Thinking Process

When the pebble is dropped from the top, a variable force called viscous force will act which increases with increase in speed. And at equilibrium this velocity becomes constant.

Answer (c) When the pebble is falling through the viscous oil the viscous force is

$$F=6\pi \eta rv$$

where $r$ is radius of the pebble, $v$ is instantaneous speed, $\eta$ is coefficient of viscosity. As the force is variable, hence acceleration is also variable so $v$ - $t$ graph will not be straight line.First velocity increases and then becomes constant known as terminal velocity.

• Option (a): This graph shows a linear increase in velocity with time, which implies a constant acceleration. However, in the case of a pebble falling through viscous oil, the acceleration is not constant due to the increasing viscous force opposing the motion. Therefore, the velocity will not increase linearly.

• Option (b): This graph shows a parabolic increase in velocity with time, which suggests an increasing acceleration. In reality, the acceleration decreases over time as the viscous force increases, eventually leading to a constant terminal velocity. Hence, this graph does not accurately represent the situation.

• Option (d): This graph shows a velocity that increases and then decreases over time. In the case of a pebble falling through viscous oil, the velocity should increase until it reaches a constant terminal velocity. The velocity should not decrease after reaching a maximum value, making this graph incorrect.

Answer (d) In a streamline flow at any given point, the velocity of each passing fluid particles remaines constant. If we consider a cross-sectional area, then a point on the area cannot have different velocities at the same time, hence two streamlines of flow cannot cross each other.

• (a) The diagram shows streamlines that are parallel and do not intersect, which is characteristic of a streamline flow. In streamline flow, the fluid particles follow smooth paths in layers, and these paths do not cross each other.

• (b) The diagram depicts streamlines that are curved but do not intersect. This is also a characteristic of streamline flow, where the fluid particles follow smooth, non-intersecting paths.

• (c) The diagram shows streamlines that are converging but do not intersect. In streamline flow, streamlines can converge or diverge, but they do not cross each other, which is consistent with streamline flow behavior.

Answer (b) As we know for a streamline flow of a liquid velocity of each particle at a particular cross-section is constant, because $Av=$ constant (law of continuity) between two cross-section of a tube of flow.

• (a) The velocity of a fluid particle does not remain constant along a streamline because the direction of the velocity can change even if the speed remains constant.

• (c) The velocity of all fluid particles at a given instant is not constant because different streamlines can have different velocities at the same instant.

• (d) The speed of a fluid particle does not necessarily remain constant along a streamline because the speed can vary depending on changes in the cross-sectional area of the flow.

Answer (a) Consider the diagram where an ideal fluid is flowing through a pipe.

As given

$d_1=$ Diameter at 1 st point is 2.5 .

$d_2=$ Diameter at 2nd point is 3.75 .

Applying equation of continuity for cross-sections $A_1$ and $A_2$.

$$\begin{array}{rl}\Rightarrow A_1{v}_1& =A_2{v}_2\\ \Rightarrow \frac{v_1}{v_2}& =\frac{A_2}{A_1}=\frac{\pi (r_2^2)}{\pi (r_1^2)}=(\frac{r_2}{r_1}{)}^2\\ & =(\frac{\frac{3.75}{2}}{\frac{2.5}{2}}{)}^2=(\frac{3.75}{2.5}{)}^2=\frac{9}{4}[\begin{array}{c}{r}_2=\frac{d_2}{2}\\ r_1=\frac{d_1}{2}\end{array}]\end{array}$$

• Option (b) $3:2$: This ratio is incorrect because it does not account for the squared relationship between the radii of the two sections of the pipe. The correct ratio of velocities is derived from the square of the ratio of the diameters, not a simple linear ratio.

• Option (c) $\sqrt{3}:\sqrt{2}$: This ratio is incorrect because it does not follow from the equation of continuity, which involves the areas (and thus the squares of the radii) of the cross-sections. The correct ratio should be the square of the ratio of the diameters, not the square roots.

• Option (d) $\sqrt{2}:\sqrt{3}$: This ratio is incorrect for the same reason as option (c). The equation of continuity requires the ratio of the areas, which involves the squares of the radii, not their square roots.

