Class 11th

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

This is a multiple choice answer as classified in NCERT Exemplar

(c) (a) work =force $*distance=$ [MLT^-2] [L]= [ML²T^-2]

Torque= force (distance)= [ML²T^-2]

(b) angular momentum = mvr = [MLT^-1] [L]= [ML²T-1]

Planck's constant= E/V= [ML²T^-2]/ [T^-1]= [ML²T^-1]

(c) tension=force= [MLT^-2]

Surface tension = force/length= [MLT^-2]/L= [ML⁰T^-2]

(d) impulse =force (time)= [MLT^-2] [T]= [MLT^-1]

linear momentum = mass (velocity)= [M] [LT^-1]

They both have different formula.

This is a multiple choice answer as classified in NCERT Exemplar

(a) Given length = (1.62 $\pm 0.1$ )cm

And breadth b= (10.1 $\pm 0.1$ )cm

Area = l $*$ b=163.62 cm²

But when we rounding off to three significant figures then it would be =164cm²

$\frac{?A}{A}=\frac{?l}{l}+\frac{?b}{b}=\frac{0.1}{16.2}+\frac{0.1}{10.1}=\frac{2.63}{163.62}$

$so?A=A*\frac{2.63}{163.62}=16.62*\frac{2.63}{163.32}$ = 2.63cm²

So by rounding it off it become 3cm²

Area A= (164 $\pm 3$ )cm²

This is a multiple choice answer as classified in NCERT Exemplar

(d) Rounding off 2.745 to 3 significant figures would be 2.74 so result after in three significant figures is 2.74.

This is a multiple choice answer as classified in NCERT Exemplar

(c) As we know that formula for density is

density = $\frac{mass}{volume}=\frac{4.237g}{2.5{cm}^3}=1.6948$  so density is 1.7g/cm³