Class 11th

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

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

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

Putting m = 1 kg and r = 1 m,

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

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'

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

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

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

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

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

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

JJ Thomson, through the cathode ray tube experiment, discovered electrons. He passes the electricity from the tube and sees rays coming from the negative end. These are tiny negatively charged particles, he named them electrons.

JJ Thomson's model of the atom proposed that an atom is like a sphere (pudding) and electrons are spread around it like a plum or chocolate chips. Basically atom is overall neutral.

The structure of an atom reflects how the atom is made, which includes the nucleus at the centre where protons and neutrons are present, and electrons revolving around the nucleus.

Boys (10)            Girls (5)

  (3)                        (3)

B₁ & B₂ should not be selected together

Total number of ways    

= (56 + 56) * 10 = 1120

$x^4-3x^3-2x^2+3x+1=0$

$\Rightarrow x=\pm 1$

and let a, b are roots of x² – 3x – 1 = 0

$\therefore \alpha +\beta =3\alpha \beta =-1$

$\therefore 1^3+{\left(-1\right)}^3+{\alpha }^3+{\beta }^3$

= 27 + 3 (3) = 36