Board Question Paper Solution: October 2014 Physics

Maharashtra State Board - HSC Physics

SECTION – I

Q.1. Attempt any THREE: [9]

i. Draw a diagram showing all components of forces acting on a vehicle moving on a curved banked road. Write the necessary equation for maximum safety speed and state the significance of each term involved in it.

Diagram:

Imagine a vehicle on a banked road inclined at angle \(\theta\).

Weight \(mg\) acts vertically downwards.
Normal reaction \(N\) acts perpendicular to the road surface.
Frictional force \(f_s\) acts downwards along the slope (preventing upward skidding at maximum speed).

Equation for Maximum Safety Speed:

v_{max} = \sqrt{rg \left( \frac{\mu_s + \tan\theta}{1 - \mu_s \tan\theta} \right)}

Significance of terms:

\(v_{max}\): Maximum safe velocity of the vehicle.
\(r\): Radius of the curved road.
\(g\): Acceleration due to gravity.
\(\mu_s\): Coefficient of static friction between tires and road.
\(\theta\): Angle of banking.

ii. Explain Maxwell distribution of molecular speed with necessary graph.

Explanation:

Maxwell derived the law of distribution of molecular speeds for a gas in thermal equilibrium. According to this distribution:

The gas molecules move with all possible speeds ranging from zero to infinity.
The number of molecules having very low speeds or very high speeds is very small.
The number of molecules increases with speed, reaches a maximum for a particular speed called the "most probable speed" (\(v_{mp}\)), and then decreases.
The area under the graph represents the total number of molecules.

Graph: A curve plotted with "Speed of molecules" on the X-axis and "Number of molecules per unit speed interval" (\(dN/dv\)) on the Y-axis. The curve is asymmetrical, starting at the origin, rising to a peak, and tapering off towards the right.

iii. Find the total energy and binding energy of an artificial satellite of mass 800 kg orbiting at a height of 1800 km above the surface of the earth. (G = 6.67 × 10^-11 S.I. units, Radius of earth : R = 6400 km, Mass of earth : M = 6 × 10²⁴ kg)

Given:

Mass of satellite, \(m = 800 \text{ kg}\)
Height, \(h = 1800 \text{ km} = 1.8 \times 10^6 \text{ m}\)
Radius of Earth, \(R = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}\)
Mass of Earth, \(M = 6 \times 10^{24} \text{ kg}\)
\(G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2\)

Orbital Radius: \(r = R + h = 6.4 \times 10^6 + 1.8 \times 10^6 = 8.2 \times 10^6 \text{ m}\)

Total Energy (T.E.):

T.E. = -\frac{GMm}{2r} \] \[ T.E. = -\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 800}{2 \times 8.2 \times 10^6} \] \[ T.E. = -\frac{32016 \times 10^{13}}{16.4 \times 10^6} \] \[ T.E. \approx -1.95 \times 10^{10} \text{ J}

Binding Energy (B.E.):

B.E. = -T.E. = +\frac{GMm}{2r} \] \[ B.E. = +1.95 \times 10^{10} \text{ J}

Answer: Total Energy = \(-1.95 \times 10^{10} \text{ J}\), Binding Energy = \(1.95 \times 10^{10} \text{ J}\).

iv. Wavelengths of two notes in the air are \(\left(\frac{70}{153}\right)\) m and \(\left(\frac{70}{157}\right)\) m. Each of these notes produces 8 beats per second with a tuning fork of fixed frequency. Find the velocity of sound in the air and frequency of the tuning fork.

Given:

\(\lambda_1 = \frac{70}{153} \text{ m}\)
\(\lambda_2 = \frac{70}{157} \text{ m}\)
Beat frequency with tuning fork = 8 Hz for both notes.

Formula: \(v = n\lambda \Rightarrow n = \frac{v}{\lambda}\)

Frequencies of the two notes:

n_1 = \frac{v}{\lambda_1} = \frac{v}{70/153} = \frac{153v}{70} \] \[ n_2 = \frac{v}{\lambda_2} = \frac{v}{70/157} = \frac{157v}{70}

Since \(\lambda_1 > \lambda_2\), \(n_1 < n_2\). Let \(N\) be the frequency of the tuning fork.
Since both produce 8 beats/sec, the frequencies are \(N \pm 8\).
Since \(n_2 > n_1\), we must have \(n_1 = N - 8\) and \(n_2 = N + 8\).
Thus, \(n_2 - n_1 = 16\).

