Model Question Paper Set - I: Physics Solutions
Physics Paper 1 This Paper Physics Paper 2 Physics Paper 3 Physics Paper 4
SECTION – I
Q. 1. Select and write the most appropriate answer from the given alternatives for each sub-question: [7 Marks]
i. A driver of a van travelling with high speed suddenly sees a herd of cows in front at a distance ‘d’. To avoid collision he applies brakes instead of taking turn side ways. His action is _______.
Answer: (C) inappropriate as centripetal force is greater than the retarding force
Explanation: To turn safely, the required centripetal force ($mv^2/r$) must be provided by friction ($\mu mg$). To stop, the work is done by friction over distance $d$. Generally, the braking distance $d = v^2/(2\mu g)$ is half the minimum turning radius $r = v^2/(\mu g)$. Thus, it requires less distance to stop than to turn. However, the question frames it around the forces. If the speed is high, the required centripetal force for a sharp turn (radius $d$) would likely exceed the maximum available frictional (retarding) force, causing a skid. Hence, turning is inappropriate because the required centripetal force is greater than the available retarding force. Note: The driver's action (braking) is actually appropriate, but option (C) correctly explains why turning would be inappropriate (required force > available force). Contextually, option (C) is the intended logic for why turning fails.
ii. The ratio of radius of gyration of hollow cylinder to solid cylinder is _______.
Answer: (B) $\sqrt{2}$
Moment of Inertia (M.I.) of a hollow cylinder $I_h = MR^2$, so radius of gyration $K_h = R$.
M.I. of a solid cylinder $I_s = \frac{1}{2}MR^2$, so $K_s = \frac{R}{\sqrt{2}}$.
Ratio $\frac{K_h}{K_s} = \frac{R}{R/\sqrt{2}} = \sqrt{2}$.
M.I. of a solid cylinder $I_s = \frac{1}{2}MR^2$, so $K_s = \frac{R}{\sqrt{2}}$.
Ratio $\frac{K_h}{K_s} = \frac{R}{R/\sqrt{2}} = \sqrt{2}$.
iii. Which of the following relationship between the acceleration a and the displacement x of a particle involve simple harmonic motion?
Answer: (C) a = − 10 x
For Simple Harmonic Motion (SHM), acceleration must be directly proportional to displacement and opposite in direction: $a \propto -x$ or $a = -\omega^2 x$. Only option (C) satisfies this linear, restoring relationship.
iv. Dimensions of compressibility are _______.
Answer: (B) [L¹M⁻¹T²]
Compressibility is the reciprocal of Bulk Modulus ($K$).
$K = \frac{\text{Pressure}}{\text{Strain}}$. Strain is dimensionless. Pressure = $Force/Area = [MLT^{-2}]/[L^2] = [ML^{-1}T^{-2}]$.
Therefore, Compressibility = $1/K = [M^{-1}L^1T^2]$.
Note: The options in the image seem to have typos in the exponents or signs. Option (B) in standard notation is closest to the inverse of pressure if interpreted as $M^{-1}L^1T^2$. Let's re-evaluate the provided options carefully.
(B) is $[L^1M^{-1}T^2]$. This matches the inverse of pressure ($[M^1L^{-1}T^{-2}]$). So (B) is correct.
$K = \frac{\text{Pressure}}{\text{Strain}}$. Strain is dimensionless. Pressure = $Force/Area = [MLT^{-2}]/[L^2] = [ML^{-1}T^{-2}]$.
Therefore, Compressibility = $1/K = [M^{-1}L^1T^2]$.
Note: The options in the image seem to have typos in the exponents or signs. Option (B) in standard notation is closest to the inverse of pressure if interpreted as $M^{-1}L^1T^2$. Let's re-evaluate the provided options carefully.
(B) is $[L^1M^{-1}T^2]$. This matches the inverse of pressure ($[M^1L^{-1}T^{-2}]$). So (B) is correct.
v. A large size drop flattens _______.
Answer: (A) to bring centre of gravity lower
Large drops are affected significantly by gravity. They flatten to minimize their gravitational potential energy by lowering their centre of gravity.
vi. The kinetic energy of hydrogen in 1 litre of volume at 2 atmospheric pressure is _______.
Answer: (D) 303.9 J
Kinetic Energy $E = \frac{3}{2}PV$.
$P = 2 \text{ atm} = 2 \times 1.013 \times 10^5 \text{ Pa}$.
$V = 1 \text{ litre} = 10^{-3} m^3$.
$E = 1.5 \times (2 \times 1.013 \times 10^5) \times 10^{-3} = 3 \times 101.3 = 303.9 \text{ J}$.
