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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2501. |
wa dropped. lake g = 10 m/ s .30. A ball is dropped from a height of 5 m onto a sandy floorand penetrates the sand up to 10 cm before coming torest. Find the retardation of the ball in sand assumingit to be uniform |
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Answer» find the difference of number 15030 in 5and 4 find the difference of number 15030 in 5and 3 |
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| 2502. |
An electron is moving in a circular path under theinfluence of a transverse magnetic field of 3.57 X102 T. If the value of e/m is 1.76 x 1011 C/kg, thefrequency of revolution of the electron is(1) 1 GHz(2) 100 Mhz(3) 62.8 MHz(4) 6.28 MHz |
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Answer» the answer could be option a |
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| 2503. |
11. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in atissue in which the speed of sound is 1.7 km/s ? The operating frequency of the scanner is 42 MHz (1 MHz106 Hz)(C.B.S.E. 2011, 2015) |
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| 2504. |
A message signal of frequency 10 kHz and peak voltage 10 V is used tomodulate a carrier of frequency 1 MHz and peak voltage 20 V. Determine(i) the modulation index,(ii) the side bands produced. |
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Answer» 1) Given that Am = 10v and Ac = 20v Where Am = Amplitude of modulated wave Ac = Amplitude of Carrier wave Modulation index is given by µ = Am/Ac we get, µ = 10/20 = 0.5 2) Side bands The side bands are, (ωᴄ - ωm) and (ωc+ωm) or we can write (fc - fm) and (fc + fm) Where (ωᴄ - ωm) is the frequency of lower side band,(ωc+ωm) is the frequency of upper side band . also ωᴄ = carrier wave frequency ωm = modulated wave frequency ∴ Frequency of lower side band = (ωᴄ - ωm) = (1000-10)= 990 KHz Frequency of upper side band = (ωᴄ + ωm)= (1000+10)= 1010KHz |
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| 2505. |
VED EXAMPLES:What is the wavelength of radiowave of frequency 530 MHz?(Speed of radio waves in air-3 x10 m/s) |
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Answer» v= c/λλ =c/vλ = 3×10^8/ 530×10^6 λ =0.005660×10² λ =0.5660 m. |
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| 2506. |
T 00111. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the wavelength of sound in atissue in which the speed of sound is 1-7 km/s? The operating frequency of the scanner is 4-2 MHz. (1 MHz106 Hz)CRS F 2011 201 |
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| 2507. |
A 400 pF capacitor is charged with a 100 V battery.After disconnecting the battery this capacitor isconnected with another 400 pF capacitor. Thenthe energy loss is:- |
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Answer» energy loss kitna h |
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| 2508. |
50. Capacitor of 1 F withstand the maximum voltage6 KV while a capacitor 2uF withstand the maximumvoltage 4 kV. What maximum voltage will thesystem of these two capacitor withstand if they areconnected in series. |
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| 2509. |
rleld.2AfFind energy stored ineach capacitor and chargeon each capacitor.3412y |
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Answer» c= q/vcharge will be 2/12, 3/12 ,5/12,equivalent c = 30/11ferad |
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| 2510. |
The surface tension of a liquid is 5 N/m. If a film isheld on a ring of area 0.02 m2, its surface energy isabout(1) 5x10-2 J(3) 2 x 10-1 J(2) 2.5 x 10-2 J(4) 3 x 10-1 J |
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Answer» Option (3) is right ans |
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| 2511. |
Find number of photons in 6.62 J of radiant energy of frequency 102 Hz. (Takeh = 6.62 x 10-34 J-s)[2] |
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Answer» Given; E = 6.62J v = 10^14 Hz Let the no.of photons be x. =) E = x* hv =) 6.62 = x * 6.62 * 10^-34 Js * 10^14 =) 6.62/6.62 = x * 10^-20 =) 1 = x * 1/ 10^20 =) 10^20 = x. Hence correct answer is 10^20 photons. |
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| 2512. |
& ube of length L and radius R is joined taanother tube of length 3 and radius . A fluidis flowing through this tube. If the pressuredifference across the first tube is P, then thepressure difference across the second tube is :3P |
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Answer» how did you calculated the value of Q/t in the first step |
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| 2513. |
2. The total kinetic energy of 1 mole of N2 at 27° C will beapproximately:(1) 1500J(3) 1500 kilo Calories(2) 1500 Calories(4) 1500 erg |
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| 2514. |
hat is the rear g matter |
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Answer» anything that has mass and occupies space matter is a substance that has inertia and occupies physical space. According to modern physics matter consists of various types of particles each with mass and size matter everything that occupies space and has mass matter is everything which have mass and or occupies space. matter is everything which have mass and it occupies space |
