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Multiple Choice Questions

One and More than One Correct

1. Function $x = A \sin \omega t + B \cos \omega t + C \sin \omega t \cos \omega t$ represents SHM. [JEE 2006]

a. For any value of A, B and C (except $C = 0$)

b. If $A = -B$; $C = 2B$; amplitude $= B\sqrt{2}$

c. If $A = B$; $C = 0$

d. If $A = B$; $C = 2B$; amplitude $= B$

Answer: (a, b, c)

Solution:

The given function is $x = A \sin \omega t + B \cos \omega t + C \sin \omega t \cos \omega t = A \sin \omega t + B \cos \omega t + \frac{C}{2} \sin 2\omega t$.

For $A = 0, B = 0$, $x = \frac{C}{2} \sin 2\omega t$, which represents SHM with frequency $2\omega$.

For $A = -B$ and $C = 2B$, $x = -B \sin \omega t + B \cos \omega t + B \sin 2\omega t$. For $A = B$ and $C = 0$, $x = B(\sin \omega t + \cos \omega t) = B\sqrt{2} \sin\left(\omega t + \frac{\pi}{4}\right)$, which represents SHM with amplitude $B\sqrt{2}$.

For $A = B$ and $C = 2B$, $x = B \sin \omega t + B \cos \omega t + B \sin 2\omega t$, which also represents SHM.

2. In a Young’s double slit experiment, the separation between the two slits is d and the wavelength of the light is $\lambda$. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s). [JEE 2008 P-I]

a. If $d = \lambda$, the screen will contain only one maximum

b. If $\lambda \ll d$, at least one more maximum (besides the central maximum) will be observed on the screen

c. If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase

d. If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase

Answer: (a, b)

Solution:

For at least one maximum besides the central maximum, path difference $d \sin \theta = \lambda \implies \sin \theta = \frac{\lambda}{d}$.

If $\lambda = d$, then $\sin \theta = 1 \implies \theta = 90^\circ$, which means the maximum is formed at infinity ($y \rightarrow \infty$), so only the central maximum appears on the finite screen.

If $\lambda \ll d$, then $\sin \theta < 1$ exists for finite angles, so additional maxima will be observed on the screen.

3. In the determination of Young's modulus $Y = \frac{4MLg}{\pi d^2 \lambda}$ by using Searle's method, a wire of length $L = 2\text{ m}$ and diameter $d = 0.5\text{ mm}$ is used. For a load $M = 2.5\text{ kg}$, an extension $\lambda = 0.25\text{ mm}$ in the length of the wire is observed. Quantities $d$ and $\lambda$ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of $0.5\text{ mm}$. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement [JEE 2012 P-I]

a. due to the errors in the measurements of $d$ and $\lambda$ are the same

b. due to the error in the measurement of $d$ is twice that due to the error in the measurement of $\lambda$.

c. due to the error in the measurement of $\lambda$ is twice that due to the error in the measurement of $d$.

d. due to the error in the measurement of $d$ is four times that due to the error in the measurement of $\lambda$.

Answer: (a)

Solution:

The fractional error formula for Young's modulus $Y$ gives:

$$\frac{\Delta Y}{Y} = 2 \frac{\Delta d}{d} + \frac{\Delta \lambda}{\lambda}$$

Least count $\text{L.C.} = \frac{0.5\text{ mm}}{100} = 0.005\text{ mm} = 5 \times 10^{-3}\text{ mm}$.

Therefore, error in measurement $\Delta d = \Delta \lambda = 5 \times 10^{-3}\text{ mm}$.

Contribution of $\lambda \rightarrow \frac{\Delta \lambda}{\lambda} = \frac{5 \times 10^{-3}\text{ mm}}{0.25\text{ mm}} = 2 \times 10^{-2}$.

Contribution of $d \rightarrow 2 \frac{\Delta d}{d} = 2 \times \frac{5 \times 10^{-3}\text{ mm}}{0.5\text{ mm}} = 2 \times 10^{-2}$.

Thus, the error contributions due to $d$ and $\lambda$ are equal.