$P = 2 \text{ atm} = 2 \times 1.013 \times 10^5 \text{ Pa}$.
$V = 1 \text{ litre} = 10^{-3} m^3$.
$E = 1.5 \times (2 \times 1.013 \times 10^5) \times 10^{-3} = 3 \times 101.3 = 303.9 \text{ J}$.
vii. A stationary wave is _______.
Answer: (D) periodic in both space and time
A stationary wave has a repeating pattern of nodes and antinodes in space (periodic in space) and particles oscillate with a time period T (periodic in time).
HSC Physics Board Papers with Solution
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Q. 2. Attempt any SIX: [12 Marks]
i. What is the angle of banking necessary for a curved road of 80 m radius for safe driving at 72 km/h? (g = 9.8 m/s²)
Given: $r = 80 \, m$, $v = 72 \, km/h = 72 \times \frac{5}{18} = 20 \, m/s$, $g = 9.8 \, m/s^2$.
Formula: $\tan \theta = \frac{v^2}{rg}$
Calculation:
$\tan \theta = \frac{20^2}{80 \times 9.8} = \frac{400}{784} \approx 0.5102$
$\theta = \tan^{-1}(0.5102) \approx 27^\circ 2'$
Formula: $\tan \theta = \frac{v^2}{rg}$
Calculation:
$\tan \theta = \frac{20^2}{80 \times 9.8} = \frac{400}{784} \approx 0.5102$
$\theta = \tan^{-1}(0.5102) \approx 27^\circ 2'$
ii. Calculate the M.I. of a solid sphere of mass 20 kg and radius 0.4 m, rotating about an axis passing through a point of a distance of 0.1 m from the centre of sphere.
Given: $M = 20 \, kg$, $R = 0.4 \, m$, distance $h = 0.1 \, m$.
Formula: Parallel Axis Theorem: $I = I_{CM} + Mh^2$.
For a solid sphere, $I_{CM} = \frac{2}{5}MR^2$.
Calculation:
$I_{CM} = \frac{2}{5} \times 20 \times (0.4)^2 = 8 \times 0.16 = 1.28 \, kg \cdot m^2$.
$Mh^2 = 20 \times (0.1)^2 = 20 \times 0.01 = 0.2 \, kg \cdot m^2$.
$I = 1.28 + 0.2 = 1.48 \, kg \cdot m^2$.
Formula: Parallel Axis Theorem: $I = I_{CM} + Mh^2$.
For a solid sphere, $I_{CM} = \frac{2}{5}MR^2$.
Calculation:
$I_{CM} = \frac{2}{5} \times 20 \times (0.4)^2 = 8 \times 0.16 = 1.28 \, kg \cdot m^2$.
$Mh^2 = 20 \times (0.1)^2 = 20 \times 0.01 = 0.2 \, kg \cdot m^2$.
$I = 1.28 + 0.2 = 1.48 \, kg \cdot m^2$.
iii. Derive differential equation of linear S.H.M. in terms of displacement and angular velocity.
1. In linear S.H.M, the restoring force $F$ is directly proportional to displacement $x$ and opposite to it.
$$F = -kx$$ (where k is force constant)
2. According to Newton's second law, $F = ma = m \frac{d^2x}{dt^2}$.
3. Equating the forces: $m \frac{d^2x}{dt^2} = -kx$.
4. Rearranging: $\frac{d^2x}{dt^2} + \frac{k}{m}x = 0$.
5. Substituting $\frac{k}{m} = \omega^2$ (where $\omega$ is angular velocity), we get:
$$\frac{d^2x}{dt^2} + \omega^2 x = 0$$
$$F = -kx$$ (where k is force constant)
2. According to Newton's second law, $F = ma = m \frac{d^2x}{dt^2}$.
3. Equating the forces: $m \frac{d^2x}{dt^2} = -kx$.
4. Rearranging: $\frac{d^2x}{dt^2} + \frac{k}{m}x = 0$.
5. Substituting $\frac{k}{m} = \omega^2$ (where $\omega$ is angular velocity), we get:
$$\frac{d^2x}{dt^2} + \omega^2 x = 0$$
iv. A sphere of mass 2 kg is attached at one end of steel wire having length 1 m and radius 4 mm. It is whirled in vertical circle with angular velocity of 120 r.p.m. Estimate the elongation of wire when sphere is at lowest point in its path. (Ysteel = 20 × 10¹⁰ N/m²)
Given: $m = 2 \, kg$, $l = r = 1 \, m$, radius of wire $r_w = 4 \, mm = 4 \times 10^{-3} m$.
