6 Apr 2023 Shift 1

According to De-Broglie, Momentum $P=\;\frac{h}{\lambda }$, where $h$ is plank's constant and $\lambda$ is wavelength.Also, relation between Kinetic energy $( {\text{KE}})$ and momentum( $(P)$ is given by: $P=\sqrt{2 {\text{mKE}}}$Now, $\lambda =\;{ {\text{h}}}/{ {\text{p}}}=\;{ {\text{h}}}/{\sqrt{2 {\text{mKE}}}}$ $\;\lambda _{ {\text{e}}}=\;{ {\text{h}}}/{\sqrt{2 {\text{m}}_{ {\text{e}}} {\text{KE}}_{ {\text{e}}}}}=\;{ {\text{h}}}/{\sqrt{2 {\text{m}}_{ {\text{e}}} \times 4 {\text{k}}}}=\;{ {\text{h}}}/{\sqrt{8 {\text{m}}_{ {\text{e}}} {\text{k}}}}$ $\;\lambda _{ {\text{p}}}=\;{ {\text{h}}}/{\sqrt{2 {\text{m}}_{ {\text{p}}} {\text{KE}}_{ {\text{p}}}}}=\;{ {\text{h}}}/{\sqrt{2 {\text{m}}_{ {\text{p}}} {\text{k}}}}$ $\;\lambda _alpha=\;{ {\text{h}}}/{\sqrt{2 {\text{m}}_alpha {\text{KE}}_alpha}}=\;{ {\text{h}}}/{\sqrt{2 {\text{m}}_alpha \cdot 4 {\text{k}}}}=\;{ {\text{h}}}/{\sqrt{2 \times 2 {\text{m}}_{ {\text{p}}} \cdot 2 {\text{k}}}}=\;{ {\text{h}}}/{\sqrt{8 {\text{m}}_{ {\text{p}}} {\text{k}}}}$From the above data, $\lambda _alpha < \lambda _p < \lambda _e$

$V_{e s c}=\sqrt{\;\frac{2 G M}{r}}={\sqrt{2 g r}}$Radius of moon is less than that of earth and acceleration due to gravity is also less on moon as compared to that on earth.Thus, $V_{\;\text{esc }\;}$ of Moon $< V_{\;\text{esc }\;}$ of EarthThis is also the reason behind escape of atmosphere from moon.

From Ist law of thermodynamics,$d Q=d U+d W$Also, we can write this as, $\;\frac{d Q}{d t}=\;\frac{d U}{d t}+\;\frac{d W}{d t}$ $\;\Rightarrow 1000 W=\;\frac{d U}{d t}+200 {\text{W}}$ $\;\Rightarrow \;\frac{d U}{d t}=800 W$

At terminal velocity, net force on the ball is Zero.$\;M g=f+B$ $\;\Rightarrow M g=f+V_{\;\text{ball }\;} \rho _o g \;\; . . .(i)$ $\;\;\text{ Volume of ball }\;=\;\frac{M}{\rho }$ $\;\;\text{ From eq (i), }\;$ $\;M g=f+\;\frac{M}{\rho } \rho _o g$ $\;\Rightarrow f=M g-\;\frac{M}{\rho } \rho _o g$ $\;\Rightarrow f=M g (1-\;\frac{\rho _o}{\rho })$

$\;k x=m \omega ^2 r$ $\;\Rightarrow k x=0.1 \times 25 \times (0.2+x)$ $\;\Rightarrow 7.5 x=2.5(0.2+x)$ $\;\Rightarrow 3 x=0.2+x$ $\;\Rightarrow 2 x=0.2$ $\;\Rightarrow x=0.1 {\text{m}}$Now, tension in the spring $=k x=7.5 \times 0.1 {\text{N}}=0.75 {\text{N}}$

