Physics NCERT Exemplar Solutions Class 11th Chapter Two

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(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.

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(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

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(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

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(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.

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(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

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

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(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

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(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.

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

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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 = $?0.2s$