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For spontaneous process   $\Delta G<0$

For Isobaric process;    $\Delta P=0$

For Isothermal process;    $\Delta T=0$

$\Delta H$ reaction =  ${\displaystyle \sum _{}^{}B.E_{\left(R\right)}.-{{\displaystyle \sum _{}^{}B.E.}}_{\left(P\right)}}$

Kindly consider the following figure

Molality itself a strength representing terms for solution which does not depend upon the temperature.

Com

m1v + m2 × 0 = m1v1 + m2v2

75v = $45\frac{v}{3}+50\times 10$ $\frac{}{}$

$\Rightarrow 75\times \frac{2v}{3}=50\times 10$

$\Rightarrow v=\frac{3\times 50\times 10}{75\times 2}=10m/sec$

Here are some details about the placement salaries and top recruiters at the top Architecture colleges in India accepting NATA:

Disclaimer: This information is sourced from the official website and may vary.

Bernoulli’s equation between (1) & (2)

$\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v_2^2}{zg}+z_2$

$\frac{P_{atm}+mg/A}{\rho g}+0+40+10^{-2}$

$\frac{P_{atm}}{\rho g}+\frac{v_2^2}{2g}+0\left[v_1\approx 0\right]$

$\frac{mg}{A\rho g}+40\times 10^{-2}=\frac{v_2^2}{2g}$

$\frac{m}{A\rho }+40\times 10^{-2}=\frac{v_2^2}{2g}$

$\Rightarrow \frac{25}{0.5\times 10^3}+40\times 10^2=\frac{V_2^2}{2\times 10}$

$\Rightarrow V_2^2=\left(5\times 10^{-2}+40\times 10^{-2}\right)20$

$V_2^2=45\times 10^{-2}\times 20$

$V_2^2=90\times 10^{-1}$

$V_2=3m/sec$

$=300cm/sec$

Yes, students can compare the fees and placement salary at the top Architecture colleges in India accepting NATA, and determine the ROI, or the Return on Investment.

Disclaimer: This information is sourced from the official website and may vary.

Given, hT = 25m, R = 6400 km

hR = 49m

dm (Maximum distance for satisfactory communication)

dm = $\sqrt[]{2R{h}_T}+\sqrt[]{2R{h}_R}$ $\text{exception 25:}\text{exception 25:}$

dm = $\sqrt[]{2\times 6400\times 10^3\times 25}+\sqrt[]{2\times 6400\times 10^3\times 49}$ $\text{exception 25:}\text{exception 25:}$

= $\sqrt[]{5\times 10^4\times 6400}+\sqrt[]{6400\times 10^3\times 49\times 2}$ $\text{exception 25:}\text{exception 25:}$

= $192\sqrt[]{5}\times 10^2$ $\text{exception 25:}{}^{}$

= $k\sqrt[]{5}\times 10^2$ $\text{exception 25:}{}^{}$

k= 192

Yes, pursuing a degree at the top Architecture colleges in India accepting NATA can be expensive, as most of the colleges have an average tuition fee of more than INR 5 Lakh. Here is a further breakdown of the fees:

The Lund University student life is described as "hectic" by one student, and "easy to adapt" by another, as part of Lund University Student Reviews curated from Shiksha Study Abroad. 

Here are some key takeways from the student experience of those who studied at Lund University:

Given below are the top Architecture colleges in India that accept NATA and are ranked on top as per NIRF 2025:

Disclaimer: This information is sourced from the Architecture and Planning section of the official NIRF website.

Following are some examples of positive degree:

·       The test is easy.

·       The dress is expensive.

·       This ride is deadly.

·       My cook is as good as yours.

There are around 20+ top Architecture colleges in India that accept NATA, out of which 16 colleges are government-owned and 10 colleges are owned by private entities. 

Here is the list of books students can refer to, for studying positive degree in English:

Avoid using comparative or superlative forms. Common errors include using "more" or "most" with positive degree adjectives, or using "-er" or "-est" endings on adjectives that are already in the positive degree, can lead to errors.

Avoid using the positive degree when a comparative or superlative is more appropriate. For instance, when comparing two things, use the comparative degree. When comparing three or more things, use the superlative degree.

J (current density) = 4 × 10⁶ Am^-2

Area between radial distance  $\frac{R}{2}toR$

A =  $\pi \left[R^2-\frac{R^2}{4}\right]=\frac{3R^2}{4}\pi$

I = AJ

=  $\frac{3R^2}{4}\pi \times 4\times 10^6$

= 3R² × 10⁶ πAmp

= 3 (4 × 10^-3)² × 10⁶ πAmp

= 3 × 16 × 10^-6 × 10⁶π Amp

= 48πA