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# JEE Main 2022 June 24 β Shift 1 Physics Question Paper with Solutions

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JEE Main 2022 June 24 β Shift 1 Physics Question Paper with Solutions

Section – A

Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

1. The bulk modulus of a liquid is 3 × 1010 Nm–2. The pressure required to reduce the volume of liquid by 2% is

(A) 3 × 108 Nm–2

(B) 9 × 108 Nm–2

(C) 6 × 108 Nm–2

(D) 12 × 108 Nm–2

Sol.

$$\begin{array}{l}\because B=\frac{\Delta P}{\left ( -\frac{\Delta V}{V} \right )}\end{array}$$
$$\begin{array}{l}\Rightarrow \Delta P =3 \times 10^{10}\times \left ( 0.02 \right )\end{array}$$

= 6 × 108 N/m2

2. Given below are two statements: One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): In an uniform magnetic field, speed and energy remains the same for a moving charged particle.

Reason (R): Moving charged particle experiences magnetic force perpendicular to its direction of motion.

(A) Both (A) and (R) are true and (R) is the correct explanation of (A).

(B) Both (A) and (R) are true but (R) is NOT the correct explanation of (A).

(C) (A) is true but (R) is false.

(D) (A) is false but (R) is true.

Sol. Magnetic force

$$\begin{array}{l}\overrightarrow{F}\perp \overrightarrow{V} \end{array}$$

Wb = 0

⇒ ΔKE = 0 and speed remains constant.

3. Two identical cells each of emf 1.5 V are connected in parallel across a parallel combination of two resistors each of resistance 20 W. A voltmeter connected in the circuit measures 1.2 V. The internal resistance of each cell is

(A) 2.5 W

(B) 4 W

(C) 5 W

(D) 10 W

Sol.

$$\begin{array}{l}\frac{1.5\times 10}{10+\frac{r}{2}}=1.2\end{array}$$

$$\begin{array}{l}\Rightarrow r = 5~\Omega\end{array}$$

4. Identify the pair of physical quantities which have different dimensions.

(A) Wave number and Rydberg’s constant

(B) Stress and Coefficient of elasticity

(C) Coercivity and Magnetisation

(D) Specific heat capacity and Latent heat

Sol.

$$\begin{array}{l}\left [ S \right ]=\frac{\left [ C \right ]}{\left [ m \right ]\times \left [ \Delta T \right ]}\end{array}$$

and,

$$\begin{array}{l}\left [ L \right ]=\frac{Q}{m}\end{array}$$

⇒ They have different dimensions

5. A projectile is projected with velocity of 25 m/s at an angle q with the horizontal. After t seconds its inclination with horizontal becomes zero. If R represents horizontal range of the projectile, the value of q will be

[use g = 10 m/s2]

(A)

$$\begin{array}{l}\frac{1}{2}\sin^{-1}\left [ \frac{5t^2}{4R} \right ]\end{array}$$

(B)

$$\begin{array}{l}\frac{1}{2}\sin^{-1}\left [ \frac{4R}{5t^2} \right ]\end{array}$$

(C)

$$\begin{array}{l}\tan^{-1}\left [ \frac{4t^2}{5R} \right ]\end{array}$$

(D)

$$\begin{array}{l}\cot^{-1}\left [ \frac{R}{20t^2} \right ]\end{array}$$

Sol.

$$\begin{array}{l}t=\frac{25\sin\theta}{g}\end{array}$$

and,

$$\begin{array}{l}R=\frac{\left ( 25 \right )^2\left ( 2\sin\theta\cos\theta \right )}{g}\end{array}$$
$$\begin{array}{l}\Rightarrow R=\frac{25 \times 25 \times 2}{g}\times \frac{gt}{25}\times \cos\theta\end{array}$$
$$\begin{array}{l}\Rightarrow R=50t\cos\theta\end{array}$$
$$\begin{array}{l}\therefore \tan\theta =\frac{gt}{25}\times \frac{50t}{R}\end{array}$$
$$\begin{array}{l}=\frac{20t^2}{R}\end{array}$$
$$\begin{array}{l}\therefore \theta =\cot^{-1}\left ( \frac{R}{20t^2} \right )\end{array}$$

