Class 12th

This is a short answer type question as classified in NCERT Exemplar

no reversibility of lens maker formula is not possible.

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill 3x\hfill & \hfill -x+y\hfill & \hfill -x+z\hfill \\ \hfill x-y\hfill & \hfill 3y\hfill & \hfill z-y\hfill \\ \hfill x-z\hfill & \hfill y-z\hfill & \hfill 3z\hfill \end{array}\right|\\ C_1\to C_1+C_2+C_3\\ =\left|\begin{array}{ccc}\hfill x+y+z\hfill & \hfill -x+y\hfill & \hfill -x+z\hfill \\ \hfill x+y+z\hfill & \hfill 3y\hfill & \hfill z-y\hfill \\ \hfill x+y+z\hfill & \hfill y-z\hfill & \hfill 3z\hfill \end{array}\right|\\ Taking\left(x+y+z\right)commonfrom{C}_1\\ =\left(x+y+z\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill -x+y\hfill & \hfill -x+z\hfill \\ \hfill 1\hfill & \hfill 3y\hfill & \hfill z-y\hfill \\ \hfill 1\hfill & \hfill y-z\hfill & \hfill 3z\hfill \end{array}\right|\\ R_1\to R_1-R_2,R_2\to R_2-R_3\\ =\left(x+y+z\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -x-2y\hfill & \hfill -x+y\hfill \\ \hfill 0\hfill & \hfill 2y+z\hfill & \hfill -y-2z\hfill \\ \hfill 0\hfill & \hfill y-z\hfill & \hfill 3z\hfill \end{array}\right|\\ Expandingalong{C}_1=\left(x+y+z\right)\left[1\left|\begin{array}{cc}\hfill -x-2y\hfill & \hfill -x+y\hfill \\ \hfill 2y+z\hfill & \hfill -y-2z\hfill \end{array}\right|\right]\\ =\left(x+y+z\right)\left[\left(-x-2y\right)\left(-y-2z\right)-\left(2y+z\right)\left(-x+y\right)\right]\\ =\left(x+y+z\right)\left(xy+2zx+2y^2+4yz+2xy-2y^2+zx-zy\right)\\ =\left(x+y+z\right)\left(3xy+3zx+3yz\right)\\ =3\left(x+y+z\right)\left(xy+yz+zx\right)\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill 0\hfill & \hfill x{y}^2\hfill & \hfill x{z}^2\hfill \\ \hfill x^{2}y\hfill & \hfill 0\hfill & \hfill y{z}^2\hfill \\ \hfill x^{2}z\hfill & \hfill z{y}^2\hfill & \hfill 0\hfill \end{array}\right|\\ Takingx{,}^2{y}^{2}and{z}^{2}commonfrom{C}_1,C_{2}and{C}_{3}respectively\\ =x^2{y}^2{z}^2\left|\begin{array}{ccc}\hfill 0\hfill & \hfill x\hfill & \hfill x\hfill \\ \hfill y\hfill & \hfill 0\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill z\hfill & \hfill 0\hfill \end{array}\right|\\ Expandingalong{R}_1=x^2{y}^2{z}^2\left[0\left|\begin{array}{cc}\hfill 0\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill 0\hfill \end{array}\right|-x\left|\begin{array}{cc}\hfill y\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill 0\hfill \end{array}\right|+x\left|\begin{array}{cc}\hfill y\hfill & \hfill 0\hfill \\ \hfill z\hfill & \hfill z\hfill \end{array}\right|\right]\\ =x^2{y}^2{z}^2\left[-x\left(0-yz\right)+x\left(yz-0\right)\right]\\ =x^2{y}^2{z}^2\left(xyz+xyz\right)=x^2{y}^2{z}^2\left(2xyz\right)=2x^3{y}^3{z}^3\end{array}$

This is a short answer type question as classified in NCERT Exemplar

m=v/u = D/f

So m = 25/10 = 2.5= 0.025m

P= 1/0.025= 40D

This is the required power of lens.

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{ccc}\hfill a+x\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill x\hfill & \hfill a+y\hfill & \hfill z\hfill \\ \hfill x\hfill & \hfill y\hfill & \hfill a+z\hfill \end{array}\right|\\ C_1\to C_1+C_2+C_3\\ =\left|\begin{array}{ccc}\hfill a+x+y+z\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill a+x+y+z\hfill & \hfill a+y\hfill & \hfill z\hfill \\ \hfill a+x+y+z\hfill & \hfill y\hfill & \hfill a+z\hfill \end{array}\right|=\left(a+x+y+z\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill y\hfill & \hfill z\hfill \\ \hfill 1\hfill & \hfill a+y\hfill & \hfill z\hfill \\ \hfill 1\hfill & \hfill y\hfill & \hfill a+z\hfill \end{array}\right|\\ \left(Takinga+x+y+zcommon\right)\\ R_1\to R_1-R_2,R_2\to R_2-R_3\\ =\left(a+x+y+z\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill -a\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill a\hfill & \hfill -a\hfill \\ \hfill 1\hfill & \hfill y\hfill & \hfill a+z\hfill \end{array}\right|\\ Expandingalong{C}_1=\left(a+x+y+z\right)\left|1\left(a^2-0\right)\right|=a^2\left(a+x+y+z\right)\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- (i) Here it is given that, an electron absorbs 2 photons each of frequency ν then ν where, v′ is the frequency of emitted electron.

Given, E_max= hv- ${?}_0$

Now, maximum energy for emitted electrons is E_max= h2v- ${?}_0=2\mathrm{h}\mathrm{v}-{?}_0$

(ii) The probability of absorbing 2 photons by the same electron is very low. Hence, such emission will be negligible

This is a short answer type question as classified in NCERT Exemplar

As the refractive index for red is less than that for blue, parallel beams of light incident on a lens will be bent more towards the axis for blue light compared to red.

