NEET-UG 2016

Given, v = At + Bt²

$\frac{dx}{dt}$ = At + Bt²

$\int dx=\int \left(At+B{t}^2\right)dt$

x = $\frac{A{t}^2}{2}+\frac{B{t}^3}{3}+C$

At t = 1, particle is at

x(t = 1) = $\frac{A}{2}+\frac{B}{3}+C$

At t = 2, particle is at

x(t = 2) = $2A+\frac{8B}{3}+C$

$\therefore$ distance travelled by the particle between 1 s and 2 s is,

= x(t = 2) - x(t = 1)

= ($2A+\frac{8B}{3}+C$) - ($\frac{A}{2}+\frac{B}{3}+C$)

= $\frac{3}{2}A+\frac{7}{3}B$

Let the two vectors are $\overrightarrow{A}$ and $\overrightarrow{B}.$

Given that, $\left|\overrightarrow{A}+\overrightarrow{B}\right|=\left|\overrightarrow{A}-\overrightarrow{B}\right|$

$\therefore$ $\sqrt{A^2+B^2+2AB\cos \theta }$ = $\sqrt{A^2+B^2-2AB\cos \theta }$

$\Rightarrow$ $4AB\cos \theta$ = 0

$\because$ $4AB$ $\ne$ 0

$\therefore$ $\cos \theta$ = 0

$\Rightarrow$ $\theta$ = 90^o

$\overrightarrow{r}=\cos \omega t\hat{x}+\sin \omega t\hat{y},$

$\therefore$ $\overrightarrow{v}=-\omega \sin \omega t\hat{x}+\omega \cos \omega t\hat{y}$

and $\overrightarrow{a}=-{\omega }^2\cos \omega t\hat{x}-{\omega }^2\sin \omega t\hat{y}$ = $-{\omega }^2\overrightarrow{r}$

$\overrightarrow{r}.\overrightarrow{v}$ = 0

$\Rightarrow$ $\overrightarrow{r}\perp \overrightarrow{v}$

As position vector $\left(\overrightarrow{r}\right)$ is directly away from the origin, so, acceleration ($-{\omega }^2\overrightarrow{r}$) is directed towards the origin.

On a banked road,
$\frac{V_{\max }^2}{Rg}=\left(\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }\right)$

Maximum safe velocity of a car on the banked road
$V_{\max }=\sqrt{Rg\left[\frac{{\mu }_s+\tan \theta }{1-{\mu }_s\tan \theta }\right]}$

To complete the loop a body must enter a vertical loop of radius R with the minimum velocity v = $\sqrt{5gR}$.

Given: Mass of particle, M = 10g $=\frac{10}{1000}kg$

radius of circle R = 6.4 cm
Kinetic energy E of particle = 8 × 10^–4J
acceleration a_t = ?

$\frac{1}{2}m{v}^2=E\Rightarrow \frac{1}{2}\left(\frac{10}{1000}\right)v^2=8\times 10^{-4}$

$\Rightarrow v^2=16\times 10^{-2}$

$\Rightarrow v=4\times 10^{-1}=0.4m/s$

Now, using
v² = u² + 2a_ts (s = 4$\pi$R)

${\left(0.4\right)}^2=0^2+2a_t\left(4\times \frac{22}{7}\times \frac{6.4}{100}\right)$

$\Rightarrow a_t={\left(0.4\right)}^2\times \frac{7\times 100}{8\times 22\times 6.4}=0.1m/s^2$

Given force $(\overrightarrow{F})=2t\hat{i}+3t^2\hat{j}$

According to Newton's second law of motion,

$m\frac{d\overrightarrow{v}}{dt}=2t\hat{i}+3t^2\hat{j}(m=1kg)$

$\Rightarrow \underset{0}{\overset{\overrightarrow{v}}{\int }}d\overrightarrow{v}=\underset{0}{\overset{t}{\int }}\left(2t\hat{i}+3t^2\hat{j}\right)dt$

$\Rightarrow v=t^2\hat{i}+t^3\hat{j}$

Power (P) = $\overrightarrow{F}.\overrightarrow{v}\left(2t\hat{i}+3t^2\hat{j}\right)\left(t^2\hat{i}+t^3\hat{j}\right)$

= (2t³ + 3t⁵)W

Moment of inertia of complete disc about point 'O'.

