Physics NCERT Exemplar Solutions Class 11th Chapter Two

'A' at static equilibrium, E _{(inside conductor)} = 0

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(b, d) We know that one light year = 9.46 $*{10}^{11}m$ = distance that light travels in 1 year with a speed 3 $*{10}^8$ m/s

1 par sec =3.08 $*{10}^{16}m$ = distance at which average radius of earth's orbit subtends an angle of par second.

Here second and year represent time.

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(a, b) we know that pressure = force/area

Pressure= $\frac{force*distance}{area*distance}=\frac{work}{volume}=\frac{energy}{volume}$

So we can say that pressure is also equal to energy per unit volume.

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(a,b,d) h=[ML²T^-1]

[c]=[LT^-1], [m_e]=M

[G]= [M^-1L³T^-2]

[e]=[AT],[m_p]=[M]

[ $\frac{hc}{G}$ ] = $\frac{\left[M{L}^2{T}^{-1}\right]\left[L{T}^{-1}\right]}{M^{-1}{L}^3{T}^{-2}}=M^2$

M= $\sqrt[]{\frac{hc}{G}}$

$\frac{h}{c}=\frac{M{L}^2{T}^{-1}}{L{T}^{-1}}=\left[ML\right]$

L= $\frac{h}{cM}=\frac{h}{c}\sqrt[]{\frac{G}{hc}}=\frac{\sqrt[]{Gh}}{c^{3/2}}$

C= LT^-1

T= $\frac{\left[L\right]}{\left[C\right]}$ = $\frac{\sqrt[]{Gh}}{c^{3/2}c}=\frac{\sqrt[]{Gh}}{c^{5/2}}$

Hence a,b and d any can be used to express L,M and T in terms of three chosen fundamental quantities.

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(b, d) as we know E= hv

So h=E/v= $\frac{ML{T}^{-2}}{T^{-1}}$ =ML²T^-1

Dimension of linear impulse = dimensions of momentum= [MLT^-1]

As we know that J= $?p$

Angular impulse =tdt= $?L$ = change in angular momentum

Hence angular momentum is [ML²T^-1] same as the dimension of h.

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(a, e) It is given that P, Q, R are having different dimensions. Hence cannot be added or subtracted, so we can say that (a) and (e) are not meaningful. We cannot say about the dimensions of product of these quantities. Hence b, c and d may be meaningful.

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(b, c) for (c) LHS = [L]

RHS= [L/T]=LT^-1

LHS $\ne$ RHS

Hence is c option is not correct.

In option (b) dimension of angle is vt i.e =L

RHS= L.L=L² and LHS = L

LHS $\ne$ RHS

So option b is also not correct

LHS and RHS in both equation B and C are not equal.

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(d) E=  $\propto {\rho }^a{A}^b{T}^c$

E= $=k{\rho }^a{A}^b{T}^c$

E= [ML²T^-2] and $\rho =\left. [ML{T}^{-1}\right]$

A= [L²] and T= [T]

E= [k] [ $\rho$ ]^a [A]^b [T]^c

ML²T^-2= [MLT^-1]^a [L²]^b [T]^c

= M^aL^2b+aT^-a+c

by principle of homogeneity we get,

ML²T^-2= M^aL^2b+aT^-a+c

After solving

A=1, 2b+a=2

So 2b+1=2

2b=1, b= ½

C=-2+a, c=-2+1, c=-1

So formula become

E= $\rho$ A^1/2T^-1

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(c) Y= 1.9*10¹¹N/m²

1N=10⁵dyne

Y= 1.9*10¹¹*10⁵dyne/100²cm²

1m= 100 cm

Y= 1.9 $*{10}^{11}*{10}^{5}dyne/{100}^{2}cm$

Y= 1.9 $*{10}^{16-4}dyne/{cm}^2$

Y= 1.9 $*$ 10¹²dyne/cm²

This is a multiple choice answer as classified in NCERT Exemplar

(a) l = 5cm

So by checking it

$?$ l₁ = 5-4.9 = 0.1cm

$?$ l₂= 5-4.805 = 0.195cm

$?$ l₃= 5.25-5 = 0.25cm

$?$ l₄= 5.4-5 = 0.4cm

now $?l$ = 5-4.9= 01cm

has the least error hence 4.9 is the most precise value.