Physics

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

Consider the diagram.

Let the density of water be ${\rho }_w$  and a cubical block of ice of side L be floating in water with x of its height L submerged in water.

Volume of the block V = L³

Mass of the block m = V $\rho$ =L^{3 $\rho$}

Weight of the block = mg= L^{3 $\rho g$}

1st case

Volume of the water displaced by the submerged part of the block= xL²

Weight of the water displaced by the block

In floating condition, x L²

Weight of the block= Weight of the water displaced by the block

L^{3 $\rho g$} = xL^{2 ${\rho }_{w}g$}

$\frac{x}{L}=\frac{\rho }{{\rho }_w}=x$

2nd case

When elevator is accelerating upward with an acceleration a, then effective acceleration

= (g+a)

Then, weight of the block =m (g+a)

= L^{3 $\rho$} (g+a)

Let x₁ fraction be submerged in water when elevator is accelerating upwards.

Now, in the floating condition, weight of the block= weight of the displaced water

L^{3 $\rho (g+a$} )= ( $x_{1}L$ ²) ${\rho }_w$ (g+a)

$\frac{x_1}{L}=\frac{\rho }{{\rho }_w}$  = x

From1^st and 2^nd case,

We see that, the fraction of the block submerged in water is independent of the acceleration of the elevator.

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

Let the pressure inside the ballon be P₁ and the outside pressure be P_o, then excess pressure is P_i-P_o =2S/r

Considering the air to be an ideal gas p_iV = n_iRT_i = where, V is the volume of the air inside the balloon, n_i is the number of moles inside and T_i is the temperature inside, and p_oV =n_oRTo where V is the volume of the air displaced and n_o is the number of moles displaced and T_o is the temperature outside.

So n_{i= $\frac{P_{i}V}{R{T}_i}=\frac{M_i}{M_A}$}

Where M_i is the mass of air inside and M_A is the molar mass of air

n_o= $\frac{P_{o}V}{R{T}_o}=\frac{M_o}{M_A}$

if w Is the load it can raise then w+M₁g=M_og

as atmosphere 21% O₂ and 79%N₂ is present

so molar mass of air

M_i=0.21 $*32+0.79*28$ = 28.84g

So weight raised by the ballon

w=(M_o-M_i)g

= $\frac{0.02884*\frac{4}{3}\pi *8^3*9.8}{8.314}\left(\frac{1.013*{10}^5}{293}-\frac{1.013*{10}^5}{333}-\frac{2*5}{8*313}\right)$

= 3044.2N

Mass lifted by the ballon =w/g =3044.2/10=304.42kg

So approximately 305kg

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

(a) L_v=540 kcal kg^-1

= 540 $*{10}^{3}cal$ kg^-1 = 540 $*{10}^3*$ 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= $\frac{M_A{L}_V}{N_A}J$

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

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

= 6.8 $*{10}^{-20}J$

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

d= 3.1 $*{10}^{-10}m$

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

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

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

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

If d is the inter molecular distance then

$d_1^3=3*1601*{10}^{-29}$ m³

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

= 36.3 $*{10}^{-10}$ 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 $*{10}^{-20}$

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

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

= 2.05 $*{10}^{-11}N$

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

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=- $\rho Adh*g$

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= $\frac{P_o}{P_{o}g}in\left(10\right)$

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

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

= 16 $*$ 10³m

(d) we know that

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