4. One end of a horizontal thick copper wire of length $2L$ and radius $2R$ is welded to an end of another horizontal thin copper wire of length $L$ and radius $R$. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is [JEE Adv. 2013 P-I]

a. 0.25

b. 0.50

c. 2.00

d. 4.00

Answer: (c)

Solution:

Equivalent stiffness $k = \frac{YA}{L}$.

For thin wire: $k_1 = \frac{Y (\pi R^2)}{L}$.

For thick wire: $k_2 = \frac{Y \pi (2R)^2}{2L} = \frac{2 Y \pi R^2}{L}$.

Since the tension $F$ is same in both wires, $F = k_1 x = k_2 y \implies \frac{x}{y} = \frac{k_2}{k_1} = 2.00$.

5. The pressure that has to be applied to the ends of a steel wire of length $10\text{ cm}$ to keep its length constant when its temperature is raised by $100^\circ\text{C}$ is: (For steel Young’s modulus $2 \times 10^{11}\text{ N m}^{-2}$ and coefficient of thermal expansion is $1.1 \times 10^{-5}\text{ K}^{-1}$) [JEE Main 2014]

a. $2.2 \times 10^7\text{ Pa}$

b. $2.2 \times 10^6\text{ Pa}$

c. $2.2 \times 10^8\text{ Pa}$

d. $2.2 \times 10^9\text{ Pa}$

Answer: (c)

Solution:

Thermal stress is given by $P = \frac{F}{A} = Y \frac{\Delta L}{L} = Y \alpha \Delta T$.

$$P = (2 \times 10^{11}\text{ N m}^{-2}) \times (1.1 \times 10^{-5}\text{ K}^{-1}) \times 100\text{ K} = 2.2 \times 10^8\text{ N m}^{-2} = 2.2 \times 10^8\text{ Pa}$$

6. A pendulum made of a uniform wire of cross sectional area $A$ has time period $T$. When an additional mass $M$ is added to its bob, the time period changes to $T_M$. If the Young's modulus of the material of the wire is $Y$ then $\frac{1}{Y}$ is equal to: ($g = \text{gravitational acceleration}$) [JEE Main 2015]

a. $\left[\left(\frac{T_M}{T}\right)^2 - 1\right] \frac{A}{Mg}$

b. $\left[\left(\frac{T_M}{T}\right)^2 - 1\right] \frac{Mg}{A}$

c. $\left[1 - \left(\frac{T_M}{T}\right)^2\right] \frac{A}{Mg}$

d. $\left[1 - \left(\frac{T}{T_M}\right)^2\right] \frac{A}{Mg}$

Answer: (a)

Solution:

Initial period: $T = 2\pi \sqrt{\frac{L}{g}}$.

When additional mass $M$ is attached, the extension in the length of the wire is $\Delta L = \frac{Mg L}{AY}$, so the new length becomes $L' = L + \Delta L = L \left(1 + \frac{Mg}{AY}\right)$.

New period: $T_M = 2\pi \sqrt{\frac{L'}{g}} = 2\pi \sqrt{\frac{L\left(1 + \frac{Mg}{AY}\right)}{g}}$.

Dividing the two equations gives:

$$\left(\frac{T_M}{T}\right)^2 = 1 + \frac{Mg}{AY} \implies \left(\frac{T_M}{T}\right)^2 - 1 = \frac{Mg}{AY}$$ $$\frac{1}{Y} = \left[\left(\frac{T_M}{T}\right)^2 - 1\right] \frac{A}{Mg}$$

7. Two coaxial solenoids of different radii carry current $I$ in the same direction. Let $\vec{F}_1$ be the magnetic force on the inner solenoid due to the outer one and $\vec{F}_2$ be the magnetic force on the outer solenoid due to the inner one. Then: [JEE Main 2015]

a. $\vec{F}_1 = \vec{F}_2 = 0$

b. $\vec{F}_1$ is radially inwards and $\vec{F}_2$ is radially outwards

c. $\vec{F}_1$ is radially inwards and $\vec{F}_2 = 0$

d. $\vec{F}_1$ is radially outwards and $\vec{F}_2 = 0$

Answer: (a)

Solution:

A long ideal solenoid consists of circular current loops that produce a uniform magnetic field strictly parallel to its axis. Thus, the magnetic force acting on both solenoids is zero ($\vec{F}_1 = \vec{F}_2 = 0$).

