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p + q = 2

 p⁴ + q⁴ = 272

${\left(p^2+q^2\right)}^2-2p^2{q}^2=272$          

  ${\left[{\left(p+q\right)}^2-2pq\right]}^2-2p^2{q}^2=272$          

Let pq = t Þ (4 – 2t)² – 2t² = 272

2t² – 16 t – 256 = 0

->t = pq = 16

Required equation x² – 2x + 16 = 0

$W=q\Delta V=20\times 15=300J$

Let the spring is in extension state and $a_B>a_A,{\overrightarrow{a}}_{AB}=-a_A\widehat{i}-\left(-a_B\widehat{i}\right)=\left(-a_A+a_B\right)\widehat{i}$ ${}_{}{}_{}{\stackrel{}{}}_{}{}_{}\stackrel{}{}{}_{}\stackrel{}{}{}_{}{}_{}\stackrel{}{}$

Hence we can say that block moves away from block B in the frame of B

F – kx = MaB . (1)

kx = Ma. (2)

$\Rightarrow a_B=\frac{F}{M}-a$

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D₁ is forward biased & D₂ is reverse biased.

$l=\frac{6}{120+130+50}=\frac{6}{300}=2\times 10^{-2}A=20\times 10^{-3}A=20mA$

A  (A -> B) -> B is a tautology

$\widehat{r}-\widehat{i}=2cos\theta \widehat{n}......\left(1\right)$

$\widehat{i}.\widehat{n}=-cos\theta ......\left(2\right)$

$\Rightarrow \widehat{r}=\widehat{i}-2\left(\widehat{i}.\widehat{n}\right)\widehat{n}$

$\Rightarrow \overrightarrow{b}=\overrightarrow{a}-2\left(\overrightarrow{a}.\overrightarrow{c}\right)\overrightarrow{c}$

$P=\frac{P_1{V}_1+P_2{V}_2}{V_1+V_2}=\frac{2\times 4.5+3\times 5.5}{10}=\frac{9+16.5}{10}=25.5\times 10^{-1}atm$

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{8-sin2x}}=asi{n}^{-1}\left(\frac{sinx+cosx}{b}\right)+c}$

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{9-{\left(sinx+cosx\right)}^2}}=}$  

Put (sin x + cos x) = t Þ (cos x – sin x) dx = dt

$={\displaystyle \int \frac{dt}{\sqrt[]{3^2-t^2}}=si{n}^{-1}\frac{t}{3}+c=si{n}^{-1}\left(\frac{sinx+cosx}{3}\right)+c}$           

= a = 1, b = 3

$\therefore \left(a,b\right)\equiv \left(1,3\right)$       

The VITMEE syllabus of chemical engineering is as follows -

• Law of thermodynamics
• Law of conservation of mass and energy
• Classification of fluids 
• Introduction to particulate solids 
• Fourier's law of heat conduction
• Differential and integral method of analysis of rate data
• Basic principles of unit operation and unit process 
• Diffusion in liquids
• Overview of industrial biochemical processes

$\beta =\frac{\lambda D}{d}=\frac{500\times 10^{-9}\times 1}{2\times 10^{-3}}=250\times 10^{-6}m=250\times 10^{-3}mm=0.25mm$

$N=mgcos30°+qEsin30°$

$a=\frac{mgsin\theta -qEcos\theta -\mu N}{m}=2.30m/s^2$

$S=ut+\frac{1}{2}a{t}^2$

$\Rightarrow t=\sqrt[]{\frac{2\mathcal{l}}{a}}=1.31sec$

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• National Overseas Scholarship for Students with Disabilities
• Scholarship for U.P. State Students
• National Fellowship for ST students
• Internship/Action research in Jharkhand Skill Development Mission
• Scholarship to eligible students of Jammu and Kashmir under the J&K Special Scholarship Scheme

Frequency increases on filing. So, initial frequency of A is 335 Hz.

f = 340 – 5 = 335 Hz

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$           

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$  

Obviously B (2, 2) satisfying condition (i)

From defined of bulk modulus

$K=-\frac{P}{\left(dV/V\right)}.......\left(1\right)$

m = $\rho v=constant$

$\rho dV+Vd\rho =0$

$\frac{dV}{V}=-\frac{d\rho }{\rho }.........\left(2\right)$

With the help of equations (1) and (2), we can write

$K=-\frac{P}{\left(-\frac{d\rho }{\rho }\right)}=\frac{\rho P}{d\rho }\Rightarrow d\rho =\frac{\rho P}{K}$

Central University of Jharkhand MBA fees is inclusive of various components such as tuition fees, caution deposits, etc. Moreover, students must note that only those who opt for hostel facilities are charged a hostel fee. Combining all, the total fees for the MBA programme amount to INR 1.62 lakh. Check the table below to learn the component-wise bifurcation of the total fees:

Note: The above-mentioned fee is as per the official sources. However, it is indicative and subject to change.

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

$Z=\sqrt[]{{\left(120\right)}^2+{\left(10-100\right)}^2}=150\Omega$

$\omega =\frac{1}{\sqrt[]{LC}}=\frac{1}{\sqrt[]{10^{-1}\times 10^{-4}}}=\sqrt[]{10^5}$

$?\omega =2\pi f$

$\Rightarrow f=\frac{10^3}{2\pi \sqrt[]{10}}=50Hz$