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Note that D^r < 0, hence given inequality (1) is true only if $N^r\ge 0$  

  i.e.   $\left(x-8\right)\left(x-2\right)\le 0and{2}^{x-3}>31$

i.e.    $2\le x\le 8and{2}^{x-3}\ge 31$

only x = 8 satisfies both the inequality.

$Q_1=C_{1}V=2V\mu C$

$Q_2=Q_3=\frac{C_2{C}_3}{C_2+C_3}V=\frac{6\times 12}{6+12}V=4V\mu C$

$Q_1:Q_2:Q_3=1:2:2$

No, candidates will not get laptop or desktop to appear in the exam. They need to appear in the exam using their own laptop or desktop.

$\boxed{\begin{array}{lll}\text{12.(D)}& {\int }_0^{\frac{\pi }{6}}??\frac{4{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \frac{2}{5}\times 2\mathrm{c}\mathrm{o}\mathrm{s}?\theta d\theta }{\left(4-4{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \right)}=\frac{16}{5\times 8}\int \frac{{\mathrm{s}\mathrm{i}\mathrm{n}}^2?\theta \mathrm{c}\mathrm{o}\mathrm{s}?d\theta }{{\mathrm{c}\mathrm{o}\mathrm{s}}^3?\theta }& x^{5/2}=2\mathrm{s}\mathrm{i}\mathrm{n}?\theta \\ & & \frac{5}{2}{x}^{3/2}dx=2\mathrm{c}\mathrm{o}\mathrm{s}?\theta d\theta \\ & & =\frac{2}{5}{\int }_0^{\frac{\pi }{6}}??{\mathrm{t}\mathrm{a}\mathrm{n}}^2?\theta d\theta =\frac{2}{5}{\int }_0^{\frac{\pi }{6}}??\left({\mathrm{s}\mathrm{e}\mathrm{c}}^2?\theta -1\right)d\theta =\frac{2}{5}[\mathrm{t}\mathrm{a}\mathrm{n}?\theta -\theta {]}_0^{\pi /6}=\frac{2}{5}\left(\frac{1}{\sqrt[]{3}}-\frac{\pi }{6}\right)\end{array}}$

Let us consider an elementary ring of radius r and thickness dr in which current is flowing.

So, No. of turns in this elementary ring

$dN=\left(\frac{N}{b-a}\right)dr$

$\therefore {\left(dB\right)}_{atcentre}=\frac{{\mu }_{0}ldN}{2r}$

$B=\frac{{\mu }_{0}lN}{2\left(b-a\right)}\mathcal{l}n\frac{b}{a}$

Diode, in forward biased condition only, will allow current to flow through it.

Pot. different across resistor is

$\Delta V=\left(10sin\omega t-3\right)volt$  

But in reverse biased condition of diode,

$\Delta V=0$ (across diode)

Clearly PM. PN $={\mathrm{O}\mathrm{P}}^2$ =OP2

i.e. $PM\cdot PN=16$ PM⋅PN=16

The NIMCET is conducted by an individual NIT on a rotational basis every year. So the NIT that is conducting the NIMCET in that year will release the NIMCET answer key. In 2025, the National Institute of Technology Tiruchirappalli conducted the NIMCET and hence, released the answer key.

The accepted modes to pay the UPESEAT 2026 application form fee is credit/debit cards, UPI and netbanking.

To download the SRMJEEE hall ticket, candidates will have to follow the steps mentioned below:

• Visit SRM portal
• Enter credentials - SRMJEEE application number and date of birth and click on Login.
• Click on "Book Exam Slot".
• Select preferred date and time of exam.
• Click on Confirm Slot or Change Test Date
• After confirmation, click on View / Print Hall-Ticket to download admit card.

Using F = MA = $m\frac{V}{T}$ $\frac{}{}$

m = FTV¹

$\Delta APC,AC=AP=12$  

$\Delta ABP,tan60°=\frac{12}{AB}$           

$AB=\frac{12}{\sqrt[]{3}}=4\sqrt[]{3}$            

$Area=4\sqrt[]{3}.4\sqrt[]{6}=48\sqrt[]{2}$            

The SRMJEEE admit card will consist of the following details - 

• Name of candidate
• Roll number
• Exam timings
• Remote mode exam guidelines
• Dos and Donts
• Post exam procedures

Load of mass will be equally distributed among the four colours so force on each columns will be 125 × 10³ N.

Cross section area of the column = $\pi \left[{\left(1\right)}^2-{\left(0.5\right)}^2\right]=2.355m^2$

Using young’s modulus : $\epsilon =\frac{\sigma }{Y}=\frac{F}{AY}=\frac{125\times 10^3}{2.355\times 2\times 10^{11}}=2.65\times 10^{-7}$  

$\left(a-1\right)x+0y+z=\alpha$        ……(1)

$x+\left(b-1\right)y+0z=\beta$                       ……(2)

$0x+y+\left(c-1\right)z=\gamma$                       ……(3)

$\left|\begin{array}{ccc}\hfill \left(a-1\right)\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill 1\hfill & \hfill \left(b-1\right)\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill \left(c-1\right)\hfill \end{array}\right|=0$           For no unique solution D = 0

$\left(a-1\right)\left(b-1\right)\left(c-1\right)+1=0$            

$\therefore a=2;b=2;c=0$            

Hence, |a + b + c| = 4

Yes, the SRMJEEE exam is not conducted at a designated exam center and candidates can appear for the test from the comfort of their home via their personal laptop or desktop. However, still they will have to download the admit card and keep it handy asit will consist of the following details - 

• Name of candidate
• Roll number
• Exam timings
• Remote mode exam guidelines
• Dos and Donts
• Post exam procedures

If you find any discrepancy in the NIMCET answer keys, you can raise objections online at nimcet.admissions.nic.in. You would need to submit your objections with supporting proof. For every objection, INR 500 has to be paid online, i.e., through net banking/credit card/debit card. Authorities will not accept the objections if the fee is not paid.

V = 10 $\left(1?e^{?t/RC}\right)$ ${}^{}$

2 = 20 $\left(1?e^{?t/RC}\right)$ ${}^{}$

$\frac{1}{10}=1?e^{?t/RC}$

$e^{t/RC}=\frac{10}{9}$

$\frac{t}{RC}=ln\left(\frac{10}{9}\right)=0.105$

$C=\frac{t}{R\times 0.105}=\frac{10^{?6}}{10\times 0.105}=0.95?F$

The GATE 2026 cutoff will be released along with the results or after a few days. IIT Guwahati will announce the GATE 2026 result on March 19. The cutoff will be different for each subject/branch and varies every year. For preparation or planning their dream college/institute, candidates must refer to the previous year's GATE cutoff. Going through the cutoff of the respective branch helps students to understand the difficulty level of the question papers.