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${\omega }_0=10^{5}rad/sec$

P = 16w, 120v at resonance

$P=\frac{v^2}{R}\Rightarrow 16=\frac{{\left(120\right)}^2}{R}\Rightarrow R=\frac{14400}{16}=900\Omega$

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As we know that

$Y=\frac{FL}{A?L}$          

$?L=0.04m=\frac{FL}{AY}...............\left(i\right)$           

If length and diameter both are doubled

$?L\text{'}=\frac{F.2L}{4A.Y}=\frac{FL}{2Y}=0.02m=2cm$       

$l={\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)dx={\left[xf\left(x\right)\right]}_0^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx=2e^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}}}$        .(A)

Put   $l_1={\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}$            .(i)

Using properties ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$  

$l_1={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(2-x\right)dx={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(x\right)dx}}$    .(ii)

Adding (i) and (ii) we get

  $2l_1=2{\displaystyle \underset{0}{\overset{2}{\int }}f\text{'}\left(x\right)dx\Rightarrow l_1={\left[f\left(x\right)\right]}_0^2}$         

f(2) – f(0) = e² – 1

From (A) l = 2e² – e² + 1 = e² + 1

The ACJ exam pattern for PG Diploma in Business and Financial Jorunalism is given here.

Let m_AB = m = 1 kg

AB = 0.4 m =  $\mathcal{l}$

d = OM = $\sqrt[]{0.16-0.04}=\sqrt[]{0.12}$

$d=2\sqrt[]{3}\times 10^{-1}$  

$I_{ABaboutO}=\frac{m{\mathcal{l}}^2}{12}+m{d}^2$           

$\therefore l_{hexagon,aboutO}=6\left[\frac{m{\mathcal{l}}^2}{12}+m{d}^2\right]=\frac{m{\mathcal{l}}^2}{2}+6m{d}^2$       

$=\frac{1\times {\left(0.4\right)}^2}{2}+6\times 1\times {\left(2\sqrt[]{3}\times 10^{-1}\right)}^2=\frac{0.16}{2}+6\times 4\times 3\times 10^{-2}$   

= 0.08 + 0.72 = 0.8 kgm² = 8 × 10^-1 kgm²

Flux through the 6 sides of square (i.e. cube)

${?}_T=\frac{q_{in}}{{\epsilon }_0}=\frac{12\times 10^{-6}C}{{\epsilon }_0}$

$\therefore$ Flux through a square

$?=\frac{{?}_T}{6}=\frac{12\times 10^{-6}}{{\epsilon }_0\times 6}=\frac{2\times 10^{-6}}{{\epsilon }_0}$  

$\Rightarrow ?=225.98\times 10^{3}N{m}^2/C?226N{m}^2/C$

No, the Asian College of Journalism will declare the exam schedule. All the candidates must appear for the exam as per the schedule. No request regarding change in the schedule of examination will be entertained.

The ACJ exam pattern for PG Diploma in Journalism Course is given here.

• Sections: English Language and General Knowledge (including Current Affairs).
• Total Marks: 200
• Slots: Two (Morning and Afternoon)
• Mode of examination: Online
• Both sections will be for 100 marks each.
• Exam Duration: 2 hours (Section I) and 90 minutes (Section II)

$T=27°C,P=1atm=10^{5}N/m^2,V_{rms}=200m/s$

As we know

$v_{rms}=\sqrt[]{\frac{3RT}{M}}\Rightarrow \frac{v_{rms}}{v{\text{'}}_{rms}}=\frac{\sqrt[]{\frac{3RT}{M}}}{\sqrt[]{\frac{3RT\text{'}}{M}}}$

$V{\text{'}}_{rms}=200\times \frac{2}{\sqrt[]{3}}=\frac{x}{\sqrt[]{3}}$

x = 400

The ACJ question paper has two sections.

• Section I: English Language
• Section II: General Knowledge

Candidates will get 2 hours to complete Section I and 1.5 hours to complete Section II. 

$x_{cm}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

$=\frac{\sigma \pi a^2\times a-\frac{\sigma \pi a^2}{4}\times \frac{3a}{2}}{\sigma \pi a^2-\frac{\sigma \pi a^2}{4}}=\frac{\frac{5}{8}\sigma \pi a^3}{\frac{3}{4}\sigma \pi a^2}=\frac{5}{8}\sigma \pi a^2\times \frac{4}{3}\sigma \pi a^2=\frac{5}{6}a$

The Asian College of Journalism will not deduct marks if a student does not attempt a question. So, students can skip a question if they are not sure of the answer.

The total marks for ACJ entrance exam is 200 (100 marks each for Section I and II).

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The placement rate recorded for 2023 stood at 84.92%. The data tabulated below provides Siliguri Institute of Technology placement highlights for 2023:

Travelling wave is represented by $f\left(t\pm \frac{x}{v}\right)$ $\frac{}{}$

Thus, y = A sin (15x – 2t) represents a travelling wave

The highest package recorded during Siliguri Institute of Technology placement season 2023 was INR 51 LPA, which was provided by Microsoft. Check the table below to know more about the top packages provided by each company:

To calculate your score using the CPGET answer key, compare your answers with the correct ones given in the key. For every correct answer, add 1 mark. Since CPGET exam pattern includes no negative marking, you do not need to deduct marks for wrong answers. The total will be your expected score in the exam.