By lens makers formula 1/f= $(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$

So f_br so focal length of blue is less than red light 

This is a Short Answer Type Questions as classified in NCERT Exemplar
Sol:

$\begin{array}{l}Let\Delta =\left|\begin{array}{cc}\hfill x^2-x+1\hfill & \hfill x-1\hfill \\ \hfill x+1\hfill & \hfill x+1\hfill \end{array}\right|\\ C_1\to C_1-C_2\\ =\left|\begin{array}{cc}\hfill x^2-2x+2\hfill & \hfill x-1\hfill \\ \hfill 0\hfill & \hfill x+1\hfill \end{array}\right|\\ =\left(x+1\right)\left(x^2-2x+2\right)-0\\ =x^3-2x^2+2x+x^2-2x+2=x^3-x^2+2\end{array}$

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}L.H.S.\\ Let\Delta =\left|\begin{array}{ccc}\hfill xa\hfill & \hfill yb\hfill & \hfill zc\hfill \\ \hfill yc\hfill & \hfill za\hfill & \hfill xb\hfill \\ \hfill zb\hfill & \hfill xc\hfill & \hfill ya\hfill \end{array}\right|\\ Expandingalong{R}_1\\ \Rightarrow xa\left|\begin{array}{cc}\hfill za\hfill & \hfill xb\hfill \\ \hfill xc\hfill & \hfill ya\hfill \end{array}\right|-yb\left|\begin{array}{cc}\hfill yc\hfill & \hfill xb\hfill \\ \hfill zb\hfill & \hfill ya\hfill \end{array}\right|+zc\left|\begin{array}{cc}\hfill yc\hfill & \hfill za\hfill \\ \hfill zb\hfill & \hfill xc\hfill \end{array}\right|\\ \Rightarrow xa\left(yz{a}^2-x^{2}bc\right)-yb\left(y^{2}ac-xz{b}^2\right)+zc\left(xy{c}^2-z^{2}ab\right)\\ \Rightarrow xyz{a}^3-x^{3}abc-y^{3}abc+xyz{b}^3+xyz{c}^3-z^{3}abc\\ \Rightarrow xyz\left(a^3+b^3+c^3\right)-abc\left(x^3+y^3+z^3\right)\\ \Rightarrow xyz\left(a^3+b^3+c^3\right)-abc\left(3xyz\right)\left[?\left(x+y+z=0\right)\left(\therefore x^3+y^3+z^3=3xyz\right)\right]\\ \Rightarrow xyz\left(a^3+b^3+c^3-3abc\right)\\ R.H.S.xyz\left|\begin{array}{ccc}\hfill a\hfill & \hfill b\hfill & \hfill c\hfill \\ \hfill c\hfill & \hfill a\hfill & \hfill b\hfill \\ \hfill b\hfill & \hfill c\hfill & \hfill a\hfill \end{array}\right|\\ R_1\to R_1+R_2+R_3\\ \Rightarrow xyz\left|\begin{array}{ccc}\hfill a+b+c\hfill & \hfill a+b+c\hfill & \hfill a+b+c\hfill \\ \hfill c\hfill & \hfill a\hfill & \hfill b\hfill \\ \hfill b\hfill & \hfill c\hfill & \hfill a\hfill \end{array}\right|\\ \Rightarrow xyz\left(a+b+c\right)\left|\begin{array}{ccc}\hfill 1\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill c\hfill & \hfill a\hfill & \hfill b\hfill \\ \hfill b\hfill & \hfill c\hfill & \hfill a\hfill \end{array}\right|\left(a+b+ccommonfrom{R}_1\right)\\ C_1\to C_1-C_{2}and{C}_2\to C_2-C_3\\ \Rightarrow xyz\left(a+b+c\right)\left|\begin{array}{ccc}\hfill 0\hfill & \hfill 0\hfill & \hfill 1\hfill \\ \hfill c-a\hfill & \hfill a-b\hfill & \hfill b\hfill \\ \hfill b-c\hfill & \hfill c-a\hfill & \hfill a\hfill \end{array}\right|\\ Expandingalong{R}_1\\ \Rightarrow xyz\left(a+b+c\right)\left[1\left|\begin{array}{cc}\hfill c-a\hfill & \hfill a-b\hfill \\ \hfill b-c\hfill & \hfill c-a\hfill \end{array}\right|\right]\\ \Rightarrow xyz\left(a+b+c\right)\left[{\left(c-a\right)}^2-\left(b-c\right)\left(a-b\right)\right]\\ \Rightarrow xyz\left(a+b+c\right)\left(c^2+a^2-2ca-ab+b^2+ac-bc\right)\\ \Rightarrow xyz\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\\ \Rightarrow xyz\left(a^3+b^3+c^3-3abc\right)\left[a^3+b^3+c^3-3abc=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\right]\\ L.H.S.=R.H.S.\\ Hence,proved.\end{array}$

This is a Short Answer Type Questions as classified in NCERT Exemplar

Explanation- $\lambda =h/\sqrt[]{2mqv}$

$\lambda \propto \frac{1}{\sqrt[]{mq}}$

$\frac{{\lambda }_p}{{\lambda }_a}=\frac{\sqrt[]{m_a{q}_a}}{\sqrt[]{m_p{q}_p}}=\frac{\sqrt[]{4m_p*2e}}{\sqrt[]{m_p*e}}$ = $\sqrt[]{8}$

So wavelength of proton is $\sqrt[]{8}$ times of alpha particle