Following is the solved question paper of Physics Model Question prepared by National Examination Board (NEB) for Grade XII students of 2079 (2023) batch.
Check other Physics Model Sets:
- NEB Class 12 Physics Board Question Paper with Solutions – 2078
- NEB Class 12 Physics Model Question Paper with Solutions – Set 1
Physics Model Question for NEB XII 2079 (2023)
Q 12 a) Define simple harmonic motion.
Solution: Simple harmonic motion is defined as a harmonic motion having constant amplitude and frequency. Example: motion of a simple pendulum, etc.
Q 12 b) Derive an expression for the time period of oscillation for a mass m attached to a vertical spring of force constant k.
Solution: $$\rm T = 2 \pi \sqrt{ \frac{m}{k}}$$
Q 12 c) What will be the time period of this system if it is taken inside of the satellite?
Solution: The time period of a spring mass system only depends upon its mass and force constant. These parameters remain the same inside the satellite as on the Earth. Hence, the time period of this system remains same if it is taken inside of the satellite.
Q 13 a) State Bernoulli’s principle.
Solution: Bernoulli’s principle states, “The total energy per unit mass of an ideal fluid flowing through a tube at each cross-section is constant.”
Q 13 b) Figure below shows a liquid of density 1200 km/m3 flowing steadily in a tube of varying cross-sections. The cross-section at point A is 1.0 cm2 and that at B is 20 mm2, points A and B are in the same horizontal plane. The speed of liquid at A is 10 cm/s. Calculate.
i) the speed at B.
Solution: Using the equation of continuity, the product of area and velocity at A and B remain same.$$\rm v_A a_A = v_B a_b$$$$\rm 10 \times 1 = v_B \times (20 \times 10^{-2})$$$$\rm v_B = 50 cm/s$$Hence, the required velocity of the liquid at point B is 50 cm/s.
ii) the difference in pressure at A and B.
Solution: It is mentioned that the points lie in same horizontal plane so, $\triangle$h = 0. By using Bernoulli’s principle, the total energy per unit mass at A is equal to the total enery per unit mass at B.$$\rm \frac{1}{2} v_A^2 + gh_1 + \frac{P_1}{\rho} = \frac{1}{2} v_B^2 + gh_2 + \frac{P_2}{\rho}$$$$\rm \frac{P_1 – P_2}{\rho} = \frac{1}{2} ( v_B^2 – v_A^2) + g(h_2 – h_1)$$$$\rm \Rightarrow P_1 – P_2 = \frac{1}{2} \times (50^2 – 10^2) \times 10^{-4} \times 1200$$$$\rm \therefore \triangle P = 144 Pa$$ Hence, the required pressure difference at A and B is 144 Pa.
Q 14 a) Draw a PV diagram of a petrol engine and explain its working based on its PV diagram.
Solution: Explain.
Q 14 b) Compare the efficiency of petrol engines with that of diesel engines based on their compression ratios.
Solution: The compression ratio of a diesel engine is much larger than that of a petrol engine. The efficiency of such internal combustion engines are computed by the formula$$\rm \eta = 1 – \left ( \frac{1}{\rho} \right )^{\gamma – 1}$$where $\rho$ is the compression ratio. Hence, the efficiency of diesel engines is higher than petrol enginges.
Q 15 a) When the wire of a sonometer is 75 cm long, it is in resonance with a tuning fork. On shortening the wire by 0.5 cm, it makes 3 beats with the same fork. The beat is the difference in frequencies. Calculate the frequency of the tuning fork.
Solution: At resonance, the length of wire is 75 cm. Considering resonance condition with the same standing wave, when length of wire decreases (75 cm – 0.5 cm = 74.5 cm), the wavelength also decreases and frequency of vibration of wire increases. In each case, the velocity of wave $(v = \lambda f)$ remains constant. So, $$\lambda f_{fork} = \lambda _{new} f_{new}$$$$\rm \Rightarrow f_{new} = \frac{75}{74.5} f_{fork}$$Also, it is given$$\rm f_{new} – f_{fork} = 3$$$$\rm \frac{75}{74.5} f_{fork} – f_{fork} = 3$$$$\rm \therefore f_{fork} = 447 Hz$$
Q 15 b) The diagram below shows an experiment to measure the speed of a sound in a string. The frequency of the vibrator is adjusted until the standing wave shown in the diagram is formed. The frequency of the vibrator is 120 Hz. Calculate the speed at which a progressive wave would travel along the string.
Solution: At resonance, three complete wavelength is formed in a length of wire 75 cm.$$\rm 3 \lambda = 75 cm \Rightarrow \lambda = 25 cm = 0.25 m$$Here, frequency of vibrator $(f = 120 Hz)$.
Hence, the velocity of the progressive wave (v) = $\rm \lambda \times f$$$\rm \Rightarrow (0.25) \times 120 = 30 m/s$$
Q 16 a) State Lenzs law in electromagnetism. Justify this law is in favor of the principle of conservation of energy.