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| 2515. |
60. A ball is dropped from the top of a building.acquires a velocity of 30 ms on reaching theground. Calculate the height of the building.Take g= 10 ms? |
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| 2516. |
e p i £k 1 ()1310 086 D हद 01 हो— 2k (3Y) 1) O'T ° |
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Answer» a) option is correct 1kg = 1kg |
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| 2517. |
The frequency and velocity of a wave arerespectively 15 x 103 Hz and 330 m/s. Findits wave length-(a) 0.4 x 10 cm (b) 3.3 cm(c) 2.2 cm(d) 0.2 x 102 cm |
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| 2518. |
3) 7 x 1020 Photons/sec 4) 9 x IU PlO2. The number of photons emitted per second by aMedium wave transmitter of 10 kW power, emituntaheradio waves of wavelength 500m2.51 x 10312) 1.51 x 10312951 x 1031e 20%3) 7.51 x 1031 |
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Answer» thank u |
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| 2519. |
1. A sound wave travelling along a string is described byy (x, t) 4 x 10-3 sin (160 x- 6 t)where x, y and t are in SI units. Calculate the (a) amplitude, (b) wavelength, (c) time periodand (d) frequency of the wave. |
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Answer» comparing this equation with y = asin(-wt+kx) ...... w = 6 k= 2π/wavelength = 160. amplitude is a = 4×10-³ = 4/1000 = 0.004 m wavelength is = 2π/k = 2π/160 = π/80 m time period is 2π/w = 2π/6 = π/3 s frequency is w/2π = 6/2π = 3/π s-¹ |
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| 2520. |
२)!(2) (37) 2(3)The distance between two consecutive crests in a wave train produced in a string is 5 cm. If 2complete waves pass through any point per second, the velocity of the wave isकसा डारा में उत्पन्न तर के दो निकटतम शीर्ष (श्रृंग) बिन्दुओं के मध्य की दरी 5 सेमी है। यदि दो पूर्ण कम्पन किसी बि से ।प्रति सैकण्ड गजरते हैं, तो ध्वनि वेग होगा।(1) 10 cm/sec(2) 2.5 cm/sec (3) 5 cm/sec(4) 15 cm/seclicht is incident on system of mirror as shown in the adjacent figure. What is the totalmes out after two reflections ? |
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| 2521. |
I. A sound wave travelling along a string is described byy (x,t) 4 x 10-3 sin (160 x 6 t)where x y and t are in SI units. Calculate the (a) amplitude, (b) wavelength, (c) time periodand (d) frequency of the wave. |
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Answer» comparing this equation with y = asin(-wt+kx) ...... w = 6 k= 2π/wavelength = 160. amplitude is a = 4×10-³ = 4/1000 = 0.004 m wavelength is = 2π/k = 2π/160 = π/80 m time period is 2π/w = 2π/6 = π/3 s frequency is w/2π = 6/2π = 3/π s-¹ |
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| 2522. |
(1) 10m102 |
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Answer» option b |
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| 2523. |
8. Between two stations a train starting from rest first accelerates uniformly, then moves with constant velocity and finallyretards uniformly to come to rest. If the ratio of the time taken be 1 8 1and the maximum speed attained be60 km/h, then what is the average speed over the whole journey?A) 48 km/hB) 52 km/hC) 54 km/hD) 56 km/h |
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Answer» Let it accelerates with say ‘a’ in time ‘x’ then it reaches a velocity ‘v’ and continues motion for time ‘8x’ abd finally then retards with ‘b’ in time ‘x’.v = 0 + ax=> v = ax=> 60 = ax=> a = 60/xAnd,0 = v – bx=> b = 60/xDistance covered in time of acceleration,S1= 0 + ½ at2=> S1= ½ ax2= ½ × 60/x × x2= 30xDistance covered while moving with constant velocity is,S2= vt = 60 × 8x = 480xDistance covered in time of retardation,02= v2- 2bS3=> S3= 602/(2 × 60/x)=> S3= 30xSo, total distance = 30x + 480x + 30x = 540xTotal time = x + 8x + x = 10xSo, average velocity = 540x/10x = 54 km/h |
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| 2524. |
31. A bullet of mass m is fired from a gun of mass MThe recoiling gun compresses a spring of forceconstant k by a distance d. Then the velocity of thebullet is :-(1) kd VM/mvkmkM(3)(4) KVN |
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| 2525. |
A bullet of mass m is fired from a gun of mass MThe recoiling gun compresses a spring of forceconstant k by a distance d. Then the velocity of thebullet is :-(1) kd M/m(2) Vkm(4) KMVd |
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| 2526. |
8. Between two stations a train starting from rest first accelerates uniformly, then moves with constant velocity and finallyretards uniformly to come to rest. If the ratio of the time taken be 1: 8 1and the maximum speed attained be60 km/h, then what is the average speed over the whole journey?A) 48 km/hB) 52 km/hC) 54 km/hD) 56 km/h |