$\omega = 120 \, rpm = 120 \times \frac{2\pi}{60} = 4\pi \, rad/s$.
$Y = 20 \times 10^{10} \, N/m^2$.
Tension at lowest point: $T = mg + mr\omega^2$.
$T = 2(9.8) + 2(1)(4\pi)^2 = 19.6 + 2(16 \times 9.87) \approx 19.6 + 315.8 \approx 335.4 \, N$.
(Using $\pi^2 \approx 9.87$. If $\pi^2 \approx 10$, $T \approx 19.6 + 320 = 339.6 N$).
Elongation: $\Delta l = \frac{Tl}{AY} = \frac{Tl}{\pi r_w^2 Y}$.
$A = \pi (4 \times 10^{-3})^2 = 16\pi \times 10^{-6} \, m^2$.
$\Delta l = \frac{335.4 \times 1}{(16\pi \times 10^{-6}) \times (20 \times 10^{10})}$.
$\Delta l = \frac{335.4}{320\pi \times 10^4} \approx \frac{335.4}{1005 \times 10^4} \approx 3.33 \times 10^{-5} \, m$.
$\omega = 120 \, rpm = 120 \times \frac{2\pi}{60} = 4\pi \, rad/s$.
$Y = 20 \times 10^{10} \, N/m^2$.
Tension at lowest point: $T = mg + mr\omega^2$.
$T = 2(9.8) + 2(1)(4\pi)^2 = 19.6 + 2(16 \times 9.87) \approx 19.6 + 315.8 \approx 335.4 \, N$.
(Using $\pi^2 \approx 9.87$. If $\pi^2 \approx 10$, $T \approx 19.6 + 320 = 339.6 N$).
Elongation: $\Delta l = \frac{Tl}{AY} = \frac{Tl}{\pi r_w^2 Y}$.
$A = \pi (4 \times 10^{-3})^2 = 16\pi \times 10^{-6} \, m^2$.
$\Delta l = \frac{335.4 \times 1}{(16\pi \times 10^{-6}) \times (20 \times 10^{10})}$.
$\Delta l = \frac{335.4}{320\pi \times 10^4} \approx \frac{335.4}{1005 \times 10^4} \approx 3.33 \times 10^{-5} \, m$.
v. Define capillary action. Give its two applications.
Definition: The phenomenon of rise or fall of a liquid inside a capillary tube when it is dipped in the liquid is called capillary action or capillarity.
Applications:
1. Oil rises up the wick of a lamp due to capillary action.
2. Sap (water and minerals) rises from roots to the top of trees/plants.
3. Blotting paper absorbs ink via capillary pores.
Applications:
1. Oil rises up the wick of a lamp due to capillary action.
2. Sap (water and minerals) rises from roots to the top of trees/plants.
3. Blotting paper absorbs ink via capillary pores.
vi. Wavelengths of two notes in air are 90/175 m and 90/173 m. Each note produces four beats per second with a third note of fixed frequency. Calculate the frequency of the third note and velocity of sound in air.
$\lambda_1 = \frac{90}{175} \approx 0.514 \, m$, $\lambda_2 = \frac{90}{173} \approx 0.520 \, m$.
Since $\lambda_1 < \lambda_2$, frequency $n_1 > n_2$.
Let $N$ be the frequency of the third note.
Given: $|n_1 - N| = 4$ and $|n_2 - N| = 4$.
Since $n_1 > n_2$, we must have $n_1 = N + 4$ and $n_2 = N - 4$.
So, $n_1 - n_2 = 8$.
Also $n = \frac{v}{\lambda}$. So $v(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}) = 8$.
$v (\frac{175}{90} - \frac{173}{90}) = 8 \Rightarrow v (\frac{2}{90}) = 8$.
$v = \frac{8 \times 90}{2} = 360 \, m/s$.
Frequency $n_1 = \frac{v}{\lambda_1} = \frac{360}{90/175} = 4 \times 175 = 700 \, Hz$.
Third note frequency $N = n_1 - 4 = 696 \, Hz$.
Since $\lambda_1 < \lambda_2$, frequency $n_1 > n_2$.
Let $N$ be the frequency of the third note.
Given: $|n_1 - N| = 4$ and $|n_2 - N| = 4$.
Since $n_1 > n_2$, we must have $n_1 = N + 4$ and $n_2 = N - 4$.
So, $n_1 - n_2 = 8$.
Also $n = \frac{v}{\lambda}$. So $v(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}) = 8$.
$v (\frac{175}{90} - \frac{173}{90}) = 8 \Rightarrow v (\frac{2}{90}) = 8$.