\frac{157v}{70} - \frac{153v}{70} = 16 \] \[ \frac{v}{70}(157 - 153) = 16 \] \[ \frac{4v}{70} = 16 \] \[ v = \frac{16 \times 70}{4} = 4 \times 70 = 280 \text{ m/s}

Now, find the frequencies:

n_1 = \frac{153 \times 280}{70} = 153 \times 4 = 612 \text{ Hz}

Since \(n_1 = N - 8\):

N = 612 + 8 = 620 \text{ Hz}

Answer: Velocity of sound = 280 m/s, Frequency of tuning fork = 620 Hz.

HSC Physics Board Papers with Solution

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Q.2. Attempt any SIX: [12]

i. Draw a diagram showing different stages (cases) of projection for artificial satellite.

This requires drawing the Earth with a launch tower. Different trajectories are shown based on horizontal velocity \(v_h\):

\(v_h < v_c\) (Critical velocity): Satellite falls back to Earth (Parabolic/Elliptical path intersecting Earth).
\(v_h = v_c\): Circular orbit.
\(v_c < v_h < v_e\) (Escape velocity): Elliptical orbit.
\(v_h = v_e\): Parabolic path (escapes).
\(v_h > v_e\): Hyperbolic path (escapes).

ii. State the law of conservation of angular momentum and explain with a suitable example.

Statement: If the resultant external torque acting on a rotating body is zero, its angular momentum remains constant.

\(L = I\omega = \text{constant}\). If \(I\) increases, \(\omega\) decreases, and vice versa.

Example: A ballet dancer or ice skater.
When the dancer folds her arms close to her body, her moment of inertia (\(I\)) decreases. To conserve angular momentum (\(L\)), her angular velocity (\(\omega\)) increases, allowing her to spin faster.

iii. Define the angle of contact and state its any ‘two’ characteristics.

Definition: The angle of contact is defined as the angle between the tangent drawn to the free surface of the liquid and the surface of the solid at the point of contact, measured within the liquid.

Characteristics:

It is constant for a given solid-liquid pair.
It depends on the nature of the liquid and the solid in contact.
It depends on the impurities present in the liquid.
It depends on the temperature of the liquid.

iv. With a neat and labelled diagram, explain Ferry’s perfectly black body.

Description:

Fery's black body consists of a double-walled hollow sphere.
The space between the walls is evacuated to minimize heat loss by conduction and convection.
The inner surface is coated with lampblack (absorptivity \(\approx 98\%\)).
There is a small aperture acting as the inlet for radiation.
A conical projection is placed directly opposite the aperture to prevent direct reflection of incident light back out of the hole.
Any radiation entering the hole undergoes multiple internal reflections and is almost completely absorbed. Thus, the aperture acts as a perfectly black body.

v. A stone of mass 5 kg, tied to one end of a rope of length 0.8 m, is whirled in a vertical circle. Find the minimum velocity at the highest point and at the midway point. (g = 9.8 m/s²)

Given: \(m = 5 \text{ kg}\), \(r = 0.8 \text{ m}\), \(g = 9.8 \text{ m/s}^2\).

1. Minimum velocity at highest point (\(v_H\)):

v_H = \sqrt{rg} = \sqrt{0.8 \times 9.8} = \sqrt{7.84} = 2.8 \text{ m/s}

2. Minimum velocity at midway (horizontal) point (\(v_M\)):

v_M = \sqrt{3rg} = \sqrt{3 \times 0.8 \times 9.8} = \sqrt{23.52} \approx 4.85 \text{ m/s}

vi. The maximum velocity of a particle performing linear S.H.M. is 0.16 m/s. If its maximum acceleration is 0.64 m/s², calculate its period.

Given: \(v_{max} = A\omega = 0.16 \text{ m/s}\), \(a_{max} = A\omega^2 = 0.64 \text{ m/s}^2\).

Calculation:

\frac{a_{max}}{v_{max}} = \frac{A\omega^2}{A\omega} = \omega \] \[ \omega = \frac{0.64}{0.16} = 4 \text{ rad/s}

Period \(T\):

T = \frac{2\pi}{\omega} = \frac{2 \times 3.142}{4} = \frac{6.142}{4} = 1.571 \text{ s}

Answer: Period = 1.57 s.

vii. Water rises to a height 3.2 cm in a glass capillary tube. Find the height to which the same water will rise in another glass capillary having half area of cross section.

