Physics 2 Marks Questions

$\text{Pa}+\text{hpg}=\text{pa}+\frac{\text{mg}}{\text{A}} $

$\Rightarrow \text{hpg}=\frac{\text{mg}}{\text{A}}$

$\Rightarrow \text{h}=\frac{\text{m}}{\text{pA}}$

$\Rightarrow \text{h}=-\frac{\text{m}}{1000\times \text{A}}$

$=\frac{4}{1000\times900\times10^{-4}}$

$=\frac{1}{2}\text{m}=50\text{cm}$

Let the volume of ice in water be a mo and the volume of ice in K. oil

be bm² Therefore, Total Volume = a + b

Density of ice = density of water at 4°C Specific Gravity.

$\therefore$ Density of the ice = 0.9 × 1000kg/m²

Density of Ice = 900kg/m²

Therefore, total mass of the ice cube = (a + b) × 900kg

Therefore, Its weight = (a + b) × 900 × 10

By the law of the Flotation, the total force of buoyancy by the water

and the Kerosene oil will be equal to the weight of the ice cube.

 $\therefore$ The total force of buoyancy = a × 1000 + Y × 800kg

(a + b) × 900 = a × 1000 + b × 800

900a + 900b = 1000a + 800b

100a = 100b

$\therefore$ a = b

Hence, the ratio of there volume is 1 : 1.

$\text{w}_1+\text{w}_2=\text{u}\cdot$

$\Rightarrow\text{mg}+\text{v}\times\rho_\text{s}\times\text{g}=\text{v}\times\rho_\text{w}\times\text{g}$

$\Rightarrow1-\rho_\text{s}=0.19$

$\Rightarrow\rho_\text{s}=1-(0.19)=0.8\text{gm}/\text{cc}$

(where $\rho_\text{s}$ = density of sphere in gm/cc)

so, specific gravity of the material is 0.8.

Let the density of the spherical body is Pg/cm

By the Principle of the Flotation,

Weight of the spherical body = Weight of the fluid displaced.

$\therefore$ Volume of the spherical body × density of spherical body × g = Volume of body displaced in water × density of water × g

Volume of the spherical body = $\frac{4}{\pi}\big(\text{R}^3-\text{r}^3\big)$

$=\frac{4}{3}\times\frac{22}{7}\times\big(8^3-6^3\big) $

$=\frac{4}{3\pi}\big(8^3-6^3\big)$

Volume of the spherical body displaced in water = Outer volume $=\frac{\text{outer}\ \text{volume}}{2}=\frac{4}{3\pi}\big(8^3\big)$

Also, density of water $=1\text{g}/\text{cm}^3.$

$\therefore\frac{4}{3\pi}\big(\text{R}^3-\text{r}^3\big)\times\rho=\frac{\Big[\frac{4}{3\pi}\big(\text{R}^3\big)\Big]}{2}\times1$

$\therefore\frac{4}{3\pi}\big(8^3-6^3\big)\times\rho=\frac{2}{3\pi}\big(8^3\big)$

$2\big(8^3-6^3\big)\times\rho=(8^3)$

$2\times296\rho=-512$

$\text{p}=0.8648\text{g}/\text{cm}^3$

$\therefore\text{p}=864.8\text{kg}/\text{m}^3.$

Hence, the density of the material of the spherical body =864.8Kg/m³.

Hope it helps.

Now, a. We know volume rate of the flow does not depends upon the the height of the pipe, since pipe. 

Velocity of the fluid will remains unchanged.

1. $\text{At}\ \text{A},$

$\text{Area}\times\text{velocity}=1\text{cm}^3/\text{s}.$

$\frac{4}{100}\text{cm}^2\times\text{velocity}=1$

$\therefore\text{velocity}=25\text{cm}/\text{seconds}.$

2. $\text{At}\ \text{B},$

$\text{Area}\times\text{velocity}=1\text{cm}^3/\text{s}.$

$\frac{2}{100}\times\text{velocity}=50\text{cm}/\text{seconds}.$

$\text{velocity}=50\text{cm}/\text{seconds}.$

3. Pressure difference, will be affected, this is because, there will be the gravitational work also which will be included.

Using the Bernoulli's equation,

Using the Bernoulli's Equation,

$\text{p}_1+\text{h}_1\rho\text{g}+\frac{1}{2}\rho\text{v}_1^2=\text{p}_2+\text{h}_2\rho\text{g}+\frac{1}{2}\rho\text{v}_2^2$

$\therefore\text{p}_1-\text{p}_2=\frac{1}{2}\rho\big(\text{v}_2^2-\text{v}_1^2\big)+\rho\text{g}(\text{h}_2-\text{h}_1)$

$\therefore\text{p} _1-\text{p}_2=\frac{1}{2}\times1\times\big(50^2-25^2\big)+1\times1000\Big(-\frac{15}{16}\Big)$

Let the pressure at the base of the back limb at A = P, and the pressure at the base of the front limb at B = Ps. The separation between the vertical limbs AB = (given). When the U- tube is accelerated at an acceleration a, the difference of pressure between horizontal points A and B is given as,

$\text{P}_\text{a}-\text{P}_\beta=\text{l}\rho\text{a}_0$

where p is the density of the liquid. Due to this pressure difference, the liquid in limb A will rise to a heighth more than in limb B. The pressure difference due to this height difference balances the difference of pressure between points A and B.

