Class 12th

This is a multiple choice answer as classified in NCERT Exemplar

(c) R= 20cm, $\mu$ = 1.5, f= $\frac{R}{\mu -1}$ = $\frac{20}{1.5-1}$ =40cm so lens acts as convex lens.

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat\\ \Rightarrow \left|\begin{array}{cc}\hfill 2x\hfill & \hfill 5\hfill \\ \hfill 8\hfill & \hfill x\hfill \end{array}\right|=\left|\begin{array}{cc}\hfill 6\hfill & \hfill -2\hfill \\ \hfill 7\hfill & \hfill 3\hfill \end{array}\right|\\ \Rightarrow 2x^2-40=18+14\Rightarrow 2x^2=32+40\\ \Rightarrow 2x^2=72\Rightarrow x^2=36\\ \therefore x\pm 6\\ Hence,thecorrect{x}^{2}optionis\left(c\right).\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(c) According to VIBGYOR, among all given sources of light, the blue light have the smallest wavelength. According to Cauchy relationship, smaller the wavelength higher the refractive index and consequently smaller the critical angle. So, corresponding to blue colour, the critical angle is least which facilitates total internal reflection for the beam of blue light. The beam of green light would also undergo total internal reflection.

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

Sol:

$\begin{array}{l}Giventhatx=sintandy=sinpt\\ Differentiatingbothsidesw.r.t.t\\ \Rightarrow \frac{dx}{dt}=costand\frac{dy}{dt}=cospt.p=p.cospt\\ \Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{p.cospt}{cost}\\ \Rightarrow \frac{dy}{dx}=\frac{p.cospt}{cost}\\ Againdifferentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=p.\frac{d}{dx}\left(\frac{cospt}{cost}\right)\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{cost.\frac{d}{dx}\left(cospt\right)-cospt.\frac{d}{dx}\left(cost\right)}{co{s}^{2}t}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{cost.\left(-sinpt\right).p\frac{dt}{dx}-cospt.\left(-sint\right).\frac{dt}{dx}}{co{s}^{2}t}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{-pcost.sinpt+cospt.sint}{co{s}^{2}t}\right]\frac{dt}{dx}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left[\frac{-pcost.sinpt+cospt.sint}{co{s}^{2}t}\right]\frac{1}{cost}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\\ Nowwehavetoprovethat\\ \left(1-x^2\right)\frac{d^{2}y}{d{x}^2}-x\frac{dy}{dx}+ p^{2}y=0\\ L.H.S.=\left(1-x^2\right)\left[p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\right]-x\left(\frac{p.cospt}{cost}\right)+ p^{2}y\\ \Rightarrow =\left(1-si{n}^{2}t\right)\left[p.\left(\frac{-pcost.sinpt+cospt.sint}{co{s}^{3}t}\right)\right]-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =co{s}^{2}t\left[\frac{-p^{2}cost.sinpt+pcospt.sint}{co{s}^{3}t}\right]-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =\frac{-p^{2}cost.sinpt+pcospt.sint}{cost}-\frac{p.sintcospt}{cost}+ p^2.sinpt\\ \Rightarrow =\frac{-p^{2}cost.sinpt+pcospt.sint-p.sintcospt+p^2.sinptcost}{cost}\\ \Rightarrow =\frac{0}{cost}=0=R.H.S.\\ Hence,proved.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(b) A passenger in an aeroplane may see a primary and a secondary rainbow like concentric circles.

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

Sol:

 $\begin{array}{l}\left(i\right)Giventhat{x}^m.y^n={\left(x+y\right)}^{m+n}\\ Takinglogonbothsides,weget,\\ \Rightarrow log{x}^m.y^n=log{\left(x+y\right)}^{m+n}\left[?logxy=logx+logy\right]\\ \Rightarrow log{x}^m+log{y}^n=\left(m+n\right)log\left(x+y\right)\\ \Rightarrow mlogx+nlogy=\left(m+n\right)log\left(x+y\right)\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow m.\frac{d}{dx}logx+n.\frac{d}{dx}logy=\left(m+n\right)\frac{d}{dx}log\left(x+y\right)\\ \Rightarrow m.\frac{1}{x}+n.\frac{1}{y}.\frac{dy}{dx}=\left(m+n\right)\frac{1}{x+y}\left(1+\frac{dy}{dx}\right)\\ \Rightarrow \frac{m}{x}+\frac{n}{y}.\frac{dy}{dx}=\frac{m+n}{x+y}.\left(1+\frac{dy}{dx}\right)\\ \Rightarrow \frac{m}{x}+\frac{n}{y}.\frac{dy}{dx}=\frac{m+n}{x+y}+\frac{m+n}{x+y}.\frac{dy}{dx}\\ \Rightarrow \frac{n}{y}.\frac{dy}{dx}-\frac{m+n}{x+y}.\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}\\ \Rightarrow \left(\frac{n}{y}-\frac{m+n}{x+y}\right).\frac{dy}{dx}=\frac{m+n}{x+y}-\frac{m}{x}\\ \Rightarrow \left(\frac{nx+ny-my-ny}{y\left(x+y\right)}\right).\frac{dy}{dx}=\left(\frac{mx+nx-mx-my}{x\left(x+y\right)}\right)\\ \Rightarrow \left(\frac{nx-my}{y\left(x+y\right)}\right).\frac{dy}{dx}=\left(\frac{nx-my}{x\left(x+y\right)}\right)\\ \Rightarrow \frac{dy}{dx}=\frac{nx-my}{x\left(x+y\right)}*\frac{y\left(x+y\right)}{nx-my}=\frac{y}{x}\\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}Hence,proved.\end{array}$