6. A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms–1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is

[use g = 9.8 ms–2]

(A) 4.9 m

(B) 9.8 m

(C) 12.5 m

(D) 19.6 m

Sol.

$$\begin{array}{l}S=\frac{u^2}{2a}=\frac{u^2}{2\left ( \mu g \right )}\end{array}$$
$$\begin{array}{l}\frac{\left ( 9.8 \right )^2}{2\times 0.5 \times\left ( 9.8 \right )}\end{array}$$
$$\begin{array}{l}\frac{9.8}{1}\end{array}$$

= 9.8 m

7. A boy ties a stone of mass 100 g to the end of a

2 m long string and whirls it around in a horizontal plane. The string can withstand the maximum tension of 80 N. If the maximum speed with which the stone can revolve is

$$\begin{array}{l}\frac{K}{\pi}\text{ rev./min}\end{array}$$
. The value of K is

(Assume the string is massless and unstretchable)

(A) 400

(B) 300

(C) 600

(D) 800

Sol. T = mw2r

$$\begin{array}{l}\Rightarrow 80=0.1\times\left ( 2\pi \times\frac{K}{\pi}\times \frac{1}{60} \right )^2\times 2\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{800}{2}=\frac{K^2}{900}\end{array}$$

K = 30 × 20 = 600

8. A vertical electric field of magnitude 4.9 × 105 N/C just prevents a water droplet of a mass 0.1 g from falling. The value charge on the droplet will be

(Given g = 9.8 m/s2)

(A) 1.6 × 10–9 C

(B) 2.0 × 10–9 C

(C) 3.2 × 10–9 C

(D) 0.5 × 10–9 C

Sol. Since the droplet is at rest

⇒ Net force = 0

mg = qE

$$\begin{array}{l}\Rightarrow q=\frac{mg}{E}=2\times 10^{-9}~\text{C}\end{array}$$

9. A particle experiences a variable force

$$\begin{array}{l}\overrightarrow{F}=\left ( 4x\hat{i}+3y^2\hat{j} \right )\end{array}$$
in a horizontal xy plane. Assume distance in meters and force is newton. If the particle moves from point (1, 2) to point (2, 3) in the xy plane; then Kinetic Energy changes by

(A) 50.0 J

(B) 12.5 J

(C) 25.0 J

(D) 0 J

Sol.

$$\begin{array}{l}W=\int{\overrightarrow{F}\cdot d\overrightarrow{r}}\end{array}$$
$$\begin{array}{l}=\int_{1}^{2}4xdx+\int_{2}^{3}3y^2dy\end{array}$$
$$\begin{array}{l}=\left [ 2x^2 \right ]_1^2+\left [ y^3 \right ]_2^3\end{array}$$

= 2 × 3 + (27 – 8)

= 25 J

10. The approximate height from the surface of earth at which the weight of the body becomes $$\frac{1}{3}$$ of its weight on the surface of earth is

[Radius of earth R = 6400 km and √
$$\begin{array}{l}\sqrt{3}=1.732\end{array}$$
]

(A) 3840 km

(B) 4685 km

(C) 2133 km

(D) 4267 km

Sol. According to the given information

$$\begin{array}{l}\frac{GM}{\left ( R+h \right )^2}=\frac{1}{3}\times\frac{GM}{R^2}\end{array}$$
$$\begin{array}{l}\Rightarrow R+h=\sqrt{3}R\end{array}$$
$$\begin{array}{l}\Rightarrow h=\left ( \sqrt{3}-1 \right )R\simeq 4685~\text{km} \end{array}$$

11. A resistance of 40 Ω is connected to a source of alternating current rated 220 V, 50 Hz. Find the time taken by the current to change from its maximum value to the rms value :