$I_{Totaldisc}=\frac{M{R}^2}{2}$

Mass of removed disc

$M_{\mathrm{Removed}}=\frac{M}{4}\left(Mass\propto area\right)$

Moment of inertia of removed disc about point 'O'.

I_Removed (about same perpendicular axis)

$=I_{cm}+m{x}^2$

$=\frac{M}{4}\frac{{\left(R/2\right)}^2}{2}+\frac{M}{4}{\left(\frac{R}{2}\right)}^2=\frac{3M{R}^2}{32}$

Therefore the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the centre,

I_{Remaing disc} = I_Total – I_Removed

$=\frac{M{R}^2}{2}-\frac{3}{32}M{R}^2=\frac{13}{32}M{R}^2$

Time taken by the body to reach the bottom when it rolls down on an inclined plane without slipping is given by

$t=\sqrt{\frac{2l\left(1+\frac{k^2}{R^2}\right)}{g\sin \theta }}$

Since g is constant and $l$, R and sin $\theta$ are same for both

$\therefore$ $\frac{t_d}{t_s}=\frac{\sqrt{1+\frac{k_d^2}{R^2}}}{\sqrt{1+\frac{k_s^2}{R^2}}}=\sqrt{\frac{1+\frac{R^2}{2R^2}}{1+\frac{2R^2}{5R^2}}}$

$\left(\because k_d=\frac{R}{\sqrt{2}},k_s=\sqrt{\frac{2}{5}}R\right)$

$\Rightarrow \sqrt{\frac{3}{2}\times \frac{5}{7}}=\sqrt{\frac{15}{14}}\Rightarrow t_d>t_s$

Given, r = 50 cm = 0.5 m, $\alpha$ = 2.0 rad s^–2 , ${\omega }_0$ = 0

At the end of 2 s,
Tangential acceleration, at
= r$\alpha$ = 0.5 × 2 = 1 m s^–2

Radial acceleration, = $a_r={\omega }^{2}r={\left({\omega }_0+\alpha t\right)}^{2}r$

= (0 + 2 × 2)² × 0.5 = 8 m s^–2

$\therefore$ Net acceleration,
$a=\sqrt{a_t^2+a_r^2}=\sqrt{1^2+8^2}=\sqrt{65}\approx 8m{s}^{-2}$

As we know, escape velocity,

$V_e=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2G}{R}.\left(\frac{4}{3}\pi R^3\rho \right)}\propto R\sqrt{\rho }$

$\therefore$ $\frac{V_e}{V_p}=\frac{R_e}{R_p}\sqrt{\frac{{\rho }_e}{{\rho }_p}}$

$\Rightarrow \frac{V_e}{V_p}=\frac{R_e}{2R_e}\sqrt{\frac{{\rho }_e}{2{\rho }_e}}$

$\therefore$ Ration $\frac{V_e}{V_p}=1:2\sqrt{2}$

As we know, gravitational potential (v) and acceleration due to gravity (g) with height

$V=\frac{-GM}{R+h}=-5.4\times 10^7$   ...(i)

and $g=\frac{GM}{{\left(R+h\right)}^2}=6$   ...(ii)

Dividing (i) by (ii)

$\frac{\frac{-GM}{R+h}}{\frac{GM}{{\left(R+h\right)}^2}}=\frac{-5.4\times {10}^7}{6}$

$\Rightarrow \frac{-5.4\times {10}^7}{\left(R+h\right)}=6$

$\Rightarrow$ R + h = 9000 km so, h = 2600 km

From question, $\left(l_2-l_1\right)$ is maintained same at all temperatures hence change in length for both rods should be same

i.e., $\Delta l_1=\Delta l_2$

As we know, coefficient of linear expansion,

$\alpha =\frac{\Delta l}{l_0\Delta T}$

$l_1{\alpha }_1\Delta T=l_2{\alpha }_2\Delta T$

$l_1{\alpha }_1=l_2{\alpha }_2$

According to Archimedes principle,

Weight of the cylinder = (upthrust)₁ + (upthrust)₂

i.e., $ALdg=\left(1-p\right)LA\rho g+\left(pLA\right)n\rho g$

$\Rightarrow d=\left(1-p\right)\rho +pn\rho$

$=\rho -p\rho +np\rho$

$=\rho +\left(n+1\right)p\rho$

$=p\left[1+(n-1)p\right]$

According to wein's displacement law, maximum amount of emitted radiation corresponding to

