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From F.B. Ds of

$9kg\to 90-T_1=9a\dots .\left(1\right)$

$6kg\to T_1-T_2=6a\dots ..\left(2\right)$

$5kg\to T_2-50=5a\dots ..\left(3\right)$

$\left(1\right)+\left(2\right)+\left(3\right)$

$40=20a?a=2m/s^2$

Bulk modulus

$B=\frac{-\Delta P}{\Delta V/V}$

$B=-V\left[\frac{\Delta P}{\Delta V}\right]$

$B=-V\left[\frac{\partial P}{\partial V}\right]$

^{$B=-V\left[\frac{\partial P}{\partial V}\right]$}

$B=-V\left[\frac{-P}{V}\right]$

$\Rightarrow B=P$

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It ranks highly as part of the overall & course-specific rankings of several prominent ranking platforms. Given below is a description of the same:

· #33 for Bachelor of Arts program

· #43 for Business & Management Studies

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· #72 for Master's in Management program

· Top 100 best business schools in the UK

For minima

$d\cos ?\theta =\left(n+\frac{1}{2}\right)\lambda$

$?\cos ?\theta =\frac{2n+1}{8}$

$?n=-4,-3,-2,-1,0,1,2,3$

$?8$

minimas

Here  $\Delta V_c=0.5V,\Delta i_c=0.05mA=0.05\times 10^{-3}A$

Output resistance is given by

$R_{out}=\frac{\Delta V_c}{\Delta i_c}=\frac{0.5}{0.5\times 10^{-3}}=10^4\Omega =10k\Omega .$

From ampere’s circuital law

$?\stackrel{?}{B}.\stackrel{?}{dl}={\mu }_0{i}_{enc}$

$={\mu }_0\left(3+1\right)=4{\mu }_0$

The relation between $?$ and $?$ is given as

$?=\frac{?}{1-?}$

${?}_1=\frac{\frac{20}{21}}{1-\frac{20}{21}}=20$

${?}_2=\frac{\frac{100}{101}}{1-\frac{100}{101}}=100$

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Class 10, 10+2 mark sheet and certificate along with transfer and migration certificate

Caste category certificate (if applicable)

Aadhar card and domicile certificate

Anti-ragging form undertaking by parents and student

Recent passport size photograph with white background

Students must be dressed in formal having name & date on photograph (6 copies)

Coloured photos of Parents (4 copies in 3.5 cm x 4.5 cm format)

In a number without decimal, trailing zeros are insignificant. If a number has a decimal, then preceding zeros are insignificant.

Hence, in 40.40 all zeros are significant

m  =l0 kg

h= 5m

By work energy theorem

$mgh=\frac{1}{2}m{v}^2$

$v^2=2$ gh 

$v=\sqrt[]{2gh}$

$v=\sqrt[]{100}$

$v=10m/s$

$kx=mg-F_{up}$

$kx=mg-\frac{md}{D}g$

$x=\frac{mg}{k}\left(1-\frac{d}{D}\right)$

The authorities are yet to announce the class commencement date for SPSAT 2025 qualified candidates. The class commencement date will be announced on the official website. 

$i_b=\frac{5-0.7}{8.6}=0.5mA$

$\Rightarrow I_c=\beta I_b=100\times 0.5 mA$

By using 

$V_{CE}=V_{CC}-I_C{R}_L=18-50\times 10^{-3}\times 100=13 V$

To confirm the admission, the candidates will have to pay the required fee and verify their documents during the SPSAT 2025 counselling. 

The charge on hole is positive.

$\stackrel{?}{F}=e{E}_0\stackrel{?}{j}$

${\stackrel{?}{a}}_y=\frac{e{E}_0}{m}\stackrel{?}{j}$

$=v_0\stackrel{?}{i}+\left(\frac{e{E}_{0}t}{m}\right)\stackrel{?}{j}$

$\left|\stackrel{?}{v}\right|=\sqrt[]{v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}=v_0\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}$

$\lambda =\frac{{\lambda }_0}{\sqrt[]{1+{\left(\frac{e{E}_{0}t}{m}\right)}^2}}$

Two successive β decays increase the charge no. by 2.

VTUEEE counselling below procedure

Visit official website

Filling all details and arrange necessary

Paying fee to secure seat

Wait for seat allotment 

Report to allotted college for admission

No, the exam authority will not share the counselling call letter through email. Candidates will have to visit the official website to download the VTUEEE call letter.