Physics Ncert Solutions Class 12th

This is a Multiple Choice Questions as classified in NCERT Exemplar

Explanation- |A+B|=|A-B|

$\sqrt[]{\left|{A|}^2+\left|{B|}^2+2\right|A\right|\left|B\right|cos\theta }=\sqrt[]{\left|{A|}^2+\left|{B|}^2-2\right|A\right|\left|B\right|cos\theta }$

$\left|{A|}^2+\left|{B|}^2+2\right|A\right|\left|B\right|cos\theta$  = $\left|{A|}^2+\left|{B|}^2-2\right|A\right|\left|B\right|cos\theta$

4|A|B|cos $\theta$ =0

|A|²+|B|²cos $\theta$ =0

A=0 or B=0 so $\theta =90$ . so A perpendicular B

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- a, b, c

Explanation- (i) Speed will constant throughout

(ii) Velocity will be tangential in the direction of motion

(iii) Centripetal acceleration will be a= v²/r, will always be towards centre of the circular path.

(iv) Angular momentum is constant in magnitude and direction out of the plane perpendicularly as well.

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.

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

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