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13401.

Q.12 An object weighting 25 N hangs from a point A with the help oftwo strings BA and CA as shown in Fig. 2.1 BA is inclined at 40' tothe vertical while CA is inclined at 60 to the horizontal. Determine theforce in the strings BA and CA

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Where is the figure?

13402.

An object is thrown up with a velocity of 20 m/s. Find(a) the maximum height attained by it.(b) the total time taken by it to return back. [Take g 10 m/s2]

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13403.

b. A raibow is the combined effect of therefraction, dispersion, and total internalreflection of light.

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A rainbow is caused by combined effects of reflection, refraction, and dispersion of light. This is a natural climatic phenomenon, often observed after a shower of rain.

Sunlight while pass-through droplets of water produce a spectrum of light that appears in the sky. It takes the form of a circular arc with multiple colour.arc.

Rainbows produced by sunlight always found to arise opposite the sun in the sky. Although rainbow is formed in full circle in the sky, we normally see only an arc formed by illuminated droplets.

13404.

53, A good plane mirror reflects about 95% of light. What is thepercentage of light reflected when total internal reflectionoccurs?

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The total internal reflection reflects all the 100 percent light. There is no loss of light due to reflection because the reflection occurs within the medium.

It reflects all the 100 %light.

13405.

2C) 980(D) 1020 mA train moving with a speed of 60 km/hr is slowed down uniformly to 30 km/hr for repair purposesduring running. After this it was accelerated uniformly to reach to its original speed. If the distancecovered during constant retardation be 2 km and that covered during constant acceleration bel km.find the time lost in the above journey(A) 1 min(B) 2 min(C)4 min(D) 5 minlogin of narticle from 1= 1 = 20 sec is:

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answer c b ok by tick

4 min is the right answer

4min is the correct answer

4 minutes is the answer that is options is correct

13406.

The velocity of sound in air is 320m/s what time it take to covers 9.6km. a) 30 secb) 1 min c) 40 secd) 20 sec

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speed=distance/time320m/s=9.6km/timetime=9.6km/320m/stime=9600m/320m/s =30 sec

13407.

example Calculate the angle of (a) 10(c) 1" (second of arc or arc sec) in radian.NCERT Solved Example(degree) (b) 1' (minute of arc or arc min) andUse 360°-2 π rad., 1°-60' and 1'-60".

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Thus 2 radians equals360 degrees. This means that 1 radian = 180/ degrees, and1 degree= /180 radians.

d(A,B) = R a /180, These formulas can be checked by noticing that the arc length is proportional to the angle, and then checking the formula for the full circle,

13408.

lIit3) With the help of a neat ray diagram, explain theDeprism.phenomenon of total internal reflection.11) Define (a

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13409.

24. Frequency of photon having energy 66 eV is(A) 8 x 10^(-15) Hz (B) 12 x 10^(- 15) Hz(C) 16 × 10^15 Hz (D) .

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this question was submitted by miss

13410.

State and explain de-Broglie's hypothesis for dualnature of matter and calculate de-Brogliewavelength of a particle.3.

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De Broglie hypothesis says that all matter has both particle and wave nature. The wave nature of a particle is quantified by de Broglie wavelength defined asλ=h/p wherepis the momentum of the particle. This was called a hypothesis because there was no evidence for it when it was proposed, only analogies with existing theories. (The wavelength-momentum relation holds exactly for photons.)

Historically, de Broglie hypothesis was the next step in quantum theory after Planck, Einstein and Bohr.

In 1900, Max Planck introduced the notion that radiation is quantized to derive the black body radiation spectrum.

In 1905, Albert Einstein used Planck's idea to explain photoelectric effect, which led to wide acceptance of the quantum nature of radiation.

In 1913, Niels Bohr used quantization of radiation along with the Bohr hypothesis (that the angular momentum of electrons is quantized) to correctly predict the line spectrum of hydrogen atom and explain .

In 1923, Lois de Broglie took this idea further and proposed that matter has wave nature as radiation has particle nature.

Bohr hypothesis comes as an immediate consequence of de Broglie hypothesis - angular momentum must be conserved if an electron in an atom is seen as a wave going in circles around a nucleus such that the electron wave interferes constructively everywhere in the orbit.

In 1926 Erwin Schrödinger published the Schrödinger equation which generalized de Broglie's concept of matter waves and put them in a more robust theoretical footing.

The direct experimental confirmation came in 1927 when Clinton Davisson and Listor Germer and independently, GP Thomson observed electron diffraction.