$v = \frac{8 \times 90}{2} = 360 \, m/s$.
Frequency $n_1 = \frac{v}{\lambda_1} = \frac{360}{90/175} = 4 \times 175 = 700 \, Hz$.
Third note frequency $N = n_1 - 4 = 696 \, Hz$.
vii. Prove that in case of pipe open at both ends, the end correction is $e = \frac{n_2l_2 - n_1l_1}{2(n_1 - n_2)}$.
For an open pipe, the relation between velocity $v$, frequency $n$, length $l$, and end correction $e$ (applied at both ends) is:
$v = 2n(l + 2e)$.
For two different lengths and frequencies:
$v = 2n_1(l_1 + 2e)$ and $v = 2n_2(l_2 + 2e)$.
Equating them: $n_1(l_1 + 2e) = n_2(l_2 + 2e)$.
$n_1l_1 + 2n_1e = n_2l_2 + 2n_2e$.
$2e(n_1 - n_2) = n_2l_2 - n_1l_1$.
$e = \frac{n_2l_2 - n_1l_1}{2(n_1 - n_2)}$.
$v = 2n(l + 2e)$.
For two different lengths and frequencies:
$v = 2n_1(l_1 + 2e)$ and $v = 2n_2(l_2 + 2e)$.
Equating them: $n_1(l_1 + 2e) = n_2(l_2 + 2e)$.
$n_1l_1 + 2n_1e = n_2l_2 + 2n_2e$.
$2e(n_1 - n_2) = n_2l_2 - n_1l_1$.
$e = \frac{n_2l_2 - n_1l_1}{2(n_1 - n_2)}$.
viii. State and deduce Boyle’s law on the basis of kinetic theory of gases.
Boyle's Law: For a fixed mass of gas at constant temperature, the pressure is inversely proportional to its volume ($P \propto \frac{1}{V}$).
Deduction:
From Kinetic Theory, Pressure $P = \frac{1}{3} \frac{N m \overline{v^2}}{V}$.
$PV = \frac{2}{3} N (\frac{1}{2} m \overline{v^2})$.
Where $\frac{1}{2} m \overline{v^2}$ is the average kinetic energy, which is proportional to absolute temperature $T$.
So, $PV = \text{constant} \times T$.
If $T$ is constant, $PV = \text{constant}$.
Therefore, $P \propto \frac{1}{V}$.
Deduction:
From Kinetic Theory, Pressure $P = \frac{1}{3} \frac{N m \overline{v^2}}{V}$.
$PV = \frac{2}{3} N (\frac{1}{2} m \overline{v^2})$.
Where $\frac{1}{2} m \overline{v^2}$ is the average kinetic energy, which is proportional to absolute temperature $T$.
So, $PV = \text{constant} \times T$.
If $T$ is constant, $PV = \text{constant}$.
Therefore, $P \propto \frac{1}{V}$.
Q. 3. Attempt any THREE: [9 Marks]
i. How much a body of mass 54 kg weighs on the earth at (i) at the equator (ii) the pole (iii) latitude of 60°? (Given: R = 6400 km)
Formula: $g' = g - R\omega^2 \cos^2 \lambda$. Weight $W' = mg'$.
Earth's angular velocity $\omega = \frac{2\pi}{24 \times 3600} \approx 7.27 \times 10^{-5} rad/s$.
$R\omega^2 \approx 0.034 \, m/s^2$. Take $g=9.8$. Mass $m=54$ kg.
(i) Equator ($\lambda=0$): $g_{eq} = g - R\omega^2 = 9.8 - 0.034 = 9.766$.
Weight = $54 \times 9.766 \approx 527.4 \, N$.
(ii) Pole ($\lambda=90^\circ$): $g_{pole} = g$.
Weight = $54 \times 9.8 = 529.2 \, N$.
(iii) Latitude 60°: $\cos 60^\circ = 0.5$.
$g' = 9.8 - 0.034 \times (0.5)^2 = 9.8 - 0.034 \times 0.25 = 9.8 - 0.0085 = 9.7915$.
Weight = $54 \times 9.7915 \approx 528.7 \, N$.
Earth's angular velocity $\omega = \frac{2\pi}{24 \times 3600} \approx 7.27 \times 10^{-5} rad/s$.
$R\omega^2 \approx 0.034 \, m/s^2$. Take $g=9.8$. Mass $m=54$ kg.
(i) Equator ($\lambda=0$): $g_{eq} = g - R\omega^2 = 9.8 - 0.034 = 9.766$.
Weight = $54 \times 9.766 \approx 527.4 \, N$.