Answer (c) According to the question, the observed meniscus is of convex figure shape. Which is only possible when angle of contact is obtuse. Hence, the combination will be of mercury-glass $({140}^{\circ })$

• Water: The angle of contact at the interface of water-glass is $0^{\circ }$, which results in a concave meniscus, not a convex one.

• Ethyl alcohol: The angle of contact at the interface of ethyl alcohol-glass is $0^{\circ }$, which also results in a concave meniscus, not a convex one.

• Methyliodide: The angle of contact at the interface of methyliodide-glass is ${30}^{\circ }$, which is still an acute angle and results in a concave meniscus, not a convex one.

Answer $(b,d)$

Consider the diagram where two molecules of a liquid are shown. One is well inside the liquid and other is on the surface. The molecule $(A)$ which is well inside experiences equal forces from all directions, hence net force on it will be zero.

And molecules on the liquid surface have some extra energy as it surrounded surraind by only lower half side of liquid molecules.

• (a) the net force on it is zero: This option is incorrect because a surface molecule experiences an imbalance in forces. Unlike a molecule well inside the liquid, which experiences equal forces from all directions resulting in a net force of zero, a surface molecule is only surrounded by liquid molecules on its lower half. This creates a net downward force on the surface molecule.

• (c) the potential energy is less than that of a molecule inside: This option is incorrect because surface molecules have higher potential energy compared to molecules inside the liquid. The molecules inside the liquid are surrounded by other molecules on all sides, leading to a lower potential energy state. In contrast, surface molecules are only partially surrounded, resulting in higher potential energy.

Answer $(b,c)$

Pressure is defined as the ratio of magnitude of component of the force normal to the area and the area under consideration.

As magnitude of component is considered, hence, it will not have any direction. So, pressure is a scalar quantity.

• Option (a) is incorrect: Pressure is not the ratio of force to area where both force and area are considered as vectors. Pressure is specifically defined as the ratio of the magnitude of the normal component of the force to the area, and not the vector quantities themselves.

• Option (d) is incorrect: Pressure can indeed depend on the size of the area chosen. For a given force, if the area is smaller, the pressure is higher, and if the area is larger, the pressure is lower. Therefore, pressure does depend on the size of the area under consideration.

Thinking Process

When any body floats in a liquid, the upthrust force acting on the body due to the displaced liquid is balanced by its weight.

Answer $(a,b)$

When the coin falls into the water, weight of the (block + coin) system decreases, which was balanced by the upthrust force earlier. As weight of the system decreases, hence upthrust force will also decrease which is only possible when $l$ decreases.

As $l$ decreases volume of water displaced by the block decreases, hence $h$ decreases.

Note As the coin falls into water, it displaces some volume of water which is very less hence, we neglect volume of the coin.

• Option (c) is incorrect because if ( l ) were to increase, it would imply that the block is displacing more water, which would require a greater upthrust force. However, since the weight of the system decreases when the coin falls off, the upthrust force required also decreases, leading to a decrease in ( l ).

• Option (d) is incorrect because if ( h ) were to increase, it would imply that the block is floating higher in the water, displacing less water. However, since the weight of the system decreases when the coin falls off, the block will float lower in the water, leading to a decrease in ( h ).

Answer (c, $d)$

For liquids coefficient of viscosity, $\eta \propto \frac{1}{\sqrt{T}}$

i.e., with increase in temperature $\eta$ decreases.

For gases coefficient of viscosity, $\eta \propto \sqrt{T}$

i.e., with increase in temperature $\eta$ increases.

• (a) gases decreases: This is incorrect because, for gases, the viscosity actually increases with an increase in temperature. This is due to the fact that as temperature rises, the kinetic energy of gas molecules increases, leading to more frequent and forceful collisions, which in turn increases the viscosity.

• (b) liquids increases: This is incorrect because, for liquids, the viscosity decreases with an increase in temperature. As temperature rises, the intermolecular forces within the liquid weaken, allowing the molecules to move more freely and thus reducing the viscosity.

Answer (b, c)

Streamline flow is more likely for liquids having low density. We know that greater the coefficient of viscosity of a liquid more will be velocity gradient hence each line of flow can be easily differentiated. Also higher the coefficient of viscosity lower will be Reynolds number, hence flow more like to be streamline.

