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  $\frac{a+20}{2}=3+7r$

a + 20 = 6 + 14r     . (i)

b = -2 + 10r           . (ii)

a = 18r – 2              . (iii)

Solving (i) and (iii) we get

20 + 18r – 2 = 6 + 14r

r = -3

$\therefore$   a = 14 + 14 (-3) = -56 and b = -2 -30 = -32

$\left|a+b\right|\left|-56-32\right|=88$

According to the KUK Pharmacy NIRF report 2025, 11 BPharma students were placed out of the total 99 graduating students during KUK BPharma placements 2024. The total number of students who graduated and were placed during Kurukshetra University placements between 2023 and 2024 is presented below:

According to the official website of Asian College of Journalism, the ACJ admissions are conducted in two rounds: online test and Personal Interview. The second round of the ACJ admission process will be conducted in online mode. Candidates who qualify the first round will be eligible to appear for the interview round. Candidates should prepare for the interview well in advance.

A -> m_A = 1 kg, P_A = P

B ->  m_B = 2 kg,   P_B = P

$?kE=\frac{P^2}{2m}$

$\therefore \frac{k{E}_A}{k{E}_B}=\frac{\left(\frac{P^2}{2mA}\right)}{\left(\frac{P^2}{2m_B}\right)}=\frac{m_B}{m_A}=\frac{2}{1}$

According to the KUK Engineering NIRF report,  the median package offered to the BPharma batch 2024 was INR 1.97 LPA. The median packages offered to BPharma students during KUK placement between 2023 and 2024 are presented below:

The examination will be held online in real time in two slots (Morning for Section I and Afternoon for Section II). Candidates applying for either or both of the programmes will be intimated, within a week of receipt of the completed application, the date and time of the online test, and date and time of the online interview to follow.

Check out the key highlights of Kurukshetra University placements for the BPharma Class of 2024 from the table below:

Note: The above-mentioned data has been taken from the KUK Pharmacy NIRF report 2025.

You must check your computer and internet connection in advance before the exam starts. If you are unable to appear for exam on the prescribed date and time, the Asian College of Journalism will not give another chance.

To access the 2025 YCMOU result, students must submit their YCMOU result PRN number. Students must submit the YCMOU result prn number on the YCMOU result portal. For other details, please visit the official website of the college. 

x – 2y = 1, x – y + kz = -2, ky + 4z = 6

 x – 2y + 0. z – 1 = 0

x – y + kz + 2 = 0

0x + ky + 4z – 6 = 0

0x + ky + 4z – 6 = 0

${\Delta }_1=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -2\hfill & \hfill 0\hfill \\ \hfill -2\hfill & \hfill -1\hfill & \hfill k\hfill \\ \hfill 6\hfill & \hfill k\hfill & \hfill 4\hfill \end{array}\right|=-\left(k+10\right)\left(k+2\right)$   

For no solution

$\Delta =0,{\Delta }_1\ne 0$       

k = 2

$7.2\times \frac{5}{18}=2m/s$ [V_A = V_B = 2m/s]

$f{\text{'}}_B=f_A\left(\frac{v+v_B}{v-v_A}\right)$

$\downarrow$     

App freq heard by car B

$f{\text{'}}_B=676\left[\frac{340+2}{340-2}\right]=676\times \frac{342}{338}=684Hz$  

Similarly, $f{\text{'}}_A=f_B\left(\frac{v+v_B}{v-v_B}\right)=684Hz$

$\therefore$ Beat frequency heard by both = 684 – 676 = 8Hz

radius r = 0.5 mm

$\sigma =5\times 10^{7}s/m$

$\overrightarrow{E}=10\times 10^{-3}V/m=10^{-2}V/m$

As we know

$J=\sigma E$

$J=\frac{i}{A}=\sigma E$

$i=\sigma EA=5\times 10^7\times 10^{-2}\times \pi \times {\left(0.5\times 10^{-3}\right)}^2=1.25\pi \times 10^{-1}A=125\pi mA=x^3\pi mA$

$$ $\therefore$  x = 5

Candidates applying for ACJ entrance exam for admission to PG Diploma in Journalism and PG Diploma in Business and Financial Journalism can pay the application fee in any of the two modes:

• Online Transfer via online payment gateway (Credit Card/ Debit Card/ Net Banking)
• Offline Transfer via Demand Draft in favour of “Media Development Foundation”, payable at Chennai

Candidates sending fee via DD must ensure that their full names, email IDs, and mobile numbers are mentioned.

f = 3GHz = 3 × 10⁹ Hz

${\lambda }_{vacuum}=\frac{v}{f}=\frac{3\times 10^{8}m/s}{3\times 10^9/s}=10^{-1}m=0.1m$

${\lambda }_{vacuum}=\frac{{\lambda }_{vacuum}}{{\mu }_{medium}}=\frac{0.1}{{\mu }_{medium}}$

${\mu }_{medium}=\sqrt[]{{\mu }_r{\in }_r}=\sqrt[]{1\times 2.25}=1.5$

$\therefore {\mu }_{medium}=\frac{0.1m}{1.5}=0.0667m=6.67cm=667\times 10^{-2}cm$

Sound level is given in dB = 10 log $\frac{{}_{}}{{}_{}}\frac{}{{}_{}}$ $\left(\frac{P_0}{P_i}\right)or10log\left(\frac{l}{l_0}\right)$

As sound level decreases 5 dB every km,

So, in 20 km sound level will decrease by 20 × 5 = 100 dB.

$\Delta \beta ={\beta }_2-{\beta }_1=10log\left(\frac{l_2}{l_1}\right)$

$-100=10log\left(\frac{l_2}{l_1}\right)$

$10^{-10}=\frac{l_2}{l_1}\Rightarrow l_2=10^{-10}{l}_1$

$P_2=10^{-10}{P}_1$

$P_2=10^{-10}\times \left(0.1\times 10^3\right)\Rightarrow P_2=10^{-8}W=10^{-x}W$

$$ $\therefore$  x = 8

No, the ACJ entrance exam registration for academic year 2026-2027 is not yet open. Once released, candidates can fill out and submit ACJ online form on official website.

$l={\displaystyle \underset{1}{\overset{3}{\int }}\left[x^2-2x+1-3\right]dx=}{\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2-3\right]dx={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-3{\displaystyle \underset{1}{\overset{3}{\int }}dx}}}$  .(A)

$l_1={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dxPut{\left(x-1\right)}^2=t}$

$l_1=\frac{1}{2}\left[0{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt[]{t}}}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{dt}{\sqrt[]{t}}}+2{\displaystyle \underset{2}{\overset{3}{\int }}\frac{dt}{\sqrt[]{t}}}+3{\displaystyle \underset{3}{\overset{4}{\int }}\frac{dt}{\sqrt[]{t}}}\right]=\frac{1}{2}\left\{{\left|\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_1^2{\left|+2\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_2^3{+3\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}]}_3^4\right\}$      = $5-\sqrt[]{2}-\sqrt[]{3}$  

Hence from (A)

= $5-\sqrt[]{2}-\sqrt[]{3}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

2^nd method       

${\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-6..............\left(A\right)}$

From (A), $l=5-\sqrt[]{2}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$