The pressure due to thhis height $\text{h}=\rho\text{gh},$ Equating we get,

$\rho\text{gh}=\text{p}_\text{a}-\text{p}_\beta=\text{l}\rho\text{a}_0$

$\Rightarrow\text{gh}=\text{la}_0$

$\text{h}=\text{a}_0\text{l}/\text{g}$

Given:

Volume rate of the flow of water = 1cc/sec.

= 1cm/sec.

Since, Volume rate of the flow of water is constant throughout the Horizontal pipe, therefore,

1. $\text{At}\ \text{A},$

$\text{Area}\times\text{velocity}=1\text{cm}^3/\text{s}.$

$\frac{4}{100}\text{cm}^2\times\text{velocity}=1$

$\therefore\text{velocity}=25\text{cm}/\text{seconds}.$

2. $\text{At}\ \text{B},$,

$\text{Area}\times\text{velocity}=1\text{cm}^3/\text{s}.$

$\frac{2}{100}\times\text{velocity}=1$

$\text{velocity}=50\text{cm}/\text{seconds}.$

3. Using the Bernoulli's equation,

$\text{p}_1+\text{h}_1\text{pg}+\frac{1}{2}\rho\text{v}_1^2 =\text{p}_2+\text{h}_2\rho\text{g}+\frac{1}{2}\rho\text{v}_2^2$

$\therefore\text{p}_1-\text{p}_2=\frac{1}{2}\rho\big(\text{v}_2^2-\text{v}_1^2\big)+\rho\text{g} \big(\text{h}_2-\text{h}_1\big)$

$\therefore\text{p}_1-\text{p}_2=\frac{1}{2}\rho\big(\text{v}_2^2-\text{v}_1^2\big)$ $\big[\because$ in case of the horizontal tube, height difference is zero.$\big]$

$\therefore\text{p}_1=\text{p}_2=\frac{1}{2}\times1\times\big(50^2-25^2\big)$

$\therefore\text{p}_1-\text{p}_2=937.5\ \text{dyne}/\text{cm}^2$

$\therefore\text{p}_1-\text{p}_2=93.75\ \text{pa}.$

1. When the system is completely Immersed in a liquid, then Upthrust and the spring force acts vertically upwards, but the weight of the object acts downwards,

Thus, In the State of the Equilibrium, we can say that,

Spring force + Upthrust = Weight

$\text{Kx}+\text{v}\rho\text{g}=\text{mg}$

$\therefore\text{Kx}=\text{mg}-\text{v}\rho\text{g}$

$\therefore\text{Kx}=0.5\times10-\frac{0.5}{800}\times1000\times10$

$\therefore\text{Kx}=5-\frac{50}{8}$

$\text{Kx}=-1.25$ [Minus sign means upward direction]

$\therefore\text{x}=0.025\text{m}$ or 2.5cm.

2. Let the block be depressed by 'a' cm.

The force on the block by the spring = 50a N

Hence the acceleration $=\frac{50\text{a}}{0.5}$

$=100\text{a}\ \text{m}/\text{s}^2$

Now, $\omega^2\text{a}=100$

$\therefore\omega=10\text{s}^{-1}$

Using the formula,

$\text{T}=\frac{2\pi}{\omega}$

$\therefore\text{T}=\frac{2\pi}{10}$

$\therefore\text{T}=\frac{\pi}{5}$ seconds or 0.63 seconds.

Hence, the time period of the S.H.M. is $\frac{\pi}{5}$ or 0.63 seconds.

1. Since, the tube is opened from the top, this means, total pressure of the gas will be equal to the pressure due to the mercury column + tmospheric Pressure.

$\therefore$ Pressure of the gas in cylinder = Atmospheric Pressure + Pressure due to the mercury column.

Now, Pressure due to the mercury column = ρg(h₂ - h₁)

where, ρ is the density of the mercury = 13.6g/cm³,

g is the acceleration due to gravity = 9.8m/s² = 980cm/s², h₂ is 8 cm and h₁ is 2cm.

$\therefore$ Total Pressure of the gas = 1 atm + (8 - 2) × 13.6 × 980

= 1.013 × 10⁵Pa + 6 × 13.6 × 980 dyne/cm²

= 101300 + 7996.8Pa.

