26 Jun 2022 Shift 2

$\because U=\;\frac{1}{2} M i^2$ \Rightarrow [M]=\;\frac{[U]}{[i^2]\}=;\frac{M L^2 T^{-2}}{A^2}$$=[M L^2 T^{-2} A^{-2}]$

From virtual work done method,$4 T \times a_1+2 T \times a_2+T \times a_3+T \times a_4=0$ $\Rightarrow 4 a_1+2 a_2+a_3+a_4=0$

${\text{W}}_{ {\text{a}}}=0, {\text{W}}_{ {\text{b}}}=+ {\text{ve}}, {\text{W}}_{ {\text{c}}}=+ {\text{ve}}> {\text{W}}_{ {\text{b}}}, {\text{W}}_{ {\text{d}}}=- {\text{ve}}$ $\Rightarrow {\text{W}}_{ {\text{c}}}> {\text{W}}_{ {\text{b}}}> {\text{W}}_{ {\text{a}}}> {\text{W}}_{ {\text{d}}}$ $\Rightarrow {\text{W}}_3> {\text{W}}_2> {\text{W}}_1> {\text{W}}_4$

$K E_R=\;\frac{1}{2} l w^2$ $=\;\frac{1}{2} \times \;\frac{2}{5} \times \omega ^2 \times (m R^2)$ $K E_{\;\text{total }\;}=\;\frac{1}{2} \times \;\frac{7}{5} \times m R^2 \times \omega ^2$ $\therefore \;{K E_R}/{K E_{\;\text{total }\;}}=\;\frac{2}{7}$

$g^{'}=g_0-\omega ^2 R \cos ^2 \theta$ $\theta=$ latitude.

$R_e=\;\frac{\rho v d}{η}$Direct formula based.

$K E_{a v g}=\;\frac{3}{2} k T$ (At lower temperature)As temperature is same for both the gases.$\Rightarrow$ Both gases will have same average kinetic energy.$\Rightarrow \;{(K E_{\;\text{avg }\;})_{\;\text{arg on }\;}}/{(K E_{\;\text{avg }\;})_{\;\text{oxygen }\;}}=\;\frac{1}{1}$

$C_{e q}=\;\frac{120}{26} \mu F$ $\Rightarrow Q_{\;\text{flown }\;} \;\text{ or }\; Q=\;\frac{13 \times 120}{26} \mu C=60 \mu C$ $\Rightarrow \;\text{ Charge on }\; 15 \mu {\text{F}} \;\text{ capacitor }\;=60 \mu {\text{C}}$As all the capacitors are in series.

Now, $\;\frac{\epsilon_0 A}{d}=4 \mu F$ $\Rightarrow \;\frac{12}{8} \;\frac{\epsilon_0 A}{d}=6 \mu F$

$q^{'}=64 q . . . . . . \;\text{ (i) }\;$ $A^{'}=16 A . . .$. (ii)Dividing (i) & (ii),$\sigma^{'}=4 \sigma$ $\Rightarrow \;\frac{\sigma^{'}}{\sigma}=\;\frac{4}{1}$

Initially $5 Ω$ and $5 Ω$ are in series and then in parallel with $10 Ω$ this pattern continues thus ${\text{R}}_{\;\text{net }\;}=5 Ω$

$U=-\vec{M} \cdot \vec{B}$So $U_f-U_i=-M B(1-\cos \theta)$ $=-14 J$So $W=-\Delta U=14 J$

Self inductances are in series but their mutual inductances are linked oppositely so equivalent self inductance$L=L_1+L_2-M-M=L_1+L_2-2 M$

$E m f=\;\frac{1}{2} B \omega l^2$ $=\;\frac{1}{2} \times 0.2 \times 10^{-4} \times 5 \times 1^2 {\text{V}}$ $=0.5 \times 10^{-4} {\text{V}}$ $=50 \mu {\text{V}}$

Wavelength of microwave is maximum then visible light then X-rays and then gamma rays so the correct order will be$\lambda _{\;\text{\gamma-ray }\;}< \lambda _{\;\text{Xray }\;}< \lambda _{\;\text{visible }\;}< \lambda _{\;\text{microwave }\;}$