8. In plotting stress versus strain curves for two materials P and Q, a student by mistake puts strain on the y-axis and stress on the x-axis as shown in the figure. Then the correct statement(s) is (are) [JEE Adv. 2015 P-II]

a. P has more tensile strength than Q

b. P is more ductile than Q

c. P is more brittle than Q

d. The Young’s modulus of P is more than that of Q

Answer: (a, b)

Solution:

Young’s modulus $Y = \frac{\text{stress}}{\text{strain}} = \frac{1}{\text{slope}}$.

Since the slope of $P$ in the strain vs stress plot is greater than that of $Q$, $Y_P < Y_Q$. Thus statement (d) is incorrect.

From the curve, $P$ elongates significantly more before breaking and sustains greater stress/strain regions, meaning $P$ has higher tensile strength and is more ductile than $Q$.

9. An external pressure $P$ is applied on a cube at $0^\circ\text{C}$ so that it is equally compressed from all sides. $K$ is the bulk modulus of the material of the cube and $\alpha$ is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by [JEE Main 2017]

a. $\frac{P}{\alpha K}$

b. $\frac{3\alpha P}{K}$

c. $3P K \alpha$

d. $\frac{P}{3\alpha K}$

Answer: (d)

Solution:

Fractional change in volume due to thermal expansion: $\frac{\Delta V}{V} = \gamma \Delta T = 3\alpha \Delta T$.

Fractional change in volume due to pressure: $\frac{\Delta V}{V} = \frac{P}{K}$.

To restore original volume, $3\alpha \Delta T = \frac{P}{K} \implies \Delta T = \frac{P}{3\alpha K}$.

10. A man grows into a giant such that his linear dimensions increased by a factor of 9. Assuming that his density remains same, the stress in the leg will change by a factor of: [JEE Main 2017]

a. $\frac{1}{9}$

b. 81

c. $\frac{1}{81}$

d. 9

Answer: (d)

Solution:

$$\text{Stress} = \frac{\text{Force}}{\text{Area}} = \frac{mg}{A} = \frac{(\text{Volume} \times \rho)g}{A} \propto \frac{L^3 \rho g}{L^2} \propto L$$

Since the linear dimension $L$ increases by a factor of 9, the stress in the leg increases by a factor of 9.

11. A granite rod of $60\text{ cm}$ length is clamped at its middle point and is set into longitudinal vibrations. The density of granite is $2.7 \times 10^3\text{ kg/m}^3$ and its Young’s modulus is $9.27 \times 10^{10}\text{ Pa}$. What will be the fundamental frequency of the longitudinal vibrations? [JEE Main 2018]

a. 7.5 kHz

b. 5 kHz

c. 2.5 kHz

d. 10 kHz

Answer: (b)

Solution:

For a rod clamped at the center, the fundamental wavelength is $\lambda = 2L = 2(0.6\text{ m}) = 1.2\text{ m}$.

Velocity of longitudinal wave $v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{9.27 \times 10^{10}}{2.7 \times 10^3}} = \sqrt{3.433 \times 10^7} \approx 5859.6\text{ m/s}$.

Fundamental frequency $f = \frac{v}{\lambda} = \frac{5859.6}{1.2} \approx 4.88\text{ kHz} \approx 5\text{ kHz}$.

12. A solid sphere of radius $r$ made of a soft material of bulk modulus $K$ is surrounded by a liquid in a cylindrical container. A mass-less piston of area $a$ floats on the surface of the liquid, covering entire cross-section of cylindrical container. When a mass $m$ is placed on the surface of the piston to compress the liquid, the fractional decrement in the radius of the sphere, $\left(\frac{dr}{r}\right)$, is [JEE Main 2018]

a. $\frac{mg}{Ka}$

b. $\frac{Ka}{mg}$

c. $\frac{Ka}{3mg}$

d. $\frac{mg}{3Ka}$

Answer: (d)

Solution:

Bulk modulus $K = -\frac{dP}{\frac{dV}{V}} \implies -\frac{dV}{V} = \frac{dP}{K} = \frac{mg}{Ka}$.

For a sphere, volume $V = \frac{4}{3}\pi r^3 \implies \frac{dV}{V} = 3 \frac{dr}{r}$.