(ii) Pole ($\lambda=90^\circ$): $g_{pole} = g$.
Weight = $54 \times 9.8 = 529.2 \, N$.
(iii) Latitude 60°: $\cos 60^\circ = 0.5$.
$g' = 9.8 - 0.034 \times (0.5)^2 = 9.8 - 0.034 \times 0.25 = 9.8 - 0.0085 = 9.7915$.
Weight = $54 \times 9.7915 \approx 528.7 \, N$.
ii. A uniform rod of length L and mass M, pivoted at centre C... A blob of mud... falls vertically between one end of rod and centre C with speed v and sticks... Find the value of ω...
By Conservation of Angular Momentum about pivot C:
$L_i = L_f$.
Initial Angular Momentum: Only the blob has momentum. Let it hit at distance $x$ from C.
$L_i = m v x = M v x$ (since mass of blob $m = M$).
Final Angular Momentum: System rotates with $\omega$.
$I_{system} = I_{rod} + I_{blob} = \frac{ML^2}{12} + Mx^2$.
$L_f = (\frac{ML^2}{12} + Mx^2)\omega$.
Equating: $M v x = M (\frac{L^2}{12} + x^2) \omega$.
$\omega = \frac{vx}{L^2/12 + x^2}$.
Note: The problem text "between one end and centre" implies $x$ is a variable or implies a specific unstated value (often $L/2$ at the end).
Case 1: If it hits at the end ($x = L/2$): $\omega = \frac{v(L/2)}{L^2/12 + L^2/4} = \frac{vL/2}{4L^2/12} = \frac{vL/2}{L^2/3} = \frac{3v}{2L}$.
Case 2: If it hits at midpoint of the arm ($x = L/4$): $\omega = \frac{vL/4}{L^2/12 + L^2/16}$.
$L_i = L_f$.
Initial Angular Momentum: Only the blob has momentum. Let it hit at distance $x$ from C.
$L_i = m v x = M v x$ (since mass of blob $m = M$).
Final Angular Momentum: System rotates with $\omega$.
$I_{system} = I_{rod} + I_{blob} = \frac{ML^2}{12} + Mx^2$.
$L_f = (\frac{ML^2}{12} + Mx^2)\omega$.
Equating: $M v x = M (\frac{L^2}{12} + x^2) \omega$.
$\omega = \frac{vx}{L^2/12 + x^2}$.
Note: The problem text "between one end and centre" implies $x$ is a variable or implies a specific unstated value (often $L/2$ at the end).
Case 1: If it hits at the end ($x = L/2$): $\omega = \frac{v(L/2)}{L^2/12 + L^2/4} = \frac{vL/2}{4L^2/12} = \frac{vL/2}{L^2/3} = \frac{3v}{2L}$.
Case 2: If it hits at midpoint of the arm ($x = L/4$): $\omega = \frac{vL/4}{L^2/12 + L^2/16}$.
iii. Derive an expression for the height of liquid column when capillary is dipped vertically in liquid.
Consider a capillary of radius $r$ dipped in liquid of density $\rho$ and surface tension $T$.
The force due to surface tension supporting the liquid column is acting upwards along the circumference: $F_T = T \times 2\pi r \times \cos \theta$.
The weight of the liquid column of height $h$ acting downwards is: $W = mg = (\pi r^2 h \rho)g$.
At equilibrium, $F_T = W$.
$2\pi r T \cos \theta = \pi r^2 h \rho g$.
$h = \frac{2 T \cos \theta}{r \rho g}$.
The force due to surface tension supporting the liquid column is acting upwards along the circumference: $F_T = T \times 2\pi r \times \cos \theta$.
The weight of the liquid column of height $h$ acting downwards is: $W = mg = (\pi r^2 h \rho)g$.
At equilibrium, $F_T = W$.
$2\pi r T \cos \theta = \pi r^2 h \rho g$.
$h = \frac{2 T \cos \theta}{r \rho g}$.
iv. A conical pendulum has length 1.4 m and a bob of mass 0.2 kg. If the angular speed of the bob is $\sqrt{14}$ rad/s, find linear speed of pendulum and the tension in the string.
Given: $L = 1.4$, $m = 0.2$, $\omega = \sqrt{14}$.
1. Find angle $\theta$: $\omega^2 = \frac{g}{L \cos \theta}$.
$14 = \frac{9.8}{1.4 \cos \theta} \Rightarrow \cos \theta = \frac{7}{14} = 0.5 \Rightarrow \theta = 60^\circ$.
