Physics NCERT Exemplar Solutions Class 11th Chapter Two

This is a short answer type question as classified in NCERT Exemplar

For dimensional consistency LHS = RHS

[L]= [A]=L

As wt-kx should be dimensionless [wt]= [kx]=1

w [T]=k [L]=1

So w= [T]^-1 and k= [L]^-1

This is a short answer type question as classified in NCERT Exemplar

The formula for solid angle is

 solid angle = area/ (distance)²

= $\frac{1{cm}^2}{5{cm}^2}=\frac{1}{25}=4*{10}^{-2}steridian$

This is a short answer type question as classified in NCERT Exemplar

As we know, $\theta =\frac{l}{r}radian$

$\theta =\frac{\pi }{6}=\frac{l}{31}cm$

L=31 $*\frac{\pi }{6}cm=\frac{31*3.14}{6}cm=16.22cm$

Rounding off to three significant figures it would be 16.2cm

This is a short answer type question as classified in NCERT Exemplar

(a) Plane angle = l/r radian its si unit is radian but no dimension formula.

(b) Strain= $\frac{?L}{L}=\frac{changeinlength}{length}$ so no dimension

And g =9.8m/s²

(c) Gravitational constant G= 6.67 $*{10}^{-11}N{m}^2/{kg}^2$

(d) Reynold's number is a number having no dimension.

This is a short answer type question as classified in NCERT Exemplar

F= [MLT^-2]=100N…….1

Length L= 10m………….2

Time t= 100s……….3

Substituting the value of L and T  in eqn 1,2,3

M $*10*{100}^{-2}=100$

$\frac{M*10}{100*100}=100$

M=100 $*1000kg$

M= 10⁵kg

This is a short answer type question as classified in NCERT Exemplar

R_me= distance of moon from earth

R_se=distance of sun from earth

Let angle made by sun and moon is .

$\theta =\frac{A_{sun}}{R_{se}^2}=\frac{A_{moon}}{R_{me}^2}$

$\theta =\frac{\pi R_s^2}{R_{se}^2}=\frac{\pi R_m^2}{R_{me}^2}$

( $\frac{R_s}{R_{se}}$ )2 = ( $\frac{R_m}{R_{me}}$ )2

( $\frac{R_s}{R_m}$ )= ( $\frac{R_{se}}{R_{me}}$ )

Hence radius of sun and moon represents their sizes respectively.

This is a short answer type question as classified in NCERT Exemplar

It is given that 50 VSD $\approx 49\mathrm{M}\mathrm{S}\mathrm{D}$

1MSD = 50/49VSD

1VSD= 49/50MSD

Minimum inaccuracy = 1MSD-1VSD= 1MSD-49/50MSD=1/50 MSD

1MSD= 0.5mm

Minimum inaccuracy = 1/50 $*$ 0.5mm=0.01mm

This is a short answer type question as classified in NCERT Exemplar

Distance of galaxy = 10²⁵m

Speed of light = 3 $*$ 10⁸ m/s

T = distance /speed = $\frac{{10}^{25}}{3*{10}^8}\approx \frac{1}{3}*{10}^{17}$

$=\frac{10}{3}*{10}^{16}s$

$3.33*{10}^{16}$ s

This is a short answer type question as classified in NCERT Exemplar

(a) A wall clock can measure time correctly upto one second .

(b) A stop watch can measure time correctly upto a fraction of a second .

(c) A digital watch upto the same precision

(d) An atomic clock measure more accurately.

This is a short answer type question as classified in NCERT Exemplar

(a)

 l=R_E; r= 60R_E

$\theta =\frac{R_E}{60R_E}=\frac{1}{60}rad=\frac{1}{60}*\frac{180}{\pi }degree=\frac{3}{\pi }\approx 1^o$

Hence angle subtend is $2^o$

(b) But in 2^nd case it is given that moon is (½)° and earth is 2^o so it become 4 times.

$\frac{diameterofearth}{diameterofmoon}=\frac{2/\pi }{1/2\pi }=4$

(c) To measure parallax sun is about 400 times the earth moon distance hence $\frac{r_{sun}}{r_{moon}}=400$

To finding ratio of

$\frac{D_{sun}}{r_{sun}}=\mathrm{}\frac{D_{sun}}{r_{moon}}$

$\frac{D_{sun}}{D_{moon}}=400$

$\frac{D_{earth}}{D_{moon}}=4$

so $\frac{D_{sun}}{D_{earth}}=100$

so $\frac{D_{sun}}{D_{earth}}=100$