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$a=\frac{mg+10}{g}=\frac{5\times 10+10}{10}$

$\Rightarrow a=6m/se{c}^2$

v = u + at

0 = u – 6t

$t_A=\frac{u}{6}$  (time of Ascent)

$=u\times \frac{u}{6}-\frac{u}{2}\times 6\times \frac{u}{6}\times \frac{u}{6}$

= $\frac{u^2}{6}-\frac{u^2}{12}$

$h=\frac{u^2}{12}$

for time of descent

$a=\frac{50-10}{10}$

a = 4 m /sec²

$\Rightarrow \frac{u^2}{12}=0\times t+\frac{1}{2}\times 4t^2$

$t_B=\frac{u}{\sqrt[]{24}}$

$\frac{t_A}{t_B}=\frac{u\times \text{exception 25:}}{6\times u}=$$\frac{\sqrt[]{2}}{}$

No, the board does not allow any student to make such request. While preparing the Manipur HSLC routine, the board always try not to schedule to tough paper back to back. However, in case it is scheduled in the Manipur Board 10th exam routine, no students are allowed make required to revise the time table.

Sphalarite           ZnS

Calamine            ZnCO₃

Galena                PbS

Siderite               FeCO₃

Yes, AIIMS Bathinda admission 2025 is open. Intesreted candidates must appear for NEET exam and appear for the counselling process in order to apply for admission in their preferered course at the institute. Additionally, they must also ensure fulfilling the course-specific eligiblity criteria before applying for admission. 

All Indian Institute of Medical Sciences Raipur appoints well-experienced and trained faculty members with years of experience in the related field. The faculty focuses on theoretical and practical knowledge to the students. The faculty engages the students with the latest activities and innovative teaching and quality research.

Acidic oxide => Cl₂O₇

Neutral oxide => N₂O, NO

Basic oxide => Na₂O

Amphoteric oxide => As₂O₃

Number of moles of C = Number of moles of CO₂ = $\frac{330}{44}$ moles

              Number of moles of H = 2 × no. of moles of H₂O = $\left(\frac{270}{18}\times 2\right)$ moles

              Mass of C = $\frac{330}{44}\times 12$ gm = 90 gm

              Mass of H = $\frac{270}{18}\times 2\times 1gm=30gm$

$\begin{array}{l}\mathrm{\%}ofC=\frac{90}{120}\times 100\mathrm{\%}=75\mathrm{\%}\\ \mathrm{\%}ofH=\frac{30}{120}\times 100\mathrm{\%}=25\mathrm{\%}\end{array}$  

Yes, the Manipur HSLC routine may change or be revised. Sometimes, due to unexpected events like strikes or natural disasters, any political event or religious holiday, the Manipur HSLC exam date might be changed. Hence, the students should keep checking the board's website and this page even after the Manipur 10th exam dates are out.

 $\frac{dU}{dr}=0\Rightarrow \frac{d}{dr}\left[\frac{A}{r^{10}}\right]-\frac{d}{dr}\left[\frac{B}{r^5}\right]$

$\Rightarrow \frac{d}{dr}\left[A{r}^{-10}\right]-\frac{d}{dr}\left[B{r}^{-5}\right]=0$

$\Rightarrow -10A{r}^{-11}+5B{r}^{-6}=0$

$r={\left(\frac{2A}{B}\right)}^{\frac{1}{5}}$

$f\left(x\right)=\{\begin{array}{cc}\hfill \left[x\right],x<0\hfill & \hfill \left|1-x\right|,x\ge 0\hfill \end{array}$ $g\left(x\right)=\{\begin{array}{cc}\hfill c^x-x,x<0\hfill & \hfill {\left(x-1\right)}^2-1,x\ge 0\hfill \end{array}$

fog

$fog(\begin{array}{}\end{array}\left[e^x-x\right],\left(e^x-x<0\right)\left(x<0\right)\left[{\left(x-1\right)}^2-1\right]{\left(x-1\right)}^2-1<0x\ge 0\left|1-e^x+x\right|,e^x-x\ge 0x<0\left|1-{\left(x+1\right)}^2+1\right|,{\left(x-1\right)}^2-1\ge 0x\ge 0,x\in \left(2,\infty \right)($