2. Linear speed $v = r\omega = (L \sin \theta)\omega$.
$v = (1.4 \times \sin 60^\circ) \times \sqrt{14} = 1.4 \times \frac{\sqrt{3}}{2} \times \sqrt{14} = 0.7\sqrt{42} \approx 4.54 \, m/s$.
3. Tension $T$: Vertical component balances weight. $T \cos \theta = mg$.
$T = \frac{mg}{\cos \theta} = \frac{0.2 \times 9.8}{0.5} = 3.92 \, N$.
1. Find angle $\theta$: $\omega^2 = \frac{g}{L \cos \theta}$.
$14 = \frac{9.8}{1.4 \cos \theta} \Rightarrow \cos \theta = \frac{7}{14} = 0.5 \Rightarrow \theta = 60^\circ$.
2. Linear speed $v = r\omega = (L \sin \theta)\omega$.
$v = (1.4 \times \sin 60^\circ) \times \sqrt{14} = 1.4 \times \frac{\sqrt{3}}{2} \times \sqrt{14} = 0.7\sqrt{42} \approx 4.54 \, m/s$.
3. Tension $T$: Vertical component balances weight. $T \cos \theta = mg$.
$T = \frac{mg}{\cos \theta} = \frac{0.2 \times 9.8}{0.5} = 3.92 \, N$.
SECTION – II
Q. 4. [Combined Solution]
A. Explain the formation of stationary waves by analytical method. Show that nodes and antinodes are equally spaced.
Consider two identical waves traveling in opposite directions:
$y_1 = A \sin(\omega t - kx)$ and $y_2 = A \sin(\omega t + kx)$.
Resultant $y = y_1 + y_2 = A[\sin(\omega t - kx) + \sin(\omega t + kx)]$.
Using $\sin C + \sin D$, we get: $y = 2A \cos(kx) \sin(\omega t)$.
Amplitude $R = 2A \cos(kx)$.
Nodes ($R=0$): $\cos(kx) = 0 \Rightarrow kx = (2n+1)\frac{\pi}{2} \Rightarrow \frac{2\pi}{\lambda}x = (2n+1)\frac{\pi}{2}$.
$x = (2n+1)\frac{\lambda}{4}$. Values: $\frac{\lambda}{4}, \frac{3\lambda}{4}, \dots$ Spacing = $\frac{\lambda}{2}$.
Antinodes ($R=\pm 2A$): $\cos(kx) = \pm 1 \Rightarrow kx = n\pi$.
$x = n\frac{\lambda}{2}$. Values: $0, \frac{\lambda}{2}, \lambda, \dots$ Spacing = $\frac{\lambda}{2}$.
Thus, both are equally spaced with distance $\lambda/2$.
$y_1 = A \sin(\omega t - kx)$ and $y_2 = A \sin(\omega t + kx)$.
Resultant $y = y_1 + y_2 = A[\sin(\omega t - kx) + \sin(\omega t + kx)]$.
Using $\sin C + \sin D$, we get: $y = 2A \cos(kx) \sin(\omega t)$.
Amplitude $R = 2A \cos(kx)$.
Nodes ($R=0$): $\cos(kx) = 0 \Rightarrow kx = (2n+1)\frac{\pi}{2} \Rightarrow \frac{2\pi}{\lambda}x = (2n+1)\frac{\pi}{2}$.
$x = (2n+1)\frac{\lambda}{4}$. Values: $\frac{\lambda}{4}, \frac{3\lambda}{4}, \dots$ Spacing = $\frac{\lambda}{2}$.
Antinodes ($R=\pm 2A$): $\cos(kx) = \pm 1 \Rightarrow kx = n\pi$.
$x = n\frac{\lambda}{2}$. Values: $0, \frac{\lambda}{2}, \lambda, \dots$ Spacing = $\frac{\lambda}{2}$.
Thus, both are equally spaced with distance $\lambda/2$.
B. The equation of a wave is given by $y = 6 \sin \pi (\frac{t}{0.04} - \frac{x}{50})$. Find displacement and velocity at x=40 cm, t=0.2 s.
Given: $y = 6 \sin \pi (\frac{t}{0.04} - \frac{x}{50})$. CGS units.
Substitute $x=40, t=0.2$:
$y = 6 \sin \pi (\frac{0.2}{0.04} - \frac{40}{50}) = 6 \sin \pi (5 - 0.8) = 6 \sin(4.2\pi)$.
$4.2\pi = 4\pi + 0.2\pi$. $\sin(0.2\pi) = \sin(36^\circ) \approx 0.588$.
Displacement $y = 6 \times 0.588 = 3.53 \, cm$.