(A) 2.5 ms

(B) 1.25 ms

(C) 2.5 s

(D) 0.25 s

Sol.I = I0cos(ωt) say

⇒ At maximum ωt1 = 0 or t1 = 0

Then at rms value

$$\begin{array}{l}l=l_0l\sqrt{2}\end{array}$$

⇒ ωt2 = π/4

⇒ ω(t2 t1) = π/4

$$\begin{array}{l}\Delta t = \frac{\pi}{4\omega}=\frac{\pi T}{4\times 2\pi}\end{array}$$
$$\begin{array}{l}=\frac{1}{400}~\text{s or } 2.5~\text{ms}\end{array}$$

⇒ Option A is right answer

12. The equations of two waves are given by :

y1 = 5 sin 2π(x – vt) cm

y2 = 3 sin 2π(x – vt + 1.5) cm

These waves are simultaneously passing through a string. The amplitude of the resulting wave is :

(A) 2 cm

(B) 4 cm

(C) 5.8 cm

(D) 8 cm

Sol.y1 = 5sin(2πx – 2πvt)

y2 = 3sin(2πx – 2πvt + 3π)

⇒ Phase difference = 3π

$$\begin{array}{l}\Rightarrow A_{\text{net}}=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\left ( 3\pi \right )}\end{array}$$

Anet = 2 cm

13. A plane electromagnetic waves travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is

4.5 × 10–2 Am–1 at a point, what will be the approximate magnitude of electric field intensity at that point?

(Given : Permeability of free space μ0 = 4π × 10–7 NA–2, speed of light in vacuum c = 3 × 108 ms–1)

(A) 16.96 Vm–1

(B) 2.25 × 10–2 Vm–1

(C) 8.48 Vm–1

(D) 6.75 × 106 Vm–1

Sol.H = 4.5 × 10–2

So B = μH

Thus

$$\begin{array}{l}E=\frac{c}{n}B\end{array}$$
(where n ⇒ refractive index)

So

$$\begin{array}{l}E=\frac{3\times 10^8 \times 4\pi \times 10^{-7}\times1.61\times 4.5 \times 10^{-2}}{\sqrt{1.61\times 6.44}}\end{array}$$

E = 8.48

14. Choose the correct option from the following options given below :

(A) In the ground state of Rutherford’s model electrons are in stable equilibrium. While in Thomson’s model electrons always experience a net-force

(B) An atom has a nearly continuous mass distribution in a Rutherford’s model but has a highly non-uniform mass distribution in Thomson’s model

(C) A classical atom based on Rutherford’s model is doomed to collapse.

(D) The positively charged part of the atom possesses most of the mass in Rutherford’s model but not in Thomson’s model.

Sol. An atom based on classical theory of Rutherford’s model should collapse as the electrons in continuous circular motion that is a continuously accelerated charge should emit EM waves and so should lose energy. These electrons losing energy should soon fall into heavy nucleus collapsing the whole atom.

15. Nucleus A is having mass number 220 and its binding energy per nucleon is 5.6 MeV. It splits in two fragments ‘B’ and ‘C’ of mass numbers 105 and 115. The binding energy of nucleons in ‘B’ and ‘C’ is 6.4 MeV per nucleon. The energy Q released per fission will be :

(A) 0.8 MeV

(B) 275 MeV

(C) 220 MeV

(D) 176 MeV

Sol.

$$\begin{array}{l}^{220}A\rightarrow ^{105}B+^{115}C\end{array}$$

Q = [105 × 6.4 + 115×6.4] – [220 × 5.6] MeV

Q = 176 MeV

16. A baseband signal of 3.5 MHz frequency is modulated with a carrier signal of 3.5 GHz frequency using amplitude modulation method. What should be the minimum size of antenna required to transmit the modulated signal?

