Physics 1 Marks Question

Given:

Radius of capillary tube r = 0.5mm = 5 × 10^-4m

Depth (where pressure is to be found) h = 5.0cm = 5 × 10^-2m

Surface tension of water T = 0.075N/m

Excess pressure at 5cm before the surface:

$\text{P}=\rho_\text{hg}=1000\times(5\times10^{-2})\times9.8=490\text{N}/\text{m}^2$

Excess pressure at the surface is given by:

$\text{P}_0=\frac{2\text{T}}{\text{r}}=\frac{2\times(0.75)}{\big(5\times10^{-4}\big)}$

$=300\text{N}/\text{m}^2$

Difference in pressure: P₀ - P = 490 = 300 = 190N/m²

Hence, the required difference in pressure is 190N/m².

1. Length of the cylinder L = 2m

Area of cross-section $\text{A}=\pi\text{r}^2\Big(\frac{2}{100}\Big)^2\text{m}^2=\frac{4\pi}{10000}\text{m}^2$

Force on the cylinder $\text{F}=100\times10\text{N}=1000\text{N}$

Hence the stress $=\frac{\text{F}}{\text{A}}=\frac{1000}{\Big(\frac{4\pi}{10000}\Big)}\text{N}/\text{m}^2$

$=\Big(\frac{1}{4\pi}\Big)\times10^7\text{N}/\text{m}^2$

$=0.0796\times10^7\text{N}/\text{m}^2$

$=7.96\times10^5\text{N}/\text{m}^2$

2. $\text{Strain}=\frac{\text{Stress}}{\text{Y}}=\frac{\Big(\frac{\text{F}}{\text{A}}\Big)}{\text{Y}}$

$=\frac{\{7.96\times10^5\}}{2.0\times10^{11}}$

$=3.98^{-6}\approx4\times10^{-6}$

3. Let the comoression of the cyilender = lm

The strain $=\frac{\text{l}}{\text{L}}=\frac{\text{l}}{2}$

$\Rightarrow4\times10^{-6}=\frac{\text{l}}{2}$

$\Rightarrow\text{l}=8\times10^{-6}\text{m}$

Given:

Face area of steel plate A = 4cm² = 4 × 10^-4m²

Thickness of steel plate d = 0.5 cm = 0.5 × 10^-2m

Applied force on the upper surface F = 10N

Rigidity modulus of steel = 8.4 × 10¹⁰Nm^-2

Let $\theta$ be the angular displacement,

Rigidity modulus $\text{m}=\frac{\text{F}}{\text{A}\theta}$

$\Rightarrow\text{m}=\Big(\frac{10}{4\times10^{-4}\theta}\Big)$

$\Rightarrow\theta=\frac{10}{4\times10^{-4}\times 8.4\times10^{10}}$

$=0.297\times10^{-6}$

$\therefore$ Lateral displacement of the upper surface with respect to the lower surface $=\theta\times\text{d}$

$\Rightarrow(0.297)\times10^{-6}\times(0.5)\times10^{-2}$

$\Rightarrow1.5\times10^{-9}\text{m}$

Hence, the required lateral displacement of the steel plate is 1.5 × 10^-9m.

It means the maximum stress in the wire $\sigma =8\times 10^8\text{N}/\text{m}^2$

Y = 2.0 × 10¹¹N/ m²

Length of the wire L = 0.50m

Let corresponding maximum elongation = lm

Strain $\in=\frac{\text{l}}{\text{L}}=\frac{\text{l}}{0.50}=2\text{l}$

we have $\frac{\sigma}{\in}=\text{Y}$

$\Rightarrow\in=\frac{\sigma}{\text{Y}}$

$\Rightarrow2\text{l}=8\times\frac{10^8}{2.0}\times10^{11}$

$\Rightarrow\text{l}=\frac{2}{100}\text{m}=2\text{mm}$

Given:

Surface tension of water T = 7.5 × 10^-2N/m

Taking $\cos\theta=1$

Radius of capillary A(r_A) = 0.5mm = 0.5 × 10^-3m

Height of water level in capillary A:

$\text{h}_\text{A}=\frac{2\text{T}\cos\theta}{\text{r}_\text{A}\rho\text{g}}$

$=\frac{2\times7.5\times10^{-2}}{0.5\times10^{-3}\times1000\times10}$

$=3\times10^{-2}\text{m}=3\text{cm}$

Radius of capillary B(r_B) =1mm = 1 × 10^-3m

Height of water level in capillary B:

$\text{h}_\text{B}=\frac{2\text{T}\cos\theta}{\text{r}_\text{B}\rho\text{g}}$

$=\frac{2\times7.5\times10^{-2}}{1\times10^{-3}\times10^3\times 10}$

$=15\times10^{-3}\text{m}=1.5\text{cm}$

Radius of capillary C(r_C) = 1.5mm = 1.5 × 10^-3m
Height of water level in capillary C:

$\text{h}_\text{C}=\frac{2\text{T}\cos\theta}{\text{r}_\text{C}\rho_\text{g}}$

$=\frac{2\times7.5\times10^{-2}}{1.5\times10^{-3}\times10^{3}\times10}$

$=\frac{15}{1.5}\times10^{-3}\text{m}=1\text{cm}$

$\text{m}_2\text{g}-\text{T}=\text{m}_2\text{a}$

$\Rightarrow\text{a}=\text{g}-\frac{\text{T}}{\text{m}_2}$

$\text{T}-\text{F}=\text{m}_1\text{a}$

$\Rightarrow\text{T}=\text{m}_1\text{a}+\text{F}$

$\Rightarrow\text{T}=\text{M}_1\text{a}+\frac{\text{m}_2\text{g}}{2}$

$\Rightarrow\text{T}=\text{m}_1\text{g}-\text{m}_1\frac{\text{T}}{\text{m}_2}+\frac{\text{m}_2\text{g}}{2}$ {Putting the value of a}