${\lambda }_m=\frac{b}{T}$

${\lambda }_m=\frac{2.88\times {10}^{6}nmK}{5760K}=500nm$



From the graph U₁ < U₂ > U₃

Energy gained by the ice during its fall E = mgh

As, $\frac{mgh}{4}=m{L}_f$

$h=\frac{m{L}_f\times 4}{mg}=\frac{L_f\times 4}{g}$

$=\frac{3.4\times {10}^5\times 4}{10}=136000m=136km$

Now $COP=\frac{L_Q}{Power}=\frac{T_L}{T_H-T_L}$

$\Rightarrow Power=\frac{L_Q\times \left(T_H-T_L\right)}{T_L}$

= (600 × 4.2) × 26/277 = 236.53 W

It is observed that the rms velocity of molecule is directly proportional to temperature, so

$v_{rms}\propto \sqrt{T}$

$\Rightarrow v_{rms}^{{}^{\prime }}=v_{rms}\sqrt{\frac{T^{\prime }}{T}}$

Hence, $v_{rms}^{{}^{\prime }}=200\times \sqrt{\frac{400}{300}}=400/\sqrt{3}$

V₁ = V, V₂ = V/2

On P-V diagram, Area under adiabatic curve > Area under isothermal curve.

So compressing the gas through adiabatic process will require more work to be done.

According to Doppler's effect in sound



Apparent frequency,

$n^{\prime }=\frac{v}{v-v_S}{n}_0$

$=\frac{330}{330-15}\left(800\right)=\frac{330\times 800}{315}=838Hz$

The frequency of sound observer hears in the echo reflected from the cliff is 838 Hz.

From figure,



First harmonic is obtained at

$l=\frac{\lambda }{4}=50cm$

Third harmonic is obtained for resonance,

$l^{\prime }=\frac{3\lambda }{4}=3\times 50=150cm$

Wavelength of pulse at the lower end

${\lambda }_1\propto velocity(v_1)=\sqrt{\frac{T_1}{\mu }}$

Similarly, ${\lambda }_1\propto v_2$ $=\sqrt{\frac{T_2}{\mu }}$

$\therefore$ $\frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{T_2}{T_1}}=\sqrt{\frac{\left(m_1+m_2\right)g}{m_{2}g}}$

$=\sqrt{\frac{m_1+m_2}{m_2}}$

From figure $\tan \theta =\frac{F_e}{mg}\simeq \theta$

$\frac{k{q}^2}{x^{2}mg}=\frac{x}{2\ell }$

$\Rightarrow$ $x^3\propto q^2$   ...(i)

$\Rightarrow x^{3/2}\propto q$

Differentiating eq. (1) w.r.t. time

$3x^2\frac{dx}{dt}\propto 2q\frac{dq}{dt}$ but $\frac{dq}{dt}$ is constant

so x²(v) $\propto$ q Replace q from eq. (2)

x²(v) $\propto$ x^3/2 $\Rightarrow$ v $\propto$ x^–1/2

Given: Charge Q = at – bt²

$\therefore$ Current $i=\frac{\partial Q}{\partial t}=a-2bt$
$\left\{fori=0\Rightarrow t=\frac{a}{2b}\right\}$

From joule's law of heating, heat produced
dH = i²Rdt

$H=\underset{0}{\overset{a/2b}{\int }}{\left(a-2bt\right)}^{2}Rdt$

$H={\frac{{\left(a-2bt\right)}^{2}R}{-3\times 2b}|}_0^{\frac{a}{2b}}=\frac{a^{3}R}{6b}$

When two cells are connected in series i.e., E₁ + E₂ the balance point is at 50 cm. And when two cells are connected in opposite direction i.e., E₁ – E₂ the balance point is at 10 cm. According to principle of potential

$\frac{E_1+E_2}{E_1-E_2}=\frac{50}{10}$

$\Rightarrow \frac{2E_1}{2E_2}=\frac{50+10}{50-10}$

$\Rightarrow \frac{E_1}{E_2}=\frac{3}{2}$

Initially, the energy stored in 2 $\mu$F capacitor is

$U_i=\frac{1}{2}C{V}^2=\frac{1}{2}\left(2\times {10}^{-6}\right)V^2=V^2\times 10^{-6}J$

Initially, the charge stored in 2 $\mu$F capacitor is
Q_i = CV = (2 × 10^–6)V = 2V × 10^–6 coulomb. When switch S is turned to position 2, the charge flows and both the capacitors share charges till a common potential V_C is reached.

