2 Marks Questions - Physics STD 12 Science Questions - Vidyadip

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Question 12 Marks

A copper wire of cross-sectional area $0.01cm^2$ is under a tension of 20N. Find the decrease in the cross-sectional area. Young's modulus of copper $= 1.1 × 10^{11}N/m^2$ and Poisson's ratio = 0.32.$\Big[$Hint: $\frac{\triangle\text{A}}{\text{A}}=2\frac{\triangle\text{r}}{\text{r}}\Big]$

Answer

Given:
Cross-sectional area of copper wire $A = 0.01cm^2= 10^{-6}m^2$
Applied tension $T = 20N$
Young modulus of copper $Y = 1.1 × 10^{11}N/m^2$
Poisson ratio $\sigma=0.32$
We know that:
$\text{Y}=\frac{\text{FL}}{\text{A}\triangle\text{L}}$
$\Rightarrow\frac{\triangle\text{L}}{\text{L}}=\frac{\text{F}}{\text{AY}}$
$=\frac{20}{10^{-6}\times1.1\times10^{11}}=18.18\times10^{-5}$
Poisson's ratio, $\sigma =\frac{\frac{\triangle\text{d}}{\text{d}}}{\frac{\triangle\text{L}}{\text{L}}}=0.32$
Where d is the transverve length
So, $\frac{\triangle\text{d}}{\text{d}}=(0.32)\times\frac{\triangle\text{L}}{\text{L}}$
$=0.32\times (18.18)\times10^{-5}=5.81\times10^{-5}$
Again, $\frac{\triangle\text{A}}{\text{A}}=\frac{2\triangle \text{r}}{\text{r}}=\frac{2\triangle \text{d}}{\text{d}}$
$\Rightarrow\triangle\text{A}=\frac{2\triangle\text{d}}{\text{d}}\text{A}$
$\Rightarrow\triangle\text{A}=2\times(5.8\times10^{-5})\times(0.01)$
$=1.165\times10^{-6}\text{cm}^2$
Hence, the required decrease in the cross-sectional area is $1.164 × 10^{-6}cm^2$.

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Question 22 Marks

A barometer is constructed with its tube having radius 1.0mm. Assume that the surface of mercury in the tube is spherical in shape. If the atmospheric pressure is equal to 76cm of mercury, what will be the height raised in the barometer tube. The contact angle of mercury with glass $= 135^\circ$ and surface tension of mercury = 0.465N/m. Density of mercury $= 13600kg/m^3$.

Answer

Given: Radius of tube $r = 1.0mm$ Atmospheric pressure = 76cm of Hg Contact angle of mercury with glass $\theta=135^\circ$ Surface tension of mercury $T = 0.465N/m$ Density of mercury $= 13600kgm^{-3}$ Let h be the rise in level in the barometer.$\text{h}=\frac{2\text{T}\cos\theta}{\text{r}\rho_\text{g}}$
$=\frac{2\times465\times\Big(\frac{1}{\sqrt{2}}\Big)}{10^{-3}\times13600\times10}$
$=0.48\text{cm}$
$\therefore$ Net rise in level in the barometer tube $= H - h$
$= 76 - 0.48$
$= 75.52cm$

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Question 32 Marks

Two large glass plates are placed vertically and parallel to each other inside a tank of water with separation between the plates equal to 1mm. Find the rise of water in the space between the plates. Surface tension of water = 0.075N/m.

Answer

Given: Surface tension of water T = 0.075N/m Separation between the glass plates $d = 1mm = 10^{-3}$ m Density of water $\rho=10^3\text{Kg}/\text{m}^3$ Applying law of conservation of energy:$\text{T}(2\text{L})=[1\times\big(10^{-3}\big)\times\text{h}]\rho_\text{g}$
$\Rightarrow\text{h}=\frac{2\times(0.075)}{10^{-3}\times10^{3}\times10}$
$=0.015\text{m}=1.5\text{cm}$
Therefore, the rise of water in the space between the plates is 1.5cm.

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Question 42 Marks

A wire forming a loop is dipped into soap solution and taken out so that a film of soap solution is formed. A loop of 6.28cm long thread is gently put on the film and the film is pricked with a needle inside the loop. The thread loop takes the shape of a circle. Find the tension in the thread. Surface tension of soap solution = 0.030N/m.