13411.

g has diameter of 1 mm, if a loadof10kg.wt.isattachedother end. What extension is produced, f poisson's ratio is 0.26? How mulateral compression is produced in it?(y 12.5x 1010 N/m2

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13412.

Example (11.1 A force of 5 N is acting onan object. The object is displacedthrough 2 m in the direction of the force(Fig. 11.2). If the force acts on the objectall through the displacement, thenwork done is 5 N × 2 m =10 N m or10 J.

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13413.

The momentum of a body of mass 5 kg is 10 kg-m/s,A force of 2 N acts on the body in the direction ofmotion for 5 sec. The increase in the kinetic energy is(1) 15 J(3) 20 J(2) 30 J(4) 40 J

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13414.

43. A car is being driven by a force of 2.5 x 1010 N. Travelling at a constant speed of 5 m/s,ittakes2minutestoreach a certain place. Calculate the work done

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13415.

11. A constant force of 2 N acts on a body for 5 secondsto change its velocity. The change in its momentum1S

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hence, change in momentum = F.t = 2*5 = 10kgm/sec

plz verify it by using 2 law of newton

This is the formula used which is derived from Newton's second law

what is dp/dt in this formula

dp is change in momentum and dt is the change in time.

13416.

Find out time period

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Atime period(denoted by 'T' ) is thetimeneeded for one complete cycle of vibration to pass in a given point. As the frequency of a wave increases, thetime periodof the wave decreases. The unit fortime periodis 'seconds'

13417.

Q12. Define the gauss's law of magnetism.

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Gauss's law for magnetism is one of the four Maxwell's equations that underlie classical electrodynamics. It states that the magnetic field B has divergence equal to zero, in other words, that it is a solenoidal vector field. It is equivalent to the statement that magnetic monopoles do not exist.

13418.

CLASS IX1. Distinguish between density and relative density of a substance. The relative density of silver is 10.8. If the density ofwater is 103 kg/m3,find the density of silver.

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The ratio between density of a substance and density of water is called relative density.

The ratio between density of a substance and density of water is called relative density.

Relative density =

Density of the substance/density of water at 40C

Units: Relative density has no units because it is the ratio between the similar physical quantities (density).

Relative density = Density relative to water Density of silver = relative density * density of water = 10.8 * 1 gm/cm³ = 10.8 gm/cm³ in SI units, 10.8 * 1000 kg/m³ = 10, 800 kg / m³

13419.

m/htoune LU rest.[Ans. 20 sec)BY A car accelerates uniformly from36 km/h in 2 seconds. Calculate a. the acceleration18b. the distance covered by the car in that time.Ans. a. a = 2.5 m/s2 b. S = 15 m)

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13420.

2) A traintravels at a speed 50 km/hr for 0.5 hr, at 30km/hr for next 0.26 hr and then 70 km/hr for next0.76 hr. What is the average speed of the train?

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13421.

6Example 8.6 A car accelerates uniformlyfrom 18 km h to 36 km h in 5 s.Calculate (i) the acceleration and (ill thedistance covered by the car in that time(8.12

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Acceleration = Change in Velocity/Time Change in Velocity = 36-18 = 18 km/h=5 m/s Time= 5 Seconds Acceleration = 5/5= 1 m/s^2

Equation of motion,s=ut+(1/2)at^2 u=18 km/h=5 m/s t=5 s a=1 m/s2 s= (5*5)+(1/2*1*5*5) s=25+12.5 i.e., s=37.5 m

13422.

A car accelerates uniformly from 18 km/h to36km/ h in 5 s. The acceleration in ms is

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Thank u

13423.

A bus travelled the first one-third distance at the speed of 10 km/h, the next one-third at 20 km/h and the leone-third at 60 km/h. The average speed of the bus is(A) 9 km/h(B) 16 km/hC 18 km/h(D) 48 km/h

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Let total distance be x

It covered x/3 at 10x/3 at 20x/3 at 60

Times taken respectively are { d / t } x / 30x / 60x / 180 respectively

total time = x/30 + x /60 + x/180= 10x / 180= x / 18

Average speed of bus (avg s )= total distance / total time

avg s = x / {x/18}avg s = 18 km / h

(C) is correct option

18 is right .

uujhjjj

13424.

