physics ncert solutions class 11th

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

This is a multiple choice answer as classified in NCERT Exemplar

(b) The sum of these numbers is 663.821. The number with least decimal places is 227.2 is correct to only one decimal places and the closest number to it 664.

This is a multiple choice answer as classified in NCERT Exemplar

(b) The zero after points are not significant. The number can be written as 0.069, Hence number of significant figure are four.

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

t₁=39.6s, t₂=39.9s and t₃=39.5s

Least count of measuring instrument =0.1s

Precision in the measurement =least count of the measuring instrument =0.1s

So mean value of 20 observation

Mean value of t= $\frac{t_1+t_2+t_3}{3}=\frac{39.6+39.9+39.5}{3}=39.7s$

Absolute error in

 t₁= t-t₁ = 39.7-39.6 =0.1s

In t₂= t-t₂ = 39.7-39.9=-0.2s

In t₃= t-t₁ = 39.7-39.5= 0.2s

Mean error = $\frac{0.1+0.2+0.2}{3}=0.17s\approx 0.2$

Accuracy in the measurement = $?0.2s$

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