Properties of Matter — Page 12 | JEE Physics

Q111

Choose the correct relationship between Poisson ratio

(\sigma)

, bulk modulus (K) and modulus of rigidity

(\eta)

of a given solid object :

A

\sigma=\dfrac{3 K+2 \eta}{6 K+2 \eta}

B

\sigma=\dfrac{3 K-2 \eta}{6 K+2 \eta}

C

\sigma=\dfrac{6 K+2 \eta}{3 K-2 \eta}

D

\sigma=\dfrac{6 K-2 \eta}{3 K-2 \eta}

Correct Answer

Option B

Solution

Poisson ratio ( $\sigma$ ), bulk modulus (K) and modulus of rigidity ( $\eta$ ) are related by

\because 2\eta(1+\sigma)=3K(1-2\sigma)

2\eta+2\eta\sigma=3K-6K\sigma

\sigma=\frac{3K-2\eta}{2\eta+6K}

Q112

A fully loaded boeing aircraft has a mass of

5.4\times10^5

kg. Its total wing area is 500 m

^2

. It is in level flight with a speed of 1080 km/h. If the density of air

\rho

is 1.2 kg m

^{-3}

, the fractional increase in the speed of the air on the upper surface of the wing relative to the lower surface in percentage will be. (

\mathrm{g=10~m/s^2}

)

A 16

B 8

C 6

D 10

Correct Answer

Option D

Solution

Velocity of aircraft = 1050 km/h = 300 m/s Now, weight of aircraft =

\Delta PA

\Delta P = {{5.4 \times {{10}^5} \times g} \over {500}} = 10800

\mathrm{Pa}

From Bernoulli's principle

\Delta P = {1 \over 2}\rho \left[ {V_{upper}^2 - V_{lower}^2} \right]

10800 = {1 \over 2} \times 1.2 \times V_{lower}^2\left[ {{{\left( {{{{V_{upper}}} \over {{V_{lower}}}}} \right)}^2} - 1} \right]

{\left( {{{{V_{upper}}} \over {{V_{lower}}}}} \right)^2} = 1 + {{10800 \times 2} \over {1.2 \times {{(300)}^2}}} = 1.2

{{{V_{upper}}} \over {{V_{lower}}}} = 1.096

$\Rightarrow$ Fractional increases = 9.6%

Q113

Surface tension of a soap bubble is

2.0 \times 10^{-2} \mathrm{Nm}^{-1}

. Work done to increase the radius of soap bubble from

3.5 \mathrm{~cm}

to

7 \mathrm{~cm}

will be: Take

\left[\pi=\dfrac{22}{7}\right]

A

18 .48 \times 10^{-4} \mathrm{~J}

B

5.76 \times 10^{-4} \mathrm{~J}

C

0.72 \times 10^{-4} \mathrm{~J}

D

9.24 \times 10^{-4} \mathrm{~J}

Correct Answer

Option A

Solution

Surface area of soap bubble $=2 \times 4 \pi \mathrm{R}^{2}$ Work done $=$ change in surface energy $\times \mathrm{T}_{\mathrm{S}}$ $=\mathrm{T}_{\mathrm{S}} \times 8 \pi \times\left(\mathrm{R}_{2}^{2}-\mathrm{R}_{1}^{2}\right)$ $=2 \times 10^{-2} \times 8 \times \dfrac{22}{7} \times 49 \times \dfrac{3}{4} \times 10^{-4}$ $=18.48 \times 10^{-4} \mathrm{~J}$

Q114

A bicycle tyre is filled with air having pressure of

270 ~\mathrm{kPa}

at

27^{\circ} \mathrm{C}

. The approximate pressure of the air in the tyre when the temperature increases to

36^{\circ} \mathrm{C}

is

A 262 kPa

B 360 kPa

C 270 kPa

D 278 kPa

Correct Answer

Option D

Solution

$\mathrm{P}_{\text{in }}=270 \mathrm{kPa}, \mathrm{T}_{\text{in }}=27^{\circ} \mathrm{C}$ $=300 \mathrm{~K}$

\mathrm{T}_{\text{final }}=36^{\circ} \mathrm{C}=309 \mathrm{~K}

Hence we can consider process to be isochoric volume constant $\therefore P \propto T$

\frac{P_{\text{in }}}{P_{f}}=\frac{T_{\text{in }}}{T_{f}} \Rightarrow P_{f}=278 ~\mathrm{kPa}

Q115

A bowl filled with very hot soup cools from 98

^\circ

C to 86

^\circ

C in 2 minutes when the room temperature is 22

^\circ

C. How long it will take to cool from 75

^\circ

C to 69

^\circ

C?