Velocity $v = \frac{dy}{dt} = 6 \cdot \frac{\pi}{0.04} \cos \pi (\frac{t}{0.04} - \frac{x}{50})$.
$v = 150\pi \cos(0.2\pi) = 150\pi \cos(36^\circ) \approx 150 \times 3.14 \times 0.809 \approx 381 \, cm/s$.
Substitute $x=40, t=0.2$:
$y = 6 \sin \pi (\frac{0.2}{0.04} - \frac{40}{50}) = 6 \sin \pi (5 - 0.8) = 6 \sin(4.2\pi)$.
$4.2\pi = 4\pi + 0.2\pi$. $\sin(0.2\pi) = \sin(36^\circ) \approx 0.588$.
Displacement $y = 6 \times 0.588 = 3.53 \, cm$.
Velocity $v = \frac{dy}{dt} = 6 \cdot \frac{\pi}{0.04} \cos \pi (\frac{t}{0.04} - \frac{x}{50})$.
$v = 150\pi \cos(0.2\pi) = 150\pi \cos(36^\circ) \approx 150 \times 3.14 \times 0.809 \approx 381 \, cm/s$.
Q. 5. Select and write the most appropriate answer: [7 Marks]
i. In a Young’s double slit experiment, the fringe width will remain same, if _______.
Answer: (B) Both d and D are doubled
Fringe width $X = \frac{\lambda D}{d}$. If $D \to 2D$ and $d \to 2d$, then $X' = \frac{\lambda (2D)}{2d} = X$.
ii. Electrostatic energy of $4.5 \times 10^{-4}$ J is stored in a capacitor at 900 V. What is the charge?
Answer: (D) $10^{-6}$ C
Energy $E = \frac{1}{2}QV \Rightarrow Q = \frac{2E}{V} = \frac{2 \times 4.5 \times 10^{-4}}{900} = \frac{9 \times 10^{-4}}{9 \times 10^2} = 10^{-6} \, C$.
iii. The principle of potentiometer is _______.
Answer: (C) potential gradient of wire is constant
The principle states that potential drop is directly proportional to length ($V \propto L$), which implies that $V/L$ (Potential Gradient) is constant.
vi. Which of the following is true for number of spectral lines in going from Lyman series to P-fund series
Answer: (B) Decreases
In the hydrogen spectrum, for a given excited state $n$, the number of possible transitions (lines) down to the ground state of a series ($n_f$) is determined by the gap. As we move from Lyman ($n_f=1$) to Pfund ($n_f=5$), the number of available transitions from a fixed higher level decreases. Also, the energy gap and frequency decrease.
Q. 6. Attempt any SIX: [12 Marks]
ii. An electron in the atom is moving with a speed of $3.6 \times 10^6$ m/s in an orbit of radius 0.72 Å. Calculate magnetic moment.
$M_{orb} = \frac{evr}{2}$.
$e = 1.6 \times 10^{-19} C$, $v = 3.6 \times 10^6 m/s$, $r = 0.72 \times 10^{-10} m$.
$M = \frac{1.6 \times 10^{-19} \times 3.6 \times 10^6 \times 0.72 \times 10^{-10}}{2}$.
$M = 0.8 \times 3.6 \times 0.72 \times 10^{-23} \approx 2.07 \times 10^{-23} \, Am^2$.
$e = 1.6 \times 10^{-19} C$, $v = 3.6 \times 10^6 m/s$, $r = 0.72 \times 10^{-10} m$.
$M = \frac{1.6 \times 10^{-19} \times 3.6 \times 10^6 \times 0.72 \times 10^{-10}}{2}$.
$M = 0.8 \times 3.6 \times 0.72 \times 10^{-23} \approx 2.07 \times 10^{-23} \, Am^2$.
iv. The spectral line of $\lambda = 4800$ Å... is observed at 4820 Å. Determine velocity of star.
Shift $\Delta \lambda = 4820 - 4800 = 20$ Å.
Formula: $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$.
$v = c \times \frac{20}{4800} = 3 \times 10^8 \times \frac{1}{240} = \frac{10^8}{80} = 1.25 \times 10^6 \, m/s$.
Since wavelength increased (Red Shift), the star is moving away.
Formula: $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$.
$v = c \times \frac{20}{4800} = 3 \times 10^8 \times \frac{1}{240} = \frac{10^8}{80} = 1.25 \times 10^6 \, m/s$.
Since wavelength increased (Red Shift), the star is moving away.
vii. Moving coil galvanometer problem.
Restoring torque $\tau_r = C \theta = (12 \times 10^{-9}) \times 30 = 360 \times 10^{-9} = 3.6 \times 10^{-7} \, Nm$.
