Physics

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

To lift the disc ther must be some electrostatic force is reqired.

F=qE……. (1), E=V/d………. (2)

But the charge required during the process is - $?$ _{0 $\frac{v}{d}?$} r², using this relation in 1 and 2

F=- $?$ _{0 $\frac{v}{d}?$} r²V/d…… (3)

Also to balance something F=mg……. (4)

From equation 3 and 4

Mg= - $?$ _{0 $\frac{v}{d}?$} r²V/d, so V = $?\frac{mgd}{?0\mathrm{?}\mathrm{r}}$ ₂ this the requird solution.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, c 

Explanation – as we know average acceleration is a_av= $\frac{?v}{?t}=\frac{v_2-v_1}{t_2-t_1}$

But when acceleration is not uniform V_av is not equal to v₁+v₂/2

So we can write $?v=\frac{?r}{?t}$

$?r=r_2-r_1$ = v₂-v₁ (t₂-t₁)

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- c

Explanation– as the given track y=x² is a frictionless track thus total energy will be same throughout the journey.

Hence total energy at A = total energy at P . at B the particle is having only Ke but at P some KE is converted to P

Hence (KE)_B = (KE)_P

Total energy at A = PE= total energy at B = KE= total energy at P

= PE+KE

Potential energy at A is converted to KE and PE at P hence

(PE)P< (PE)A

Hence (height)P= (height)A

As height of p < height of A

Hence path length AB > path length BP

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

To lift the disc ther must be some electrostatic force is reqired.

F=qE……. (1), E=V/d………. (2)

But the charge required during the process is - $\in$ _{0 $\frac{v}{d}\pi$} r², using this relation in 1 and 2

F=- $\in$ _{0 $\frac{v}{d}\pi$} r²V/d…… (3)

Also to balance something F=mg……. (4)

From equation 3 and 4

Mg= - $\in$ _{0 $\frac{v}{d}\pi$} r²V/d, so V = $\surd \frac{mgd}{\in 0\mathrm{\pi }\mathrm{r}}$ ₂ this the requird solution.

NCERT Exemplar goes beyond the NCERT textbook as it contains questions based on the textbook concepts. While practicing the short answer type questions, long answer type questions, and multiple choice questions, students learn how to use these concepts for solving problems. It will help students improve their analytical thinking and problem-solving skills and develop a deeper understanding of concepts. It prepares students for CBSE Board exams and competitive exams like NEET and JEE.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a,b,c

Explanation – H= $\frac{u^2{sin}^2\theta }{2g}$

H₁=V_o²sin^{2 $\theta$} ₁/2g  , H₂=V_o²sin^{2 $\theta$} ₂/2g

H₁>H₂

V_o²sin^{2 $\theta$} ₁/2g= V_o²sin^{2 $\theta$} ₂/2g

Sin^{2 $\theta$} ₁>sin^{2 $\theta$} ₂

Sin^{2 $\theta$} ₁ – sin² $\theta$ ₂>0

(Sin $\theta$ ₁ – sin $\theta$ ₂)( Sin $\theta$ ₁ + sin $\theta$ ₂)>0

Sin $\theta$ ₁>sin $\theta$ ₂ or ₁ >₂

T= $\frac{2usin\theta }{g}=\frac{2v_{o}sin\theta }{g}$

T₁= $\frac{2v_{o}sin{\vartheta }_1}{g}$   , T₂= $\frac{2v_{o}sin{\vartheta }_2}{g}$

T₁> T₂

R= $\frac{u^{2}sin2\theta }{g}=\frac{v_o^{2}sin2\theta }{g}$

Sin $\theta$ ₁>sin $\theta$ ₂

Sin2 $\theta$ ₁> sin2 $\theta$ ₂

$\frac{R_1}{R_2}=\frac{\mathrm{S}\mathrm{i}\mathrm{n}2{\theta }_1}{sin2{\theta }_2}1$

R₁>R₂

Total energy for the first particle

U₁=K.E+P.E=1/2m_{1 $v_o^2$}

U₂= K.E+P.E= 1/2m_{2 $v_o^2$}

Total energy for the second particle

So m₁= m₂ then U₁=U₂

So m₁>m₂ then U₁>U₂

So m₁2 then U1

While applying Kirchhoff's Current Law (KCL) at junctions, the total current leaving is equal to the total current entering. For Kirchhoff's Voltage Law (KVL) in loops, the total of the voltage gains and drops around any closed loop is zero. To avoid mistakes, students must assign a direction to currents which can be positive for gains and negative for drops, and be consistent with the signs while solving the equations.

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

$\frac{1}{4\pi ??}\frac{q}{(\sqrt[]{x^2+{(d/2)}^2+h^2}}$ - $\frac{1}{4\pi ??}\frac{q}{(\sqrt[]{x^2-{(d/2)}^2+h^2}}$

If net potential is zero then,

$\frac{1}{(\sqrt[]{{\mathit{x}}^2+{(\mathit{d}/2)}^2+{\mathit{h}}^2}\mathit{}}$ = $\frac{1}{(\sqrt[]{{\mathit{x}}^2+{(\mathit{d}/2)}^2+{\mathit{h}}^2}\mathit{}}$

$x^2+{(d/2)}^2+h^2$ = $x^2-{(d/2)}^2+h^2$

2dx= 0 , x=0

X=0 , y-z plane.

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

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b

Explanation - |A+B|= |A|or |A+B|²=|A|²

|A|² +|B|²+2|A|B|cos $\theta$ = |A|²

|B| (|B|+2|A|cos $\theta$ )= 0

|B|=0 or |B|+2|A|cos $\theta$ =0

Cos $\theta$ = $\frac{-\left|B\right|}{2\left|A\right|}$

If A and B are antiparallel then $\theta$ =180

-1= $\frac{\left|B\right|}{2\left|A\right|}=\left|B\right|=2\left|A\right|$