Answer

Given: Surface tension of soap solution $T = 0.030N/m^{-1}$ Let the radius of the thread loop be r.$\Rightarrow2\pi\text{r}=6.28\text{cm}$
$\Rightarrow\text{r}=\frac{6.28}{2\times3.14}=1\text{cm}$
The excess pressure inside the loop is expressed as follows:$\triangle\text{P}=\frac{4\text{T}}{\text{r}}$
Tension in the thread:$\text{T}'=\triangle\text{P}\times(\text{area}\ \text{of}\ \text{loop})$
$\Rightarrow\text{T}=\frac{4\text{T}}{\text{r}}\times\pi\text{r}^2$
$\Rightarrow\text{T}'=4\pi\text{r}$
$=4\times0.030\times3.14\times10^{-2}\text{N}$
$=3.8\times10^{-3}\text{N}$

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Question 52 Marks

A uniform vertical tube of circular cross-section contains a liquid. The contact angle is $90^\circ$. Consider a diameter of the tube lying in the surface of the liquid. The surface to the right of this diameter pulls the surface on the left of it. What keeps the surface on the left in equilibrium?

Answer

As the angle of contact is 0, there is no force between the surface of the tube and the liquid. The diameter of the liquid surface is pulled on both sides by equal and opposite forces of surface tension. This results in no net force remaining on the surface of the liquid. Hence, the liquid stays in equilibrium.

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Question 62 Marks

If water in one flask and castor oil in other are violently shaken and kept on a table, which will come to rest earlier?

Answer

Castor oil will come to rest more quickly because it has a greater coefficient of viscosity than water.
Castor oil has a higher viscosity than water. It will therefore, lose kinetic energy and come to rest faster than water.

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Question 72 Marks

The contact angle between water and glass is $0^\circ$. When water is poured in a glass to the maximum of its capacity, the water surface is convex upward. The angle of contact in such a situation is more than $90^\circ$. Explain.

Answer

When water is poured in a glass, it reaches the brim and rises further. The edge of the glass lies below the water level. In this case, the force of attraction due to molecules of the glass surface is not perpendicular to the solid. Here, the contact angle can be greater than the standard contact angle for a pair of substances.

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Question 82 Marks

Estimate the speed of vertically falling raindrops from the following data. Radius of the drops $=0.02 \mathrm{~cm}$, viscosity of air $=1.8 \times 10^{-4}$ poise, $g=9.9 \mathrm{~m} / \mathrm{s}^2$ and density of water $=1000 \mathrm{~kg} / \mathrm{m}^3$.

Answer

Given:
Radius of the drops $\mathrm{r}=0.02 \mathrm{~cm}=2 \times 10^{-4} \mathrm{~m}$
Viscosity of air $\eta=1.8 \times 10^{-4}$ poise $=1.8 \times 10^{-5}$ decapoise
Acceleration due to gravity $\mathrm{g}=9.9 \mathrm{~m} / \mathrm{s}^2$
Density of water $\rho=1000\text{Kg}/\text{m}^3$
Let v be the terminal velocity of a drop.
The forces acting on the drops are

The weight mgmg acting downwards.
The force of buoyance, i.e., $\big(\frac{4}{3}\Big)\pi\text{r}^3\ \rho\text{g}$ acting upwards.
The force of viscosity, i.e., $6\pi\eta\text{r}\text{v}$ acting upwards.

Because the density of air is very small, the force of buoyance can be neglected.
From the free body diagram:

$6\pi\eta\text{r}\text{v}=\text{mg}$

$6\pi\eta\text{rv}=\frac{4}{3}\pi\text{r}^3\rho_\text{g}$

$\text{v}=2\text{r}^2\frac{\rho\text{g}}{9\eta}$

$=2\big(0.02\times10^{-2}\big)^2\times1000\times\frac{(9.9)}{9}\times\big(1.8\times10^{-5}\big)$

$=5\text{m}/\text{s}.$

Hence, the required vertical speed of the falling raindrops is 5m/s.

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Question 92 Marks

When rubber sheets are used in a shock absorber, what happens to the energy of vibration?

Answer

The energy of vibration dissipates as heat from the shock absorber.