$V_C=\frac{\mathrm{total}\mathrm{charge}}{totalcapacitance}$

$=\frac{2V\times {10}^{-6}}{\left(2+8\right)\times {10}^{-6}}=\frac{V}{5}volt$

Finally, the energy stored in both the capacitors

$U_f=\frac{1}{2}\left[\left(2+8\right)\times {10}^{-6}\right]\left(\frac{V}{5}\right)$

$=\frac{V^2}{5}\times 10^{-6}J$

% loss of energy, $\Delta U=\frac{U_i-U_f}{U_i}\times 100\%$

$=\frac{\left(V^2-V^2/5\right)\times {10}^{-6}}{V^2\times {10}^{-6}}\times 100\%=80\%$

Force on arm AB due to current in conductor XY is

$F_1=\frac{{\mu }_0}{4\pi }\frac{2IiL}{\left(L/2\right)}=\frac{{\mu }_{0}Ii}{\pi }$

acting towards XY in the plane of loop.

Force on arm CD due to current in conductor XY is

$F_2=\frac{{\mu }_0}{4\pi }\frac{2IiL}{3\left(L/2\right)}=\frac{{\mu }_{0}Ii}{3\pi }$

acting away from XY in the plane of loop.

$\therefore$ Net force on the loop = F₁ – F₂

$=\frac{{\mu }_{0}Ii}{3\pi }\left[1-\frac{1}{3}\right]=\frac{2}{3}\frac{{\mu }_{0}Ii}{\pi }$

For points inside the wire i.e., (r $\le$ R)

Magnetic field $B=\frac{{\mu }_{0}Ir}{2\pi R^2}$

For points outside the wire (r $\ge$ R)

Magnetic field, $B^{\prime }=\frac{{\mu }_{0}I}{2\pi R}$

$\therefore \frac{B}{B^{\prime }}=\frac{\frac{{\mu }_{0}I\left(a/2\right)}{2\pi a^2}}{\frac{{\mu }_{0}I}{2\pi \left(2a\right)}}=1:1$

Magnetic susceptibility is negative for diamagnetic material only.

Total flux linked with the solenoid, $\varphi$ = N$\varphi$₀

= 1000 $\times$ 4 $\times$ 10^-3 Wb = 4 Wb

$\therefore$ Self-inductance of solenoid

L = $\frac{\varphi }{I}=\frac{4}{4}$ = 1 H

X_L = $\omega$L = 340 $\times$ 20 $\times$ 10^-3 = 6.8 $\Omega$

X_C = $\frac{1}{340\times 50\times {10}^{-6}}$ = 58.82 $\Omega$

Average power in impedance

Z = $\sqrt{R^2+{\left(X_C-X_L\right)}^2}$

= $\sqrt{{40}^2+{\left(58.82-6.8\right)}^2}$ = 65.62 $\Omega$

Power loss in A.C. circuit,

= V_rms I_rms cos$\varphi$

= $\frac{1}{2}$V₀ I₀ cos$\varphi$

= $\frac{1}{2}\times 10\times \frac{10}{65.62}\times \frac{40}{65.62}$

= 0.46 W

In ideal capacitor, I leads from voltage by 90^o and capacitor does not consume energy.

For producing electromagnetic waves, accelerating charged particle is required.

i = $\frac{{\delta }_m+A}{2}$

$\Rightarrow$ 45^o = $\frac{{\delta }_m+60^{\circ }}{2}$

$\Rightarrow$ ${\delta }_m=30^{\circ }$

sin45º = $\mu$sin30º

$\Rightarrow$ $\frac{1}{\sqrt{2}}=\frac{\mu }{2}$

$\Rightarrow$ $\mu$ = $\sqrt{2}$

Magnification in the mirror, m = $-\frac{v}{u}$

m = –2 $\Rightarrow$ v = 2u

As v and u have same signs so the mirror is concave and image formed is real.

$m=-\frac{1}{2}$ $\Rightarrow$ v = $\frac{u}{2}$

$\Rightarrow$ Concave mirror and real image.

m = + 2 $\Rightarrow$ v = –2u


As v and u have different signs but magnification is 2 so the mirror is concave and image formed is virtual.