A 2 minutes

B 0.5 minute

C 1.4 minutes

D 1 minute

Correct Answer

Option C

Solution

From Newton's law of cooling.

\frac{d T}{d t}=-k\left(T-T_{s}\right)

Case $\mathrm{I}: d T=12^{\circ} \mathrm{C}, d t=2 \min$

\frac{12}{2}=-k\left[92-22^{\circ}\right]=-k 70

Case II : $d T=6^{\circ} \mathrm{C}$

\frac{6}{d t}=-k[72-22]=-k 50

From (1) and (2)

d t=1.4 \mathrm{~min}

Q116

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R Assertion A : Steel is used in the construction of buildings and bridges. Reason R : Steel is more elastic and its elastic limit is high. In the light of above statements, choose the most appropriate answer from the options given below

A Both A and R are correct and R is the correct explanation of A

B A is correct but R is not correct

C Both A and R are correct but R is NOT the correct explanation of A

D A is not correct but R is correct

Correct Answer

Option A

Solution

Assertion A states that steel is used in the construction of buildings and bridges, which is true.

Steel is a widely used material for construction due to its high strength, durability, and resistance to corrosion.

Reason R states that steel is more elastic and has a higher elastic limit compared to other common construction materials like concrete, and this is why it is preferred in construction.

This statement is also true.

Steel's high elasticity and high elastic limit make it an ideal material for use in construction as it can withstand greater stress before it becomes permanently deformed.

Therefore, both Assertion A and Reason R are true statements.

However, to determine the most appropriate answer, we need to see if Reason R explains Assertion A.

Reason R does indeed explain why steel is used in the construction of buildings and bridges.

Its high elasticity and high elastic limit allow it to withstand greater stress than other common construction materials before becoming permanently deformed, making it a preferred material for construction.

So, the correct answer is Both A and R are correct, and R is the correct explanation of A.

Q117

The frequency (

\nu

) of an oscillating liquid drop may depend upon radius (

r

) of the drop, density (

\rho

) of liquid and the surface tension (s) of the liquid as

\nu=r^a\rho^b s^c

. The values of a, b and c respectively are

A

\left( {{3 \over 2},{1 \over 2}, - {1 \over 2}} \right)

B

\left( { - {3 \over 2}, - {1 \over 2},{1 \over 2}} \right)

C

\left( {{3 \over 2}, - {1 \over 2},{1 \over 2}} \right)

D

\left( { - {3 \over 2},{1 \over 2},{1 \over 2}} \right)

Correct Answer

Option B

Solution

$[v]=\left[\mathrm{T}^{-1}\right]$

\begin{aligned} & {[r]=\mathrm{L} \quad[s]=\left[\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}}\right]} \\\\ & {[\rho]=\left[\frac{\mathrm{M}}{\mathrm{L}^{3}}\right]=\left[\mathrm{ML}^{-3}\right]} \\\\ & \Rightarrow v=r^{a} \rho^{b} \mathrm{~s}^{c} \\\\ & \Rightarrow \mathrm{T}^{-1}=\mathrm{L}^{a} \mathrm{M}^{b} \mathrm{~L}^{-3 b} \mathrm{M}^{c} \mathrm{~T}^{-2 c} \\\\ & \Rightarrow \mathrm{T}^{-1}=\mathrm{M}^{(b+c)} \mathrm{L}^{(a-3 b)} \mathrm{T}^{-2 c} \\\\ & -2 c=-1 \Rightarrow c=\frac{1}{2} \\\\ & b+c=0 \\\\ & \Rightarrow b=-\frac{1}{2} \\\\ & a-3 b=0 \Rightarrow 3 b=a \Rightarrow a=-\frac{3}{2} \\\\ & (a, b, c)=\left(-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}\right) \end{aligned}

Q118

A 100 m long wire having cross-sectional area

\mathrm{6.25\times10^{-4}~m^2}

and Young's modulus is

\mathrm{10^{10}~Nm^{-2}}

is subjected to a load of 250 N, then the elongation in the wire will be :