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Question 102 Marks

Consider an ice cube of edge 1.0cm kept in a gravity free hall. Find the surface area of the water when the ice melts. Neglect the difference in densities of ice and water.

Answer

Given: Edge of the ice cube a = 1.0cm The water that is formed due to the melting of ice acquires a spherical surface. In the absence of gravity, let the radius of the spherical surface be r. Volume of ice cube = volume of spherical surface of water$\Rightarrow\text{a}^3=\frac{4}{3}\pi\text{r}^3$
$\Rightarrow\text{r}=\Big[\frac{3\text{a}^3}{4\pi}\Big]^{\frac{1}{3}}$
Surface area of spherical water surface $=4\pi\text{r}^2$$=4\pi\Big[\frac{3\text{a}^3}{4\pi}\Big]^{\frac{2}{3}}$
$=(36\pi)^{\frac{1}{3}}\text{cm}^2$

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Question 112 Marks

In the upper wire is made of steel and the lower of copper. The wires have equal cross-section. Find the ratio of the longitudinal strains developed in the two wires.

Answer

Since both wires have equal cross-sections, the stress in both wires will be the same. Let the elongation in steel wire be l and in copper wire = l'. Since both the wires have the same length, say L, the strain in steel wire $=\frac{\text{l}}{\text{L}}$ and the strain in copper wire $=\frac{\text{l}'}{\text{L}}.$ If the modulus of elasticity of steel wire is Y and the modulus of elasticity of copper = Y', then strain in steel $\frac{\text{wire}}{\text{steel}}$ in copper wire

$=\frac{\Big(\frac{\text{stress}\ \text{in}\ \text{steel}\text{wire}}{\text{Y}}\Big)}{\Big(\frac{\text{stress} \ \text{in} \ \text{copper}}{\text{Y}'}\Big)}$

$=\frac{\text{Y'}}{\text{Y}}$

$=\frac{2.0\times10^{11}}{1.3\times10^{11}}$ {Taking the values from the previous problem}

$=\frac{2.0}{1.3}$

$=\frac{20}{13}$

$=1.54$

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Question 122 Marks

The contact angle between pure water and pure silver is $90^\circ$. If a capillary tube made of silver is dipped at one end in pure water, will the water rise in the capillary?

Answer

No, the water will neither rise nor fall in the silver capillary. According to Jurin's law, the level of water inside a capillary tube is given by$\text{h}=\frac{2\text{T}\cos\theta} {\text{r}\rho\text{g}}$
Here, $\theta=90^\circ$$\Rightarrow\text{h}=\frac{2\text{T}\cos90^\circ}{\text{r}\rho\text{g}}$
$\Rightarrow\text{h}=0$
Thus, the water level neither rises nor falls.

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Question 132 Marks

The yield point of a typical solid is about 1%. Suppose you are lying horizontally and two persons are pulling your hands and two person are pulling your legs along your own length. How much will be the increase in your length if the strain is 1%? Do you think your yield point is 1% or, much less than that?

Answer

Let my length = L Let the increase in length = l Strain$=\frac{\text{l}}{\text{L}}=\frac{1}{100}$
$\Rightarrow\text{l}=\frac{\text{L}}{100}$
So, the increase in length will be $\frac{\text{L}}{100}.$ Yes, the yield point is much less than the 1% strain because the human body consists of joints and not one uniform solid structure.

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Question 142 Marks

It is said that a liquid rises or is depressed in a capillary due to the surface tension. If a liquid neither rises nor depresses in a capillary, can we conclude that the surface tension of the liquid is zero?

Answer

No, we cannot conclude the surface tension to be zero solely by the fact that the liquid neither rises nor falls in a capillary. The height of the liquid inside a capillary tube is given by.$\text{h}=\frac{2\text{T}\cos\theta}{\text{r}\rho\text{g}}$
From the equation, we see that the height (h) of the liquid may also be zero if the contact angle $\theta$ between the liquid and the capillary tube is 90º or 270º

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Question 152 Marks

The lower end of a capillary tube is immersed in mercury. The level of mercury in the tube is found to be 2cm below the outer level. If the same tube is immersed in water, upto what height will the water rise in the capillary?