$\begin{array}{l}\left(ii\right)Giventhat\frac{dy}{dx}=\frac{y}{x}\\ Differentiatingbothsidesw.r.t.x\\ \Rightarrow \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(\frac{y}{x}\right)\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{x.\frac{dy}{dx}-y.1}{x^2}\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{x.\frac{y}{x}-y}{x^2}\left[?\frac{dy}{dx}=\frac{y}{x}\right]\\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{y-y}{x^2}=\frac{0}{x^2}=0\\ Hence,\frac{d^{2}y}{d{x}^2}=0Hence,proved.\end{array}$

This is a multiple choice answer as classified in NCERT Exemplar

(c) When an object approaches a convergent lens from the left of the lens with a uniform speed of 5 m/s, the image away from the lens with a non-uniform acceleration.

This is a multiple choice answer as classified in NCERT Exemplar

(d) Since v ∝ λ, the light of red colour is of the highest wavelength and therefore of the highest speed. Therefore, after travelling through the slab, the red colour emerge first.

This is a Long Answer Type Questions as classified in NCERT Exemplar $\begin{array}{l}Giventhatf\left(x\right)=\{\begin{array}{cc}\hfill x^2+3x+p,\hfill & \hfill x\le 1\hfill \\ \hfill qx+2,\hfill & \hfill x>1\hfill \end{array}atx=1.\\ L.H.L.f\text{'}\left(c\right)=\underset{x\to 1^{-}}{lim}\frac{f\left(x\right)-f\left(c\right)}{x-c}\\ f\text{'}\left(1\right)=\underset{x\to 1^{-}}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}\\ =\underset{x\to 1^{-}}{lim}\frac{\left(x^2+3x+p\right)-\left(1+3+p\right)}{x-1}\\ =\underset{h\to 0}{lim}\frac{\left[{\left(1-h\right)}^2+3\left(1-h\right)+p\right]-\left(1+3+p\right)}{1-h-1}\\ =\underset{h\to 0}{lim}\frac{\left[1+h^2-2h+3-3h+p\right]-\left(4+p\right)}{-h}\\ =\underset{h\to 0}{lim}\frac{\left[h^2-5h+4+p\right]-\left[4+p\right]}{-h}\\ =\underset{h\to 0}{lim}\frac{h^2-5h+4+p-4-p}{-h}\\ =\underset{h\to 0}{lim}\frac{h^2-5h}{-h}=\underset{h\to 0}{lim}\frac{h\left[h-5\right]}{-h}=5\\ R.H.L.f\text{'}\left(1\right)=\underset{x\to 1^{+}}{lim}\frac{f\left(x\right)-f\left(1\right)}{x-1}\\ =\underset{x\to 1^{+}}{lim}\frac{\left(qx+2\right)-\left(1+3+p\right)}{x-1}\\ =\underset{h\to 0}{lim}\frac{\left[q\left(1+h\right)+2\right]-\left[4+p\right]}{1+h-1}\\ =\underset{h\to 0}{lim}\frac{q+qh+2-4-p}{h}=\underset{h\to 0}{lim}\frac{q+qh-2-p}{h}\\ Forexistingthelimit\\ q-2-p=0\Rightarrow q-p=0\dots \left(1\right)\\ \underset{h\to 0}{\Rightarrow lim}\frac{qh-0}{h}=q\\ IfL.H.L.f\text{'}\left(1\right)=R.H.L.f\text{'}\left(1\right)thenq=5.\\ Nowputt\end{array}$