V_A=v {𝑗}^{\^}And V_B=-v {𝑖}^{\^}Time to reach from ${\text{A}}$ to ${\text{B}}=\;\frac{2 \pi R}{4} \times \;\frac{1}{v}=\;\frac{\pi R}{2 v}$Displacement from ${\text{A}}$ to ${\text{B}}=R {\sqrt{2}}$Now, Average velocity from ${\text{A}}$ to ${\text{B}}=\;\frac{\;\text{ Displacement }\;}{\;\text{ Time }\;}=\;{R {\sqrt{2}}}/{\frac{\pi R}{2 v}}=\;{2 {\sqrt{2}} v}/{\pi}$Instantaneous velocity at $B$ is -v {𝑖}^{\^}According to question, $\;\frac{\;\text{ instantaneous velocity }\;}{\;\text{ average velocity }\;}=\;{\pi}/{x {\sqrt{2}}}$ $\;\;{v}/{{2 {\sqrt{2}} v}/{\pi}}=\;{\pi}/{x {\sqrt{2}}}$ $\;\Rightarrow \;{\pi}/{2 {\sqrt{2}}}=\;{\pi}/{x {\sqrt{2}}}$ $\;\Rightarrow x=2$

In parallel combination, $\;\frac{1}{R_{e q}}=\;\frac{1}{R_1}+\;\frac{1}{R_2}$ $\Rightarrow \;\frac{1}{R_{e q}}=\;\frac{1}{10}+\;\frac{1}{15}=\;\frac{5}{30}=\;\frac{1}{6}$Now, for error calculation,$\;\;\frac{d R_{e q}}{R_{e q}^2}=\;\frac{d R_1}{R_1^2}+\;\frac{d R_2}{R_2^2}$ $\;\Rightarrow \;\frac{d R_{e q}}{36}=\;\frac{0.5}{100}+\;\frac{0.5}{225}$ $\;d R_{e q}=36 \times 0.5 \times (\;\frac{13}{900}) =18 \times \;\frac{13}{900}=\;\frac{26}{100}=0.26$Now, $\;\frac{d R_{e q}}{R_{e q}} \times 100=\;\frac{0.26}{6} \times 100=\;\frac{26}{6}=4.33$

For $r \le R, V=\;\frac{K Q}{R}$, i.e., Constant everywhere inside.For $r > R, V=\;\frac{K Q}{r}$, i.e., Decreases with ${\text{r}}$.

It is a case of solid infinite current carrying wire. Using ampere circuital law,CASE I: if ${\text{r}} \le {\text{R}}$ $B=\;\frac{\mu _0 i}{2 \pi R^2} r$CASE II: $r > {\text{R}}$ $B=\;\frac{\mu _0 i}{2 \pi r}$

New range is given by ${\sqrt{2 R(h+0.21 h)}}$ $\;={\sqrt{2 R h \times 1.21}}$ $\;=1.1 {\sqrt{2 R h}}$It means new range increases by $10 \%$.

Collision frequency is given by $Z=n \pi d^2 V_{\;\text{avg }\;}$, where ${\text{n}}$ is number of molecules per unit volume.$\;\frac{z_1}{z_2}=\;\frac{n_1}{n_2}=\;\frac{3}{12}=\;\frac{1}{4}=0.25$

$\;E=\;\frac{h c}{\lambda }$ $\;\Rightarrow \lambda =\;\frac{h c}{E}$For $\lambda _{\min }$, E must be maximum. And ${\text{E}}$ is maximum for ${\text{D}}$.

In EM waves, average electric energy density is equal to average magnetic energy density.$\;\frac{1}{4} E_0 E_0^2=\;\frac{1}{4 \mu _0} B_0^2$

Average Density of planet $=$ average density of earth$\;\;\frac{M_e}{\frac{4}{3} \pi R_e^3}=\;\frac{M_p}{\frac{4}{3} \pi R_p^3}$ $\;\Rightarrow \;\frac{M_e}{R_e^3}=\;\frac{2 M_e}{R_p^3}$ $\;\Rightarrow R_p=2^{\;\frac{1}{3}} R_e-------(i)$ $\;\;\text{ Now, }\; g=\;\frac{G M}{R^2}$ $\;\;\frac{g_e}{g_p}=\;\frac{M_e}{R_e^2} \times \;\frac{R_p^2}{2 M_e}=2^{\;\frac{2}{3}-1}=2^{-\;\frac{1}{3}}$ $\;\Rightarrow g_p=2^{\;\frac{1}{3}} g_e$ $\;\Rightarrow W_p=2^{\;\frac{1}{3}} W_e$

A semiconductor starts conduction more as the temperature increases. It means resistance decreases with increase in temperature. So, if temperature increases, its resistivity decreases.Also, $\rho =\;\frac{m}{n e^2 \tau }$As Temperature increase, $\tau$ decreases but ${\text{n}}$ increases and ${\text{n}}$ is dominant over $\tau$.