Solution: In electromagnetism, Lenzs’ law states that the induced current sets itself in a conductor in a way that opposes the cause that produces it.
Yes, this law favours the principle of conservation of energy. Mechanical work done spent in changing magnetic flux is converted into electromotive force.
Q 16 b) A magnet is quickly moved in the direction indicated by an arrow between two coils $C_1$ and $C_2$ as shown in the figure. What will be the direction of the induced current in each coil as indicated by the movement of magnet? Explain.
Solution: Owing to Lenz’s law, the induced current will tend to oppose the effect of moving magnet. In coil $C_1$, current will induce in a way to attract the magnet so, south pole will be near the North of the magnet. In coil $C_2$, magnetic flux is increasing so, South pole will induce near the South of the magnet to reduce it. Accordingly, the direction of current can be shown in a figure. In both coils, the direction of induced current with be clockwise, as observed from the magnet.
Q 17 a) State the principle of the Potentiometer. A potentiometer is also called a voltmeter of infinte resistance, why?
Solution: Potentiometer works on the principle, “The potential difference through a wire is proportional to its length provided that a wire of uniform cross-section is used and steady current is flown through it.”
A potentiometer is also called a voltmeter of infinte resistance because it does not permit any current through it.
Q 17 b) In the meter bridge experiment, the balance point was observed at J with l = 20 cm.
i) The values of R and X were doubled and then interchanged. What would be the new position of balance point?
Solution: In a meter bridge,$$\rm X \times l = R \times (100 – l)$$. It is given that the balance point lenth $\rm l = 20 cm$. So, $\rm (100 – l) = (100 – 20) = 80 cm$.$$\rm \Rightarrow \frac{X}{R} = \frac{80}{20} = 4$$When R’ = 2X and X’ = 2R, let l’ be the new balancing length then, $$\rm (2R) \times l \prime = 2X \times (100 – l \prime)$$$$\rm \frac{l \prime}{100 – l \prime} = \frac{X}{R}$$$$\rm \frac{l \prime}{100 – l \prime} = 4$$$$\rm l \prime = 400 – 4 l \prime$$$$\rm 5 l \prime = 400$$$$\rm l \prime = 80 cm$$Hence, the new position of balance point would be 80 cm.
ii) If the galvanometer and battery are interchanged at the balance position, how will the balance point get affected?
Solution: The balance point does not change by interchanging the position of galvanometer and battery at the balance point.
Q 17 a) State the two Kirchoff’s laws for electrical circuits.
Solution: Kirchoff’s laws for electrical circuits are stated below:
Kirchoff’s first law: According to this law, the algebraic sum of currents entering any junction is equal to the algebraic sum of currents leaving the junction. It is based on the principle of conservation of charge.$$\rm \sum I = 0$$Kirchoff’s second law: According to this law, the algebraic sum of potential difference across any conducting loop is zero. It is based on the principle of conservation of energy.$$\rm \sum E = 0$$
Q 17 b) In Meter Bridge shown below, the null point is found at a distance of 60.0 cm from A. If now a resistance of 5 ohm is connected in series with S, the null point occurs at 50 cm. Determine the values of R and S.
Solution: Let the balancing length be denoted by l. In balanced condition in a meter bridge,$$\rm R \times (100 – l) = S \times l$$When l = 50 cm,$$\rm R \times (100 – 60) = S \times 60$$$$\Rightarrow \rm R \times 40 = S \times 60$$$$\Rightarrow \rm S = \frac{2}{3} R$$When a resistance of 5 ohm is connected in series with S, the new resistance at S can be written as (S + 5). If l = 50 cm, then$$\rm R \times (100 – 50) = (S + 5) \times 50$$$$\rm \Rightarrow R = S + 5$$$$\rm \Rightarrow R = \frac{2}{3} R + 5$$$$\rm \therefore R = 15 \Omega$$Similiarly, S = (2R)/3 = 15 $\rm \Omega$
Q 18) The graph below shows the maximum kinetic energy of the emitted photoelectrons as the frequency of incident radiation on a sodium plate is varied.
a) From the graph determine the maximum frequency of radiation that can cause a photoelectric effect?
Solution: From the graph, the maximum frequency of radiation that can cause a photoelectric effect is $\rm 10 \cdot 10^{14} Hz$. However, any frequency higher than or equal to $\rm 5.6 \cdot 10^{14}$ Hz (threshold frequency) can cause the photoelectric effect.
b) Calculate the work function for sodium.
Solution: Work function of a metal ($\rm \phi$) is defined as the minimum energy required to just pull its valence electron from the orbit.$$\rm \phi = hf_o$$Here, $\rm f_o = 5.6 \times 10^{14} Hz$.
We know, $\rm h = 6.64 \times 10^{-34} Js$
Then$$\rm \phi = 6.64 \times 10^{-34} \times 5.6 \times 10^{14}$$
c) Use the graph to calculate the value of the Planck constant in Js.
Solution: The photoelectric equation $\rm E = hf – hf_o$ represents the graph given in the equation. From the equation, it is clear that the slope of the graph gives the value of Planck’s constant. To express it in Js, enery must be represented in Joules. Now,$$\rm h = slope$$$$\rm h = \frac{\triangle E}{\triangle f}$$$$\rm h = \frac{(1 – 0) eV}{(8 – 5.6) \times 10^{14} Hz}$$$$\rm h = \frac{1}{2.4} \times (1.6 \times 10^{-19} \times 10^{-14})$$$$\rm h = 6.66 \times 10^{-34} Js$$
Q 19 a) Figure below shows the experimental setup of Millikan’s oil drop experiment. Find the expression for a charge of an oil drop of radius r moving with constant velocity v in a downward direction using a free body diagram.
Solution:
Q 19 b) What will be the expression for the charge of an oil drop if the electric force is greater than its weight?
Solution:
Q 19 c) Determine the electric field supplied when the electric force applied between the two horizontal plates just balances an oil drop with 4 electrons attached to it and mass of oil drop is 1.3 $\times$ 10$^{-14}$ kg.
Solution:
Q 19 a) A diode can be used as a rectifier. What characteristic of a diode is used in rectification?
Solution: An ideal diode functions as short circuit in forward bias and behaves as an open circuit in reverse bias. This characteristic of a diode is used in rectification.
Q 19 b) Draw a circuit diagram for full wave rectifier.
Solution: Refer to any textbook for the circuit diagram.
Q 19 c) A NOR gate ‘opens’ and gives an output only if both inputs are ‘low’, but an OR gate ‘closes’. An AND gate ‘opens’ only if both outputs are ‘high’, but a NAND gate ‘closes’. Construct a truth table for the circuit shown below including the states of E, F and G.
Solution:
Q 20 a) What is the significance of the negative energy of the electron in an orbit?
Solution: The negative energy of an electron in an orbit signifies that it is bound to the nucleus. It cannot leave the orbit unless sufficient external energy is supplied for the purpose.
Q 20 b) The energy levels of an atom are shown in Fig. below. Which one of these transitions will result in the emission of a photon of wavelength 275 nm? Explain with calculation. (h = 6.64 $\times$ 10$^{-34}$ Js, c = 3.0 $\times$ 10$^8$ m/s).
Q 20 c) Find the expression for the wavelength of radation emitted from a hydrogen atom when an electron jumps from igher energy level $n_2$ to the lower energy level $n_1$.
Solution:
Q 21 a) A student is trying to make an accurate measurement of the wavelength of green light from a mercury lamp ($\lambda = 546 nm$). Using a double slit of separation of 0.50 mm, he finds he can see ten clear fringes on a screen at a distance of 0.80 m from the slits. He then tries an alternative experiment using a diffraction grating that has 3000 lines/cm.
i) What will be the width of the ten fringes that he can measure in the first experiment?
Solution:
ii) What will be the angle of the second-order maximum in the second experiment?
Solution:
iii) Suggest which experiment you think will give the more accurate measurement of wavelength ($\lambda$).
Solution: The diffraction grating experiment will give a more accurate measurement of the wavelength.
Q 21 b) A physics student went to buy polaroid sunglass. The shopkeeper gave him two similar-looking sunglasses. In what way he can differentiate between polaroid sunglasses and non-polaroid sunglasses?
Solution:
Q 21 c) At what angle of incidence wll the light reflected from water ($\mu = 1.3$) be completely polarized?
Solution: Using Brewster’s law, light incident ona surface at a polarizing angle ($\theta_p$) will be completely polarized in the given condition$$\rm \mu = tan \theta_p$$$$\rm tan \theta_p = 1.3$$$$\rm \theta_p = 52.43 ^o$$
Q 21 a) In what way does the intensity of sound heard by an observer change if the distance with the source changes by four times?
Solution: $$\rm I \propto \frac{1}{r^2}$$$$\rm I_2 = \frac{r_1^2}{r_2^2} \times I_1$$$$\rm I_2 = \frac{I_1}{16}$$
Q 21 b) A train is traveling at 30 m/s in still air. The frequency of the note emitted by the train whistle is 262 Hz. What frequency is heard by a passenger on the other train moving in the opposite direction to the first at 18 m/s
(i) when approaches the first?
Solution:
(ii) when receding from the first?
Solution:
Q 21 c) In a sinosoidal sound wave of moderate loudness, the maximum pressure variations are of the order of 3.0 $\times$ 10$^{-2}$ Pa above and below the atmospheric pressure. Find the corresponding maximum displacement if the frequency is 1000 Hz in air at atmospheric pressure and density. The speed of sound is 344 m/s and the bulk modulus of the medium is 11.42 $\times$ 10$^5$ Pa.
Solution:
Q 22 a) What is choke coil?
Solution: A choke coil is an inductor that is used to moderate current in an AC circuit.
Q 22 b) Why is it preferred over a resistor in ac circuit?
Solution: An ideal choke coil has zero resistance so, there is no energy loss in the coil.
Q 22 c) In figures (a), (b), and (c), three ac circuits with equal currents have been shown.
Leave a Reply