(A) 42.8 m

(B) 42.8 mm

(C) 21.4 mm

(D) 21.4 m

Sol.

$$\begin{array}{l}\nu_c=3.5\times 10^9 \text{ Hz}\end{array}$$

$$\begin{array}{l}\therefore \lambda=\frac{c}{\nu_c}=\frac{3\times 10^8}{3.5\times 10^9}\end{array}$$

∴ Size of antenna

$$\begin{array}{l}=\frac{\lambda}{4}\end{array}$$
$$\begin{array}{l}=\frac{8.57 \times 10^{-2}}{4}\end{array}$$

= 21.4 mm

17. A Carnot engine whose heat sinks at 27°C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?

(A) Increases by 18°C

(B) Increases by 200°C

(C) Increases by 120°C

(D) Increases by 73°C

Sol. Initially :

$$\begin{array}{l}\frac{1}{4}=1-\frac{300}{T_H}\end{array}$$

TH = 400 K

Finally : Efficiency becomes

$$\begin{array}{l}\frac{1}{2}\end{array}$$
$$\begin{array}{l}\therefore \frac{1}{2}=1-\frac{300}{T_H}\end{array}$$
$$\begin{array}{l}\therefore T_H=600~\text{K}\end{array}$$

⇒ Temperature of the source increases by 200°C.

18. A parallel plate capacitor is formed by two plates each of area 30p cm2 separated by 1 mm. A material of dielectric strength 3.6 × 107 Vm–1 is filled between the plates. If the maximum charge that can be stored on the capacitor without causing any dielectric breakdown is 7 × 10–6 C, the value of dielectric constant of the material is :

$$\begin{array}{l}\left [\text{Use } \frac{1}{4\pi\epsilon_0}=9\times 10^9~\text{Nm}^2\text{ C}^{-2} \right]\end{array}$$

(A) 1.66

(B) 1.75

(C) 2.25

(D) 2.33

Sol. Field inside the dielectric

$$\begin{array}{l}=\frac{\sigma}{k\epsilon_0}\end{array}$$

According to the given information,

$$\begin{array}{l}\frac{\sigma}{k\epsilon_0}=3.6\times 10^7\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{\frac{Q}{A}}{k\epsilon_0}=3.6\times 10^7\end{array}$$

k = 2.33

19. The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance

$$\begin{array}{l}\frac{r}{2}\end{array}$$
from the centre is :

(A)

$$\begin{array}{l}\frac{B}{2}\end{array}$$

(B) 2B

(C)

$$\begin{array}{l}\left ( \frac{2}{\sqrt{5}} \right )^3B\end{array}$$

(D)

$$\begin{array}{l}\left ( \frac{2}{\sqrt{3}} \right )^3B\end{array}$$

Sol.

$$\begin{array}{l}B=\frac{\mu_0l}{2r}\end{array}$$
$$\begin{array}{l}B_a=\frac{\mu_0lr^2}{2\left ( r^2+\frac{r^2}{4} \right )}\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{B_a}{B}=\left ( \frac{2}{\sqrt{5}} \right )^3\end{array}$$
$$\begin{array}{l}\Rightarrow B_a=\left ( \frac{2}{\sqrt{5}} \right )^3 B\end{array}$$

20. Two metallic blocks M1 and M2 of same area of cross-section are connected to each other (as shown in figure). If the thermal conductivity of M2 is K then the thermal conductivity of M1 will be :

(A) 10K

(B) 8K

(C) 12.5K

(D) 2K

Sol. Thermal current is same so

$$\begin{array}{l}\frac{dQ}{dt}=\frac{\Delta T_1}{\frac{l_1}{K_1A}}=\frac{\Delta T_2}{\frac{l_2}{K_2A}}\end{array}$$
$$\begin{array}{l}\text{or }\frac{20}{16}\times K’ =\frac{80}{8}\times K\end{array}$$

K’ = 8 K

Section – B

Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

1. 0.056 kg of Nitrogen is enclosed in a vessel at a temperature of 127°C. The amount of heat required to double the speed of its molecules is ___ k cal.

(Take R = 2 cal mol–1 K–1)

Sol. Because the vessel is closed, it will be an isochoric process.