$\;1 \times sin \;45=\mu sin \;30$ $\;\Rightarrow \;{1}/{{\sqrt{2}}}=\mu \times \;\frac{1}{2}$ $\;\Rightarrow \mu ={\sqrt{2}}----(i)$ $\;\;\text{ Now, }\; \;\frac{\mu _1}{\mu _2}=\;\frac{V_2}{V_1}=\;\frac{\lambda _2}{\lambda _1}----(i i)$ $\;\;\text{ Using eq (i) and (ii), }\;$ $\;\lambda _2=\;{1}/{{\sqrt{2}}} \lambda _1$ $\;\;\text{ And }\; V_2=\;{1}/{{\sqrt{2}}} V_1$Now, for relation between frequencies,Frequency, $v=\;\frac{v}{\lambda }$ $\; {\text{Or}} \;\frac{v_1}{v_2}=\;\frac{v_1}{v_2} \times \;\frac{\lambda _2}{\lambda _1}=1$ $\;v_1=v_2$

Both the springs are in parallel.$\;K_{e q}=K_1+K_2$ $\;T=2 \pi \sqrt{\;\frac{m}{K_{e q}}}=2 \pi \sqrt{\;\frac{m}{K_1+K_2}}$

NOR gate

Induced emf can be induced in a coil by changing magnetic flux.And $\varphi =\vec{B} \cdot \vec{d A}$By rotating coil, angle between coil and magnetic field changes and hence flux changes.By changing area, magnetic flux changes.

For a ground to ground projectile, Horizontal range is given by $R=\;\frac{u^2 sin \;2 \theta}{g}$ And for $R_{\max }, sin \;2 \theta$ must be maximum.

\;{B_{\;\text{net }\;}}^{\to }=\;\frac{\mu _0 i}{2 r} {𝑖}^{\^}+\;\frac{\mu _0 i}{2 r} {𝑗}^{\^} $\;\Rightarrow B_{\;\text{net }\;}=\;\frac{\mu _0 i}{2 r} {\sqrt{2}}=4 \pi \times 10^{-7} \times {\sqrt{2}} \times {\sqrt{2}} \times \;\frac{1}{2 \times 0.2}=2 \times 3.14 \times 10^{-6}=628 \times 10^{-8} {\text{T}}$

$\;Y=\;\frac{\;\text{ stress }\;}{\;\text{ strain }\;}$ $\;\Rightarrow \;\text{ strain }\;=\;\frac{\;\text{ stress }\;}{Y}=\;\frac{F}{A Y}=\;\frac{62.8 \times 1000}{\pi r^2 \times 2 \times 10^{11}}=\;\frac{62.8 \times 1000}{3.14 \times 400 \times 10^{-6} \times 2 \times 10^{11}}=\;\frac{200}{8} \times 10^{-5}=25 \times 10^{-5}$

$\;R=\;\frac{\rho l}{A}$ $\;R^{'}=\;\frac{\rho \times 1.2 l}{0.96 A}=\;\frac{10}{8} \times R=1.25 R$It means $25 \%$ increase in Resistance.

$r_n=\;\frac{0.51 n^2}{z} A^0$For ${\text{Li}}^{++}, z=3$.So $r_5=0.51 \times \;\frac{25}{3} \times 10^{-10} {\text{m}}=17 \times 25 \times 10^{-12} {\text{m}}=425 \times 10^{-12} {\text{m}}$