$\lambda _{o b s}=\lambda _{\;\text{source }\;} \sqrt{\;\frac{1+\frac{v}{C}}{1-\;\frac{v}{C}}}$For $v< < C$,$\;\frac{670.7}{670}=1+\;\frac{v}{C}$ $\Rightarrow v=\;\frac{0.7}{670} \times 3 \times 10^8 {\text{m}} / {\text{s}}$ $\Rightarrow v ≃ 3.13 \times 10^5 {\text{m}} / {\text{s}}$

$\;\frac{h c}{\lambda }-\varphi =K E . . . . . . . .( {\text{i}})$ $R=\;\frac{m v}{B q}=\;\frac{\sqrt{2 m(K E)}}{B q} . . . . . .(ii)$Putting the values,$\varphi ≃ 1.36 {\text{eV}}$

$\;\frac{d A}{d t}-(-\lambda _1 A)+(-\lambda _2 A)$ $\Rightarrow \;\frac{d A}{d t}=-(\lambda _1+\lambda _2) A$ $\Rightarrow \lambda _{e f f}=\lambda _1+\lambda _2$ $\Rightarrow \;\frac{\ln 2}{(t_{1 / 2})_{e f f}}=\;\frac{\ln 2}{(t_{1 / 2})_1}+\;\frac{\ln 2}{(t_{1 / 2})_2}$ $\Rightarrow (t_{1 / 2})_{e f f}=\;\frac{4.5 \times 3}{7.5}$ hours $=1.8$ hours

When the amplifier connects with positive feedback, it acts as the oscillator the feedback here is positive feedback which means some amount of voltage is given to the input.

Modulate signal $s(t) \equiv [1+20 \sin (1000 \pi t)] \sin (10^7 \pi t)$ $\equiv \sin (10^7 \pi t)+10 \cos (10^7 \pi t-10^3 \pi t)+10 \cos (10^7 \pi t+10^3 \pi t)$ $\Rightarrow$ Required amplitude $=10$

Let they meet at ${\text{t}}= {\text{t}}$So first ball gets $t$ sec.$\& 2^{\;\text{nd }\;}$ gets $( {\text{t}}-2)$ sec. & they will meet at sameheight${\text{h}}_1=50 {\text{t}}-\;\frac{1}{2} {\text{gt}}^2$ ${\text{h}}_2=50( {\text{t}}-2)-\;\frac{1}{2} {\text{g}}( {\text{t}}-2)^2$ ${\text{h}}_1= {\text{h}}_2$ $50 {\text{t}}-\;\frac{1}{2} {\text{gt}}^2=50( {\text{t}}-2)-\;\frac{1}{2} {\text{g}}( {\text{t}}-2)^2$ $100=\;\frac{1}{2} g[t^{2-}(t-2)^2]$ $100=\;\frac{10}{2}[4 t-4]$ $5= {\text{t}}-1$ $t=6 {\sec} \;\text{. }\;$

Impulse $=$ change in momentum$= {\ m}[ {\ v}-(- {\ v})]=2 {\ mv}$ $=2 \times 0.4 \times 15=12 {\ Ns}$

At given instant$a_{\;\text{sys }\;}=\;\frac{6 m \times g}{10 m}=\;\frac{6 g}{10}$ $\therefore T_{78}=(3 m) \times a_{\;\text{sys }\;}$ $=(3 m) \times (\;\frac{6 g}{10})$ $=\;\frac{3 \times 2 \times 6 \times 10}{10}=36 {\text{N}}$

$Q=m s \Delta T$ $\;\frac{d Q}{d t}=(\;\frac{d m}{d t})_{\;\text{water }\;} S \Delta T=(\;\frac{d m}{d t})_{\;\text{oil }\;} C$ $\Rightarrow 2 \times 4.2 \times 10^3 \times 40=(\;\frac{d m}{d t})_{o i l} \times 8 \times 10^6$ $\Rightarrow (\;\frac{d m}{d t})_{o i l}=\;\frac{8 \times 4.2 \times 10^4}{8 \times 10^6} {\text{kg}} / {\text{minute}}$ $=42 {\text{g}} / {\min}$

$(1-\;\frac{324}{T_H})=\;\frac{300-180}{300}$ $1-\;\frac{2}{5}=\;\frac{324}{T_H}$ $T_H=\;\frac{324 \times 5}{3}=540$

