Physics

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

Explanation – for equilibrium balance net torque should be zero

Mgl= Wcoill

500gl = Wcoill

Wcoil= 500 $*$ 9.8N

Taking moment of force about mid point then magnetic field

Mgl+mgl=Wcoill+IBlsin90

Mgl=BILl

M= $\frac{0.2*4.9*1*{10}^{-2}}{9.8}=10$ -3kg= 1g

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.

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