$m=+\frac{1}{2}$ $\Rightarrow$ v = $-\frac{u}{2}$

As v and u have different signs with magnification $\frac{1}{2}$ so the mirror is convex and image formed is virtual.

Given: Focal length of objective, f₀ = 40cm

Focal length of eye–piece f_e = 4 cm

image distance, u₀ = 200 cm

Using lens formula for objective lens

$\frac{1}{v_0}-\frac{1}{u_0}=\frac{1}{f_0}$

$\frac{1}{v_0}-\frac{1}{-200}=\frac{1}{40}$

$\Rightarrow$ $\frac{1}{v_0}=\frac{1}{40}-\frac{1}{200}$ = $\frac{4}{200}$

$\Rightarrow$ v₀ = 50 cm

Tube length($l$) = Distance between lenses = v_o + f_e = 50 + 4 = 54 cm

For the first minima, the path difference between extreme waves

sin $\theta$ = $\frac{\lambda }{a}$

$\Rightarrow$ $\frac{1}{2}$ = $\frac{\lambda }{a}$

$\Rightarrow$ $a$ = 2$\lambda$

For first secondary maximum, the path difference between extreme waves

$\sin {\theta }^{\prime }=\frac{3\lambda }{2a}$ = $\frac{3}{2}\left(\frac{1}{2}\right)$

$\Rightarrow$ ${\theta }^{\prime }$ = ${\sin }^{-1}\left(\frac{3}{4}\right)$

Path difference S₂P – S₁P = $\sqrt{{\left(50\lambda \right)}^2+{\left(5\lambda \right)}^2}$ - 50$\lambda$ = 0.25$\lambda$

S₂P – S₁P = $\frac{\lambda }{4}$

Phase difference,

$\Delta$$\varphi$ = $\frac{2\pi }{\lambda }\times \frac{\lambda }{4}$ = $\frac{\pi }{2}$

So, resultant intensity at the desired point 'P' is

I = I₀cos²$\frac{\varphi }{2}$ = I₀cos²$\frac{\pi }{4}$ = $\frac{I_0}{2}$

At closest distance of approach, the kinetic energy of the particle will convert completely into electrostatic potential energy.

Kinetic energy = $\frac{1}{2}m{v}^2$

Potential energy = $\frac{KQq}{r}$

$\therefore$ $\frac{1}{2}m{v}^2$ = $\frac{KQq}{r}$

$\Rightarrow$ r $\propto$ $\frac{1}{m}$

The wave number of the last line of the Balmer series in hydrogen spectrum is given by

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

= $\frac{R}{4}$ = $\frac{{10}^7}{4}$ = 0.25 $\times$ 10⁷ m^$-$1

For electron De-Broglie wavelength,

$\lambda$_e = $\frac{h}{\sqrt{2mE}}$

For photon of energy, E = h$\nu$ = $\frac{hc}{{\lambda }_p}$

$\Rightarrow$ $\lambda$_p = $\frac{hc}{E}$

$\therefore$ $\frac{{\lambda }_e}{{\lambda }_p}$ = $\frac{h}{\sqrt{2mE}}\times \frac{E}{hc}$ = $\frac{1}{c}{\left(\frac{E}{2m}\right)}^{\frac{1}{2}}$

According to Einstein's photoelectric effect,

eV = $\frac{hc}{\lambda }$ - $\frac{hc}{{\lambda }_0}$ .......(1)

e$\frac{V}{4}$ = $\frac{hc}{2\lambda }$ - $\frac{hc}{{\lambda }_0}$ .......(2)

Dividing equation (i) by (ii) by,

4 = $\frac{\frac{1}{\lambda }-\frac{1}{{\lambda }_0}}{\frac{1}{2\lambda }-\frac{1}{{\lambda }_0}}$

$\Rightarrow$ $\lambda$₀ = 3$\lambda$

Here, the p-n junction diode is forward biased, hence it offers zero resistance.