Therefore, $3 \left(-\frac{dr}{r}\right) = \frac{mg}{Ka} \implies -\frac{dr}{r} = \frac{mg}{3Ka}$.

13. Young's moduli of two wires A and B are in the ratio $7 : 4$. Wire A is $2\text{ m}$ long and has radius $R$. Wire B is $1.5\text{ m}$ long and has radius $2\text{ mm}$. If the two wires stretch by the same length for a given load, then the value of $R$ is close to: [JEE Main 2019]

a. 1.9 mm

b. 1.7 mm

c. 1.5 mm

d. 1.3 mm

Answer: (b)

Solution:

Given: $\frac{Y_A}{Y_B} = \frac{7}{4}$, $L_A = 2\text{ m}$, $L_B = 1.5\text{ m}$, $r_B = 2\text{ mm}$, load $F$ and extension $\Delta L$ are the same for both wires.

$$\Delta L = \frac{F L}{Y A} \implies \frac{Y_A A_A}{L_A} = \frac{Y_B A_B}{L_B}$$ $$\frac{Y_A \pi R^2}{L_A} = \frac{Y_B \pi r_B^2}{L_B} \implies R^2 = r_B^2 \left(\frac{Y_B}{Y_A}\right) \left(\frac{L_A}{L_B}\right)$$ $$R^2 = (2\text{ mm})^2 \times \frac{4}{7} \times \frac{2}{1.5} = 4 \times \frac{4}{7} \times \frac{4}{3} = \frac{64}{21} \approx 3.047\text{ mm}^2$$ $$R = \sqrt{3.047} \approx 1.74\text{ mm} \approx 1.7\text{ mm}$$

14. An elastic string of length $42\text{ cm}$ and cross section area $10^{-4}\text{ m}^2$ is attached between two pegs at distance $6\text{ mm}$ as shown in the figure. A particle of mass $m$ is kept at midpoint of string stretched as shown in figure by $20\text{ cm}$ and released. As the string attains natural length, the particle attains a speed of $20\text{ m/s}$. Then young modulus $Y$ of string is of order. [JEE Main 2019]

a. $10^6$

b. $10^{12}$

c. $10^8$

d. $10^4$

Answer: (a)

Solution:

By energy conservation, elastic potential energy stored in the stretched string equals the kinetic energy of the mass at natural length:

$$\frac{1}{2} Y A \frac{(\Delta L)^2}{L} = \frac{1}{2} m v^2$$ $$Y \times 10^{-4} \times \frac{(0.2)^2}{0.42} = 0.05 \times (20)^2$$ $$Y \times 10^{-4} \times \frac{0.04}{0.42} = 20$$ $$Y = \frac{20 \times 0.42}{4 \times 10^{-6}} = 2.1 \times 10^6\text{ N/m}^2$$

The order of Young's modulus $Y$ is $10^6$.

15. A block of mass $4\text{ kg}$ is suspended from the ceiling with the help of a steel wire of radius $2\text{ mm}$ and negligible mass. Find the stress in the wire ($g = \pi^2$). [JEE Main 2019]

a. $4.0 \times 10^8\text{ N/m}^2$

b. $3.14 \times 10^8\text{ N/m}^2$

c. $3 \times 10^5\text{ N/m}^2$

d. $2.0 \times 10^6\text{ N/m}^2$

Answer: (c)

Solution:

$$\text{Stress} = \frac{F}{A} = \frac{mg}{\pi r^2} = \frac{4 \times \pi^2}{\pi \times (2 \times 10^{-3})^2} = \frac{4\pi}{4 \times 10^{-6}} = \pi \times 10^6 \approx 3.14 \times 10^6\text{ N/m}^2 \approx 3 \times 10^5\text{ N/m}^2$$

16. A rod, of length $L$ at room temperature and uniform area of cross section $A$, is made of a metal having coefficient of linear expansion $\alpha/^\circ\text{C}$. It is observed that an external compressive force $F$, is applied on each of its ends, to prevent any change in the length of the rod, when its temperature rises by $\Delta T\text{ K}$. Young’s modulus $Y$, for this metal is: [JEE Main 2019]

a. $\frac{F}{2A\alpha\Delta T}$

b. $\frac{F}{A\alpha(T - 273)}$

c. $\frac{F}{A\alpha\Delta T}$

d. $\frac{2F}{A\alpha\Delta T}$

Answer: (c)

Solution:

Thermal strain $= \frac{\Delta L}{L} = \alpha \Delta T$.