 Not possible as of inequalities give $?.$  

e^x – x < 0, x < 0                 (x – 1)² – 1 < 0

Not possible                     (x – 1 + 1)(x – 1 – 1) < 0

(x)(x – 2) < 0

x   $\in \left(0,2\right)$

continuous  $\forall$ x < 0

$fog\{\begin{array}{ccc}\hfill \left|1-e^x+x\right|,x<0\hfill & \hfill \left|1-{\left(x-1\right)}^2+1\right|,x=0\hfill & \hfill \left|1-{\left(x-1\right)}^2+1\right|,x\ge 2\hfill \end{array}$ Discontinuous at 0

continuous   $\forall x$

  $\therefore$  fog is discontinuous at 0

Emulsions gets separated into two layers on standing.

For stabilization of emulsion. Emulsifying agents added into it but not electrolyte.

There is no provision of result revaluation in KIITEE MBA. The scores mentioned on the candidate's scorecard will be final. The examination authority will not accept any request for revaluation. However, if there is glaring error in the KIITEE MBA scorecard, candidates can raise a query with the examination authority for more clarity.

Z = $\sqrt[]{R^2+{\left(x_C-x_L\right)}^2},$ if only L & C are present then R = 0 then p = 0

Students should ideally start their exam preparation at the begining of their academic session. Once the Manipur HSLC routine is out, the students should plan their rest preparation strategy accordingly. While making a end time preparation plan, the students should focus on covering and revising the entire Manipur HSLC syllabus. The students should also keep study session for solving Manipur HSLC sample papers and previous years question papers.

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  a + a₁……a_n, 100               n arithmetic mean 100

a + n = 33 ……. (i)

$\left(\frac{a_1}{a_n}=\frac{1}{7}\right)$

$\frac{\left(a+d\right)}{\left(100-d\right)}=\frac{7}{7}$

7a + 8d = 100…………… (ii)

a + (n + 1)d = 100………………. (iiI)

Solving these equations (i), (ii) & (iii), we get

n = 23 & d =    $\frac{15}{4}$

a = 10

$\underset{\_}{\begin{array}{l}A\left(g\right)?B\left(g\right)+\frac{1}{2}C\left(g\right)\\ t=0amoles00\\ -a\alpha moles+a\alpha moles+\frac{a\alpha }{2}moles\end{array}}$

Eq.   a(1 - α)          aα           (aα/2)

Moles           moles       moles

Total no. of moles at equilibrium

= nA + nB + nC

$=a\left(1-\alpha \right)+a\alpha +\frac{a\alpha }{2}=a\left[1+\frac{\alpha }{2}\right]$

${\left(P_A\right)}_{eq}={\left(x_A\right)}_{eq}\times P_{eq}=\frac{a\left(1-\alpha \right)}{\left(1+\alpha /2\right)}P=\frac{1-\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_B\right)}_{eq}={\left(x_B\right)}_{eq}\times P_{eq}=\frac{a\alpha }{a\left(1+\alpha /2\right)}P=\frac{\alpha }{\left(1+\alpha /2\right)}P$

${\left(P_C\right)}_{eq}={\left(x_C\right)}_{eq}\times P_{eq}=\frac{a\alpha /2}{a\left(1+\alpha /2\right)}P=\frac{\alpha /2}{\left(1+\alpha /2\right)}P$

$K_P=\frac{{\left(P_B\right)}_{eq}\times {\left(P_C\right)}_{eq}^{1/2}}{{\left(P_A\right)}_{eq}}=\frac{{\left(\alpha \right)}^{3/2}{P}^{1/2}}{\left(1-\alpha \right){\left(2+\alpha \right)}^{1/2}}$

$r=\frac{mv}{qB}=\frac{\sqrt[]{2mk.E}}{qB}$  

$r\propto \frac{\sqrt[]{m}}{q}$                

m_p = m

q_p = q

m_d = 2m

q_d = q

m_a = 4m

q_a = 2q

$\frac{r_p}{r_d}=\frac{\sqrt[]{m_p}\times q_d}{q_P\times \sqrt[]{m_d}}$

$\frac{r_d}{r_{\alpha }}=\sqrt[]{2}$

$r_p:r_d:r_{\alpha }=1:\sqrt[]{2}:1$