Given: \(h_1 = 3.2 \text{ cm}\). Area \(A_2 = \frac{A_1}{2}\).

Relation between Area and Radius: \(A = \pi r^2\).
Since \(A_2 = \frac{A_1}{2} \Rightarrow \pi r_2^2 = \frac{\pi r_1^2}{2} \Rightarrow r_2 = \frac{r_1}{\sqrt{2}}\).

Capillary rise formula: \(h = \frac{2T\cos\theta}{r\rho g} \Rightarrow h \propto \frac{1}{r}\).

\frac{h_2}{h_1} = \frac{r_1}{r_2} \] \[ \frac{h_2}{3.2} = \frac{r_1}{r_1 / \sqrt{2}} = \sqrt{2} \] \[ h_2 = 3.2 \times 1.414 = 4.5248 \text{ cm}

Answer: Height = 4.52 cm.

viii. A 36 cm long sonometer wire vibrates with frequency of 280 Hz in fundamental mode, when it is under tension of 24.5 N. Calculate linear density of the material of wire.

Given:

\(L = 36 \text{ cm} = 0.36 \text{ m}\)
\(n = 280 \text{ Hz}\)
\(T = 24.5 \text{ N}\)

Formula: \(n = \frac{1}{2L}\sqrt{\frac{T}{m}}\), where \(m\) is linear density.

n^2 = \frac{1}{4L^2} \frac{T}{m} \Rightarrow m = \frac{T}{4L^2 n^2} \] \[ m = \frac{24.5}{4 \times (0.36)^2 \times (280)^2} \] \[ m = \frac{24.5}{4 \times 0.1296 \times 78400} \] \[ m = \frac{24.5}{40642.56} \] \[ m \approx 6.028 \times 10^{-4} \text{ kg/m}

Answer: Linear density \(m \approx 6.03 \times 10^{-4} \text{ kg/m}\).

Q.3. Select and write the most appropriate answer from the given alternatives for each sub-questions: [7]

i. A thin wire of length L and uniform linear mass density ρ is bent into a circular coil. Moment of inertia of the coil about tangential axis in its plane is _______.

(A) \( \frac{3\rho L^2}{8\pi^2} \)
(B) \( \frac{8\pi^2}{3\rho L^3} \)
(C) \( \frac{3\rho L^3}{8\pi^2} \)
(D) \( \frac{8\pi}{3\rho L^2} \)

Calculation:
Mass \(M = \rho L\). Length \(L = 2\pi R \Rightarrow R = \frac{L}{2\pi}\).
M.I. about diameter \(I_d = \frac{1}{2}MR^2\).
M.I. about tangent in plane \(I_t = I_d + MR^2 = \frac{3}{2}MR^2\).
\(I_t = \frac{3}{2} (\rho L) \left(\frac{L}{2\pi}\right)^2 = \frac{3\rho L^3}{8\pi^2}\).

ii. The average displacement over a period of S.H.M. is _______. (A = amplitude of S.H.M.)

(A) 0
(B) A
(C) 2A
(D) 4A

Reason: Displacement is a vector. The particle returns to the starting point after one period, so net displacement is 0.

iii. In which of the following substances, surface tension increases with increase in temperature?

(A) Copper
(B) Molten copper
(C) Iron
(D) Molten iron

Reason: Generally surface tension decreases with temperature. However, for molten copper and molten cadmium, surface tension increases with temperature over a certain range.

iv. The ratio of diameters of two wires of the same material and length is n : 1. If the same load is applied to both the wires then increase in the length of the thin wire is (n > 1) _______.

(A) \(n^{1/4}\) times
(B) \(n^{1/2}\) times
(C) n times
(D) \(n^2\) times

Calculation: \(Y = \frac{FL}{A\Delta l} \Rightarrow \Delta l \propto \frac{1}{A} \propto \frac{1}{d^2}\).
\(d_{thick}/d_{thin} = n/1\).
\(\frac{\Delta l_{thin}}{\Delta l_{thick}} = \left(\frac{d_{thick}}{d_{thin}}\right)^2 = n^2\).

v. The co-efficient of reflection of an opaque body is 0.16. Its co-efficient of emission is _______.