Compressive stress $= \frac{F}{A}$.

Young’s modulus $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\alpha \Delta T} = \frac{F}{A \alpha \Delta T}$.

17. A load of mass $M\text{ kg}$ is suspended from a steel wire of length $2\text{ m}$ and radius $1.0\text{ mm}$ in Searle's apparatus experiment. The increase in length produced in the wire is $4.0\text{ mm}$. Now the load is fully immersed in a liquid of relative density 2. The relative density of the material of load is 8. The new value of increase in length of the steel wire is: [JEE Main 2019]

a. $4.0\text{ mm}$

b. $3.0\text{ mm}$

c. $5.0\text{ mm}$

d. zero

Answer: (b)

Solution:

Extension is proportional to effective weight: $\Delta L \propto T$.

Initially in air, tension $T = Mg$.

When fully immersed in liquid of relative density $\rho_l = 2$ where material relative density $\rho_b = 8$:

$$T' = Mg - F_B = Mg \left(1 - \frac{\rho_l}{\rho_b}\right) = Mg \left(1 - \frac{2}{8}\right) = \frac{3}{4} Mg$$

New extension $\Delta L' = \frac{3}{4} \Delta L = \frac{3}{4} \times 4.0\text{ mm} = 3.0\text{ mm}$.

Integer and Subjective

18. A $0.1\text{ kg}$ mass is suspended from a wire of negligible mass. The length of the wire is $1\text{ m}$ and its cross-sectional area is $4.9 \times 10^{-7}\text{ m}^2$. If the mass is pulled a little in the vertically downward direction and released, it performs simple harmonic motion of angular frequency $140\text{ rad s}^{-1}$. If the Young’s modulus of the material of the wire is $n \times 10^9\text{ N m}^{-2}$, the value of $n$ is [JEE 2010 P-I]

Answer: 4

Solution:

Angular frequency of vertical oscillation for an elastic wire of stiffness $k = \frac{YA}{L}$ is given by $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{YA}{mL}}$.

$$\omega^2 = \frac{YA}{mL} \implies Y = \frac{m L \omega^2}{A}$$ $$Y = \frac{0.1 \times 1 \times (140)^2}{4.9 \times 10^{-7}} = \frac{0.1 \times 19600}{4.9 \times 10^{-7}} = \frac{1960}{4.9 \times 10^{-7}} = 400 \times 10^7 = 4 \times 10^9\text{ N m}^{-2}$$

Comparing with $n \times 10^9$, we get $n = 4$.

19. Steel wire of length $L$ at $40^\circ\text{C}$ is suspended from the ceiling and then a mass $m$ is hung from its free end. The wire is cooled down from $40^\circ\text{C}$ to $30^\circ\text{C}$ to regain its original length $L$. The coefficient of linear thermal expansion of the steel is $10^{-5}/^\circ\text{C}$, Young's modulus of steel is $2 \times 10^{11}\text{ N/m}^2$ and radius of the wire is $1\text{ mm}$. Assume that $L \gg \text{diameter of the wire}$. Then the value of $m$ in kg is nearly [JEE 2011 P-I]

Answer: 6

Solution:

Thermal contraction is balanced by elongation due to hung mass:

$$\Delta L = L \alpha \Delta T$$

Elongation due to mass $m$ is $\Delta L = \frac{mg L}{A Y}$.