• (a) high density: High density increases the inertia of the fluid, making it more difficult to maintain a smooth, orderly flow. This can lead to turbulence rather than streamline flow.

• (d) low viscosity: Low viscosity means the fluid has less internal resistance to flow, which can result in higher velocities and a greater likelihood of turbulent flow rather than streamline flow.

Answer No, surface tension is a scalar quantity.

Surface tension $=\frac{\text{ Work done }}{\text{ Surface area }}$, where work done and surface area both are scalar quantities.

Answer Given, density of ice $({\rho }_{\text{ice }})=0.917g/c{m}^3$

Density of water $({\rho }_w)=1g/c{m}^3$

Let $V$ be the total volume of the iceberg and $V^{\prime }$ of its volume be submerged in water. In floating condition.

Weight of the iceberg $=$ Weight of the water displaced by the submerged part by ice

or

$$\begin{array}{rl}V{\rho }_{ice}g& =V^{\prime }{\rho }_{w}g\\ \frac{V^{\prime }}{V}& =\frac{{\rho }_{ce}}{{\rho }_w}=\frac{0.917}{1}=0.917(\because \text{ Weight }=mg=v\rho g)\end{array}$$

Answer Consider the diagram,

The scale is adjusted to zero, therefore, when the block suspended to a spring is immersed in water, then the reading of the scale will be equal to the thrust on the block due to water.

Thrust $=$ weight of water displaced

$=V{\rho }_{w}g$ (where $V$ is volume of the block and ${\rho }_w$ is density of water)

$$=\frac{m}{\rho }{\rho }_{w}g=(\frac{{\rho }_w}{\rho })mg$$

$(\because .$ Density of the block $.\rho =\frac{\text{ mass }}{\text{ volume }}=\frac{m}{V})$

Thinking Process

As the elevator is accelerating upward the net acceleration of the block with respect to the elevator can be calculate by the concept of pseudo force.

Answer 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.

$$\begin{array}{rl}\text{ Volume of the block }(V)& =L^3\\ \text{ Mass of the block }(m)& =V\rho =L^3\rho \\ \text{ Weight of the block }& =mg=L^3\rho g\end{array}$$

1st case

Volume of the water displaced by the submerged part of the

$\therefore$ Weight of the water displaced by the block

In floating condition, $x{L}^2{\rho }_{w}g$

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

$$\begin{array}{cc}{L}^3\rho g& =x{L}^2{\rho }_{w}g\\ \text{ or }& \frac{x}{L}=\frac{\rho }{{\rho }_w}=x\end{array}$$

2nd case

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

Then, weight of the block

$$\begin{array}{rl}& =(g+a)\\ & =m(g+a)\\ & =L^3\rho (g+a)\\ & L^3\rho (g+a)=(x_1{L}^2){\rho }_w(g+a)\\ & \frac{x_1}{L}=\frac{\rho }{{\rho }_w}=x\end{array}$$

( $\because$ Pseudo force is downward)

Let $x_1$ fraction be submerged in water when elevator is accelerating upwards.

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

or

From 1st and 2nd case,

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

Note We should not confuse with the concept of pseudo force, i.e., pseudo force is downward, hence fraction will change due to increased force.

Answer Given, radius $(r)=2.5\times {10}^{-5}m$

Surface tension $(S)=7.28\times {10}^{-2}N/m$

Angle of $contact(\theta )=0^{\circ }$

The maximum height to which sap can rise in trees through capillarity action is given by

$$\begin{array}{rl}h& =\frac{2S\cos \theta }{r\rho g}\text{ where }S=\text{ Surface tension, }\rho =\text{ Density, }r=\text{ Radius }\\ & =\frac{2\times 7.28\times {10}^{-2}\times \cos 0^{\circ }}{2.5\times {10}^{-5}\times 1\times {10}^{-3}\times 9.8}=0.6m\end{array}$$

This is the maximum height to which the sap can rise due to surface tension. Since, many trees have heights much more than this, capillary action alone cannot account for the rise of water in all trees.

Thinking Process

As the tanker starts accelerating, free surface of the tanker will not be horizontal because pseudo force acts.

Answer Consider the diagram where a tanker is accelerating with acceleration a.

Consider an elementary particle of the fluid of mass $dm$.

