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

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

If there was no cut, then the object would have been at a height of 0.5 cm from the principal axis OO’.

Applying lens formula, we have

1/v-1/u=1/f $\frac{1}{\mathit{D}-\mathit{v}}+\frac{1}{\mathit{v}}=\frac{1}{\mathit{f}}$ Z

= $\frac{1}{-50}+\frac{1}{25}$

V= 50cm

Magnification m = v/u= 50/-50=-1

So coordinates of image are (50cm, -1cm)

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

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill 1\hfill & \hfill cosC\hfill & \hfill cosB\hfill \\ \hfill cosC\hfill & \hfill 1\hfill & \hfill cosA\hfill \\ \hfill cosB\hfill & \hfill cosA\hfill & \hfill 1\hfill \end{array}\right|\\ Expandingalong{C}_1\\ =1\left|\begin{array}{cc}\hfill 1\hfill & \hfill cosA\hfill \\ \hfill cosA\hfill & \hfill 1\hfill \end{array}\right|-cosC\left|\begin{array}{cc}\hfill cosC\hfill & \hfill cosB\hfill \\ \hfill cosA\hfill & \hfill 1\hfill \end{array}\right|+cosB\left|\begin{array}{cc}\hfill cosC\hfill & \hfill cosB\hfill \\ \hfill 1\hfill & \hfill cosA\hfill \end{array}\right|\\ =1\left(1-co{s}^{2}A\right)-cosC\left(cosC-cosAcosB\right)+cosB\left(cosAcosC-cosB\right)\\ =si{n}^{2}A-co{s}^{2}C+cosAcosBcosC+cosAcosBcosC-co{s}^{2}B\\ =si{n}^{2}A-co{s}^{2}B-co{s}^{2}C+2cosAcosBcosC\\ =-cos\left(A+B\right).cos\left(A-B\right)-co{s}^{2}C+2cosAcosBcosC\left[?si{n}^{2}A-co{s}^{2}B=-cos\left(A+B\right).cos\left(A-B\right)\right]\\ =-cos\left(-C\right).cos\left(A-B\right)+cosC+\left(2cosAcosB-cosC\right)\left[?A+B+C=0\right]\\ =-cosC\left(cosAcosB+sinAsinB\right)+cosC+\left(2cosAcosB-cosC\right)\\ =-cosC\left(cosAcosB+sinAsinB-2cosAcosB+cosC\right)\\ =-cosC\left(-cosAcosB+sinAsinB+cosC\right)\\ =cosC\left(cosAcosB-sinAsinB-cosC\right)\\ =cosC\left[cos\left(A+B\right)-cosC\right]\\ =cosC\left[cos\left(-C\right)-cosC\right]\left[?A+B=-C\right]\\ =cosC\left[cosC-cosC\right]=cosC.0=0R.H.S.\\ L.H.S.=R.H.S.Henceproved.\end{array}$

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

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill a^2+2a\hfill & \hfill 2a+1\hfill & \hfill 1\hfill \\ \hfill 2a+1\hfill & \hfill a+2\hfill & \hfill 1\hfill \\ \hfill 3\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right|\\ R_1\to R_1-R_2,R_2\to R_2-R_3\\ =\left|\begin{array}{ccc}\hfill a^2-1\hfill & \hfill a-1\hfill & \hfill 0\hfill \\ \hfill 2a-2\hfill & \hfill a-1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill \left(a+1\right)\left(a-1\right)\hfill & \hfill a-1\hfill & \hfill 0\hfill \\ \hfill 2\left(a-1\right)\hfill & \hfill a-1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right|\\ Taking\left(a-1\right)commonfrom{C}_{1}and{C}_2\\ =\left(a-1\right)\left(a-1\right)\left|\begin{array}{ccc}\hfill a+1\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 2\hfill & \hfill 1\hfill & \hfill 0\hfill \\ \hfill 3\hfill & \hfill 3\hfill & \hfill 1\hfill \end{array}\right|\\ Expandingalong{C}_3\\ ={\left(a-1\right)}^2\left[1\left|\begin{array}{cc}\hfill a+1\hfill & \hfill 1\hfill \\ \hfill 2\hfill & \hfill 1\hfill \end{array}\right|\right]\\ ={\left(a-1\right)}^2\left(a+1-2\right)={\left(a-1\right)}^2\left(a-1\right)={\left(a-1\right)}^{3}R.H.S.\\ L.H.S.=R.H.S.Henceproved.\end{array}$

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

Explanation- p=100W

_{$\lambda$ 1}= 1nm

_{$\lambda$ 2}= 500nm

Also n₁and n₂ represent number of photon of x-rays and visible light emitted from the two sources per sec