Answer

Let T be the surface tension, r be the inner radius of the capillary tube and $\rho$ be the density of the liquid. For $\cos\theta=1$ height (h) of the liquid level is given as:$\text{h}=\frac{2\text{T}\cos\theta}{\text{r}\rho\text{g}}$
Now, for mercury:$\text{h}_\text{Hg}=\frac{2\text{T}_{\text{Hg}}}{\text{r}\rho_\text{Hg}\ \text{g}}\ \cdots(1)$
For water:$\text{h}_\text{w}=\frac{2\text{T}_\text{w}}{\text{r}\rho_\text{w}\text{g}}\ \cdots(2)$
$\frac{\text{h}_\text{w}}{\text{h}_\text{Hg}}=\frac{\text{T}_\text{w}}{\text{T}_\text{Hg}}\times\frac{\rho_\text{Hg}}{\rho_\text{w}}$
$=\big(\frac{0.075}{0.465}\Big)\times(13.6)$
$=2.19$
Height of the water level: $h_w= 2 × 2.19 = 4.38cm$ Hence, the required rise in the water level in the capillary tube is 4.38cm.

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Question 162 Marks

A load of 10kg is suspended by a metal wire 3m long and having a cross-sectional area $4mm^2$. Find.

The stress.
The strain and.
The elongation. Young's modulus of the metal is $2.0 × 10N^{11}N/ m^2.$

Answer

Length of wire L = 3m,
Load $F = 10 × 10N = 100N, $ $\{$Taking $g = 10m/s^2\} $

Area of cross-section $A = 4mm^2$

$\Rightarrow\text{A}=4\times10^{-6}\text{m}^2$

The stress = Load (force) on unit area of cross-section $=\frac{\text{F}}{\text{A}}$

$=\frac{100}{4}\times10^{-6}\text{N}/\text{m}^2$

$=25\times10^6\text{N}/\text{m}^2$

$=2.5\times10^7\text{N}/\text{m}^2$

Let the elongation of the wire under this stress bel, The strain $=\frac{\text{l}}{\text{L}},$ Young's modulus of the metal $Y = 2.0 × 10^{11}N/m^2$ We

have $\frac{\text{Stress}}{\text{Strain}}=\text{Y}$ (constant)

$\Rightarrow\frac{\Big(\frac{\text{F}}{\text{A}}\Big)}{\text{Strain}}= \text{Y}$

$\Rightarrow\text{Strain}= \frac{\Big(\frac{\text{F}}{\text{A}}\Big)}{\text{Y}}= 2.5 \times\frac{10^7}{2.0}\times10^{11}$

$\Rightarrow\text{strain=1.25}\times10^{-4}\text{m}$

strain $=\frac{\text{l}}{\text{L}}=1.25\times10^{-4}$

$\Rightarrow\text{l}=3.0\times1.25\times10^{-4}\text{m}$

$=3.75\times10^{-4}\text{m}$

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Question 172 Marks

A steel wire and a copper wire of equal length and equal cross-sectional area are joined end to end and the combination is subjected to a tension. Find the ratio of.

The stresses developed in the two wires.
The strains developed. Y of steel $= 2 \times 10^{11}N/m^2.$ Y of copper $= 1.3 \times 10^{11}N/m^2.$

Answer

Given:Young's modulus of steel $= 2 \times 10^{11} N m^{-2}$
Young's modulus of copper $= 1.3 \times 10^{11} N m^{-2}$
Both wires are of equal length and equal cross-sectional area. Also, equal tension is applied on them.
As per the question:
$L_{steel} = L_{cu}$
$A_{steel} = A_{cu}$
$F_{cu} = F_{steel}$
Here: $L_{steel}$ and $L_{Cu}$ denote the lengths of steel and copper wires, respectively. $A_{steel}$ and $A_{Cu}$ denote the cross-sectional areas of steel and copper wires, respectively.
$F_{steel}$ and $F_{Cu}$ denote the tension of steel and cooper wires, respectively.

