Ask & Answer: India's Largest Education Community

A = [1]
[0 1]
Now A² = [1] [1] = [1 2]
[0 1] [0 1] [0 1]
A³ = A².A = [1 2] [1] = [1 3]
[0 1] [0 1] [0 1]
Similarly A²? ¹¹ = [1 2011]
[0 1]

If charge (-Q) at origin is replaced by (+Q), then electric field at the centre of the cube is zero. Thus, electric field at the centre of the cube is as if only (-2Q) charge is present at the origin.

$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\frac{\left(-2Q\right)}{{\left(\frac{\sqrt[]{3}}{2}a\right)}^2}.\frac{1}{\sqrt[]{3}}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)=\frac{|-2Q}{3\sqrt[]{3}\pi {\epsilon }_0{a}^2}\left(\widehat{x}+\widehat{y}+\widehat{z}\right)$

x = t² - t + 1 . (i)
y = t² + t + 1 . (ii)
(i) + (ii)
x + y = 2 (t² + 1) . (iii)
(i) – (ii)
x - y = -2t . (iv)
from (iii) and (iv)
x² – 2xy + y² – 2x – 2y + 4 = 0
Here H² = ab and Δ≠ 0
This is parabola, so e = 1

$F_{1}cos45°+F_{2}cos45°+F_3=\frac{m{v}^2}{r}$

$\Rightarrow \frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(\sqrt[]{2}r\right)}^2}.\frac{1}{\sqrt[]{2}}+\frac{G{m}^2}{{\left(2r\right)}^2}=\frac{m{v}^2}{r}$  

  $\Rightarrow v=\sqrt[]{\frac{Gm}{4r}\left(2\sqrt[]{2}+1\right)}$            

Putting m = 1 kg and r = 1 m,

$v=\frac{1}{2}\sqrt[]{G\left(1+2\sqrt[]{2}\right)}$            

Length of normal
k = y √ (1 + (dy/dx)²)

⇒ ∫ (y dy) / √ (k² - y²) = ∫ dx
Let k² – y² = t²
-2y dy = 2t dt
-√ (k² - y²) = x + c
It passes through (0, k)
x² + y² = k²

| 1+x² |
| x 1+x | = − (x - α? ) (x − α? ) (x – α? ) (x – α? )
| x² x 1+x |

Put x = 0
| 1 0 |
| 0 1 0 | = - (-α? ) (-α? ) (-α? ) (-α? )
| 0 1 |
α ⇒? α? α? α? = −1

Velocity of block in equilibrium, in first case,

$v=A\omega =A.\sqrt[]{\frac{k}{M}}$            

Velocity of block in equilibrium, is second case,

$v\text{'}=A\text{'}\omega \text{'}=A\text{'}\sqrt[]{\frac{k}{M+m}}$                  

From conservation of momentum,

Mv = (M + m) v’

$\Rightarrow MA\sqrt[]{\frac{k}{M}}=\left(M+m\right)A\text{'}\sqrt[]{\frac{k}{M+m}}\Rightarrow A\text{'}=A\sqrt[]{\frac{M}{M+m}}$          

(1 + x)? = C? + C? x + C? x² + . C? x?
Put x = 1
2? = C? + C? + C? + C? + … C?
Put x = ω
(1 + ω)? = C? + C? ω + C? ω² + . C? (ω)?
Put x = ω²
(1 + ω²)? = C? + C? ω² + C? ω² + … C? (ω²)?
∴ 1 + ω + ω² = 0
Now add all three equation (s)
2? + (1 + ω)? + (1 + ω²)? = 3 [C? + C? + C? + …]
C? + C? + C? + C? + …
= (2? + (1+ω)? + (1+ω²)? ) / 3
= (2? + (1+ω)? + (-ω)? ) / 3 = (2? - (-1)? ) / 3

The total number of seats for the MBBS course at NTRUHS is offered through the university and its affiliated colleges. The total number of seats offered in the course is around 3,240. To get a seat in the programme, applicants must complete their Class 12 from  Physics, Chemistry, Biology (Botany, Zoology)/Biotechnology and English with a minimum of 50% in the Science subjects. Further, appear in NEET and counselling process.   

Since, $\frac{x^2}{\alpha kT}$ should be dimensionless.

So, dimension of  $\alpha ,\left[\alpha \right]=\frac{L^2}{M{L}^2{T}^{-2}}=M^{-1}{T}^2$

Dimension of  $\alpha {\beta }^2$ should be that of W.

So, $\left[\alpha {\beta }^2\right]=M{L}^2{T}^{-2}$

$\Rightarrow \left[{\beta }^2\right]=\frac{M{L}^2{T}^{-2}}{M^{-1}{T}^2}=M^2{L}^2{T}^{-4}\Rightarrow \left[\beta \right]=ML{T}^{-2}$

Negation of (p → q) ∨ (p ∨ q)
~ [ (p → q) ∨ (p ∨ q)]
≡ ~ (p → q) ∧ ~ (p ∨ q)
[∴ ~ (p → q) ≡ p ∧ ~q]
≡ (p ∧ ~q) ∧ (~p ∧ ~q)
= F (contradiction)

$\frac{dq}{dt}=t=20t+8t^2$

$\Rightarrow {\displaystyle \underset{0}{\overset{q}{\int }}dq={\displaystyle \underset{0}{\overset{15}{\int }}\left(20t+8t^2\right)dt}}$

$\Rightarrow q=20\times \frac{15^2}{2}+8.\frac{15^3}{3}=11250C$  

I = ∫? ² [log? x] dx
By using graph
I = ∫? 0 dx + ∫? ² 1 dx
I = 0 + e² - e

A : Series limit of Lyman series

B : Third line of Balmer series

C : Second line of Paschen series

According to question
aα² + bα + c = 0 . (i)
−aβ² + bβ + c = 0 . (ii)
a/2 γ² + bγ + c = 0 . (iii)
bα + c = -aα² . (iv)
bβ + c = aβ² . (v)
bγ + c = -a/2 γ² . (vi)
By observing (v) and (vi), we get β > γ
By observing (iv) and (vi), we get α > γ
So α < γ < β

Sister Nivedita University courses are offered to students at the UG, PG,  and PhD levels. The university is mainly concerned with streams such as Science, Humanities, Engineering, Social Sciences, Agricultural Sciences, etc. SNU offers a BTech programme to students at the UG level for a duration of 4 years. The fee structure of the Sister Nivedita University BTech programme ranges between INR 6.6 Lacs and INR 9 lakh. 

The total tuition fee for the BA programme ranges between INR 7,400 and INR 15,600 depending on the affiliated college and specialisation. The tuition fee is typically paid as per the university's schedule. Candidates will need to pay the first installment of the Kashmir University fees to confirm their admission into the BA programme.