(A) 0.94
(B) 0.84
(C) 0.74
(D) 0.64

Calculation: Opaque \(\Rightarrow t = 0\).
\(a + r = 1 \Rightarrow a = 1 - r = 1 - 0.16 = 0.84\).
By Kirchhoff's law, \(a = e\), so \(e = 0.84\).

vi. Let velocity of a sound wave be ‘v’ and ‘ω’ be angular velocity. The propagation constant of the wave is _______.

(A) \(\sqrt{\frac{\omega}{v}}\)
(B) \(\sqrt{\frac{v}{\omega}}\)
(C) \(\frac{\omega}{v}\)
(D) \(\frac{v}{\omega}\)

Reason: Propagation constant \(k = \frac{2\pi}{\lambda}\). Since \(v = \frac{\omega}{k}\), we get \(k = \frac{\omega}{v}\).

vii. The value of end correction for an open organ pipe of radius ‘r’ is _______.

(A) 0.3 r
(B) 0.6 r
(C) 0.9 r
(D) 1.2 r

Reason: The end correction \(e\) at one open end is approx \(0.6r\). While total correction for an open pipe is \(2e = 1.2r\), the phrase "value of end correction" typically refers to the constant \(e\) itself.

Q.4. Attempt any ONE: [7]

Option 1: Distinguish between forced vibrations and resonance. Draw neat, labelled diagrams for the modes of vibration of a stretched string in second harmonic and third harmonic.

Problem: The area of the upper face of a rectangular block is 0.5 m × 0.5 m and the lower face is fixed. The height of the block is 1 cm. A shearing force applied at the top face produces a displacement of 0.015 mm. Find the strain and shearing force. (Modulus of rigidity : η = 4.5 × 10¹⁰ N/m²)

Distinguish Forced Vibrations vs Resonance:

Forced Vibration: Body vibrates with frequency of external periodic force, not its natural frequency. Amplitude is generally small.
Resonance: Special case of forced vibration where external frequency matches natural frequency. Amplitude becomes maximum.

Diagrams:

2nd Harmonic: String vibrates in 2 loops. (Nodes at ends and center, Antinodes in between).
3rd Harmonic: String vibrates in 3 loops.

Problem Solution:

Given: \(A = 0.5 \times 0.5 = 0.25 \text{ m}^2\), \(h = 1 \text{ cm} = 0.01 \text{ m}\), \(x = 0.015 \text{ mm} = 1.5 \times 10^{-5} \text{ m}\), \(\eta = 4.5 \times 10^{10} \text{ N/m}^2\).

1. Shearing Strain (\(\theta\)):

\theta = \frac{x}{h} = \frac{1.5 \times 10^{-5}}{0.01} = 1.5 \times 10^{-3}

2. Shearing Force (\(F\)):

\eta = \frac{\text{Shear Stress}}{\text{Shear Strain}} = \frac{F/A}{\theta} \] \[ F = \eta A \theta = (4.5 \times 10^{10}) \times (0.25) \times (1.5 \times 10^{-3}) \] \[ F = 4.5 \times 0.25 \times 1.5 \times 10^7 \] \[ F = 1.6875 \times 10^7 \text{ N}

OR - Option 2: Define phase of S.H.M. Show variation of displacement, velocity and acceleration with phase...
Problem: A body starts rotating from rest. Due to a couple of 20 Nm it completes 60 revolutions in one minute. Find the moment of inertia of the body.

Problem Solution:

Given: \(\omega_0 = 0\), Torque \(\tau = 20 \text{ Nm}\).

Displacement \(\theta = 60 \text{ rev} = 60 \times 2\pi = 120\pi \text{ rad}\).

Time \(t = 1 \text{ min} = 60 \text{ s}\).

Using kinematic equation: \(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\)

120\pi = 0 + \frac{1}{2}\alpha (60)^2 \] \[ 120\pi = 1800\alpha \] \[ \alpha = \frac{120\pi}{1800} = \frac{\pi}{15} \text{ rad/s}^2

Using \(\tau = I\alpha\):

I = \frac{\tau}{\alpha} = \frac{20}{\pi/15} = \frac{300}{\pi} \approx 95.49 \text{ kg m}^2

SECTION – II

Q.5. Attempt any THREE: [9]

i. In a biprism experiment... distance between 4th bright band on one side and 4th dark band on the other side...

Given: \(\lambda = 4800 \text{ Å} = 4.8 \times 10^{-7} \text{ m}\). \(d = 3 \text{ mm} = 3 \times 10^{-3} \text{ m}\).