So I_AB = $\frac{4-\left(-6\right)}{1\times {10}^3}$ = 10^-2 A

Voltage gain = Current gain × Resistance gain

= 0.96 $\times$ $\frac{800}{192}$ = 4

Power gain = [Current gain] × [Voltage gain]

= 0.96 × 4 = 3.84

Output of the circuit, Y = (A + B)·C

Y = 1 if C = 1 and A = 0, B = 1

or A = 1, B = 0 or A = B = 1

For any two electrons occupying the same orbital values of n, l and m_l are same but m_s is different [ + $\frac{1}{2}$ and - $\frac{1}{2}$]

Let the number of moles of each gas = x

Fraction of hydrogen escaped = $\frac{1}{2}x$

$\frac{r_{O_2}}{r_{H_2}}=\sqrt{\frac{M_{H_2}}{M_{O_2}}}$

$\Rightarrow \frac{n_{O_2}/t}{\frac{x}{2}/t}=\sqrt{\frac{2}{32}}=\sqrt{\frac{1}{16}}=\frac{1}{4}$

$\Rightarrow n_{O_2}=\frac{1}{8}x$

Then, fraction of oxygen escaped = $\frac{1}{8}$

This is Clausius-Clapeyron equation.

$\frac{d\ln P}{dT}=\frac{\Delta H_v}{R{T}^2}$

MY ⇌ M⁺ + Y^–
   s        s       s

K_sp = s.s = s²

$\Rightarrow$ s = $\sqrt{6.2\times {10}^{-13}}$

= 7.87 × 10^–7 mol L^–1

NY₃ ⇌ N⁺ + 3Y^–
   s        s       3s

K_sp = s.(3s)³ = 27s⁴

6.2 × 10^–13 = 27s⁴

$\Rightarrow$ s = 3.89 × 10^–4 mol L^–1

Hence, molar solubility of MY in water is less than that of NY₃.

Given that

w_s = 6.5 g, w_A = 100 g

p_s = 732 mm of Hg

k_b = 0.52, T^o_b = 100^oC

p^o = 760 mm of Hg

$\frac{p^o-p_s}{p^o}=\frac{n_2}{n_1}$

$\Rightarrow$ $\frac{760-732}{760}=\frac{n_2}{\frac{100}{18}}$

$\Rightarrow$ n₂ = 0.2046 mol

$\Delta$T_b = K_b × m

T_b - T^o_b = $k_b\times \frac{n_2\times 1000}{w_{A\left(g\right)}}$

$\Rightarrow$ T_b - 100^oC = $\frac{0.52\times 0.2046\times 1000}{100}$ = 1.06

$\Rightarrow$ T_b = 101.06 ^oC

p_Benzene = x_Benzene. p^o_Benzene

p_Toluene = x_Toluene. p^o_Toluene

For an ideal 1 : 1 molar mixture of benzene and toluene

x_Benzene = $\frac{1}{2}$ and x_Toluene = $\frac{1}{2}$

p_Benzene = $\frac{1}{2}$p^o_Benzene = $\frac{1}{2}\times 12.8$ = 6.4 kPa

p_Toluene = $\frac{1}{2}$p^o_Toluene = $\frac{1}{2}\times 3.85$ = 1.925 kPa

Thus, the vapour will contain a high percentage of benzene as the partial vapour pressure of benzene is higher as compared to that of toluene.

$\Delta G=\Delta H-T\Delta S$

If $\Delta$H < 0 and $\Delta$S > 0
   $\Delta G=(-ve)-T(+ve)$

then at all temperatures, $\Delta G$ = -ve, spontaneous reaction.

If $\Delta$H < 0 and $\Delta$S = 0
   $\Delta G=(-ve)-T(0)$

then at all temperatures, $\Delta G$ = -ve at all temperatures.

2H⁺ (aq) + 2e^– $\to$ H₂(g)

E_cell = E^o_cell - $\frac{0.0591}{2}\log \frac{p_{H_2}}{{\left[H^{+}\right]}^2}$

$\Rightarrow$ 0 = 0 - $\frac{0.0591}{2}\log \frac{p_{H_2}}{{\left[{10}^{-7}\right]}^2}$

$\Rightarrow$ $\frac{p_{H_2}}{{\left[{10}^{-7}\right]}^2}=1$

$\Rightarrow$ P_H2 = 10^–14 atm

For the first order reaction

A $\to$ Product

Rate $\propto$ [A]

k = $\frac{2.303}{t_2-t_1}\log \frac{{\left(rate\right)}_1}{{\left(rate\right)}_2}$

= $\frac{2.303}{20-10}\log \frac{0.04}{0.03}$

= 0.0287 sec^-1

$t_{\frac{1}{2}}=\frac{0.693}{k}=\frac{0.693}{0.0287}$ = 24.14 sec

A catalyst provides an alternate path to the reaction which has lower activation energy.