The acting forces on the particle with respect to the tanker are shown above .

Now, balancing forces (as the particle is in equilibrium) along the inclined direction component of weight $=$ component of pseudo force $dmg\sin \theta =dma\cos \theta$ (we have assumed that the surface is inclined at an angle $\theta$ ) where, $dm$ is pseudo force

$$\begin{array}{ccc}\Rightarrow & g\sin \theta & =a\cos \theta \\ \Rightarrow & & a& =g\tan \theta \\ \Rightarrow & & \tan \theta & =\frac{a}{g}=\text{ slope }\end{array}$$

Thinking Process

In this process, conservation of mass will be applied. Before and after collapse, mass and hence, volume of the system remains conserved.

Answer Consider the diagram.

Radii of mercury droplets

$$\begin{array}{rl}& r_1=0.1cm=1\times {10}^{-3}m\\ & r_2=0.2cm=2\times {10}^{-3}m\end{array}$$

Surface tension $(T)=435.5\times {10}^{-3}N/m$

Let the radius of the big drop formed by collapsing be $R$.

$\therefore$ Volume of big drop $=$ Volume of small droplets

or

$$\begin{array}{rl}\frac{4}{3}\pi R^3& =\frac{4}{3}\pi r_1^3+\frac{4}{3}\pi r_2^2\\ R^3& =r_1^3+r_2^3\\ & =(0.1{)}^3+(0.2{)}^3\\ & =0.001+0.008\\ & =0.009\end{array}$$

$$R^3=r_1^3+r_2^3$$

or

$$R=0.21cm=2.1\times {10}^{-3}m$$

$\therefore$ Change in surface area

$$\mathrm{\Delta }A=4\pi R^2-(4\pi r_1^2+4\pi r_2^2)$$

$$=4\pi [R^2-(r_1^2+r_2^2)]$$

$$\begin{array}{rl}\therefore \text{ Energy released }=& T\cdot \mathrm{\Delta }A(\text{ where }T\text{ is surface tension of mercury) }\\ & =T\times 4\pi [R^2-(r_1^2+r_2^2)]\\ & =435.5\times {10}^{-3}\times 4\times 3.14[(2.1\times {10}^{-3}{)}^2.\\ & ..=435.5\times 4\times 3.14[4.41-5]\times {10}^{-6}\times {10}^{-3}+4\times {10}^{-6})]\\ & =-32.23\times {10}^{-7}\text{ (Negative sign shows absorption) }\end{array}$$

Therefore, $3.22\times {10}^{-6}J$ energy will be absorbed.

Note In this process, energy is not conserved. Energy is Ist due to the collapsing, in the form of radiations.

Answer When a big drop of radius $R$, breaks into $N$ droplets each of radius $r$, the volume remains constant.

$$\begin{gathered} \therefore \text{Volume of big drop}=N\times \text{volume of each small drop} \\ \frac{4}{3}\pi R^3=N\times \frac{4}{3}\pi r^3 \\ \text{or}{R}^3=N{r}^3 \\ \text{or}N=\frac{R^3}{r^3} \\ \text{Now},\text{change in surface area}=4\pi R^2-N4\pi r^2 \\ =4\pi (R^2-n{r}^2) \\ \text{Energy released}=T\times \mathrm{\Delta }A=S\times 4\pi (R^2-N{r}^2)[T=\text{Surface tension}] \end{gathered}$$

Due to releasing of this energy, the temperature is lowered.

If $\rho$ is the density and $s$ is specific heat of liquid and its temperature is lowered by $\mathrm{\Delta }\theta$, then energy released $=ms\mathrm{\Delta }\theta$ [s $=$ specific heat $\mathrm{\Delta }\theta =$ change in temperature]

$$\begin{array}{rl}T\times 4\pi (R^2-N{r}^2)& =(\frac{4}{3}\times R^3\times \rho )s\mathrm{\Delta }\theta \\ \mathrm{\Delta }\theta & =\frac{T\times 4\pi (R^2-N{r}^2)}{\frac{4}{3}\pi R^3\rho \times s}\\ & =\frac{3T}{\rho s}[\frac{R^2}{R^3}-\frac{N{r}^2}{R^3}]\\ & =\frac{3T}{\rho s}[\frac{1}{R}-\frac{(R^3/r^3)\times r^2}{R^3}]\\ & =\frac{3T}{\rho s}[\frac{1}{R}-\frac{1}{r}]\end{array}$$

Answer Given, surface tension of water

$$(S)=7.28\times {10}^{-2}N/m$$

Vapour pressure $(p)=2.33\times {10}^{3}Pa$

The drop will evaporate, if the water pressure is greater than the vapour pressure.