$\frac{\text{Stress of cu}}{\text{Stress of steel}}=\frac{\text{F}_\text{Cu}}{\text{A}_\text{Cu}}\frac{\text{A}_{\text{steel}}}{\text{F}_{\text{steel}}}=1$
$\frac{\text{Strain of cu}}{\text{Strain of steel}}=\frac{\frac{\Delta\text{L}_\text{steel}}{\text{L}_\text{steel}}}{\frac{\Delta\text{L}_\text{Cu}}{\text{L}_\text{Cu}}}=\frac{\text{F}_{\text{steel}}\text{L}_{\text{steel}}\text{A}_{\text{Cu}}\text{Y}_{\text{Cu}}}{\text{A}_{\text{steel}}\text{Y}_{\text{steel}}\text{F}_{\text{Cu}}\text{L}_{\text{Cu}}}$

$\Big(\text{Using} \ \frac{\Delta\text{L}}{\text{L}}=\frac{\text{F}}{\text{AY}}\Big)$
$\Rightarrow\frac{\text{Strain of cu}}{\text{Strain of steel}}=\frac{\text{Y}_\text{Cu}}{\text{Y}_\text{steel}}=\frac{1.3\times10^{11}}{2\times10^{11}}$
$\Rightarrow\frac{\text{Strain of cu}}{\text{Strain of steel}}=\frac{13}{20}$
$\Rightarrow\frac{\text{Strain of steel}}{\text{Strain of Cu}}=\frac{20}{13}$
Hence, the required ratio is $20 : 13.$

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Question 182 Marks

A metal sphere of radius 1mm and mass 50mg falls vertically in glycerine. Find:

The viscous force exerted by the glycerine on the sphere when the speed of the sphere is 1cm/s.
The hydrostatic force exerted by the glycerine on the sphere and.
The terminal velocity with which the sphere will move down without acceleration. Density of glycerine = $1260kg/m^3$ and its coefficient of viscosity at room temperature = 8.0 poise.

Answer

Given: Radius of metallic sphere $r = 1 mm = 10^{-3}m$ Speed of the sphere $v = 10^{-2}m/ s$ Coefficient of viscosity $\eta =8$ poise = 0.8 decapoise Mass $m = 50 mg = 50 \times 10^{-3}kg$ Density of glycerin $\sigma = 1260\text{kg}/ \text{m}^3$

Viscous force exerted by glycerine on the sphere $\text{F}=6\pi\eta\text{rv}$

$\Rightarrow \text{F}=6\times (3.14)\times(0.8)\times 10^{-3}\times (10^{-2})$
$=1.50\times 10^{-4}\text{N}$

Let V be the volume of the sphere.

Hydrostatic force exerted by glycerin on the sphere $\text{F}'=\text{V}\sigma\text{g}$
$\Rightarrow \text{F}'=\frac{4}{3}\pi\text{r}^2\sigma \text{g}$
$=\Big(\frac{4}{3}\Big)\times (3.14)\times (10^{-6})\times 1260\times 10$
$=5.275\times 10^5\text{N}$

Let the terminal velocity of the sphere be v'.

The forces acting on the drops are.

The weight mg acting downwards.
The force of buoyance, i.e., $\frac{4}{3}\pi \text{r}^3\sigma \text{g}$ acting upwards.
The force of viscosity, i.e., $6\pi \eta\text{rv}'$ acting upwards.

From the free body diagram:

$6\pi \eta \text{ev}'+\frac{4}{3}\pi\text{r}^3\sigma\text{g}=\text{mg}$
$\Rightarrow \nu =\frac{\text{mg}-\frac{4}{3}\pi\text{r}^2\sigma\text{g}}{6\pi\eta\text{r}}$
$=\frac{50\times10^{-3}-\frac{4}{3}\times3.14\times10^{-6}\times1260\times10}{6\times3.14\times0.8\times10^{-3}}$
$=\frac{500-\frac{4}{3}\times3.14\times10^{-3}\times1260\times10}{6\times3.14\times0.8}$
$=2.3\text{cm}/\text{s}$

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Question 192 Marks

If a compressed spring is dissolved in acid, what happened to the elastic potential energy of the spring?

Answer

When a compressed spring dissolves in an acid, the acid molecules leave the sold lattice of the spring faster than the uncompressed spring. This in turn increases the kinetic energy of the solution. As a result, the temperature of the acid also increases. However, this temperature increase will be very small because the mechanical energy content in the spring is lesser than its chemical energy content.

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2 Marks Questions - Physics STD 12 Science Questions - Vidyadip