$\;\;\text{ Given, }\; a=-2 x$ $\;\Rightarrow \;\frac{v d v}{d x}=-2 x$ $\;\Rightarrow v d v=-2 x \text{dx}$ $\;\Rightarrow int_{v_1}^{}{v_2} v d v=-2 int_{0}^{x} x \text{dx}$ $\;\Rightarrow \;\frac{v_2^2}{2}-\;\frac{v_1^2}{2}=-\;\frac{2 x^2}{2}$ $\;\Rightarrow \;\frac{m v_1^2}{2}-\;\frac{m v_2^2}{2}=m x^2=\;\frac{10}{1000} x^2=10^{-2} x^2= (\;\frac{10}{x}) ^{-2}$ $\; {\text{n}}=2 .$

$\;\;\frac{V_s}{V_p}=\;\frac{N_s}{N_p}$ $\;\Rightarrow \;\frac{120}{12000}=\;\frac{N_s}{N_p}$ $\;\Rightarrow \;\frac{N_s}{N_p}=\;\frac{1}{100}---(i)$For an ideal transformer, input power $=$ Output power And power is given by $P=i V$ $\;i_p V_p=i_s V_s=60000 {\text{W}}$ $\;i_p=\;\frac{60000}{12000}=5$Now, $R_p=\;\frac{V_p}{i_p}=\;\frac{12000}{5}=2400 Ω$ $R_s=\;\frac{V_s}{i_s}=\;\frac{120}{60000 / 120}=120 \times \;\frac{120}{60000}=\;\frac{120}{500}=0.240 Ω=240 {\text{m}} Ω$

$C_1=$ $\;{\;\frac{E_0 A}{\frac{2 d}{3}} \times \frac{4 E_0 A}{\frac{d}{3}}}/{\frac{E_0 A}{2 d / 3}+\frac{4 E_0 A}{d / 3}}$ $=\;\frac{18}{\frac{3}{2}+12} \;\frac{E_0 A}{d}=18 \times \;\frac{2}{27} \;\frac{E_0 A}{d}=\;\frac{4}{3} \;\frac{E_0 A}{d}$ According to ${\text{qn}}, \;\frac{4}{3} \;\frac{E_0 A}{d}=2 \Rightarrow \;\frac{E_0 A}{d}=\;\frac{3}{2}----(i)$Now, $C_2=$ $\;{\;\frac{E_0 A}{\frac{d}{3}} \times \frac{4 E_0 A}{\frac{2 d}{3}}}/{\frac{E_0 A}{d / 3}+\frac{4 E_0 A}{2 d / 3}}$ $=\;\frac{18}{3+6} \;\frac{E_0 A}{d}=2 \times \;\frac{E_0 A}{d}=2 \times \;\frac{3}{2}=3$

Using parallel axix theorem,$\;I_{\;\text{sys }\;}= (\;\frac{2}{5} m r^{2+}m d^2) \times 2$ $\;\Rightarrow I_{\;\text{sys }\;}= (\;\frac{2}{5} \times 2 \times 0.01+2 \times 0.04) \times 2=(0.008+0.08) \times 2=0.088 \times 2=176 \times 10^{-3}$

$\;f^{'}=f_0+40$ $\;\Rightarrow f_0 (\;\frac{330+15}{330-15}) =f_0+40$ $\;\Rightarrow f_0 \times \;\frac{345}{315}=f_0+40$ $\;\Rightarrow f_0 \times \;\frac{30}{315}=40$ $\;\Rightarrow f_0=40 \times \;\frac{315}{30}=420 {\text{Hz}}$

$\; sin \;60=\;\frac{4}{3} sin \;r$ $\;\Rightarrow sin \;r=\;\frac{3}{4} \times \;{{\sqrt{3}}}/{2}=\;{3 {\sqrt{3}}}/{8}---(i)$ $\;\cos r=\sqrt{1-\;\frac{27}{64}}=\;{{\sqrt{37}}}/{8}=0.75$ $\;\Rightarrow {\text{tanr}}=\sqrt{\;\frac{27}{37}}$ $\;\Rightarrow \;\frac{x}{1.5}=0.85$ $\;\Rightarrow x=0.85 \times 1.5=1.275 {\text{m}}$ $\;\tan 30=\;\frac{y}{2.15-1.275}=\;\frac{y}{0.875}$ $\;y=\;\frac{0.875}{1.732}=0.50$So length of pole above water surface $=0.50 {\text{m}}=50 {\text{cm}}$