Distances: Slit to Biprism = 15 cm. Biprism to Eyepiece = 85 cm.
Total distance \(D = 15 + 85 = 100 \text{ cm} = 1 \text{ m}\).

Fringe width \(X = \frac{\lambda D}{d}\).

Position of 4th bright band from center: \(y_{B4} = 4X\).

Position of 4th dark band (on other side): \(y_{D4} = (4 - 0.5)X = 3.5X\).

Total distance = \(y_{B4} + y_{D4} = 4X + 3.5X = 7.5X\).

\text{Distance} = 7.5 \times \frac{4.8 \times 10^{-7} \times 1}{3 \times 10^{-3}} \] \[ \text{Distance} = 2.5 \times 4.8 \times 10^{-4} \] \[ \text{Distance} = 12 \times 10^{-4} \text{ m} = 1.2 \text{ mm}

ii. Six capacitors of capacities 5, 5, 5, 5, 10 and X µF are connected as shown... Find X if balanced, and resultant between A and C.

Analysis: The circuit is a bridge. Outer arms are AB, BC, AD, DC. The central arm is BD.
From diagram: \(C_{AB}=5, C_{BC}=5, C_{AD}=5, C_{BD}=5\).

The arm DC contains two capacitors in series: 10 µF and X µF. Let equivalent be \(C_{DC}\).

a. Value of X if balanced:

Condition: \(C_{AB} / C_{AD} = C_{BC} / C_{DC}\) (assuming A and C are input/output).

\frac{5}{5} = \frac{5}{C_{DC}} \Rightarrow C_{DC} = 5 \mu\text{F}

Since \(C_{DC}\) is series of 10 and X: \(\frac{10X}{10+X} = 5\).

10X = 50 + 5X \Rightarrow 5X = 50 \Rightarrow X = 10 \mu\text{F}

b. Resultant capacitance between A and C:

If balanced, the central branch (BD) is ignored. The circuit becomes two parallel branches (ABC and ADC).

Upper branch (ABC): 5 and 5 in series \(\to 2.5 \mu\text{F}\).
Lower branch (ADC): 5 and \(C_{DC}\) (which is 5) in series \(\to 2.5 \mu\text{F}\).

Resultant \(C_{eq} = 2.5 + 2.5 = 5 \mu\text{F}\).

iii. Show that the current flowing through a moving coil galvanometer is directly proportional to the angle of deflection of coil.

Derivation:

Deflecting Torque: \(\tau_d = NIAB\) (where N=turns, I=current, A=area, B=magnetic field). The radial field ensures \(\sin\theta=1\).
Restoring Torque: \(\tau_r = k\alpha\) (where k=torsional constant, \(\alpha\)=deflection).
At equilibrium: \(\tau_d = \tau_r \Rightarrow NIAB = k\alpha\).
Therefore, \(I = \left(\frac{k}{NAB}\right)\alpha\).
Since k, N, A, B are constants, \(I \propto \alpha\).

Q.6. Attempt any SIX: [12]

v. A red light of wavelength 6400 Å in air has wavelength 4000 Å in glass. If the wavelength of violet light in air is 4400 Å, find its wavelength in glass. (Assume that µr ≈ µv)

Solution:

Refractive index for red: \(\mu_r = \frac{\lambda_{air}}{\lambda_{glass}} = \frac{6400}{4000} = 1.6\).

Given \(\mu_r \approx \mu_v\), so \(\mu_v = 1.6\).

For violet light: \(\lambda_{g,v} = \frac{\lambda_{a,v}}{\mu_v} = \frac{4400}{1.6}\).

\lambda_{g,v} = \frac{44000}{16} = 2750 \text{ Å}

vi. The magnetic moment of a magnet of dimensions 5 cm × 2.5 cm × 1.25 cm is 3 Am². Calculate the intensity of magnetization.

Solution:

Volume \(V = 5 \times 2.5 \times 1.25 \text{ cm}^3 = 15.625 \text{ cm}^3 = 15.625 \times 10^{-6} \text{ m}^3\).

Intensity \(I = \frac{M}{V} = \frac{3}{15.625 \times 10^{-6}}\).

I = \frac{3 \times 10^6}{15.625} = 1.92 \times 10^5 \text{ A/m}

vii. An A.C. circuit consists of inductor of inductance 125 mH connected in parallel with a capacitor of capacity 50 µF. Determine the resonant frequency.