To double the speed, temperature must be 4 times

$$\begin{array}{l}\left ( v \propto \sqrt{T} \right )\end{array}$$

So Tf = 1600 K, Ti = 400 K

number of moles are $$\frac{56}{28}=2$$

so

$$\begin{array}{l}Q=nCv \Delta T = 2\times \frac{5}{2}\times 2 \times 1200\end{array}$$

= 12000 cal = 12 K cal

2. Two identical thin biconvex lenses of focal length 15 cm and refractive index 1.5 are in contact with each other. The space between the lenses is filled with a liquid of refractive index 1.25. The focal length of the combination is ___ cm.

Sol.

$$\begin{array}{l}\frac{1}{f_1}=\left ( \frac{\mu_e}{\mu_m}-1 \right )\left ( \frac{1}{R_1}-\frac{1}{R_2} \right )\end{array}$$

here |R1| = |R2| = R

$$\begin{array}{l}\Rightarrow \frac{1}{f_{l_1}}=\left ( 1.5-1 \right )\left ( \frac{2}{R} \right )=\frac{1}{15}\end{array}$$
$$\begin{array}{l}\Rightarrow \frac{1}{R}=\frac{1}{15} \text{ or } R = 15 \text{ cm}\end{array}$$

for the concave lens made up of liquid

$$\begin{array}{l}\frac{1}{f_{l_2}}=\left ( 1.25-1 \right )\left ( -\frac{2}{R} \right )=-\frac{1}{30} \text{ cm}\end{array}$$

now for equivalent lens

$$\begin{array}{l}\frac{1}{f_e}=\frac{2}{f_{l_1}}+\frac{1}{f_{l_2}}\end{array}$$
$$\begin{array}{l}=\frac{2}{15}-\frac{1}{30}=\frac{3}{30}=\frac{1}{10}\end{array}$$

or fe = 10 cm

3. A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10 mA and the collector current changes by 1.5 mA. The load resistance is 5 kW. The voltage gain of the transistor will be ___.

Sol.

$$\begin{array}{l}R_B=\frac{10\times 10^-3}{10\times 10^{-6}}\end{array}$$

= 103 Ω

$$\begin{array}{l}\therefore A_v=\left ( \frac{\Delta l_C}{\Delta l_B} \right )\times \left (\frac{R_C}{R_B} \right )\end{array}$$
$$\begin{array}{l}=\frac{1.5\times 10^{-3}}{10\times 10^{-6}}\times\frac{5\times 10^3}{1\times 10^3}\end{array}$$
$$\begin{array}{l}=\frac{1.5\times 5}{10}\times\left ( 1000 \right )\end{array}$$

= 750

4. As shown in the figure an inductor of inductance 200 mH is connected to an AC source of emf 220 V and frequency 50 Hz. The instantaneous voltage of the source is 0 V when the peak value of current is

$$\begin{array}{l}\frac{\sqrt{a}}{\pi}\end{array}$$
A. The value of a is ___.

Sol.

$$\begin{array}{l}l_{rms}=\frac{V_{rms}}{z}\end{array}$$

z = X2 = ω2

$$\begin{array}{l}=2\pi \times 50 \times \frac{200}{1000}\end{array}$$

= 20 p

$$\begin{array}{l}\therefore l_{rms}=\frac{220}{20\pi}=\frac{11}{\pi}\end{array}$$
$$\begin{array}{l}\therefore l_{\text{peak}}=\sqrt{2}\times\frac{11}{\pi}\end{array}$$
$$\begin{array}{l}=\frac{\sqrt{2\times 121}}{\pi}\end{array}$$
$$\begin{array}{l}=\frac{\sqrt{242}}{\pi}\end{array}$$

5. Sodium light of wavelengths 650 nm and 655 nm is used to study diffraction at a single slit of aperture 0.5 mm. The distance between the slit and the screen is 2.0 m. The separation between the positions of the first maxima of diffraction pattern obtained in the two cases is _____ × 10–5 m.

