Units and Measurements

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 $\frac{D_{sun}}{D_{earth}}=100$

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

Because they cannot be derived from one another as they are independent quantities.

Also we can say that all other quantities in mechanics can be expressed in terms of length mass and time.

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

Since f ( $\theta )$ is a sum of different powers of $\theta$  an we know that it is a dimensionless quantity. By principle of homogeneity as RHS is dimensionless. Hence LHS should also be dimensionless.