Let a water droplet or radius $R$ can be formed without evaporating.

Vapour pressure $=$ Excess pressure in drop.

$$\begin{array}{cc}\therefore & p=\frac{2S}{R}\\ \text{ or }R& =\frac{2S}{p}=\frac{2\times 7.28\times {10}^{-2}}{2.33\times {10}^3}\\ & =6.25\times {10}^{-5}m\end{array}$$

Thinking Process

As we are going up in the atmosphere, thickness of the gases above us decreases hence, pressure also decreases.

Answer (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.

$\begin{array}{rlrlr}& \text{ i.e., }(p+dp)A-pA=-\rho gAdh(\because \text{ Weight }=\text{ Density }\times \text{ Volume }\times g)& =-\rho \times Adh\times g& \Rightarrow dp=-\rho gdh.(\rho =\text{ density of air })& \end{array}$

Negative sign shows that pressure decreases with height.

(b) Let $p_0$ be the density of air on the surface of the earth.

As per question, pressure $\propto$ density

$$\begin{array}{rl}& \Rightarrow \frac{p}{p_0}=\frac{\rho }{{\rho }_0}\\ & \Rightarrow \rho =\frac{{\rho }_0}{p_0}p\\ & \therefore dp=-\frac{{\rho }_{o}g}{p_0}pdh[\because dp=-\rho gdh]\\ & \Rightarrow \frac{dp}{p}=-\frac{{\rho }_{0}g}{p_0}dh\end{array}$$

$$\Rightarrow {\int }_{p_0}^p\frac{dp}{p}=-\frac{{\rho }_{0}g}{p_0}{\int }_0^{h}dh[\begin{array}{l}\because \text{ at }h=0,r=p_0\\ \text{ and }\text{ at }h=h,p=p\end{array}]$$

$$\begin{array}{rl}& \Rightarrow \ln \frac{p}{p_0}=-\frac{{\rho }_{0}g}{p_0}h\\ & \text{ By removing log, }p=p_{0}e(-\frac{{\rho }_{0}gh}{p_0})\end{array}$$

(c) As $p=p_0{e}^{-\frac{{\rho }_{0}gh}{p_0}}$,

$$\begin{array}{ccc}\Rightarrow & \ln \frac{p}{p_0}& =-\frac{{\rho }_{0}gh}{P_0}\\ & \text{ By question, }& p& =\frac{1}{10}{p}_0\\ \Rightarrow & \ln (\frac{\frac{1}{10}{p}_0}{p_0})& =-\frac{{\rho }_{0}g}{p_0}h\\ \Rightarrow & \ln \frac{1}{10}& =-\frac{{\rho }_{0}g}{p_0}h{\rho }_0\end{array}$$

(d) We know that

$$\begin{array}{rl}h& =-\frac{p_0}{{\rho }_{o}g}\ln \frac{1}{10}=-\frac{p_o}{p_{0}g}\ln (10{)}^{-1}=\frac{p_o}{p_{0}g}\ln 10\\ & =\frac{p_0}{{\rho }_{o}g}\times 2.303[\because \ln (x)=2.303{\log }_{10}(x)]\\ & =\frac{1.013\times {10}^5}{1.22\times 9.8}\times 2.303=0.16\times {10}^{5}m\\ & =16\times {10}^{3}m\end{array}$$

Temperature $(T)$ remains constant only near the surface of the earth, not at greater heights.