= 181268Pa.

= 1.8 × 10⁵Pa.

2. The pressure of the mercury at the bottom of the U -tube.

= The pressure of mercury column on the open side + Atmospheric pressure.

= 13.6 × 10⁶/1000 × 9.8 × 0.08 + 1.013 × 10⁵

= 1.12 × 10⁵Pa.

1. The volume of the boat immersed in water will be that volume of water which weight = 50kg. Since the density of water = 1000kg/m?, this volume will be 

$=\frac{50}{1000}\text{m}^3=\frac{1}{20}\text{m}^3\cdot$

Hence the fraction of the volume of the boat immersed in water $=\frac{\big(\frac{1}{20}\text{m}^3\big)}{\big(1\text{m}^3\big)}=\frac{1}{20}\cdot$

2. In the leak condition, when the outer water level comes to the top of the side of the boat, the total volume of water displaced is equal to the internal volume of the

boat = 1m³. The force of buoyancy = weight of this water = 1000kg.

This force balances the weight of the boat plus the weight of the water inside. If the weight of the water inside be Xkg, then X + 50 = 1000Kg

$\Rightarrow\text{x}=950\text{Kg}$

Hence the volume of inside water$=\frac{50}{1000}\text{m}^3=0.950\text{m}^3$

The inside water volume fraction $=\frac{\big(0.950\text{m}^3\big)}{1\text{m}^3}$

$=\frac{19}{20}\cdot$

Given conditions,

density (p) = 1000Kg/m².

acceleration due to the gravity (g) = 10m/s².

height (h) = 4m

Using the Formula,

P = pgh.

$\therefore$ P = 1000 × 10 × 4N/m².

P = 40000N/m².

Hence, the pressure is 40,000

Let the hole be at a height h from the bottom. The height of the water above the hole = H - h. The speed of the water at the hole

$\text{v}=\sqrt{\{2\text{g}(\text{h}-\text{h})\}}$

Let the time taken by the water to strike the floor is t. Then for the vertical fall,

$\text{h}=0\times\text{t}+\frac{1}{2}\text{gt}^2$

$\Rightarrow\text{t}^2=2\text{h}/\text{g}$

$\Rightarrow\text{t}=\sqrt{(2\text{h}/\text{g})}$

Let the distance of the striking point from the hole be X. Then,

$\text{X}=\text{v}\times\text{t}=\sqrt{2\text{g}(\text{H}-\text{h})}\times\sqrt{(2\text{h}/\text{g})}$

$=\sqrt{\{2^2\times(\text{H}-\text{h})\text{h}\}}=2\sqrt{(\text{H}-\text{h})}\text{h}$

$\Rightarrow\text{X}=2\sqrt{(\text{H}-\text{h})}\times\sqrt{\text{h}}$ For maximum $\text{x},\frac{\text{dx}}{\text{dh}}=0$

$2\times\Bigg[\sqrt{(\text{H}-\text{h})}\big]\times\frac{\text{d}\sqrt{\text{h}}}{\text{dh}}+_\sqrt{\text{h}}\times\frac{\text{d}\sqrt{(\text{H}-\text{h})}}{\text{dh}}\Bigg]=0$

$\Rightarrow\sqrt{(\text{H}-\text{h})}\times\Big\{\frac{1}{2\sqrt{\text{h}}}\Big\}+\sqrt{\text{h}}\times\frac{1}{\Big\{2\sqrt{(\text{H}-\text{h})\times(-1)}\Big\}}=0$

$\Rightarrow\sqrt{(\text{H}-\text{h})}\times\Big\{\frac{1}{2\sqrt{\text{h}}}\Big\}+\sqrt{\text{h}}\times\frac{1}{\big\{2\sqrt{(\text{H}-\text{h})}\times(-1)\big\}}=0$

$\Rightarrow\frac{\sqrt{(\text{H}-\text{h})}}{(\sqrt{\text{h}})}=\frac{\sqrt{\text{h}}}{\big\{\sqrt{(\text{H}-\text{h})}\big\}}$

$\Rightarrow\text{H}-\text{h}=\text{h}$

$\Rightarrow2\text{h}=\text{H}$

$\Rightarrow\text{h}=\frac{\text{H}}{2}$

Hence the required height of the hole from the bottom for the greatest horizontal distance of water to strike is half the height of the water level.