${\text{Y}}: {\text{CCP}} \Rightarrow 4 {\text{Y}}$ ${\text{X}}=1 / 3 {\text{THV}}=1 / 3 \times 8 \Rightarrow 8 / 3 {\text{x}}$ $\therefore \;\text{ Formula }\;: {\text{X}}_{8 / 3} {\text{Y}}_4 \;\text{ or }\; {\text{X}}_2 {\text{Y}}_3$

Nitrogen detection by lassaigne's method ${\text{Na}}+ {\text{C}}+ {\text{N}} \to {\text{NaCN}}$ $6 {\text{NaCN}}+ {\text{FeSO}}_4 \to {\text{Na}}_4 [ {\text{Fe}}( {\text{CN}})_6] + {\text{Na}}_2 {\text{SO}}_4$ {\text{Na}}_4 [ {\text{Fe}}( {\text{CN}})_6] + {\text{Fe}}^{3+} \to {\text{Fe}}_4 [ {\text{Fe}}( {\text{CN}})_6] _3_{(\text{Prussian} \text{blue})}Sulphur detection by Sodium nitroprusside ${\text{Na}}_2 [ {\text{Fe}}( {\text{CN}})_5 {\text{NO}}] + {\text{Na}}_2 {\text{S}} \to {\text{Na}}_4 [ {\text{Fe}}( {\text{CN}})_5 {\text{NOS}}] _{[\text{Purple}]}$Phosphorus detection by ammonium molybdate ${\text{Na}}_3 {\text{PO}}_4+3 {\text{HNO}}_3 \to {\text{H}}_3 {\text{PO}}_4+3 {\text{NaNO}}_3$ ${\text{H}}_3 {\text{PO}}_4+12 ( {\text{NH}}_4) _2 {\text{MoO}}_4+21 {\text{HNO}}_3 \to$ ${( {\text{NH}}_4) _3 {\text{PO}}_4 \cdot 12 {\text{MoO}}_3}_{ (\text{canary} \text{yellow}) }+21 {\text{NH}}_4 {\text{NO}}_3+12 {\text{H}}_2 {\text{O}}$ Halogen give specific coloured ppt with ${\text{AgNO}}_3( {\text{aq}})$ {\text{NaCl}}+ {\text{AgNO}}_3( {\text{aq}}) \to {\text{AgCl}}+ {\text{NaNO}}_3 _{( {\text{White}}) } {\text{NaBr}}+ {\text{AgNO}}_3( {\text{aq}}) \to {\text{AgBr}}+ {\text{NaNO}}_3 _{( {\text{Pale}} \;\text{ yellow }\;) } {\text{NaI}}+ {\text{AgNO}}_3( {\text{aq}}) \to \;\text{ AgI }\;+ {\text{NaNO}}_3 _{ (\;\text{ Yellow }\;)}

The standard electrode potential ($E^\circ$) of the ${\text{M}}^{+} / {\text{M}}$ couple describes the overall process: ${\text{M}}(s) \rightarrow {\text{M}}^{+}(aq) + e^{-}$. This process can be analyzed using a Born-Haber cycle, which breaks it down into distinct thermodynamic steps. These steps are the sublimation of the solid metal ({\text{M}}(s) \rightarrow {\text{M}}(g)$), the ionization of the resulting gaseous atom ({\text{M}}(g) \rightarrow {\text{M}}^{+}(g) + e^{-}$), and the hydration of the gaseous ion ({\text{M}}^{+}(g) \rightarrow {\text{M}}^{+}(aq)$). The electrode potential is a function of the enthalpy changes of these three steps. Ionization energy is specifically defined as the energy required to remove an electron from an isolated gaseous atom, not a solid one. Therefore, the "ionization of a solid metal atom" is not a defined step in this cycle.