Answer (a) Given , $L_v=540kcalk{g}^{-1}$

$$=540\times {10}^{3}calk{g}^{-1}=540\times {10}^3\times 4.2Jk{g}^{-1}$$

$\because$ Energy required to evaporate $1kg$ of water $=L_{V}kcal$

$$\therefore M_{A}kg\text{ of water requires }{M}_A{L}_{V}kcal[\because Q=mL]$$

Since, there are $N_A$ molecules in $M_{A}kg$ of water the energy required for 1 molecule to evaporate is

$$\begin{array}{rl}U& =\frac{M_A{L}_V}{N_A}J[\text{ where }{N}_A=6\times {10}^{26}=\text{ Avogadro number }]\\ & =\frac{18\times 540\times 4.2\times {10}^3}{6\times {10}^{26}}J\\ & =90\times 18\times 4.2\times {10}^{-23}J\\ & =6.8\times {10}^{-20}J\end{array}$$

(b) Let the water molecules to be points and are separated at distance $d$ from each other. Volume of $N_A$ molecule of water $=\frac{M_A}{{\rho }_w}$

$$[\because V=\frac{M}{\rho }]$$

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

The volume around one molecule is

$$\begin{array}{rl}{d}^3& =(M_A/N_A{\rho }_w)\\ \therefore & d=(\frac{M_A}{N_A{\rho }_w}{)}^{1/3}=(\frac{18}{6\times {10}^{26}\times {10}^3}{)}^{1/3}\\ (30\times {10}^{-30}{)}^{1/3}m& \approx 3.1\times {10}^{-10}m\end{array}$$

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

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

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

$\therefore 1$ molecule occupies $\frac{18\times 1601\times {10}^{-3}}{6\times {10}^{26}}{m}^3$

If $d$ is the inter- molecular distance, then

$$\begin{array}{rl}{d}_1^3& =(3\times 1601\times {10}^{-29})m^3\\ d_1& =(30\times 1601{)}^{1/3}\times {10}^{-10}m\\ & =36.3\times {10}^{-10}m\end{array}$$

$$\therefore d_1=(30\times 1601{)}^{1/3}\times {10}^{-10}m$$

(d) Work done to change the distance from $d$ to $d_1$ is $=F(d_1-d)$

This work done is equal to energy required to evaporate 1 molecule.

$$\begin{array}{rl}\therefore F(d_1-d)& =6.8\times {10}^{-20}\\ \text{ or }F& =\frac{6.8\times {10}^{-20}}{d_1-d}\\ & =\frac{6.8\times {10}^{-20}}{(36.3\times {10}^{-10}-3.1\times {10}^{-10})}\\ & =2.05\times {10}^{-11}N\end{array}$$

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

Thinking Process

Pressure inside the curved surface will be greater than of outside pressure.

Answer Let the pressure inside the balloon be $p_i$ and the outside pressure be $p_0$, then excess pressure is $p_i-p_0=\frac{2S}{r}$. where, $S=$ Surface tension

$r=$ radius of balloon

Considering the air to be an ideal gas $p_{j}V=n_{j}R{T}_j$ 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 $p_{0}V=n_{0}R{T}_0$ where $V$ is the volume of the air displaced and $n_0$ is the number of moles displaced and $T_0$ 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

and

$$n_0=\frac{p_{0}V}{R{T}_0}=\frac{M_0}{M_A}$$

where, $M_O$ is the mass of air outside that has been displaced. If $w$ is the load it can raise, then $w+M_{i}g=M_{o}g$

$\Rightarrow W=M_{0}g-M_{j}g$

As in atmosphere $21$ and $79$-is present

$\therefore$ Molar mass of air

$$M_i=0.21\times 32+0.79\times 28=28.84g$$

$\therefore$ Weight raised by the balloon

$$\begin{array}{rl}W& =(M_0-M_i)g\\ \Rightarrow W& =\frac{M_{A}V}{R}(\frac{p_0}{T_0}-\frac{p_i}{T_i})g\\ & =\frac{0.02884\times \frac{4}{3}\pi \times 8^3\times 9.8}{8.314}(\frac{1.013\times {10}^5}{293}-\frac{1.013\times {10}^5}{333}-\frac{2\times 5}{8\times 313})\\ & =\frac{0.02884\frac{4}{3}\pi \times 8^3}{8.314}\times 1.013\times {10}^5(\frac{1}{293}-\frac{1}{333})\times 9.8\\ & =3044.2N\end{array}$$

$\therefore$ Mass lifted by the balloon $=\frac{w}{g}=\frac{3044.2}{10}\approx 304.42kg$.

$$\approx 305kg\text{. }$$