Sol. Position of 1st maxima is

$$\begin{array}{l}\frac{3}{2}\frac{\lambda D}{a}\end{array}$$

⇒ According to given values, required separation

$$\begin{array}{l}=\frac{3}{2}\times\left ( 655 \text{ nm}-650 \text{ nm} \right )\times\frac{2m}{0.5 \text{ nm}}\end{array}$$

⇒ Required separation = 3 × 10–5 m.

6. When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is v1. When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes v2. If v2 = x v1, the value of x will be ________.

Sol. Let us say the work function is f

$$\begin{array}{l}\Rightarrow 2\phi = \phi + \frac{1}{2}mv_1^2~~~~\cdots\left ( 1 \right )\end{array}$$

and

$$\begin{array}{l}\Rightarrow 5\phi = \phi + \frac{1}{2}mv_2^2~~~~\cdots\left ( 2 \right )\end{array}$$

From (1) and (2)

$$\begin{array}{l}\frac{v_2^2}{v_1^2}=\frac{4}{1} \text{ or } \frac{v_2}{v_1}=2\end{array}$$

7. From the top of a tower, a ball is thrown vertically upward which reaches the ground in 6 s. A second ball thrown vertically downward from the same position with the same speed reaches the ground in 1.5 s. A third ball released, from the rest from the same location, will reach the ground in ____ s.

Sol. Based on the situation

$$\begin{array}{l}h=-ut_1+\frac{1}{2}gt_1^2~~~~~~~~~\rightarrow \text{ throwing up}\cdots\left ( i \right )\end{array}$$
$$\begin{array}{l}h=-ut_2+\frac{1}{2}gt_2^2~~~~~~~~~\rightarrow \text{ throwing up}\cdots\left ( ii \right )\end{array}$$
$$\begin{array}{l}h=\frac{1}{2}gt^2~~~~~~~~~\rightarrow \text{ dropping up}\cdots\left ( iii \right )\end{array}$$

and

$$\begin{array}{l}0=u\left (t_1-t_2 \right )-\frac{1}{2}g\left ( t_1-t_2 \right )^2~~~~~~\cdots\left ( iv \right )\end{array}$$

solving above equations

$$\begin{array}{l}t=\sqrt{t_1t_2}\end{array}$$
$$\begin{array}{l}\Rightarrow t=\sqrt{6\times1.5}=3\text{ s}\end{array}$$

8. A ball of mass 100 g is dropped from a height h =10 cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance

$$\begin{array}{l}\frac{h}{2}\end{array}$$
. The spring constant is ______ Nm–1.

(Use g = 10 ms–2)

Sol.

$$\begin{array}{l}mg\left ( h+\frac{h}{2} \right )=\frac{1}{2}k\left ( \frac{h}{2} \right )^2\end{array}$$
$$\begin{array}{l}\Rightarrow 0.1 \times 10 \times \left ( 0.15 \right )=\frac{1}{2}k\left ( 0.05 \right )^2\end{array}$$

k =120 N/m

9. In a potentiometer arrangement, a cell gives a balancing point at 75 cm length of wire. This cell is now replaced by another cell of unknown emf. If the ratio of the emf’s of two cells respectively is 3 : 2, the difference in the balancing length of the potentiometer wire in above two cases will be _____ cm.

Sol. At balancing point, we know that emf is proportional to the balancing length. i.e.,

emf ∝ balancing length

Now, let the emf’s be 3ε and 2ε.

3ε = k(75) …(1)

and 2ε = k(l) …(2)

l = 50 cm

Difference is (75 – 50) cm = 25 cm.

10. A metre scale is balanced on a knife edge at its centre. When two coins, each of mass 10 g are put one on the top of the other at the 10.0 cm mark the scale is found to be balanced at 40.0 cm mark. The mass of the metre scale is found to be x × 10–2 kg. The value of x is _________.

Sol.

If λ is the mass per unit length of the scale then

0.02 × (30) × 10 + l40 × 20 × 10 = l60 × 30 × 10

0.006 = λ10

Or 100 λ = 0.06 kg

= 6 × 10–2 kg

x = 6

s