A car moves from A to B with speed 20 km/h and back to A with speed 30 km/h. The averagespeed during the whole journey isA) 25 km/h(B) 24 km/h(C) 50 km/h(D) 35 km/hS-nou

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(A) 25km/his a correct answer

25km/his a correct answer

(A)25km/his a correct answer

A. 25km/h is a correct answer .

13425.

Two slits are made one millimetre apart and theused one meter away What is the fringe separation whenlight of wavelength 500 nm is used?

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13426.

A car travels from A to B at a speed of 20 km h-and returns at a speed of 30 km h-'. The averagespeed of the car for the whole journey is(A) 5 km h-(B) 24 km h-(C) 25 km h-(D) 50 km h-10.

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C) is the correct answer is

( d). 50 is the correct answer

Hii frnd..

the speed average is

Total average speed is simply = Total distance/Total timeLets say,D = distance travelled by the car in EACH directiont1 = time spent on onward tript2 = time spent on return tripThus, the total distance travelled by the car = D+D= 2DAnd, by the formula, Speed = Distance/TimeS1 = D/t1 => t1 = D/S1S2 = D/t2 => t2 = D/S2Total average speed = Total Distance/Total time = 2D/(t1+t2) = 2D/(D/S1+D/S2) = 2S1*S2/(S1+S2)

Remember this general formula for a total average speed problems:Total average speed = 2S1*S2/(S1+S2)

So in the question S1 = 20km/hr

&S2 = 30km/hr

So av. speed = 2 x 20 x 30/20+30

= 24km/hr

24km / hr is the right answer

please accept best

25 is the right answer

option b is right answer

13427.

A. Ca(OH)2C CaCOsWhat is chemical formula of quick lime?A. Ca(OH)2C. CaCO3B. CaOD. Ca(HCO3)2B. CaOD. Ca(HCO3)2

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Option B is correct.

13428.

0 TIiIl, 2.0 Inim.7. A beam of monochromatic light of wavelength500 nm falls on two parallel slits. The distancebetween the slits is 0.15 mm. Determine the width ofthe interference fringes on a screen placed at a distanceof 1.5 m from the slits.Ans. 5 mm.

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fringe width is wavelength x distance of screen/slit differencehence=500*10^-9*1.5/0.15*10^-3=5mm

13429.

69:9. Abus traveling along a straight highway covers one-third of the total distance between two places with avelocity 20 km h-. The remaining part of the dis-tance was covered with a velocity of 30 km h- forthe first half of the remaining time and with velocity50 km h- for the next half of the time. Find theaverage velocity of the bus for its whole journey.

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Let distance be xx/3 at 20 kmphremaining 2x/31/2 of remaining is 1/2 of 2x/3 = x/3 at 30 kmphand remaining ( x - x/3 - x/3 = x/3) at 50 kmphfirst part takes (x/3)/20 hours2nd part takes (x/3)/30 hours3rd part takes (x/3)/50 hoursTotally time taken for a distance of x km = (x/3) ( 1/20 + 1/30 + 1/50) hoursTotally time taken for a distance of x km = (x/30)( 15 + 10 + 6)/30 = (x/30)(31/30) = 31x/900

Average speed = total distance/total time = x/ ( 31x/900) = 900/31 kmph = 29.03 kmph

13430.

bcaboe Caca + ab , bc + ab : bc-caIf a,b,c are in А.Р., then prove thatare in H.P

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13431.

The work function of metal is 1 eV. Light of wavelength3000 A is incident on this metal surface. The velocityof emitted photo-electrons will be(1) 10 m/sec (2) 1 x103 m/sec(3) 1x104m/sec (4) 1 x 10 m/sec

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13432.

Assuming human pupl to have a radius of 0.25 em and a comfortable viewingdistance of 25cm, the minimum separation between two objects that humaneye can resolve at 500 nm wavelength is2 30 um3) 100 um4) 300 um

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13433.

Example 4 (a) The work function for the surface of aluminium is 4-2 eV. How much potentialdifference will be required to stop the emission of maximum energy electrons emitted by light of2000 À wavelength ?(b) What will be the wavelength of that incident light for which stopping potential will be zero ?h = 66 x 10-34 Js, c = 3 108 ms-1.

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13434.

4. How can three resistors of resistances 2 Ω. 3 Ω, and 6 Ω be connectedto give a total resistance of (a) 4 Ω、 (b) 1 Ω?tel en İstance that can be secured

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Thank you

13435.

3. Three resistors of resistances 25Ω, 50Ω, and 75Ω respectively are connected inseries in a circuit. What is the effective resistance of the combination of the threeresistors?