A

\mathrm{6.25\times10^{-6}~m}

B

\mathrm{4\times10^{-3}~m}

C

\mathrm{4\times10^{-4}~m}

D

\mathrm{6.25\times10^{-3}~m}

Correct Answer

Option B

Solution

Elongation in wire $\delta=\dfrac{\mathrm{F} \ell}{\mathrm{AY}}$

\begin{aligned} & \delta=\frac{250 \times 100}{6.25 \times 10^{-4} \times 10^{10}} \\\\ & \delta=4 \times 10^{-3} \mathrm{~m} \end{aligned}

Q119

A wire of length '

L

' and radius '

r

' is clamped rigidly at one end. When the other end of the wire is pulled by a force

f

, its length increases by '

l

'. Another wire of same material of length '

2 \mathrm{~L}

' and radius '

2 r

' is pulled by a force '

2 f

'. Then the increase in its length will be :

A

2 l

B

4 l

C

l

D

l / 2

Correct Answer

Option C

Solution

Let $A$ be the cross-sectional area of the first wire, and let $Y$ be its Young's modulus.

The strain in the wire is given by $\epsilon = \dfrac{l}{L}$ , where $l$ is the increase in length.

The stress in the wire is given by $\sigma = \dfrac{f}{A}$ .

According to Hooke's law, the stress is proportional to the strain, so we have $\sigma = Y \epsilon$ .

Solving for $f$ , we get $f = \dfrac{YA}{l}$ .

The second wire has twice the length and four times the cross-sectional area of the first wire, so its cross-sectional area is $4A$ and its Young's modulus is still $Y$ .

When a force of $2f$ is applied to this wire, the stress in the wire is $\sigma = \dfrac{2f}{4A} = \dfrac{f}{2A}$ .

Using Hooke's law again, we have $\sigma = Y \epsilon$ .

Solving for $\epsilon$ , we get $\epsilon = \dfrac{\sigma}{Y} = \dfrac{f}{2AY}$ .

The increase in length of the second wire is given by $\Delta l = \epsilon \cdot 2L = \dfrac{f \cdot 2L}{2AY}$ .

Substituting the expression for $f$ that we derived earlier, we get $\Delta l = \dfrac{YL \cdot 2A \cdot l}{2AY \cdot L} = \boxed{l}$ .

Therefore, the increase in length of the second wire is the same as the increase in length of the first wire, which is $l$ .

Q120

Given below are two statements: one is labelled as Assertion

\mathbf{A}

and the other is labelled as Reason

\mathbf{R}

Assertion A : A spherical body of radius

(5 \pm 0.1) \mathrm{mm}

having a particular density is falling through a liquid of constant density. The percentage error in the calculation of its terminal velocity is

4 \%

. Reason R : The terminal velocity of the spherical body falling through the liquid is inversely proportional to its radius. In the light of the above statements, choose the correct answer from the options given below

A A is false but

\mathbf{R}

is true

B

\mathrm{A}

is true but

\mathbf{R}

is false

C Both

\mathbf{A}

and

\mathbf{R}

are true but

\mathbf{R}

is NOT the correct explanation of

\mathbf{A}

D Both

\mathbf{A}

and

\mathbf{R}

are true and

\mathbf{R}

is the correct explanation of

\mathbf{A}

Correct Answer

Option B

Solution

The terminal velocity

v_t

of a spherical body falling through a viscous fluid is given by Stokes' Law, which states that:

v_t = \frac{2}{9}\frac{(\rho_s - \rho_f)gr^2}{\eta}

where: $\rho_s$ is the density of the sphere $\rho_f$ is the density of the fluid $g$ is the acceleration due to gravity $r$ is the radius of the sphere $\eta$ is the dynamic viscosity of the fluid As per Stokes' Law, the terminal velocity is proportional to the square of the radius of the sphere (since the radius term

r^2

is in the numerator).

Note that Reason R states that the terminal velocity is "inversely proportional" to its radius, which is contrary to the relationship presented by Stokes' Law.

Therefore, Reason R is false.

Moving on to Assertion A, we can consider the percentage error in the radius to determine the percentage error in the terminal velocity.

If the radius

r

has an error of

\pm 0.1 \mathrm{mm}

at

5 \mathrm{mm}

, then the relative error in the radius is:

\frac{0.1}{5} = 0.02 \text{ or } 2\%

Since the terminal velocity varies with the square of the radius, the percentage error in the terminal velocity would be twice the percentage error in the radius.

\text{Percentage error in } v_t = 2 \times \text{ (Percentage error in } r)

\text{Percentage error in } v_t = 2 \times 2\% = 4\%

This is in agreement with Assertion A, making it true.

Given this analysis, the correct statement is: Option B: A is true but

\mathbf{R}

is false.