Solution:

\(L = 125 \text{ mH} = 0.125 \text{ H}\), \(C = 50 \mu\text{F} = 50 \times 10^{-6} \text{ F}\).

Formula: \(f_r = \frac{1}{2\pi\sqrt{LC}}\).

f_r = \frac{1}{2\pi\sqrt{0.125 \times 50 \times 10^{-6}}} \] \[ f_r = \frac{1}{2\pi\sqrt{6.25 \times 10^{-6}}} \] \[ f_r = \frac{1}{2\pi \times 2.5 \times 10^{-3}} \] \[ f_r = \frac{1000}{5\pi} = \frac{200}{\pi} \approx 63.66 \text{ Hz}

viii. Calculate the de Broglie wavelength of an electron moving with 1/3rd of the speed of light in vacuum.

Solution:

\(v = \frac{c}{3} = 10^8 \text{ m/s}\). Mass \(m = 9.1 \times 10^{-31} \text{ kg}\). \(h = 6.63 \times 10^{-34}\).

\lambda = \frac{h}{mv} = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^8} \] \[ \lambda = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-23}} \] \[ \lambda = 0.728 \times 10^{-11} \text{ m} = 0.0728 \text{ Å}

Q.7. Select and write the most appropriate answer: [7]

i. (B) limit of resolution decreases (since Limit = \(\lambda/2NA\))
ii. (B) \(2\pi \times 10^{-5}\) T (Calculated: \(B = \mu_0 n I\), \(n=10\), \(I=5\))
iii. (C) energy and charge
iv. (C) \(\frac{(E_1 - E_2)\lambda_1\lambda_2}{c(\lambda_2 - \lambda_1)}\)
v. (C) the band gap of the material of semiconductor
vi. (D) space wave
vii. (A) \(\frac{qV}{dm}\) (Acc = \(F/m = qE/m = q(V/d)/m\))

Q.8. Attempt any ONE: [7]

Option 1: Potentiometer Problem... resistance in series to get potential gradient \(10^{-3}\) V/cm.

Solution:

Given: \(L = 4 \text{ m}\), \(R_{wire} = 5 \Omega\). Cell \(E = 2 \text{ V}, r = 1 \Omega\).

Required Gradient \(K = 10^{-3} \text{ V/cm} = 0.1 \text{ V/m}\).

Potential difference across wire \(V_{AB} = K \times L = 0.1 \times 4 = 0.4 \text{ V}\).

Current in circuit \(I = \frac{V_{AB}}{R_{wire}} = \frac{0.4}{5} = 0.08 \text{ A}\).

Also \(I = \frac{E}{R_{wire} + r + R_{ext}}\).

0.08 = \frac{2}{5 + 1 + R} \] \[ 0.08(6 + R) = 2 \] \[ 6 + R = \frac{2}{0.08} = 25 \] \[ R = 19 \Omega

Answer: Series resistance = 19 \(\Omega\).

Option 2: Photoelectric Effect Problem... Threshold 230 nm, Incident 180 nm. Find KEmax.

Solution:

\(\lambda_0 = 230 \text{ nm}\), \(\lambda = 180 \text{ nm}\).

\(KE_{max} = h c \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right)\).

Value of \(hc \approx 1240 \text{ eV}\cdot\text{nm}\).

KE_{max} = 1240 \left( \frac{1}{180} - \frac{1}{230} \right) \text{ eV} \] \[ KE_{max} = 1240 \left( \frac{230 - 180}{180 \times 230} \right) \] \[ KE_{max} = 1240 \left( \frac{50}{41400} \right) \] \[ KE_{max} = \frac{62000}{41400} \approx 1.5 \text{ eV}

In Joules: \(1.5 \times 1.6 \times 10^{-19} = 2.4 \times 10^{-19} \text{ J}\).

Maharashtra HSC Physics Board Question Paper Solution October 2014

Board Question Paper Solution: October 2014 Physics

SECTION – I

Q.1. Attempt any THREE: [9]

HSC Physics Board Papers with Solution

Q.2. Attempt any SIX: [12]

Q.3. Select and write the most appropriate answer from the given alternatives for each sub-questions: [7]

Q.4. Attempt any ONE: [7]

SECTION – II

Q.5. Attempt any THREE: [9]

Q.6. Attempt any SIX: [12]

Q.7. Select and write the most appropriate answer: [7]

Q.8. Attempt any ONE: [7]