E/t=P=n₁hc/ $\lambda$ ₁₌n₂hc/ $\lambda$ ₂

So n₁/ $\lambda$ ₁ = n₂/ $\lambda$ ₂

So n₁/n₂= 1/500

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

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill y+z\hfill & \hfill z\hfill & \hfill y\hfill \\ \hfill z\hfill & \hfill z+x\hfill & \hfill x\hfill \\ \hfill y\hfill & \hfill x\hfill & \hfill x+y\hfill \end{array}\right|\\ C_1\to C_1-\left(C_2+C_3\right)\\ =\left|\begin{array}{ccc}\hfill 0\hfill & \hfill z\hfill & \hfill y\hfill \\ \hfill -2x\hfill & \hfill z+x\hfill & \hfill x\hfill \\ \hfill -2x\hfill & \hfill x\hfill & \hfill x+y\hfill \end{array}\right|\\ Taking-2commonfrom{C}_1\\ =-2\left|\begin{array}{ccc}\hfill 0\hfill & \hfill z\hfill & \hfill y\hfill \\ \hfill x\hfill & \hfill z+x\hfill & \hfill x\hfill \\ \hfill x\hfill & \hfill x\hfill & \hfill x+y\hfill \end{array}\right|\\ Expandingalong{C}_1\\ =-2\left[x\left|-zy-zy\right|\right]=-2\left(-xyz\right)=4xyzR.H.S.\\ L.H.S.=R.H.S.Henceproved.\end{array}$

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

Since, the object distance is u. Let us consider the two ends of the object be at distance

So u₁= u+L/2 and u₂=u+L/2 respectively so that|u₁- u₂| =L .

 Let the image of the two

ends be formed at v1 and v2, respectively so that the image length would be

L’= |v₁-v₂|……… (1)

Applying mirror formula 1/v+1/u=1/f or v= fu/u-f

On solving two positions of image v₁= $\frac{F(u-L/2)}{u-f-L/2}$ v₂= $\frac{F(u+L/2)}{u-f+L/2}$

For length substituting in equation 1

L’=  |v₁-v₂|= $\frac{F^{2}L}{{(u-F)}^2\times \frac{L^2}{4}}$

The objects is short and kept away from focus we have

$\frac{L^2}{4}$ << ( ${u-f)}^2$

Hence L’= $\frac{F^2}{{(u-f)}^2}L$

This is required expression.

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

Explanation- In photoelectric effect, we can observe that most electrons get scattered into the metal by absorbing a photon.Thus, all the electrons that absorb a photon doesn't come out as photoelectron. Only a few come out of metal whose energy becomes greater than the work function of metal.

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

$\begin{array}{l}L.H.S.=\left|\begin{array}{ccc}\hfill y^2{z}^2\hfill & \hfill yz\hfill & \hfill y+z\hfill \\ \hfill z^2{x}^2\hfill & \hfill zx\hfill & \hfill z+x\hfill \\ \hfill x^2{y}^2\hfill & \hfill xy\hfill & \hfill x+y\hfill \end{array}\right|\\ Taking{R}_1\to x{R}_1,R_2\to y{R}_2,R_3\to z{R}_{3}anddividingthedeterminantbyxyz.\\ =\frac{1}{xyz}\left|\begin{array}{ccc}\hfill x{y}^2{z}^2\hfill & \hfill xyz\hfill & \hfill xy+zx\hfill \\ \hfill y{z}^2{x}^2\hfill & \hfill yzx\hfill & \hfill yz+xy\hfill \\ \hfill z{x}^2{y}^2\hfill & \hfill zxy\hfill & \hfill zx+zy\hfill \end{array}\right|\\ Takingxyzcommonfrom{C}_{1}and{C}_2\\ =\frac{xyz.xyz}{xyz}\left|\begin{array}{ccc}\hfill yz\hfill & \hfill 1\hfill & \hfill xy+zx\hfill \\ \hfill zx\hfill & \hfill 1\hfill & \hfill yz+xy\hfill \\ \hfill xy\hfill & \hfill 1\hfill & \hfill zx+zy\hfill \end{array}\right|\\ C_3\to C_3+C_1\\ =xyz\left|\begin{array}{ccc}\hfill yz\hfill & \hfill 1\hfill & \hfill xy+yz+zx\hfill \\ \hfill zx\hfill & \hfill 1\hfill & \hfill xy+yz+zx\hfill \\ \hfill xy\hfill & \hfill 1\hfill & \hfill xy+yz+zx\hfill \end{array}\right|\\ Taking\left(xy+yz+zx\right)commonfrom{C}_3\\ =\left(xyz\right)\left(xy+yz+zx\right)\left|\begin{array}{ccc}\hfill yz\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill zx\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill xy\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right|\\ =\left(xyz\right)\left(xy+yz+zx\right)\left|\begin{array}{ccc}\hfill yz\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill zx\hfill & \hfill 1\hfill & \hfill 1\hfill \\ \hfill xy\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right|=0\left[?C_{2}and{C}_{3}areidentical\right]\\ L.H.S.=R.H.S.Henceproved.\end{array}$

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

Prism for is given by $\mu =\frac{sin\left\{\frac{A+D_m}{2}\right\}}{sin\left(\frac{A}{2}\right)}$

Given that D_m= A 

After substituting the values, $\mu$ = $\frac{sinA}{sin\frac{A}{2}}$

On solving, $\mu$ = $\frac{2sin\frac{A}{2}cos\frac{A}{2}}{sin\frac{A}{2}}$ 2 cos A/2

So cos A/2 = $\sqrt[]{3}/2$

So A= 60⁰ so this is the required prism angle