${\text{Cl}}$ has the most negative $\Delta {\text{H}}_{ {\text{eg}}}$ among all the elements and ${\text{Ne}}$ has the most positive $\Delta {\text{H}}_{ {\text{eg}}}$

This question requires matching specific enzymatic reactions with the correct enzyme catalyst. Let's analyze each pair:

1. Sucrose $\to$ Glucose and Fructose: This hydrolysis is catalyzed by the enzyme Invertase. So, A matches with III.
2. Glucose $\to$ ethyl alcohol and ${\text{CO}}_2$: This fermentation process is carried out by the enzyme complex Zymase, found in yeast. So, B matches with I.
3. Starch $\to$ Maltose: The breakdown of starch into the disaccharide maltose is catalyzed by Diastase. So, C matches with IV.
4. Proteins $\to$ Amino acids: The digestion of proteins into amino acids is initiated by proteases like Pepsin in the stomach. So, D matches with II.

Based on these pairings, the correct combination is A-III, B-I, C-IV, and D-II.

Photochemical smog occurs in warm, dry and sunny climate. The main components come from the action of sunlight on unsaturated hydrocarbon and nitrogen oxides produced by automobiles and factories.

The setting of cement is a hydration process. One of the main constituents of cement, tricalcium aluminate ($C_3A$), reacts very rapidly with water, which can cause the cement to set almost instantly. This phenomenon is called a "flash set," making the cement unworkable. To control this rapid reaction, Gypsum (calcium sulfate dihydrate, $CaSO_4 \cdot 2H_2O$) is added during the final grinding of cement clinker. Gypsum reacts with $C_3A$ to form an insoluble layer of ettringite, which coats the $C_3A$ particles. This layer slows down the hydration process, thereby increasing the initial setting time and allowing for proper mixing, transport, and placement.

The Assertion (A) is correct. When a hydrogen atom loses its single electron, only the nucleus, which is a proton, remains. The size of a proton is on the order of $10^{-15}$ m. The given size, $1.5 \times 10^{-3}$ pm, converts to $1.5 \times 10^{-3} \times 10^{-12}$ m = $1.5 \times 10^{-15}$ m, which is a physically correct order of magnitude for a proton's size.

The Reason (R) is also correct. A bare proton ($H^+$) has an extremely high charge density due to its small size. This makes it highly unstable and reactive, so it immediately combines with surrounding molecules (like water to form $H_3O^+$) and does not exist independently.

However, the Reason (R) explains the chemical reactivity and existence of a proton, which is a consequence of its small size and high charge. It does not explain why the proton has that specific physical size as stated in the Assertion (A). Therefore, both statements are correct, but R is not the correct explanation of A.

${\text{HVZ}}$ reactions $= {\text{Br}}_2 /$ red ${\text{P}}$Iodoform reaction $= {\text{NaOH}}+ {\text{I}}_2$Etard reaction $=$ (i) ${\text{CrO}}_2 {\text{Cl}}_2, {\text{CS}}_2$ (ii) ${\text{H}}_2 {\text{O}}$Gatterman-Koch Reaction $= {\text{CO}}, {\text{HCl}}$, Anhydrous, ${\text{AlCl}}_3$

$[ {\text{Fe}}( {\text{CN}})_6] ^{3-}$Unpaired electron $=1$ $\mu =\sqrt{ {\text{n}}( {\text{n}}+2)}={\sqrt{1 \times 3}}=1.74 \;\text{ B.M. }\;$ $[ {\text{Fe}} ( {\text{H}}_2 {\text{O}}) _6] ^{3+}$ No pairing because ${\text{H}}_2 {\text{O}}$ is WFLNumber of unpaired electrons $=5, \mu =5.92 {\text{BM}}$Assertion is true, Reason is true but not correct explanation.

This question requires matching vitamins with their respective deficiency diseases. Let's analyze each pair.

• Vitamin A is essential for vision. Its severe deficiency leads to Xerophthalmia, a condition where the eye fails to produce tears, resulting in dryness and hardening of the cornea. Thus, A matches with III.
• Thiamine is Vitamin B1. A deficiency in thiamine causes Beri-Beri, a disease that affects the nervous system and cardiovascular system. Thus, B matches with I.
• Ascorbic acid is Vitamin C. Its deficiency causes Scurvy, characterized by bleeding gums, weakness, and anemia. Thus, C matches with IV.
• Riboflavin is Vitamin B2. Its deficiency can lead to Cheilosis, which involves inflammation and cracking at the corners of the mouth. Thus, D matches with II.
  Combining these gives the sequence A-III, B-I, C-IV, D-II.