Physics

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

When no current flows in the circuit, the total potential difference provided by a source is known as the Electromagnetic force (EMF). On the other hand, when current flows in the circuit, the actual voltage across the terminals of the cell is called the terminal voltage. The formula that depicts the relation is:

Where r is the internal resistance and I is the current.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- b, d

Explanation – given A+B+C = 0

B $*\left(A+B+C\right)=B*0=0$

B $*A$ +B $*B+B*C$ =0

B $*A+0+B*C=0$

B $*A=-B*C$

A $*B=B*C$

(A $*B$ ) $*C=\left(B*C\right)*C$

It cannot be zero

(b) (A $*B$ ).C= (B $*AC$ ).C=0 . if b|C then B $*C$ =0 then (B $*C$ ) $*C=0$

(c) (A $*B$ )=X=ABsin $?X$ . The direction of X is perpendicular to the plane containing A and B (A $*B$ ) $*C=X*C.$

(d) if c²= A²+B², then angle between A and B is 90⁰

(A $*B$ ).C= (AB sin90⁰X).C=AB (X.C)

= ABC cos90⁰= 0

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer-b, d

Explanation- as given motion is two-dimensional motion and given that instantaneous speed v_o is positive constant. Acceleration is rate of change of velocity. hence it will also be in the plane of motion.