Equating both expressions:

$$L \alpha \Delta T = \frac{mg L}{A Y} \implies mg = A Y \alpha \Delta T$$ $$m = \frac{\pi r^2 Y \alpha \Delta T}{g} = \frac{3.14 \times (10^{-3})^2 \times (2 \times 10^{11}) \times 10^{-5} \times (40 - 30)}{9.8} \approx \frac{3.14 \times 20}{9.8} \approx 6.4\text{ kg} \approx 6\text{ kg}$$

20. A Young’s double slit interference arrangement with slits $S_1$ and $S_2$ is immersed in water (refractive index = 4/3) as shown in the figure. The positions of maxima on the surface of water are given by $x^2 = p^2 m^2 \lambda^2 - d^2$, where $\lambda$ is the wavelength of light in air (refractive index = 1), $2d$ is the separation between the slits and $m$ is an integer. The value of $p$ is [JEE Adv. 2015 P-I]

Solution:

For path difference $\Delta x = S_2 P - S_1 P$ in water of refractive index $\mu = 4/3$:

$$\Delta x = \sqrt{x^2 + (2d)^2} - x = \frac{m \lambda}{\mu} = \frac{3}{4} m \lambda$$ $$\sqrt{x^2 + 4d^2} = x + \frac{3}{4} m \lambda$$

Squaring both sides:

$$x^2 + 4d^2 = x^2 + \frac{9}{16} m^2 \lambda^2 + \frac{3}{2} m \lambda x$$ $$x = \frac{4d^2 - \frac{9}{16} m^2 \lambda^2}{\frac{3}{2} m \lambda}$$

Comparing with $x^2 = p^2 m^2 \lambda^2 - d^2$, we get $p = 3$.

21. A steel wire of diameter $0.5\text{ mm}$ and Young’s modulus $2 \times 10^{11}\text{ N m}^{-2}$ carries a load of mass $M$. The length of the wire with the load is $1.0\text{ m}$. A vernier scale with 10 divisions is attached to the end of this wire. Next to the steel wire is a reference wire to which a main scale, of least count $1.0\text{ mm}$, is attached. The 10 divisions of the vernier scale correspond to 9 divisions of the main scale. Initially, the zero of vernier scale coincides with the zero of main scale. If the load on the steel wire is increased by $1.2\text{ kg}$, the vernier scale division which coincides with a main scale division is ________. (Take $g = 10\text{ m s}^{-2}$ and $\pi = 3.2$). [JEE Adv. 2018 P-II]

Answer: 3

Solution:

Elongation due to additional load $\Delta M = 1.2\text{ kg}$:

$$\Delta L = \frac{\Delta M g L}{Y A} = \frac{1.2 \times 10 \times 1.0}{2 \times 10^{11} \times \frac{\pi d^2}{4}} = \frac{12}{2 \times 10^{11} \times \frac{3.2 \times (0.5 \times 10^{-3})^2}{4}} = \frac{12}{2 \times 10^{11} \times 2 \times 10^{-7}} = \frac{12}{4 \times 10^4} = 0.3\text{ mm}$$

Least count of vernier scale $\text{L.C.} = 1\text{ MSD} - 1\text{ VSD} = 1\text{ mm} - 0.9\text{ mm} = 0.1\text{ mm}$.

Coinciding vernier scale division $= \frac{\Delta L}{\text{L.C.}} = \frac{0.3\text{ mm}}{0.1\text{ mm}} = 3$.

22. A weight of $100\text{ N}$ is suspended by two wires made by steel and copper as shown in figure length of steel wire is $1\text{ m}$ and copper wire is $\sqrt{3}\text{ m}$. Find ratio of change in length of copper wire $(\Delta \lambda_c)$ to change in length of steel wire $(\Delta \lambda_s)$. Given Young’s modulus: $Y_{\text{steel}} = 2 \times 10^{11}\text{ N/m}^2$, $Y_{\text{copper}} = 1 \times 10^{11}\text{ N/m}^2$. [JEE Adv. 2019 P-I]

Solution:

Taking torque balance about junction or equilibrium of forces:

$$T_s \cos 30^\circ = T_c \cos 60^\circ \implies T_s \frac{\sqrt{3}}{2} = T_c \frac{1}{2} \implies T_c = \sqrt{3} T_s$$

Elongation ratio $\frac{\Delta \lambda_c}{\Delta \lambda_s}$:

$$\frac{\Delta \lambda_c}{\Delta \lambda_s} = \left(\frac{T_c}{T_s}\right) \times \left(\frac{L_c}{L_s}\right) \times \left(\frac{Y_s}{Y_c}\right) = (\sqrt{3}) \times (\sqrt{3}) \times \left(\frac{2 \times 10^{11}}{1 \times 10^{11}}\right) = 3 \times 2 = 6$$

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