Given $v_{20}=2 v_1$Also $v_{20}=4 \times 19+v_1$So $v_{20}=152 {\text{Hz}}$

$\;\frac{d F}{d l}=\;\frac{\mu _0 i_1 i_2}{2 \pi d}$So $\;\frac{2 \times 10^{-7} \times 5 \times 5}{d}=\;\frac{10^{-5}}{10 \times 10^{-2}}$ $d=\;\frac{2 \times 10^{-7} \times 5 \times 5}{10^{-4}}$ $=50 {\text{mm}}$ $=5 {\text{cm}}$

So $r=h \;\frac{\sin i_c}{\sqrt{1-\sin ^2 i_c}}$So $A=\pi r^2$ $=\;\frac{\pi h^2 \sin ^2 i_c}{1-\sin ^2 i_c}$ $=\;\frac{\pi 7 \times \frac{9}{16}}{1-\;\frac{9}{16}}=\;\frac{\pi \times 7 \times 9}{7}=9 \pi$

$40 {\text{M}}=1 {\text{cm}}$ $\Rightarrow {\text{M}}=0.025 {\text{cm}} . . . . . . . . .(1)$Also, $50 {\text{V}}=49 {\text{M}}$ $\Rightarrow$ Least count $= {\text{M}}- {\text{V}}= {\text{M}}-\;\frac{49}{50} {\text{M}}=\;\frac{M}{50}$ $\Rightarrow {\text{LC}}=\;\frac{0.025}{50} {\text{cm}}$ $=\;\frac{250}{50} \times 10^{-6} {\text{m}}$ $\Rightarrow {\text{LC}}=5 \times 10^{-6} {\text{m}}$

$\;\frac{h c}{\lambda }-\varphi =K E=e V_0$ $\Rightarrow \;\frac{6.63 \times 10^{-34} \times 3 \times 10^8}{6630 \times 10^{-10}}-6.63 \times 10^{-34} f_{t h}=1.6 \times 10^{-19} \times 0.4$ $\Rightarrow f_{t h} ≃ 35.11 \times 10^{13} H$

We know,Radial nodes $= {\text{n}}- {\text{l}}-1$and Angular nodes $=1$For 4d orbital,$n=4$ $I=2$ $\therefore$ Radial nodes $=4-2-1=1$Angular nodes $=2$

(A) Invertase $\to$ Cane sugar into glucose(B) Zymase $\to$ Glucose into ethanol(C) Diastase $\to$ Starch into maltose(D) Maltase $\to$ Maltose into glucose

${\text{Sc}}, {\text{Pb}}, {\text{Bi}}$ are metalsTe is a metalloid

Depressant prevent one component from coming to the froth.For eg., in Galena ore, the depressant (NaCN) prevents impurity $( {\text{ZnS}})$ from coming to the froth.

${\text{Ca}}( {\text{HCO}}_3)_2 {\longrightarrow }^{\Delta } {\text{CaCO}}_3 ↓+ {\text{H}}_2 {\text{O}}+ {\text{CO}}_2 ↑$ ${\text{Mg}}( {\text{HCO}}_3)_2 {\longrightarrow }^{ {\text{A}}} {\text{Mg}}( {\text{OH}})_2+2 {\text{CO}}_2 ↑$

Beryllium does not give flame test because of its small size and high ionization energy the energy of flame is not sufficient to excite the electrons to higher energy level.

The oxide of nitrogen which contains odd electron is ${\text{NO}}_2$

${\text{MnO}}_4^{2-}$ is an intermediate oxidation state and is converted into compounds having higher and lower oxidation states.

${\text{Ce}} \to [ {\text{Xe}}] 4 f^1 5 d^1 6 s^2$ ${\text{Ce}}^{+4} \to [x {\text{e}}] 4 {\text{f}}^{\circ} 5 {\text{d}}^{\circ} 6 {\text{s}}^{\circ}$Cerium in $+4$ oxidation state acquires inert gas configuration.

Highly polluted water should have BOD value of 17 ppm or more

$\;{ {\text{NH}_3}}_{\;\text{ Acid }\;} \longrightarrow \;{ {\text{NH}_2^{-}}}_{\;\text{ Conjugate base }}+ {\text{H}}^{+}$ Conjugate base of acid is always a stronger nucleophile.