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13436.

32.Ultraviolet light of6.2 eV falls on aluminium surface (workfunction 4.2 eV). The kinetic energy (in joule) of thefastest electron emitted is approximately:(a) 3 x 1041(c) 3 x 10-17(b)3 x 10-19(d)3 x 10-15

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the pic is not clear

13437.

4) x2 x 6 x 10300. The sun radiradiations received on the earth surface from the sun is1.4kW/m2. The average earth- sun distance is 1.5x 10" nm.The mass lost by the sun per day is (1 day 86400 sec)

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13438.

Calculate the equivalent resistance of two resistors of 1 Ω and 10 Ω connected in parallel

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Given R1 = 1 ohm, R2 = 10 ohm

Equivalent resistance when R1 and R2 connected in parallel= 1/R1 + 1/R2= 1/1 + 1/10= (10 + 1)/10= 11/10= 1.1 ohm

13439.

61022Three resistors are connected as shown in thediagram. Through the resistor 5 ohm, a currentof 1 ampere is flowing.(i) What is the current through the other wother 1 amp.two resistors ?(in) What is the p.d. across AB and acrossAC ?51522

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like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like

11 ohm is the answer

11 is the best answer me

13440.

8.If energy (E), velocity (v) and time (T) are chosenas the fundamental quantities, the dimensionalformula of surface tension will beA. [Ev-2 r']B. (Ev-1r?]C. [Ev-2T-2jD. [E-2-1T-]Alif eandlatu-L---

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c is the snswer. SEEJAN MAITY. Am i m write or wrong.

13441.

4. Assuming that the moon completes one revolutionin a circular orbit around the earth in 27.3 days,calculate the acceleration of the moon towardsthe earth. The radius of the circular orbit can betaken as 3.85 × 10° km. (Ans. 2.73 ×10-3 ms-2)ting along a

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time period 2πr/vso calculate v firstthen calculate V^2/rit gives acceleration

13442.

Two isdentical resistors each of resistance 2s are connected in tuno) in series (i) in paralle, to abuttery of 12 V. Cakculate the ratio of power consumed in the two cases

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13443.

A projectile can have the same range Rfor two angles of projection. Ift, and tbe the times of flight in the two cases,then the product of the twotimes offlight is directly proportional to

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13444.

4) If the height of a satellite completing one revolution around the earth in T seconds is hmeter, then what would be the height of a satellite taking 2v2 T seconds in onerevolution.

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13445.

1. Find the energy of a photon in electron volt of microwaves having wavelength3 mm. Given h 6.6 x 1034 Js and 1 eV 1.6 x10-19 J

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13446.

Find the ratio of the potential differences that must be applied across the parallel and seriescombination of two capacitors C, and C, with their capacitances in the ratio 1 :2 so that theenergy stored in the two cases becomes the same151

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C2 = 2 C1.

parallel combination: C_eq = 3 C1 Energy E = 1/2 * C_eq * V1² = 3/2 * C1 V1²

Series combination: C_eq = C1 C2 /(C1 +C2) = 2 C1/ 3 Energy E =1/2 * 2/3 * C1 * V2² = 1/3 * C1 V2²

Equating energies, we get V1 : V2 =√2 : 3

Plot of energy vs Capacitance: This will be linear if Voltage applied is assumed to be constant.

13447.

IS2. A set of 24 tuning forks is arranged in a series increasing frequencies if each forks gives 4beats per second with the preceding one and if the frequency of the last is twice that of thefirst, find the frequency of the first and the last fork.

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13448.

21. The Young's modulus of the material of a wire is 1.1 x 1011 N/mp. If the wire undergoes a longitudinalstrain of 0.1%, then the energy stored per unil volume is(a) 5.5 x 103) (b) 5.5 x 10*(©) 11 x 10"] (d) 11 x 10

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13449.

A pteridophyte called Horse-tail is(A) Equisetum157.(B) Lycopodium(C) Marsilea(D) Selaginella

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Answer : A) EquisetumExplanation :The termshorse tail,horsetailorhorse's tailmay refer to the following: Thetailof ahorse. Equisetales, the order ofpteridophytesthat includes extinct species known in the fossil record. Equisetum, a genus in the Equisetales that includes: Equisetum arvense, or "field horsetail"Like if you find it useful

13450.

Two wires of the same material and length butdiameters in the ratio 1 : 2 are stretched by thesame force. The potential energy per unit volumefor the two wires when stretched will be in the ratio.

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