Ask & Answer: India's Largest Education Community

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = $\frac{1}{2\pi \sqrt[]{LC}}=\frac{1}{2\pi \sqrt[]{\left(0.5\times 10^{-3}\right)\times \left(200\times 10^{-6}\right)}}$

$\Rightarrow f=\frac{10^4}{2\pi \sqrt[]{10}}\approx 5\times 10^{2}Hz$ $\left[Taking\pi =\sqrt[]{10}\right]$

According to Ellingham diagram, the metal oxide with lower $\Delta G^o$ is more stable than metal oxide of higher $\Delta G^o$ at same temperature.

According to Young’s double slit experiment, we can write

$\beta =\frac{\lambda D}{d}\Rightarrow \Delta \beta ={\beta }_2-{\beta }_1=\frac{\lambda }{d}\Delta D$

$\Rightarrow \lambda =\frac{d\Delta \beta }{\Delta D}=\frac{1\times 10^{-3}\times 3\times 10^{-5}}{5\times 10^{-2}}=60\times 10^{-8}m=600nm$

In boron family lower O.S (+1) is more stabilize down the group due to inert pair effect.

According to Nuclear activity, we can write

$N_0\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{2}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{4}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{8}\stackrel{t_{\frac{1}{2}}}{\to }\frac{N_0}{16}=\left(0.0625\right)N_0$         

Time required = 4 × $t_{\frac{1}{2}}=20yrs$  

$\frac{dE}{dT}$ is the temperature Co-efficient of cell. The cell having less variation of EMF, with respect to temperature have high efficiency.

According to question, we can write

$\mathcal{l}=2\pi r\Rightarrow r=\frac{\mathcal{l}}{2\pi }$

$I_1=\frac{m{\mathcal{l}}^2}{3},and{I}_2=\frac{m{r}^2}{2}=\frac{m{\mathcal{l}}^2}{8{\pi }^2}$

$\Rightarrow \frac{I_1}{I_2}=\frac{m{\mathcal{l}}^2}{3}\times \frac{8{\pi }^2}{m{\mathcal{l}}^2}=\frac{8{\pi }^2}{3}$

According to definition of displacement current, we can write

$I_d={\epsilon }_0\frac{d?}{dt}={\epsilon }_0\frac{d\left(ES\right)}{dt}={\epsilon }_0\frac{d\left(\frac{V}{\mathcal{l}}S\right)}{dt}=\frac{{\epsilon }_{0}S}{\mathcal{l}}\left(\frac{dV}{dt}\right)$

$\Rightarrow \mathcal{l}=\frac{8.85\times 10^{-12}\times 40\times 10^{-4}\times 10^6}{4.425\times 10^{-6}}8\times 10^{-3}m$

$Bondorder=\frac{e^{-}ofBondingMO-e^{-}ofABMO}{2}$  

Hence Bond order for  $O_2^{+}=\frac{10-5}{2}=2.5$  

$O_2=\frac{10-6}{2}=2.0$

$O_2^{-}=\frac{10-7}{2}=1.5$

$O_2^{--}=\frac{10-8}{2}=1.0$  

According to Work energy theorem, we can write

$K_f-K_i=W_{ElectricForce}\Rightarrow \frac{1}{2}m{v}^2-\frac{1}{2}m{v}_0^2=-eV\Rightarrow v^2=v_0^2-\frac{2eV}{m}$

$\Rightarrow v^2={\left(6.0\times 10^5\right)}^2-\frac{2\times 1.6\times 10^{-19}}{9\times 10^{-31}}=\frac{324-128}{9}\times 10^{10}=\frac{196}{9}\times 10^{10}\Rightarrow V=\frac{14}{3}\times 10^{5}m/s$

According to Kirchhoff’s Law, we can write

-20 + 2000I + 600 × 5I = 0 $\Rightarrow I=\frac{20}{5000}A$  

Reading of voltmeter = 2000I = 2000 × $\frac{20}{5000}=8volt$  

Radius of Bohr’s orbit $R_n=0.529\stackrel{o}{A}\times \frac{n^2}{z}$  

Radius of Bohr’s orbit for hydrogen,   $R_n=0.529\stackrel{o}{A}\times n^2$

For third orbit (R₃) = $0.529\stackrel{o}{A}\times 9$ = r₃ and

Fourth orbit (R₄) = $0.59\stackrel{o}{A}\times 16=r_4$

$\therefore r_4=\frac{16r_3}{9}$  

Wt of liquid = 135 – 40 = 95 gm

Volume of liquid =  $\frac{wt.}{density}=\frac{95}{0.95}=100ml$  

Hence volume of vessel = 100 ml = 0.1 lit from ideal gas equation,

$M=\frac{dRT}{P}=\frac{0.5}{0.1}\times \frac{0.0821\times 250}{0.82}$  

$\therefore M=125g/mol$  

Brunel University MSc in Management course duration is one year. The programme is accredited by AACSB. MSc in Management focuses on management principles and leadership skills. The skills consist of problem solving, decision-making and how to make an effective communication which will also help students who want to explore other industries.

$Molality=\frac{molesofsolute\times 1000}{wtofsolvent\left(gm\right)}$  

$Molality=\frac{35\times 1000\times 1.46}{36.5\times 100}$ = 14.0 M

According to Concept of resonance tube, we can write

$\frac{\lambda }{4}+e={\mathcal{l}}_{1}and\frac{3\lambda }{4}+e={\mathcal{l}}_2\Rightarrow \frac{\lambda }{2}={\mathcal{l}}_2-{\mathcal{l}}_1\Rightarrow \frac{V}{2\nu }={\mathcal{l}}_2-{\mathcal{l}}_1$

$\Rightarrow {\mathcal{l}}_2=\frac{v}{2\nu }+{\mathcal{l}}_1=\frac{336}{400}+0.20=0.84+0.20=1.04m=104cm$

Since velocity does not change, so acceleration will be zero.

mg = FB + Fv $\Rightarrow \frac{4\pi }{3}{r}^3\rho g=\frac{4\pi }{3}{r}^3\sigma g+6\pi \eta rv$

$\Rightarrow v=\frac{2r^2\left(\rho -\sigma \right)g}{9\eta }=\frac{2\times 0.1\times 0.1\times 10^{-6}\times \left(10^4-10^3\right)\times 10}{9\times 1.0\times 10^{-5}}$

$\Rightarrow h=\frac{400}{2g}=20m$

Least  count of Vernier = 0.1mm

Reading of Vernier Scale = 5 × 0.1 = 0.5mm

The corrected diameter of sphere = Main Scale Reading + Vernier Scale reading + Zero correction = 1.7 + 0.05 + 0.05 = 1.8cm = 180 × 10^-2 cm.

According to question, we can write

$R=\frac{u^{2}sin2\theta }{g}\Rightarrow u^2=g{R}_{max},where\theta =45°$

$\Rightarrow H_{max}=\frac{u^2}{2g}=\frac{g{R}_{max}}{2g}=\frac{R_{max}}{2}=\frac{100}{2}=50m$

$9^n-8n-1={\left(1+8\right)}^n-8n-1$  

$=(1+8n{+}^n\text{?}{C}_2\cdot 8^2{+}^n\text{?}{C}_3\cdot 8^3+....]-8n-1$  

So, a = ⁿC₂ + ⁿC₃8 + ⁿC₄8² +….

Similarly, b = ⁿC₂ + ⁿC₃ 5 + ⁿC₄ 5² +….

a - b = ⁿC₃ (8 – 5) + ⁿC₄ (8² – 5²) +….