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$\mathrm{}\tau \propto \frac{1}{\sqrt[]{\mathrm{T}}}$

$A=A_0{e}^{-\lambda t}$

$\begin{array}{rr}& 500=700{\mathrm{e}}^{-\lambda \mathrm{t}}\\ & \lambda \mathrm{t}=\mathrm{l}\mathrm{n}?\frac{7}{5}\\ & \frac{\mathrm{l}\mathrm{n}?2}{{\mathrm{t}}_{1/2}}\times 30=\mathrm{l}\mathrm{n}?\frac{7}{5}\\ & ?\mathrm{}{\mathrm{t}}_{1/2}=\frac{\mathrm{l}\mathrm{n}?2\times 30}{\mathrm{l}\mathrm{n}?\frac{7}{5}}\end{array}$

${\mathrm{t}}_{1/2}=61.8\mathrm{m}\mathrm{i}\mathrm{n}\approx 62\mathrm{m}\mathrm{i}\mathrm{n}$ .

For $2^{\text{nd}}$ ${}^{}$ minima

$\mathrm{d}\mathrm{s}\mathrm{i}\mathrm{n}?\theta =2\lambda$

$\frac{\text{exception 25:}}{}$ $\mathrm{s}\mathrm{i}\mathrm{n}?\theta =\frac{\sqrt[]{3}}{2}$   (given)

$\Rightarrow \mathrm{}\frac{\lambda }{\mathrm{d}}=\frac{\sqrt[]{3}}{4}$

So for $1^{\text{st}}$ ${}^{}$ minima is

$\mathrm{d}\mathrm{s}\mathrm{i}\mathrm{n}?\theta =\lambda$

$\frac{}{}$ $\mathrm{s}\mathrm{i}\mathrm{n}?\theta =\frac{\lambda }{d}=\frac{\sqrt[]{3}}{4}$   (from equation (i)

$$ $\theta ={25.65}^{?}\mathrm{}$   (from sin table)

${}^{}$  $\theta \approx {25}^{?}$ .

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Magnitude of electric field is constant & the surface is equipotential.

  $\mathrm{}I=I_{0}t-I_0{t}^2$ $\mathrm{}{}_{}{}_{}{}^{}$

$?=\mathrm{B}\mathrm{A}$

$?={\mu }_0\mathrm{n}\mathrm{I}\mathrm{A}$

$V_R=\frac{d?}{dt}=-{\mu }_0{\mathrm{n}\mathrm{A}\mathrm{l}}_0(1-2\mathrm{t})$

${}_{}$ $V_R=0$ at $\frac{}{}$  $t=\frac{1}{2}$ and $I_R=\frac{V_R}{\text{Resistance of loop}}$ ${}_{}\frac{{}_{}}{}$

$\frac{\text{Energy}}{\text{Volume}}=\frac{{\mathrm{M}\mathrm{L}}^2{\mathrm{T}}^{-2}}{{\mathrm{L}}^3}=\left({\mathrm{M}\mathrm{L}}^{-1}{\mathrm{T}}^{-2}\right)$

The distance between two successive bright fringes is fringe width  $\left(\beta \right)$ .

$\beta =\frac{\lambda D}{d}=\frac{589\times {10}^{-9}\times 1.5}{0.15\times {10}^{-3}}=5.9\mathrm{m}\mathrm{m}$

Relation between De-Broglie wavelength and K.E. is

$\lambda =\frac{\mathrm{h}}{\sqrt[]{2\left(\mathrm{K}\mathrm{E}\right){\mathrm{m}}_{\mathrm{e}}}}\Rightarrow \lambda \propto \frac{1}{\sqrt[]{\mathrm{K}\mathrm{E}}}$

$\begin{array}{rr}& \frac{{\lambda }_{\mathrm{A}}}{{\lambda }_{\mathrm{B}}}=\frac{\sqrt[]{{\mathrm{K}\mathrm{E}}_{\mathrm{B}}}}{\sqrt[]{\mathrm{K}\mathrm{E}}}\Rightarrow \frac{1}{2}=\sqrt[]{\frac{{\mathrm{T}}_{\mathrm{A}}-1.5}{{\mathrm{T}}_{\mathrm{A}}}}\\ & \Rightarrow {\mathrm{T}}_{\mathrm{A}}2\mathrm{e}\mathrm{V}\\ & \therefore {\mathrm{K}\mathrm{E}}_{\mathrm{B}}=2-1.5=0.5\mathrm{e}\mathrm{V}\\ & {?}_{\mathrm{B}}=4.5-0.5=4\mathrm{e}\mathrm{V}\end{array}$