$\Rightarrow\text{T}\frac{(\text{m}_1+\text{m}_2)}{\text{m}_2}=\frac{(2\text{m}_1+\text{m}_2)\text{g}}{2}$

$\Rightarrow\text{T}=\text{m}_2\frac{(2\text{m}_1+\text{m}_2)\text{g}}{2(\text{m}_1+\text{m}_2)}$

Length of the rod = L = 2m

The length increased during the night = 1 = 0.1cm = 0.001m

The strain developed during the night $=\frac{\text{l}}{\text{L}}=\frac{0.001}{2}=0.0005$

If the tension developed = TN

The stress developed $=\frac{\text{T}}{\text{A}}\text{N}/\text{m}^2$

$=\frac{\text{T}}{0.0004}\text{N}/\text{m}^2$

But stress = strain × y

$\frac{\text{T}}{0.0004}=0.0005\times1.9\times10^{11}\text{N}$

$\Rightarrow\text{T}=0.0004\times0.0005\times1.9\times10^{11}\text{N}$

$\Rightarrow\text{T}=2\times10^{-7}\times1.9\times10^{11}\text{N}$

$\Rightarrow\text{T}=3.8\times10^4\text{N}$

1. Excess pressure inside mercury drop:

$\text{P}=\frac{2\text{T}_\text{Hg}}{\text{r}}$

$=\frac{0.465\times2}{2\times10^{-3}}=465\text{N}/\text{m}^2$

2.  Excess pressure inside the soap bubble:

$\text{P}=\frac{4\text{T}_\text{S}}{\text{r}}$

$=\frac{4\times0.03}{4\times10^{-3}}=30\text{N}/\text{m}^2$

3. Excess pressure inside the air bubble:

$\text{P}=\frac{4\text{T}_\text{a}}{\text{r}}$

$=\frac{2\times0.076}{4\times10^{-3}}=38\text{N}/\text{m}^2$

1. Force exerted by air on the surface area:

$\text{F}=\text{P}_0\text{A}$

$\Rightarrow\text{F}=1.0\times10^5\times10^{-6}=0.1\text{N}$

2. Force exerted by mercury below the surface area:

Pressure $\text{P}'=\text{P}_0+\frac{2\text{T}}{\text{r}}$

$\text{F}=\text{P}'\text{A}=\Big(\text{P}_0+\frac{2\text{T}}{\text{r}}\Big)\text{A}$

$=\big(0.1+\frac{2\times0.465}{4\times10^{-3}}\Big)\times10^{-6}$

$=1+0.00023=0.10023\text{N}$

3. Force exerted by mercury surface in contact with it:

$\text{P}=\frac{2\text{T}}{\text{r}}$

$\text{F}=\text{PA}=\frac{2\text{T}}{\text{r}}\text{A}$

$=\frac{2\times0.465}{4\times10^{-3}}\times10^{-6}=0.00023\text{N}$

Given:

Bulk modulus of water B = 2.1 × 10⁹Nm^-2

In order to decrease the volume (V) of a water sample by 0.01%, let the increase in pressure be P.

$\frac{\text{v}\times0.1}{100}=\triangle\text{V}$

$\Rightarrow\frac{\triangle\text{V}}{\text{V}}=10^{-4}$

From B $=\frac{\text{P}\text{V}}{\triangle\text{V}},$ We have:

$\Rightarrow\text{P}= \text{B}\Big(\frac{\triangle\text{V}}{\text{V}}\Big)$

$=2.1\times10^{9}\times10^{-4}$

$=2.1\times10^{5}\text{N}/\text{m}^2$

Hence, the requred increase in pressure is 2.1 × 10⁵Nm^-2.

1. Let T be the surface tension and ρ be the density of the liquid.

Then, for $\cos\theta=1,$ height (h) of liquid level:

$\text{h}=\frac{2\text{T}}{\text{r}\rho_\text{g}}\ \cdots(1)$

where g is the acceleration due to gravity:

$\Rightarrow\text{h}=\frac{2\times(0.076)}{10^{-3}\times10\times100}$

$=1.52\text{cm}$

$=1.52\times10^{-2}\text{m}$

$=1.52\text{cm}$

2. Let the new length of the tube be h'.

$\text{h}'=\frac{2\text{T}\cos\theta}{\text{r}\rho\text{g}}$

$\cos\theta=\frac{\text{h'}\text{r}\rho\text{g}}{2\text{T}}$

Using equation (1), we get:

$\cos\theta=\frac{\text{h}'}{\text{h}}=\frac{1}{2}$ $\Big($Because $\text{h}'=\frac{\text{h}}{2}\Big)$

$\Rightarrow\theta=\cos^{-1}\Big(\frac{1}{2}\Big)=60^\circ$

The water surface in the capillary makes an angle of 60^∘ with the wall.

$\Big(\frac{4}{3}\Big)\pi\text{R}^3=\Big(\frac{4}{3}\big)\pi\text{r}^3\times8$

$\text{r}=\frac{\text{R}}{2}=10^{-3}\text{m}$

$\therefore$ Increase in surface energy = TA' - TA

$=(8\times4\pi\text{r}^2-4\pi\text{R}^2)\text{T}$

$=4\pi\text{T}\Big[8\times\Big(\frac{\text{R}^2}{4}\Big)-\text{R}^2\Big]$

$=4\pi\text{T}\text{R}^2$

$=4\times(3.14)\times(0.465)\times\big(4\times10^{-6}\big)$

$=23.36\times10^{-6}$

$=23.4\mu\text{j}$