Oxymercuration-DemercurationAddition of ${\text{H}}_2 {\text{O}}$Markovnikov's addition without rearrangement

Electrophile will attack at ortho and para position with respect to better electron releasing group (ERG)Para position with respect to $-\mathrm{OH}(+\mathrm{R})$ group and it will be meta position with respect to $-\mathrm{CH}_3$ group.

\text{CH}_3\;\text{CH}_2\;-\text{C}_{||}_{O}-\text{CH}_3 {\to }^{\text{HCN}} \text{CH}_3\text{CH}_2-{\text{C}}^{|\;\;}^{\text{CN}}_{|\;\;}_{\text{OH}}-\text{CH}_3{\to }^{95% \; \text{H}_2\text{SO}_4}_{\text{Heat}} CH_3-CH=C^{|\;\;}^{\text{CH}_3}-COOH

Soaps are sodium or potassium salts of long-chain fatty acids (carboxylic acids), having the general structure $R-COO^{-}Na^{+}$. Synthetic detergents are cleansing agents that are not soaps. They are broadly classified into anionic, cationic, and non-ionic detergents.

Let's analyze the given options:

• A. Sodium lauryl sulphate is an anionic detergent.
• B. ${\text{CH}}_3-( {\\text{CH}}_2)_{16}- {\\text{COO}}^{-} {\\text{Na}}^{+}$ is Sodium Stearate. This is the sodium salt of a long-chain fatty acid (stearic acid), which by definition is a soap.
• C. Cetyltrimethylammonium bromide is a cationic detergent.
• D. This is a polyoxyethylene ester, which is a type of non-ionic detergent.

Since soap is not a synthetic detergent, Sodium Stearate is the correct answer.

B−complex (vitamins B group) is a water-soluble vitamin. Also, vitamin C is a water-soluble vitamin.Vitamin A, D, E and K are fat soluble vitamins.

{}_{\text{Initially}:}_{}_{}_{\text{finally}:}{{\text{CH}}_4}_{6.25}_{}_{6.25-6.5/2}+2 {{\text{O}}_2}_{6.5} _{}_{0}\longrightarrow {\text{CO}}_2_{0}_{}_{6.5/2}+2 {{\text{H}}_2 {\text{O}}}_{0}_{}_{6.5} Initially,Mass of ${\text{CNG}}( {\text{CH}}_4)=100 {\text{gm}}$ $\therefore$ Moles of ${\text{CH}}_4=\;\frac{100}{16}=6.25$ molesMass of ${\text{O}}_2=208 {\text{gm}}$ $\therefore$ Moles of ${\text{O}}_2=\;\frac{208}{32}=6.5$ molesFrom reaction you can see,1 mole of ${\text{CH}}_4$ react with 2 mole of ${\text{O}}_2$ $\therefore 6.25$ mole of ${\text{CH}}_4$ react with $6.25 \times 2=12.5$ moles of ${\text{O}}_2$But here only $6.5$ moles of ${\text{O}}_2$ presentSo, ${\text{O}}_2$ will act as limiting reagent.$\therefore$ Produced ${\text{CO}}_2$ will depends on moles of ${\text{O}}_2$.From 2 moles of ${\text{O}}_2 1$ mole of ${\text{CO}}_2$ $\therefore$ From $6.5$ moles of ${\text{O}}_2 \;\frac{6.5}{2}$ moles of ${\text{CO}}_2$ produced.$\therefore$ Weight of ${\text{CO}}_2=\;\frac{6.5}{2} \times 44=143 {\text{gm}}$

$A \to 4-(2 \times \;\frac{1}{2})=3$ ${\text{B}} \to 12 \times \;\frac{1}{4}+1 \times 1=4$So, Compound is ${\text{A}}_3 {\text{B}}_4$The value of $x$ is 3 .

Number of lone pair on central atom for ${\text{H}}_2 {\text{O}}$ and ${\text{XeF}}_4$ is equal to 2.

${\text{H}}_2 {\text{O}}(l) \to {\text{H}}_2 {\text{O}}( {\text{g}})$ ${\text{n}}=\;\frac{36}{18}=2 {\text{mol}}$ $\Delta {\text{U}}=\Delta {\text{H}}-\Delta {\text{n}}_{ {\text{g}}} {\text{RT}}$ $=41.1-\;\frac{1 \times 8.31 \times 373}{1000} {\text{kJ}} / {\text{mol}}$ $=38 {\text{kJ}} / {\text{mol}}$