$\mathrm{}\stackrel{?}{E}=\frac{\sigma }{2{\epsilon }_0}\mathrm{c}\mathrm{o}\mathrm{s}?{60}^{?}(-\hat{x})+\left. [\frac{\sigma }{2{\epsilon }_0}-\frac{\sigma }{2{\epsilon }_0}\mathrm{s}\mathrm{i}\mathrm{n}?{60}^{?}\right]\left(\hat{y}\right)$

$\stackrel{?}{\mathrm{E}}=\frac{\sigma }{2{\epsilon }_0}\left. [\left(1-\frac{\sqrt[]{3}}{2}\right)\stackrel{\mathrm{ˆ}}{\mathrm{y}}-\frac{1}{2}\stackrel{\mathrm{ˆ}}{\mathrm{x}}\right]$

Magnetic energy per unit volume $=\frac{{\mathrm{B}}^2}{2{\mu }_0}$

The BIT MCA 2026 exam date has not been announced yet. In 2025, the exam was conducted on May 31, 2025; hence, for 2026, the exam is expected to be conducted in May 2026 only. The entrance examination was held as a proctored-based online test for all registered candidates. For admission into the MCA courses offered by the institutes, the candidates had to qualify for the entrance examination. More details about BIT MCA 2026 will be announced on the official website.

$\mathrm{}I=I_0{\mathrm{c}\mathrm{o}\mathrm{s}}^2?\theta$

$\frac{{\mathrm{I}}_0}{10}={\mathrm{I}}_0{\mathrm{c}\mathrm{o}\mathrm{s}}^2?\theta$

$\mathrm{c}\mathrm{o}\mathrm{s}?\theta =\frac{1}{\sqrt[]{10}}=0.31<\frac{1}{\sqrt[]{2}}$ which is 0.707

So, $\theta >{45}^{?}$ ${}^{}$ and $90-\theta <{45}^{?}$ ${}^{}$ so only one option is correct i.e. ${18.4}^{?}$ ${}^{}$

Angle rotated should be $={90}^{?}-{71.6}^{?}={18.4}^{?}$ ${}^{}{}^{}{}^{}$ .

BIT Mesra has released the placements for the batch passing out in 2025. The key highlights of BIT Mesra placements for the Class of 2025 are tabulated below:

$\mathrm{}{W}_{\text{equator}}=W_{\text{pole}}-m{\omega }^{2}R$

$\begin{array}{rr}& =196-19.6\times {\left(\frac{2\pi }{24\times 60\times 60}\right)}^2\times 6400\times {10}^3\\ & =195.33\mathrm{N}\end{array}$

$\left(\mathrm{0,3}\right)$

$\begin{array}{rr}& {\stackrel{?}{r}}_{\mathrm{c}\mathrm{m}}=\frac{1\times \left(\frac{\hat{i}}{2}+\hat{j}\right)+1\times \left(\hat{i}+\frac{5\hat{j}}{2}\right)}{2}\\ & {\stackrel{?}{r}}_{\mathrm{c}\mathrm{m}}=\frac{3}{4}\hat{i}+\frac{7}{4}\hat{j}\end{array}$

$\left(\mathrm{0,0}\right)$

BIT Mesra has a good record of placements for courses like BTech, MTech, MBA, MCA, etc. The the highest and average package offered during BIT Mesra placements stood at INR 145 LPA and INR 11.77 LPA, respectively. The key highlights of BIT Mesra placements for the Class of 2025 are tabulated below:

BIT Mesra is a well-known institute in Gorakhpur for offering placement opportunities to its Engineering, Management and Pharma students. The placement percentage is yet to be released on the official website. Meanwhile, check out the key highlights of BIT Mesra placements for the Class of 2025 in the table below: