67486 + Interview Questions in Secondary School in Physics

1.	What is inertia and second law of motion
Answer» The INHERENT property of all objects that they do not change their state rest or uniform MOTION along a straight line path unless we can not apply EXTERNAL FORCE called inertia. The rate of change of MOMENTUM of body is equal to the external force applied on the baody nd yhe change in momentum always on the direction of force called Newton secnd law of motion..

Discussion

2.	A stone thrown vertically upwards with initial velocity u reached a height h before coming down. Show that the time taken to go up in same as the time taken to came down.
Answer» When the ball is in your HAND it's U is 0 when u THROW it it catches some speed that is v1 then at a certain point is came back, on its certain point it BECAME again 0 and RETURNED with some speed v2 ; now our first v1 is U and second is v2 is V the result will be 0 if their distance is equal

Discussion

3.	a scooter acquires a velocity of 36 km/h in 1/6seconds just after the start. Calculate the acceleration of the scooter.
Answer» Speed in m/s =36*5/18= 10m/s According to Newton's FIRST law of motion, a = (V - U)/t =10/(1/6) =60m/s^2

Discussion

4.	Give difference between law of gravitatrion and acceleration due to grsavity
Answer» It sets the strength of the gravitational INTERACTION in the SENSE that if it were doubled, so would all the gravitational forces. But 'g' is gravity or acceleration of gravity. this is 'acceleration' due to gravity that an object feels near the surface of the earth. Hope it's help you FRIEND

Discussion

5.	In nth second displacement covered by an object is 4n-3 metre. Find displacement and velocity in 100 seconds
Answer» In this QUESTION we have been given the function of the displacement. Through differentiation, we can get the velocity function of the moving OBJECT. Let displacement function be S then : S = 4n — 3 V is THUS : ds / dn = 4 At t = 100 S = 4 × 100 — 3 = 397 m The velocity is 4m/s

5.

In nth second displacement covered by an object is 4n-3 metre. Find displacement and velocity in 100 seconds

Answer»

In this QUESTION we have been given the function of the displacement.

Through differentiation, we can get the velocity function of the moving OBJECT.

Let displacement function be S then :

S = 4n — 3

V is THUS :

ds / dn = 4

At t = 100

S = 4 × 100 — 3 = 397 m

The velocity is 4m/s

Discussion

6.	Draw a graph that acceleration is in uniform motion but initial velocity is not equal to zero
Answer» It is a VELOCITY TIME time GRAPH

Discussion

7.	It is easier to move a heavier box when it is fitted with wheels
Answer» because it is in rolling friction so it is EASY to MOVE a HEAVIER box when it is FITTED with WHEELS

Discussion

8.	A rectangular brick is kept on a table with a part of its length projecting out. It remains at rest if the length projected is slightly less than slightly half the total length but it falls down if the length projected is slightly more than half the total length. Give reason.
Answer» Correct REASON ⇒ The weight or the Force of the Gravity of the projected part produces a torque that will tend to overturn the brick about the edge of the table while the weight of the part on the table sometimes produces a torque that will tend to restore the brick from overturning. Let us Assuming that the uniform density of the brick in the first case, the weight of projected part will be less than the weight of the part on the table. Also, the distance of the line of action of the weight of the projected part from the edge of the table is also less in COMPARISON to the other part. Thus, the overturning torque, in this case, will be less than the restoring torque. Thus, the brick remains at rest. The situation is just reverse in the second case where the overturning torque is GREATER than the restoring torque, so the brick overturns and falls. HOPE it helps.

8.

A rectangular brick is kept on a table with a part of its length projecting out. It remains at rest if the length projected is slightly less than slightly half the total length but it falls down if the length projected is slightly more than half the total length. Give reason.

Answer»

Correct REASON ⇒ The weight or the Force of the Gravity of the projected part produces a torque that will tend to overturn the brick about the edge of the table while the weight of the part on the table sometimes produces a torque that will tend to restore the brick from overturning.

Let us Assuming that the uniform density of the brick in the first case, the weight of projected part will be less than the weight of the part on the table.

Also, the distance of the line of action of the weight of the projected part from the edge of the table is also less in COMPARISON to the other part.

Thus, the overturning torque, in this case, will be less than the restoring torque. Thus, the brick remains at rest.

The situation is just reverse in the second case where the overturning torque is GREATER than the restoring torque, so the brick overturns and falls.

HOPE it helps.

Discussion

9.	A body is in transitional equilibrium under the actions of coplanar forces. If the torque of these forces is zero about a point, is it necessary that it will also be zero about any other point?
Answer» Answer ⇒ Yes, It is necessary, i.e., it is always necessary that the torque will be zero about any other point if the RESULTANT of the torque acting on the BODY is zero about the point. If the body will be in the Transnational Equilibrium then it will move with he constant VELOCITY thus zero acceleration at every point. By the Inertial and the EQUIVALENT Frame of the Reference ,i.e the Position of the OBSERVATIONS, it will be Zero. Note ⇒ Torque is the turning effects produced in the Body when the force is applied and the body covers some distance. Torque is the Vector Quantity and also an Vector Product or the Cross Product of two vectors. Hope it helps.

9.

A body is in transitional equilibrium under the actions of coplanar forces. If the torque of these forces is zero about a point, is it necessary that it will also be zero about any other point?

Answer»

Answer ⇒ Yes, It is necessary, i.e., it is always necessary that the torque will be zero about any other point if the RESULTANT of the torque acting on the BODY is zero about the point.

If the body will be in the Transnational Equilibrium then it will move with he constant VELOCITY thus zero acceleration at every point.

By the Inertial and the EQUIVALENT Frame of the Reference ,i.e the Position of the OBSERVATIONS, it will be Zero.

Note ⇒ Torque is the turning effects produced in the Body when the force is applied and the body covers some distance.

Torque is the Vector Quantity and also an Vector Product or the Cross Product of two vectors.

Hope it helps.

Discussion

10.	If the resultant torque of all the forces acting on a body is zero about a point, is it necessary that it will be zero about any other point?
Answer» Answer ⇒ Yes, It is necessary, i.e., it is always necessary that the torque will be ZERO about any other point if the resultant of the torque ACTING on the body is zero about the point. By the Inertial and the Equivalent Frame of the Reference ,i.e the Position of the Observations it will be Zero. NOTE ⇒ Torque is the TURNING effects produced in the Body when the force is applied and the body covers some distance. Torque is the Vector Quantity and also an Vector Product or the CROSS Product of two vectors. Hope it helps.

10.

If the resultant torque of all the forces acting on a body is zero about a point, is it necessary that it will be zero about any other point?

Answer»

Answer ⇒ Yes, It is necessary, i.e., it is always necessary that the torque will be ZERO about any other point if the resultant of the torque ACTING on the body is zero about the point.

By the Inertial and the Equivalent Frame of the Reference ,i.e the Position of the Observations it will be Zero.

NOTE ⇒ Torque is the TURNING effects produced in the Body when the force is applied and the body covers some distance.

Torque is the Vector Quantity and also an Vector Product or the CROSS Product of two vectors.

Hope it helps.

Discussion

11.	If the sum of all the forces acting on a body is zero, is it necessarily in equilibrium? If the sum of all the forces on a particle is zero, is it necessarily in equilibrium?
Answer» *Answer ⇒* *No, It the sum of all the FORCES acting on the body is zero then it not necessarily that the body will be in the state of the Equilibrium.* *Explanation⇒* The sum of all the forces acting on the body is zero then it not necessarily that the body will be in the state of the Equilibrium. For the explanation let us TAKE an example. Let us assume that the two equal and OPPOSITE forces having their line of actions at a CERTAIN distance apart act on a body, then the body in this case will remain stationary at the point but will have not have the rotation due to the torque produced. Also, we can say that If the sum of all the forces acting on a particle is zero then it is necessarily in equilibrium because there will not be a distance between anti-parallel forces. Hope it HELPS.

11.

If the sum of all the forces acting on a body is zero, is it necessarily in equilibrium? If the sum of all the forces on a particle is zero, is it necessarily in equilibrium?

Answer»

Answer ⇒ No, It the sum of all the FORCES acting on the body is zero then it not necessarily that the body will be in the state of the Equilibrium.

Explanation⇒ The sum of all the forces acting on the body is zero then it not necessarily that the body will be in the state of the Equilibrium. For the explanation let us TAKE an example.

Let us assume that the two equal and OPPOSITE forces having their line of actions at a CERTAIN distance apart act on a body, then the body in this case will remain stationary at the point but will have not have the rotation due to the torque produced.

Also, we can say that If the sum of all the forces acting on a particle is zero then it is necessarily in equilibrium because there will not be a distance between anti-parallel forces.

Hope it HELPS.

Discussion

12.	A bus increases its speed for 32 km/h to 72 km/h in 10 secs then its acceleration
Answer» U = 36 km/h = 10 m/s V = 72 km/h = 72× 1000 / 60 × 60 = 72000/ 3600 = 20 m/s t = 10 second a = v- u / t = 20 - 10 / 10 = 1 m/s ^2

Discussion

13.	A block weight 100n is pushed by a force F on horizontal rough plane moving with acceleration 1m/s ehen force is doubled acceleration becomes 10 then coefficient of friction is
Answer» Take g = 10m/s^2 Final ANSWER : $\mu = 0.8$ Steps: 1) Weight = 100 N => MG = 100N => m = 10 Kg Let the External Force be F and friction force be f. Then, according to the Question : Normal ,N= W = 100 N $F - f = ma \\ => F - \mu N = ma \\ => F - 100 \mu = 101 --(1)$ 2) We know, When Force is doubled, a becomes 10 units. $2F - 100 \mu = ma \\ => 2 F - 100 \mu = 1010 \\ => 2 F - 100 \mu = 100 ---(2)$ 3) Solving above equations we get, $\mu =0.8$ Hence, COEFFICIENT of friction is 0.8 .

Discussion

14.	Find the maximum speed with which the car can strike the barrier
Answer» The MAXIMUM SPEED of the CAR DEPENDS on the car DESIGN

Discussion

15.	Question 8.14: Given below are some famous numbers associated with electromagnetic radiations in different contexts in physics. State the part of the electromagnetic spectrum to which each belongs. (a) 21 cm (wavelength emitted by atomic hydrogen in interstellar space). (b) 1057 MHz (frequency of radiation arising from two close energy levels in hydrogen; known as Lamb shift). (c) 2.7 K [temperature associated with the isotropic radiation filling all space-thought to be a relic of the ‘big-bang’ origin of the universe]. (d) 5890 Å - 5896 Å [double lines of sodium] (e) 14.4 keV [energy of a particular transition in 57 Fe nucleus associated with a famous high resolution spectroscopic method (Mössbauer spectroscopy)].Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-287
Answer» (a) we know, $\bf{\nu=\frac{c}{\lambda}}$ so, frequency = 3 × 10^8/(21 × 10^-2) = 0.1438 × 10^10 Hz = 1438 × 10^6 Hz = 1438 MHz (b) The frequency of radiation arising from the two close energy levels in hydrogen known as Lamb shift i.e. 1057 MHz is radio waves as it belongs to the short wavelength end of the ELECTROMAGNETIC spectrum. (c) Using Wein's law the wavelength of the radiation mostly emitted at given temperature of heavenly BODIES can he calculated as $\lambda_m\times T=b=0.29cmK\\or,\lambda_m=\frac{b}{T}=0.29/2.7=0.11cm$ this wavelength correspond to microwave region of electromagnetic waves. (d) 5890 A° - 5896 A° [double lines of sodium] belongs to the yellow light of the VISIBLE spectrum of electromagnetic waves. (e) The frequency of radiation can be calculated as E = HV, v = E/h = (14.4 × 10^3 × 1.6 × 10^-19)/(6.626 × 10^-34) v = 3.49 × 10^18 Hz this frequency correspond to X - RAY region of electromagnetic waves.

Discussion

16.	Question 8.13: Use the formula λm T= 0.29 cm K to obtain the characteristic temperature ranges for different parts of the electromagnetic spectrum. What do the numbers that you obtain tell you?Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-287
Answer» According to Wein's displacement LAW, λm × T = 0.29 cm K or λm = (0.29/T) cm where T is the temperature and λm is the maximum wavelength of the WAVE. For different values of maximum wavelengths such as λm = 10-4 cm, 5 × 10-5 cm and 10-6 cm, the values for temperatures can be CALCULATED as FOLLOWS: T = (0.29/λm) K case 1 :- When, λm = 10-4 cm T = 0.29/10-4 cm K T = 2900 K case 2 :- When, λm = 5× 10-5 cm T = $\frac{0.29}{5\times10^{-5}}$ cm K T = 5800 K case 3 :- When, λm = 10-6 cm T = 0.29/10-6 cm K T = 290000 K The above data shows that temperature is required to get radiations of different parts of electromagnetic spectrum and the temperature increases with decrease in the wavelength.

Discussion

17.	Question 8.12: About 5% of the power of a 100 W light bulb is converted to visible radiation. What is the average intensity of visible radiation (a) at a distance of 1 m from the bulb? (b) at a distance of 10 m? Assume that the radiation is emitted isotropically and neglect reflection.Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-287
Answer» The bulb as a point source, radiate light in all DIRECTION uniformly and it is given that only 5% of power is converted to visible radiations. (a) let a SPHERE of RADIUS , r = 1 m surface area , A = 4πr² = 4π(1)² = 4π m² now, intensity = power/area given, Power = 5 % of 100W = 5W and area = 4π m² so, intensity = 5/4π = 5/(12.42) = 0.4 W/m² (b) let a sphere of radius, R = 10m surface area , A' = 4πR² = 4π(10)² = 400π m² intensity = power/area = 5% of 100W/400π = 5/400π = 0.004 W/m²

Discussion

18.	Question 8.9: The terminology of different parts of the electromagnetic spectrum is given in the text. Use the formula E = hν (for energy of a quantum of radiation: photon) and obtain the photon energy in units of eV for different parts of the electromagnetic spectrum. In what way are the different scales of photon energies that you obtain related to the sources of electromagnetic radiation?Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-286
Answer» The PHOTONS of given we energy are released during transition between energy levels in the atomic and the EMITTED photon energy is EQUAL to difference of energies of energy levels , among which transition has taken place. For example photon energy for, $\lambda$ =1m can be calculated. then, $E = \frac{hc}{\lambda}$ = (6.63 × 10^-34 × 3 × 10^8)/1 J = 19.88 × 10^-26/1.6 × 10^-19 eV = 12.425 × 10^-7 eV so, for emission of radio waves the energy difference between energy levels should be 12.425 × 10^-7 eV SIMILARLY, photon energy for other wavelength can be calculated as shown in figure.

Discussion

19.	Question 8.8: Suppose that the electric field amplitude of an electromagnetic wave is E0 = 120 N/C and that its frequency is ν = 50.0 MHz. (a) Determine, B0, ω, k, and λ. (b) Find expressions for E and B.Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-286
Answer» Given, electric field amplitude , $E_0$ =120N/C frequency of source, $\nu$ =50MHz speed of light , c = 3 × 10^8 m/s (a) magnitude of magnetic field Strength is given as $B_0=\frac{E_0}{c}=\frac{120}{3\times10^8}\\B_0=4\times10^{-7}T$ Angular frequency of the source is given as $\omega=2\pi\nu=2\pi\times50\times10^6rad/s$ propagation constant is given as $\kappa=\frac{\omega}{c}$ k = 3.14 × 10^8/3 × 10^8 = 1.05 rad/m now, WAVELENGTH of wave is given as $\lambda=\frac{c}{\nu}$ $\lambda$ = 3 × 10^8/5 × 10^7 = 6 m (b) SUPPOSE the wave is propagating in positive x direction.then electric field vector will be in the positive y direction and the magnetic field vector will be positive z direction. this is because all the PLANES are mutually PERPENDICULAR to each other. equation of electric field vector is given as $E=E_0sin(kx-\omega t)\hat{i}\\E=120sin(1.05x-3.14\times10^8t)\hat{i}$ and magnetic field vector is given as $B=B_0sin(kx-\omega t)\hat{k}\\B=4\times10^{-7}sin(1.05x-3.14\times10^8t)\hat{k}$

Discussion

20.	Question 8.7: The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is B0 = 510 nT. What is the amplitude of the electric field part of the wave?Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-286
Answer» Amplitude of MAGNETIC field of an electromagnetic WAVE in a VACCUM , $B_0=510nT$ speed of LIGHT in vacuum,c = 3 × 10^8 m/s amplitude of electric field of the electromagnetic wave is given by the relation, $E_0=cB_0=3\times10^8\times510\times10^{-9}\\E_0=153N/C$ therefore electric field part of the wave is 153N/C

Discussion

21.	Question 8.6: A charged particle oscillates about its mean equilibrium position with a frequency of 10 9 Hz. What is the frequency of the electromagnetic waves produced by the oscillator?Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-286
Answer» The frequency of the electromagnetic WAVES PRODUCED by the oscillator is 10^9 Hz. This is because the frequency of the electromagnetic waves produced by the oscillator is EQUAL to the frequency of the electromagnetic waves produced by the oscillator.

Discussion

22.	Question 8.5: A radio can tune in to any station in the 7.5 MHz to 12 MHz band. What is the corresponding wavelength band?Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-286
Answer» A radio can tune to minimum frequency, v1=7.5MHz = 7.5 × 10^6 Hz MAXIMUM frequency,v2=12MHz=12 × 10^6 Hz SPEED of light , C = 3 × 10^8 m/s use the relation , wavelength = speed of light /frequency so, minimum wavelength = speed of light/maximum frequency = 3 × 10^8/12 × 10^6 = 25 m and maximum wavelength = speed of light/minimum wavelength = 3 × 10^8/7.5 × 10^6 = 300/7.5 = 40 m hence, range of wavelength is 25m - 40m

Discussion

23.	Question 8.3: What physical quantity is the same for X-rays of wavelength 10 −10 m, red light of wavelength 6800 Å and radiowaves of wavelength 500 m?Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-286
Answer» Answer :- speed of light in vaccum. The PHYSICAL quantity is the same for X-rays of wavelength 10^ -10 m. red light of wavelength 6800 Å and radio WAVES of wavelength 500 m as we know, speed of light in vacuum equals to 3×108 m/s. the speed of light in vacuum REMAINS same for all wavelengths. This is because speed of light is INDEPENDENT of wavelength in the vacuum.

Discussion

24.	Question 8.2: A parallel plate capacitor (Fig. 8.7) made of circular plates each of radius R = 6.0 cm has a capacitance C = 100 pF. The capacitor is connected to a 230 V ac supply with a (angular) frequency of 300 rad s −1 . (a) What is the rms value of the conduction current? (b) Is the conduction current equal to the displacement current? (c) Determine the amplitude of B at a point 3.0 cm from the axis between the plates.Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-286
Answer» Given, radius of each circular plate , R = 6cm = 0.06 m capacitance of parallel plate capacitor , C = 100pF = 100 × 10^-12 F = 10^-10 F angular FREQUENCY , $\omega$ =300rad/s (a) RMS VALUE of conduction CURRENT, $I=\frac{V}{X_C}$ where, $X_C$ =capacitative reactance. $X_C=\frac{1}{\omega C}$ $I=V\omega C=230V\times300rad/s\times10^{-10}F\\I=6.9\times10^{-6}A$ (b) yes, conduction current is equal to displacement current. (c) magnetic field is given as, $B=\frac{\mu_0r}{2πR^2}I_0$ here, $I_0$ is the maximum current = √2I r is the DISTANCE between the plates from the axis.e.g., r = 3cm = 0.03 so, B = (4π × 10^-7 × 0.03)/(2π × 0.06²) × √2 × 6.9 × 10^-6 T = 1.63 × 10^-11 T hence, magnetic field is 1.63 × 10^-11 T

Discussion

25.	Question 8.1: Figure 8.6 shows a capacitor made of two circular plates each of radius 12 cm, and separated by 5.0 cm. The capacitor is being charged by an external source (not shown in the figure). The charging current is constant and equal to 0.15 A. (a) Calculate the capacitance and the rate of charge of potential difference between the plates. (b) Obtain the displacement current across the plates. (c) Is Kirchhoff’s first rule (junction rule) valid at each plate of the capacitor? Explain.Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-285
Answer» Given, radius of each CIRCULAR plate , r = 12cm = 0.12m distance between the plate, d = 5cm = 0.05m charging current, I = 0.15A now, capacitance between the two parallel plats is given by $C = \frac{\epsilon_0A}{d}$ where , A= area of each plate = πr² = π(0.12)² = 0.0144π m² now, C = (8.85 × 10^-12 × 0.0144π)/0.05 = 8 × 10^-12 F = 8pF now, charge on each plate, q = CV C is capacitance e.g., constant now, difference q = cV with respect to time. dq/dt = cdV/dt as you know, dq/dt = i= current so, dV/dt = i/C = 0.15/8 × 10^-12 = 1.87 × 10^10 V/s therefore the CHANGE in potential difference between the plates is 1.87 × 10^10 V/s (B) displacement current across the plates is same as conduction current. hence, displacement current i = 0.15A (c)yes , kirchoff's rule is valid in each plate of capacitor provided that we take the SUM of conduction and displacement current.

Discussion

26.	Why is easy to push nail by pointed end?
Answer» Because it increase the PRESSURE and DECREASE the AREA.... Hope it HELPS...

Discussion

27.	Question 7.8: Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?Class 12 - Physics - Alternating Current Alternating Current Page-266
Answer» According to question 7.7, Capacitance ( C ) = 30 цf 1 цf = 10⁻⁶ f => 30 цf = 30 * 10⁻⁶ f Inductance ( L ) = 27 mH 1 mH = 10⁻³ H => 27 mH = 27 * 10⁻³ H Charge of the conductor ( Q ) = 6 mC 1 mC = 10⁻³ C => 6 mC = 6 * 10⁻³ C We know that, ENERGY = Q² / 2C => Energy = ( 6 * 10⁻³ )² / 2 * 30 * 10⁻⁶ => Energy = 36 * 10⁻⁶ C / 60 * 10⁻⁶ F => Energy = 36 / 60 = 6 / 1 0= 0.6 J Hence energy stored is 0.6 J. This energy will be same in LATER time also as it is shared between Inductor and capacitor.

27.

Question 7.8: Suppose the initial charge on the capacitor in Exercise 7.7 is 6 mC. What is the total energy stored in the circuit initially? What is the total energy at later time?Class 12 - Physics - Alternating Current Alternating Current Page-266

Answer»

According to question 7.7,

Capacitance ( C ) = 30 цf

1 цf = 10⁻⁶ f

=> 30 цf = 30 * 10⁻⁶ f

Inductance ( L ) = 27 mH

1 mH = 10⁻³ H

=> 27 mH = 27 * 10⁻³ H

Charge of the conductor ( Q ) = 6 mC

1 mC = 10⁻³ C

=> 6 mC = 6 * 10⁻³ C

We know that,

ENERGY = Q² / 2C

=> Energy = ( 6 * 10⁻³ )² / 2 * 30 * 10⁻⁶

=> Energy = 36 * 10⁻⁶ C / 60 * 10⁻⁶ F

=> Energy = 36 / 60 = 6 / 1 0= 0.6 J

Hence energy stored is 0.6 J.

This energy will be same in LATER time also as it is shared between Inductor and capacitor.

Discussion

28.	Name a mirror which can form virtual image
Answer» By PLANE MIRROR the IMAGE ALWAYS VIRTUAL

Discussion

29.	A stone of mass 0.5 kg is thrown with a velocity of 10 m/s across the frozen surface of a lake and it comes to rest after travelling a distance of 100 m. What is the magnitude of force of friction between the stone and ice?Choose one:3 N4 N0.25 N0.75 Ni will mark the brainliest whoever answers it fist
Answer» We will USE the FOLLOWING equation of motion : V² = U² - 2as Where V is the final SPEED and u the INITIAL velocity and S is the displacement In this case our U is 10m/s and V is 0. The S is 100. 0 = 100 - 2 × 100 × a 0 = 100 - 200a 100 = 200a a = 0.5m/s² Force = mass × Acceleration Force = 0.5 × 0.5 = 0.25 N

Discussion

30.	What is known as parallex error??
Answer» MEASUREMENTS MADE by viewing the position of some MARKER relative to something to be measured are subject to PARALLAX error if the marker is some distance away from the object of measurement and not viewed from the CORRECT position

Discussion

31.	The resistance r=v/i where v=(100+5)or(100-5)v and i=(10+0.2) or (10-0.2). Find percentage error
Answer» From the given data V = 100. and ∆V = 5 I =10. and ∆I= 0.2 ∆R/R% = ? R = V/I So ∆R/R = ∆V/V + ∆I/I ∆R/R = 5/100 + 0.2/10 =(5+2)/100 =7/100 ∆R/R =0.07 ∆R/R% = 0.07×100 = 7%

Discussion

32.	Arrange the speed of time in solid,liquid and gas in decreasing orderas well as in measuring order
Answer» GAS liquid solid are in DECREASING ORDER

Discussion

33.	A heavy particle of mass m falls freely near the earth's surface. What is the torque acting on this particle about a point 50 cm east to the line of motion? Does this torque produce any angular acceleration in the particle?
Answer» ANSWER ⇒ The Torque acing on the Particles about the point 50 cm east to the line of the MOTION is 4.9m N-meter. where m -s the mass of the Particle. This Torque cannot produce any angular acceleration in the Particles. Explanation ⇒ For First Answer, Distance of the line of the motion (r) = 50 cm = 0.5 m. Let the mass of the Particle be m. Acceleration due to gravity(g) = 9.8 m/s² ∴ Force acting on the body = m × g = 9.8m N Now, Using the Formula, Torque = Force × Distance of the line of the motion. = 9.8 m × 0.5 = 4.9m N-m. Hence, the torque is 4.9m N-m. Second answer, This torque cannot produce any angular acceleration in the particle because there is no force in the OPPOSITE direction of the weight at 50 cm east of the line of motion. For a torque to be effective through a force, there MUST be equal and opposite non-linear force acting on the particle. Hope it HELPS.

33.

A heavy particle of mass m falls freely near the earth's surface. What is the torque acting on this particle about a point 50 cm east to the line of motion? Does this torque produce any angular acceleration in the particle?

Answer»

ANSWER ⇒ The Torque acing on the Particles about the point 50 cm east to the line of the MOTION is 4.9m N-meter.

where m -s the mass of the Particle.

This Torque cannot produce any angular acceleration in the Particles.

Explanation ⇒

For First Answer,

Distance of the line of the motion (r) = 50 cm = 0.5 m.

Let the mass of the Particle be m.

Acceleration due to gravity(g) = 9.8 m/s²

∴ Force acting on the body = m × g

= 9.8m N

Now, Using the Formula,

Torque = Force × Distance of the line of the motion.

= 9.8 m × 0.5

= 4.9m N-m.

Hence, the torque is 4.9m N-m.

Second answer,

This torque cannot produce any angular acceleration in the particle because there is no force in the OPPOSITE direction of the weight at 50 cm east of the line of motion.

For a torque to be effective through a force, there MUST be equal and opposite non-linear force acting on the particle.

Hope it HELPS.

Discussion

34.	A ball is whirled in a circle by attaching it to a fixed point with a string. Is there an angular rotation of the ball about its center? If yes, is this angular velocity equal to the angular velocity of the ball about the fixed point?
Answer» Answer ⇒ Yes, there is the ANGULAR ROTATION of the ball about its centre. The Angular Velocity of the ball about its center is equal to the angular velocity of the ball about the fixed POINT. Explanation ⇒ Yes the ball has an angular rotation about its center. If we will be noticing the POSITION of the point on the ball where it is tied with the string, we can EASILY see that it is moving about its center as its position changes. This point again comes to the same position when the ball makes a complete revolution around the fixed point. Thus the ball rotates about its center through 360° in the same time in which the ball revolves through 360° around the fixed point. Hence the angular velocity of the ball about its center is equal to the angular velocity of the ball about the fixed point. Hope it helps.

34.

A ball is whirled in a circle by attaching it to a fixed point with a string. Is there an angular rotation of the ball about its center? If yes, is this angular velocity equal to the angular velocity of the ball about the fixed point?

Answer»

Answer ⇒ Yes, there is the ANGULAR ROTATION of the ball about its centre.

The Angular Velocity of the ball about its center is equal to the angular velocity of the ball about the fixed POINT.

Explanation ⇒ Yes the ball has an angular rotation about its center. If we will be noticing the POSITION of the point on the ball where it is tied with the string, we can EASILY see that it is moving about its center as its position changes.

This point again comes to the same position when the ball makes a complete revolution around the fixed point.

Thus the ball rotates about its center through 360° in the same time in which the ball revolves through 360° around the fixed point.

Hence the angular velocity of the ball about its center is equal to the angular velocity of the ball about the fixed point.

Hope it helps.

Discussion

35.	A simple pendulum is a point mass suspended by a light thread from a fixed point. The particle is displaced towards one side and then released. It makes small oscillations. Is the motion of such a simple pendulum a pure rotation? If yes where is the axis of rotation?
Answer» Answer ⇒ YES, the motion of such a simple pendulum is a pure rotation. Explanation ⇒ As we know that in a pure rotation all the particles of a body always moves in concentric CIRCLES, thus such motion of a simple pendulum is always a pure rotation but only the direction of motion reverses periodically. The axis of rotation in this case is the line perpendicular to the plane of rotation and passing through the fixed POINT where the thread of the pendulum is tied.Since the oscillations are small, for all practical purposes the movement of the BALL can be assumed on a straight line and also the motion can be taken as periodic and Simple HARMONIC motion can be assumed. Hence, we can finally say that the motion of such a simple pendulum a pure rotation. Hope it helps.

35.

A simple pendulum is a point mass suspended by a light thread from a fixed point. The particle is displaced towards one side and then released. It makes small oscillations. Is the motion of such a simple pendulum a pure rotation? If yes where is the axis of rotation?

Answer»

Answer ⇒ YES, the motion of such a simple pendulum is a pure rotation.

Explanation ⇒ As we know that in a pure rotation all the particles of a body always moves in concentric CIRCLES, thus such motion of a simple pendulum is always a pure rotation but only the direction of motion reverses periodically.

The axis of rotation in this case is the line perpendicular to the plane of rotation and passing through the fixed POINT where the thread of the pendulum is tied.Since the oscillations are small, for all practical purposes the movement of the BALL can be assumed on a straight line and also the motion can be taken as periodic and Simple HARMONIC motion can be assumed.

Hence, we can finally say that the motion of such a simple pendulum a pure rotation.

Hope it helps.

Discussion

36.	Can an object be in pure translation as well as in pure rotation?
Answer» Answer ⇒ No, The OBJECT be in pure translation as well as in pure ROTATION. Explanation ⇒ In pure translation, each particle of the body always moves in the parallel lines while in the case of the pure rotation the particles of a body always move in CONCENTRIC circles. THUS, Both the conditions cannot be TRUE at the same time. Hence, the answer is no, i.e., the object can not in Pure Rotation and Pure Transnational Motion at the same time. In Transnational motion body moves in the straight line whereas in Rotational motion body moves in the Rotational Path. Hope it helps.

36.

Can an object be in pure translation as well as in pure rotation?

Answer»

Answer ⇒ No, The OBJECT be in pure translation as well as in pure ROTATION.

Explanation ⇒ In pure translation, each particle of the body always moves in the parallel lines while in the case of the pure rotation the particles of a body always move in CONCENTRIC circles.

THUS, Both the conditions cannot be TRUE at the same time.

Hence, the answer is no, i.e., the object can not in Pure Rotation and Pure Transnational Motion at the same time.

In Transnational motion body moves in the straight line whereas in Rotational motion body moves in the Rotational Path.

Hope it helps.

Discussion

37.	Derivation of 2rd equation of motion mathematically
Answer» Before I start my discussion here I would like to state the first two laws of Newton in their most correct forms: 1)1st Law: There EXIST frames of reference with respect to which every body continues to be at rest or in uniform motion in a straight line until compelled by an external force to CHANGE that state. Such frames are called inertial frames. 2)2nd Law: In an inertial frame, if external force be applied on a body to change its dynamic state, then rate of change of momentum of the body is directly proportional to the applied force. This constant of proportionality can be taken as 1, 2, 3 or anything as the laws are not experimental but axiomatic in the purview of classical mechanics. Newton took it as 1 reducing the proportionality to equality. Now many textbooks on physics derive Newton's First Law from his Second Law in the following manner: " From Newton's second law, considering mass of the body to be a non-zero constant: Putting: Which is our first law, no force means no change of dynamic state." But had it been so, Newton wouldn't have enunciated his first law as a separate and INDEPENDENT one (though Newton's WORKS, mathematically speaking, were absolutely non-rigorous and not precise at all, and I view them as a concoction of vague definitions, nice intuitions and distorted mathematics). Newton's three laws are form the axiomatic basis of classical mechanics and axioms cannot be proved from axioms within the same formal system (thanks to Godel for the proof of this STATEMENT). Many books justify this fact by saying that during that time three was a favorite number for the physicists. Galileo's three laws of falling bodies, Kepler's three laws of planetary motion etc. The books argue that Newton's laws were three in number for the same reason. But this is absolutely not a healthy explanation. Notice that Newton's First Law asserts the existence of Inertial Frames. It is these frames where the Second Law holds. So the first law talks of the inertial frames. What is an inertial frame? Well, many books say that it is a frame where Newton's laws hold. But this somewhat cyclic. If instead, we define inertial frames as those which are at relative rest with respect to a free particle (i.e. a particle with no real forces being acted upon) then the Second Law axiomatically holds in such frames. However, using the Second Law, we cannot actually prove the EXISTENCE of inertial frames as talked of in the First Law. And hence using Newton's Second Law we CANNOT prove his First Law.

Discussion

38.	Two forces F1 and F2 are acting on a body of mass 5 kg as shown in the figure. If F1 = 20 N and F2 = 5 N, then what will be the acceleration of the body ?i will mark the brainliest whoe ever answers it first
Answer» There is no figure in the question. So I am TAKING both forces acting in OPPOSITE direction. F1 = 20 N and F2 = 5 N m = 5Kg Net force = F1 - F2 = 15N F= ma a = F/m = 15/5 = 3

Discussion

39.	Pls answer fast................
Answer» 1 OHM is the ANSWER BRO.

Discussion

40.	The nearest star to our solar system
Answer» ALPHA Centauri is the STAR NEAREST to SOLAR SYSTEM

Discussion

41.	What is quantum efficiency
Answer» It SI the ratio of the number of carrier collected by the solar CELL to the number of photons of the given ENERGY incident on the solar cell.. Hope this HELPS...

Discussion

42.	Does the speed remain same in case when a body is drped from a height
Answer» No it is FALSE STATEMENT

Discussion

43.	Please don't delete my question sir ke bal kyu pakte hai what is a scientific reason behind this
Answer» By my experience of my LIFE my MOTHER has TOLD me that when yu not eat HEALTHY food or when yu got anger by some reason then yur nerve cell response to the btain

Discussion

44.	Under which condition the displacement of a moving body will be zer
Answer» When the FINAL and INITIAL point of a body is same , then the DISPLACEMENT is zero.

Discussion

45.	What is centrifugal force
Answer» The FORCE EXERTED from the centre is cetrifugal force ex. stone is tight on STRING which is HUNG on CEILING n rotated

Discussion

46.	Derive a formula for four equivalent resistances connected in series
Answer» R=R1+R2+R3+R4 MAY be it HELPED you

Discussion

47.	What is uniform circular motion
Answer» Thatmean a BODY is MOVING EQUAL CIRCLES during equal INTERVALS

Discussion

48.	A property of sound which is most closely associated with pitch of a musical note is
Answer» Pitch of a MUSICAL note is most closely associated with the FREQUENCY of that musical note.

Discussion

49.	Define the term velocity ratio for a simple machine and how is it related to the mechanical advantage of an ideal machine
Answer» This can be calculated from the geometry of the machine. For EXAMPLE, the mechanical ADVANTAGE and distance ratio of the lever is equal to the ratio of its lever arms. ... If the output FORCE is less than the input, but the distance MOVED by the load is greater than the distance moved by the input force.

Discussion

50.	Define: what are the two ways in which the physical states of matter can be changed
Answer» PHYSICAL STATE of matter can be changed by HEATING the body . The body changes its state on heating . And its OPPOSITE, .i.e cooling . Cooling the body too leads in CHANGE of physical state . there are many other ways of changing the state of matter.

50.

Define: what are the two ways in which the physical states of matter can be changed

Answer»

PHYSICAL STATE of matter can be changed by HEATING the body . The body changes its state on heating . And its OPPOSITE, .i.e cooling . Cooling the body too leads in CHANGE of physical state .

there are many other ways of changing the state of matter.

Discussion

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