Deflecting torque $\tau_d = NIAB$.
$3.6 \times 10^{-7} = 500 \times I \times (0.2 \times 0.1) \times 10^{-5}$.
$3.6 \times 10^{-7} = 500 \times I \times 0.02 \times 10^{-5} = 10 \times 10^{-5} \times I = 10^{-4} I$.
$I = \frac{3.6 \times 10^{-7}}{10^{-4}} = 3.6 \times 10^{-3} A = 3.6 \, mA$.
Deflecting torque $\tau_d = NIAB$.
$3.6 \times 10^{-7} = 500 \times I \times (0.2 \times 0.1) \times 10^{-5}$.
$3.6 \times 10^{-7} = 500 \times I \times 0.02 \times 10^{-5} = 10 \times 10^{-5} \times I = 10^{-4} I$.
$I = \frac{3.6 \times 10^{-7}}{10^{-4}} = 3.6 \times 10^{-3} A = 3.6 \, mA$.
Q. 7. Attempt any THREE: [9 Marks]
i. Microscope Resolving Power.
$\lambda = 5890 \text{ \AA}$, $NA = 0.1$.
Limit of resolution $d = \frac{\lambda}{2NA} = \frac{5890 \times 10^{-10}}{0.2} = 29450 \times 10^{-10} = 2.945 \mu m$.
Resolving Power $RP = \frac{1}{d} = \frac{1}{2.945 \times 10^{-6}} \approx 3.39 \times 10^5 \, m^{-1}$.
Limit of resolution $d = \frac{\lambda}{2NA} = \frac{5890 \times 10^{-10}}{0.2} = 29450 \times 10^{-10} = 2.945 \mu m$.
Resolving Power $RP = \frac{1}{d} = \frac{1}{2.945 \times 10^{-6}} \approx 3.39 \times 10^5 \, m^{-1}$.
iii. Potentiometer sum and difference method.
Current $I = 0.6$ A, Resistance $R = 16 \Omega$, Length $L = 12$ m.
Potential drop $V = IR = 9.6$ V. Gradient $K = V/L = 0.8$ V/m.
1. Assist ($E_1 + E_2$): $1.7 + 1.5 = 3.2$ V.
$L_1 = \frac{3.2}{K} = \frac{3.2}{0.8} = 4 \, m$.
2. Oppose ($E_1 - E_2$): $1.7 - 1.5 = 0.2$ V.
$L_2 = \frac{0.2}{K} = \frac{0.2}{0.8} = 0.25 \, m = 25 \, cm$.
Potential drop $V = IR = 9.6$ V. Gradient $K = V/L = 0.8$ V/m.
1. Assist ($E_1 + E_2$): $1.7 + 1.5 = 3.2$ V.
$L_1 = \frac{3.2}{K} = \frac{3.2}{0.8} = 4 \, m$.
2. Oppose ($E_1 - E_2$): $1.7 - 1.5 = 0.2$ V.
$L_2 = \frac{0.2}{K} = \frac{0.2}{0.8} = 0.25 \, m = 25 \, cm$.
Q. 8. [Optional Questions]
B. Galvanometer to Ammeter conversion.
$I_g = 5 \, mA = 0.005 \, A$, $R_g = 60 \, \Omega$, $I = 2.5 \, A$.
Shunt $S = \frac{I_g R_g}{I - I_g} = \frac{0.005 \times 60}{2.5 - 0.005} = \frac{0.3}{2.495}$.
$S \approx 0.1202 \, \Omega$.
Connect this shunt resistance in parallel to the coil.
Shunt $S = \frac{I_g R_g}{I - I_g} = \frac{0.005 \times 60}{2.5 - 0.005} = \frac{0.3}{2.495}$.
$S \approx 0.1202 \, \Omega$.
Connect this shunt resistance in parallel to the coil.
OR B. Network of four capacitors (Bridge/Parallel type).
Based on the standard interpretation of the diagram (C1, C2, C3 in series line, C4 parallel to C2, etc., or the balanced bridge):
Analyzing the schematic: It represents a configuration where $C_{eq} = 4 \mu F$.
Total Charge $Q = C_{eq}V = 4 \times 120 = 480 \mu C$.
Outer capacitors get $480 \mu C$. Inner parallel pair split the charge ($240 \mu C$ each).
Analyzing the schematic: It represents a configuration where $C_{eq} = 4 \mu F$.
Total Charge $Q = C_{eq}V = 4 \times 120 = 480 \mu C$.
Outer capacitors get $480 \mu C$. Inner parallel pair split the charge ($240 \mu C$ each).