The height difference $=2\text{cm}=0.02\text{m}$

The difference of pressure between A and B $=\text{p}_\text{a}-\text{p}_\beta=\rho\text{gh}$

$=\frac{1}{2}\rho \text{Va}^2=\text{p}_\beta+\frac{1}{2}\rho \text{V}\beta^2$

$=1000\times10\times0.02=200\text{N}/\text{m}^2$

Let the rate of flow of water be Qcc/s i.e. $=\text{Q}\times10^{-6}\text{m}^3/\text{s}$

Area of the cross-section at $\text{A}=4\text{cm}^2=\frac{4}{10000}\text{m}^2$

The speed at $\text{A}=\text{v}_\text{a}=\frac{\text{Q}\times10^{-6}}{\Big(\frac{4}{10000}\Big )}\text{m}/\text{s}=\frac{\text{Q}}{400}\text{m}/\text{s}$

Area of the cross-section at $\text{B}=2\text{cm}^2=\frac{2}{10000}\text{m}^2$

The speed at $\text{B}=\text{v}_\beta=\frac{\text{Q}\times10^{-6}}{\Big(\frac{2}{10000}\Big)}\text{m}/\text{s}=\frac{\text{Q}}{200}\text{m}/\text{s}$

Since the height of both the points are the same, from Bernoulli's theorem,

$\text{p}_\text{a}+\frac{1}{2}\rho\text{V}\text{a}^2=\text{p}\beta+\frac{1}{2}\rho\text{V}\beta^2$

$\Rightarrow\text{p}_\text{a}-\text{p}_\beta=\frac{1}{2}\rho\big(\text{V}\beta^2-\text{V}\text{a}^2\big)$

$=\frac{1}{2}\times1000\times\text{Q}^2\Big\{\Big(\frac{1}{2000}\Big)^2-\Big(\frac{1}{400}\Big)^2\Big\}$

$\Rightarrow\text{p}_\text{a}-\text{p}_\beta=\Big(\frac{1}{20}\Big)\times\text{Q}^2\Big\{\frac{1}{4}-\frac{1}{16}\Big\}=\frac{3\text{q}^2}{320}$

$\Rightarrow200=\frac{3\text{Q}^2}{320}$

$\Rightarrow\text{Q}^2=200\times\frac{320}{3}=21333$

$\Rightarrow\text{Q}=146\text{cc}/\text{s}$

The Discharge $\text{Q}=500\text{cm}^3/\text{s},$

The speed at the wide section $\text{V}_\text{a}=\frac{500}{5}=100\text{cm}/\text{s}=1\text{m}/\text{s}$

The speed at the narrow section $\text{V}_\beta=\frac{500}{2}=250\text{cm}/\text{s}=2.5\text{m}/\text{s}$

Let the difference in height levels of mercury $=\text{h}$

The pressure difference at the two points $=\rho'\text{gh}$

If the pressure at the wide section $=\text{p}_\text{a}$ and at the narrow section $=\text{p}\beta$ then from the Bernoulli's theorem

$\text{p}_\text{a}+\frac{1}{2}\rho\text{V}\text{a}^2=\text{p}_\beta+\frac{1}{2}\rho\text{V}\beta^2$

$\Rightarrow\frac{1}{2}\rho\big(\text{V}\beta^2-\text{V}\text{a}^2\big)=\text{p}_\text{a}-\text{p}_\beta$

$\Rightarrow\frac{1}{2}\times1000\times\big(2.5^2-1^2\big)=\rho'\text{gh}=13600\times9.8\times\text{h}$

$\Rightarrow\text{h}=10\times\frac{(6.25-1)}{(2\times136\times9.8)}\text{m}$

$\Rightarrow\text{h}=52.5\times\frac{100}{2665.6}\text{cm}$

$\Rightarrow\text{h}=\frac{5250}{2665.6}$

$=1.97\text{cm}$

Given mass of the metal piece, m = 160g = 0.160kg

Density of metal = 8000kg/m³

Here we can say that the weight of the metal piece = buoyant force(FB) + Normal force

i.e. weight of the metal piece = mg

Buoyant force, F_B = Volume of water displaced by metal × P_water × g

mg = FB + Normal force

Normal force = mg - F_B

= m × g - V × P_water × g

Volume of water displaced by metal. = $=\frac{\text{m}}{\text{pwater}}=\frac{0.160}{8000}=0.00002$ 

Normal force = (0.160 × 9.8) - (0.00002 × 1000 × 9.8)

= 1.568 - 0.196

= 1.372N

= 1.4N

The Pressure will be same whether the mercury is used or water is

used, there is only a difference in density of liquid and height of

Fluid.

For mercury, height is 76cm.

$\therefore\text{presure}=\text{h}\rho\text{g}$

$=7\times13.6\times\text{g}$

where 13.6g/cm² is the density of Mercury.

Now, let the height of the water column be h m.

$\therefore\text{P}=\text{h} \times\rho\text{g},$ where density of water is 1g/cm².

$\therefore\text{p}=\text{h}\times76\times13.6$

$\Rightarrow\text{h}=1033.6\text{cm}$

$\therefore\text{h}=10.336\text{m}$