Class 12th

We know that  ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx}=\underset{\begin{array}{l}h\to 0\\ n\to \infty \end{array}}{lim}h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)+......+f\left(a+\left(n-1\right)h\right)\right]$

where  $nh=b-a$

Here, a = a, b= b and f(x) = x

$\therefore {\displaystyle \underset{a}{\overset{b}{\int }}xdx}=\underset{h\to 0}{lim}h\left[a\left(a+h\right)+\left(a+2h\right)+......+\left(a+\left(n-1\right)h\right)\right]$

$\Rightarrow {\displaystyle \underset{a}{\overset{b}{\int }}xdx}=\underset{h\to 0}{lim}h\left[na+\left(1+2+3+.....+\left(n-1\right)\right)\right]$

$\Rightarrow {\displaystyle \underset{a}{\overset{b}{\int }}xdx}=\underset{h\to 0}{lim}h\left[anh+h\frac{n\left(n-1\right)}{2}\right]$

$=\underset{h\to 0}{lim}h\left[anh+\frac{nh\left(nh-h\right)}{2}\right]$

$=\underset{h\to 0}{lim}h\left[a\left(b-a\right)+\frac{\left(b-a\right)\left(b-a-h\right)}{2}\right]\left[?nh=b-a\right]$

$\begin{array}{l}=\left[a\left(b-a\right)+\frac{\left(b-a\right)\left(b-a\right)}{2}\right]\\ =\left(b-a\right)\left[a+\frac{b-a}{2}\right]\\ =\left(b-a\right)\left[\frac{2a+b-a}{2}\right]\\ =\frac{\left(b-a\right)\left(b+a\right)}{2}\\ =\frac{b^2-a^2}{2}\end{array}$

This is a Multiple Choice Questions as classified in NCERT Exemplar

Answer- (c)

Explanation- magnetic field outside the toroid is inversely proportional to distance

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Kindly go through the solution

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

Explanation- in the given figure let C be the amperian loop then,

 

${\int }_Q^{p}H.dl$ = ${\int }_Q^p\frac{B}{m_0}$ .dl

 between B and dl is less than 90 degree so

${\int }_Q^{p}H.dl$ = . ${\int }_Q^p\frac{B}{{\mu }_0}$ dl>0

So magnetic field from south pole to north pole inside the bar magnet

So according to ampere law

${\int }_{pQp}^{}H.dl$ =0

${\int }_{pQp}^{}H.dl$ = $\int H.dl$  +  $\int H.dl$  = 0

${\int }_Q^{p}H.dl$ >0, so ${\int }_p^{Q}H.dl$ <0,

It will be so if the angle between H and dl is more than 90 degree so cos $\theta$  is negative . it means H must run from north to south pole .

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Kindly go through the solution

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

Explanation- I= moment of inertia of the bar magnet

m= mass of bar magnet

l= length of magnet about an any passing through its centre and perpendicular to its

length

M= magnetic moment of the magnet

B= uniform magnetic field in which magnet Is oscillating

T= 2 $\pi \sqrt[]{\frac{I}{MB}}$

I= ml²/12

When magnet is cut into two equal pieces

I¹= $\frac{m{(l/2)}^2}{2\left(12\right)}$ = ml²/96= I/8

Magnetic dipole moment M^'= M/2

Its time period oscillation is T'= 2 $\pi \sqrt[]{\frac{I}{MB}}$ = 2 $\pi \sqrt[]{\frac{I/8}{(M/2)B}}$ = $\frac{2\pi }{2}\sqrt[]{\frac{I}{MB}}$

Kindly go through the solution

Kindly go through the solution