $\frac{hc}{{\lambda }_0}$

Since, E = E₀

$\frac{h^2}{2m{\lambda }^2}=\frac{hc}{{\lambda }_0}$

$\Rightarrow$ $\lambda$₀ = $\frac{2mc{\lambda }^2}{h}$

Stopping potential is the voltage which is needed to stop energetic photo electron for reaching towards cathode.

Stopping potential, $\varphi$ = E – K_max

2 eV = 5 eV – $\varphi$ or $\varphi$ = 3 eV

Now eV₀ = E' – $\varphi$ = 6 eV – 3 eV = 3 eV

Hence stopping potential is –3V

Applying De Morgan’s law:

Output Y = [(A ⋅ B) ⋅ C]' = A' + B' + C'

When A, B, C are 0 $\Rightarrow$ Y = 1

When A, B, C are 1 $\Rightarrow$ Y = 0

Current will not flow through D₁ as it is reversed biased.

Current will flow through resistor R₁, diode D₂ and resistor R₃

Now current i = 10/(2 + 2) = 2.5 A

Voltage gain, A = $\beta \frac{R_C}{R_B}$ = $100\times \frac{2000}{1000}$ = 200

Also, A = $\frac{V_0}{V_i}$

$\Rightarrow$ $V_i=\frac{V_0}{A}$ = $\frac{4}{200}$ = 20 mA

Assume the atomic weight of element X is x and element Y is y.

For XY₂, $n=\frac{w}{Mol.wt.}$

$0.1=\frac{10}{x+2y}$

$\Rightarrow x+2y=\frac{10}{0.1}=100$    ...(i)

For X₃Y₂, $n=\frac{w}{Mol.wt.}$

$0.05=\frac{9}{3x+2y}$

$\Rightarrow 3x+2y=\frac{9}{0.05}=180$    ...(ii)

On solving equations (i) and (ii),

we get y = 30 g mol^-1

Put the value of y in equation. (i)

x + 2(30) = 100

$\Rightarrow$ x = 100 - 60 = 40 g mol^-1

d_{x² - y²} and d_z² orbitals have electron density along the axes.

For $n=3$ and $l=1$:

• $n=3$ means the electron is in the third shell

• $l=1$ means it is a $p$-subshell

Now, for a $p$-subshell:

• the possible values of $m_l$ are $-1,0,+1$

• so, number of orbitals $=3$

Each orbital can hold a maximum of $2$ electrons.

So total electrons in the $3p$ subshell:

$$3\times 2=6$$

Therefore, the correct answer is Option B: 6.

CaF₂ + H₂SO₄ $\to$ CaSO₄ + 2HF

Here, the oxidation state of every atom remains the same so, it is not a redox reaction.

C₅H₅N + H₂O → C₅H₅N⁺H + OH^–

So, the amount of pyridine that forms pyridinium ion is $\alpha$.

Now, $\alpha$ = $\sqrt{\frac{K_b}{conc.ofpyridine}}$

= $\sqrt{\frac{1.7\times {10}^9}{0.10}}$

= 1.30 $\times$ 10^-4

So, percentage of pyridine that forms pyridinium ion

= 1.30 × 10^–4 × 100

= 1.30 × 10^–2 = 0.013%