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

(a) L_v=540 kcal kg^-1

= 540 $\times {10}^{3}cal$ kg^-1 = 540 $\times {10}^3\times$ 4.2jkg^-1

Energy required to evaporate 1kg of water = L_v kcal

And M_A kg of water requires M_AL_V kcal

Since there are N_A molecules in M_A kg of water the energy required for 1 molecule to evaporate

Is

U= $\frac{M_A{L}_V}{N_A}J$

= $\frac{18\times 540\times 4.2\times {10}^3}{6\times {10}^{26}}J$

=90 $\times 18\times 4.2\times {10}^{-23}J$

= 6.8 $\times {10}^{-20}J$

(b) Let the water molecules to be points and are separated at a distance d from each other

volume of N_A molecule of water = $\frac{M_A}{{\rho }_w}$

thus the volume of one molecule is = $\frac{M_A}{N_A{\rho }_w}$

the volume around one molecule is d³= $\frac{M_A}{N_A{\rho }_w}$

d= $({\frac{M_A}{N_A{\rho }_w})}^3=({\frac{18}{6\times {10}^{26}\times {10}^3})}^{1/3}$

d= 3.1 $\times {10}^{-10}m$

(c) 1 kg of vapour occupies volume =1601 $\times {10}^{-3}$ m³

18 kg of vapour occupies 18 $\times 1601\times {10}^{-3}$ m³

6 $\times {10}^{26}$ molecules occupies 18 $\times 1601\times {10}^{-3}$ m³

1 molecule occupies = 18 $\times 1601\times {10}^{-3}$ /6 $\times {10}^{26}$

If d is the inter molecular distance then

$d_1^3=3\times 1601\times {10}^{-29}$ m³

d₁= (30 $\times 1601)$ ^{1/3 $\times {10}^{-10}m$}

= 36.3 $\times {10}^{-10}$ m

(d) Work done to change the distance from d to d₁ is = F(d₁-d)

this work done is equal to energy required to evaporate 1 molecule

F(d₁-d)=6.8 $\times {10}^{-20}$

F= $\frac{6.8\times {10}^{-20}}{\mathrm{d}1-\mathrm{d}}$

= $\frac{6.8\times {10}^{-20}}{36.3\times {10}^{-10}-3.1\times {10}^{-10}}$

= 2.05 $\times {10}^{-11}N$

(e) Surface tension =f/d= $\frac{2.05\times {10}^{-11}}{3.1\times {10}^{-10}}=6.6\times {10}^{-2}N/m$

For CBSE 10th Maths, one must focus on NCERT book. It includes all the topics mentioned in the CBSE 10th Maths syllabus and covers them thoroughly so as to clear the concepts. After finishing the NCERT textbook, if you wish to refer other reference books for practising more problems, you can refer R.D. Sharma and R.S. Aggarwal. 

Here's a more detailed breakdown-

The NCERT textbook is the primary source for learning and understanding concepts. It covers all the topics of the syllabus. 

RD Sharma offers detailed solutions and provides numerous pratice problems which are very useful in clearing doubts

RS Aggarwal book offers a wide range of problems and clear solutions, making it useful fo practice and understanding concepts.

Apart from these, students must solve and practice the CBSE 10th sample papers and previous years' question papers for familiarity with the exam pattern.

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

(a) consider a horizontal parcel of air with cross section A and height dh

Let the pressure on the top surface and bottom surface be P and p+dp. If the parcel is in equilibrium , then the net upward force must be balanced by the weight

(P+dP)-PA=- $\rho Adh\times g$

dP= - $\rho gdh$

negative sign shows that pressure decreases with height.

(b) let $\rho$ _o be the density of air on the surface of earth.

As per question , pressure $\propto$  density

$\frac{P}{P_o}=\frac{\rho }{{\rho }_0}$

$\rho =\frac{{\rho }_o}{P_o}P$

dP= - $\frac{{\rho }_{o}g}{P_o}Pdh$

$\frac{dP}{p}=-\frac{{\rho }_{o}gdh}{P_o}$

${\int }_{P_o}^P\frac{dP}{P}=-\frac{{\rho }_{o}g}{P_o}{\int }_0^{h}dh$

In $\frac{P}{P_o}=-\frac{{\rho }_{o}gh}{P_o}$

P=P_oe(- $\frac{{\rho }_{o}gh}{P_o}$ )

(c) as P =P_{o $e^{\frac{-{\rho }_{o}gh}{P_o}}$}

in $\frac{P}{P_o}=-\frac{{\rho }_{o}gh}{P_o}$

p=1/10 P_o

in( $\frac{\frac{1}{10}{P}_o}{P_o}$ ) =- $\frac{-{\rho }_{o}gh}{P_o}$

in1/10 =- $\frac{{\rho }_{o}g}{P_o}h{\rho }_0$

h=- $\frac{P_o}{{\rho }_{o}g}$ in1/10= - $\frac{P_o}{P_{o}g}in\left(10\right)$ ^-1= $\frac{P_o}{P_{o}g}in\left(10\right)$

= $\frac{P_o}{P_{o}g}\times 2.303$

= $\frac{1.013\times {10}^5}{1.22\times 9.8}\times 2.303=0.16\times {10}^{5}m$

= 16 $\times$ 10³m

(d) we know that

P $\propto \rho$  , temperature remains constant only near the surface of the earth , not at greater heights.

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