Units and Measurements

This is a multiple choice answer as classified in NCERT Exemplar

(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.

This is a multiple choice answer as classified in NCERT Exemplar

(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.

This is a multiple choice answer as classified in NCERT Exemplar

(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.

This is a multiple choice answer as classified in NCERT Exemplar

(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.

This is a multiple choice answer as classified in NCERT Exemplar

(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

This is a multiple choice answer as classified in NCERT Exemplar

(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.

This is a multiple choice answer as classified in NCERT Exemplar

(a) All the given value are correct upto two decimal places. Here 5.00mm has the smallest unit and the error in 5.00mm is least so the smallest value has less error so 5.00mm is the most precise value.

This is a multiple choice answer as classified in NCERT Exemplar

(d) According to question

A= 1.0m $\pm$ 0.2m, B= 2 $\pm 0.2$ m

Y= $\sqrt[]{AB}=\sqrt[]{1.0\left(2.0\right)}=1.414m$

After rounding off 1.4m

= $\frac{?Y}{Y}=\frac{1}{2}\left. [\frac{?A}{A}+\frac{?B}{B}\right]=\frac{1}{2}\left. [\frac{0.2}{1.0}+\frac{0.2}{2.0}\right]=\frac{0.6}{2*2.0}$

 = $\frac{0.6Y}{2\left(2.0\right)}=0.212$

 After rounding off it becomes 0.2m

Thus the correct value of $\sqrt[]{AB}$ =r+dr

 so its value become 1.4 $\pm$ 0.2m

This is a multiple choice answer as classified in NCERT Exemplar

(a) As it is given that

A=2.5ms^{-1 $\pm$} 0.5ms^-1,

B= 0.10s $\pm$ 0.01s

X= AB= (2.5) (0.10)=0.25m

$\frac{?x}{x}=\frac{0.5}{2.5}+\frac{0.01}{0.10}$

$=\frac{0.05+0.025}{0.25}=\frac{0.075}{0.25}$

After rounding off two significant figures

AB= (0.25 $\pm 0.08$ )m