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Answer» Let it accelerates with say ‘a’ in time ‘x’ then it reaches a velocity ‘v’ and continues motion for time ‘8x’ abd finally then retards with ‘b’ in time ‘x’.v = 0 + ax=> v = ax=> 60 = ax=> a = 60/xAnd,0 = v – bx=> b = 60/xDistance covered in time of acceleration,S1= 0 + ½ at2=> S1= ½ ax2= ½ × 60/x × x2= 30xDistance covered while moving with constant velocity is,S2= vt = 60 × 8x = 480xDistance covered in time of retardation,02= v2- 2bS3=> S3= 602/(2 × 60/x)=> S3= 30xSo, total distance = 30x + 480x + 30x = 540xTotal time = x + 8x + x = 10xSo, average velocity = 540x/10x = 54 km/h |
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| 2527. |
32For a body traveling with uniform acceleration, its final velocity is v - KVX,where k is constant and x is the distance traveled by the body, Then initialvelocity and acceleration isA)0,-/2 m/s2 B)k,k3/2 m/s 0,2k2 m/s D) Im/sk/2 |
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| 2528. |
1MULTIPLE CHOICErk, Power & EnergyWork is done only when(a) applied force is strong(b) applied force generates motion(c) applied force is normal to the displacement |
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Answer» option c might be the answer |
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| 2529. |
An engine applied 20N force to obtain velocity 7.3m/s. find the power of the engine |
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Answer» Power = force x velocity 7.3*20146Nm/s |
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| 2530. |
If a force F is applied on a body and it move with velocity v, then power will be :- |
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Answer» Power = Work/ Time. Work = Force x displacement So, now Power = Force x displacement / Time. But displacement / Time = velocity According to the question, Force is F and V is velocity so, Power = F x v But power is a scalar quantity so, we have dot product for power so...... Power = Fv or F.v option 1 is the answer |
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| 2531. |
12.The velocity of a body at any instant is 10 m/s. After 5 sec, velocity of the particle is 20 m/s. The velocityat 3 seconds before that instant is(1) 8 m/sec1142) 4 m/sec(3) 6 m/sec(4) 7 m/sec |
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Answer» The velocity of a body at an instant is 10 m/s Acceleration = (v - u) / t= (20 - 10) / 5 = 2 m/s²Acceleration of body is 2 m/s² Now,v = u + at10 = u + (2 × 3)u = 10 - 6u = 4 m/sVelocity 3 seconds before was 4 m/s (2) is correct option 2 is the right answer The velocity of a body at an instant is 1.m/s Acceleration =(v-u)/t =(20-10)/5 = 2 m/s*sAcceleration of a body is 2m/s*sNow, applying Newton first law of motion, v =u +at10=u+(2*3)10-6 =uu = 4m/s |
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| 2532. |
ur)circuit, shown here, the current isobserved to lead the applied voltage. An additionalitor C', when joined with the capacitorCresent in the circuit, makes the power factor ofthe circuit unity. The capacitor C', must have beenFor the LCRshowconnected in:JEE mains Online April 11, 2015] |
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| 2533. |
A man is riding on a cycle with velocity 7.2 km/hr up a hill having a slope 1 in 20. The total mass of the manand cycle is 100 kg. The power of the man is(A) 200 W(B)175W(C) 125 W(D) 98 W |
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| 2534. |
60. The energy of the electron, in the groundstate of hydrogen, is -13.6 eV. Calculatethe energy of the photon that would beemitted, if the electron were to make atransition corresponding to the emissionof the first line of the(G) Lyman series(ii) Balmer series of the hydrogenspectrum. All India 2009C |
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Answer» Energy of an electron in hydrogen atom in ground state is -13.6 eV. En= -Z² 13.6 / n² electron volt----------1 Energy levels higher than ground energy state in an atom iscalled as excited energy state/ level. Energy of electron in second excited state means : Z=1, n=3On substituting these values in equation 1 we get . En=- (1)²x 13.6/3² electron voltEn=- 1x 13.6/9En=-1.5 electron volt. ∴Energy of the electron in the second excited state in hydrogen atom is -1.5 electron volt. |
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| 2535. |
Mass=4kg , Time=5sec ,. Initial velocity=10m/sFinal velocity=20m/scalculate the power |
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| 2536. |
A car travels a distance 100 m with a constantacceleration and average velocity of 20 m s. Thefinal velocity acquired by the car is 25 m s.Find : (i) the initial velocity and (ii) accelerationof car.Ans. (i) 15 m - (ii) 2 m s-2 |
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Answer» Average velocity=(initial velocity + final velocity)/220 = (initial velocity + 25)/2 (1) Initial velocity = 40 - 25 = 15 m/s (2) using v = u + at Where, v = final velocity u = initial velocity t = time taken a = acceleration 25 = 15 + atat = 25 - 15 = 10.........(1) Now, using s = ut + 1/2* at^2 s = Distance travelled 100 = 15t + 1/2 *10*t100 = 15t + 5t20t = 100t = 100/20 = 5 s Put value of t in eq(1)at = 10a = 10/5 = 2 m/s^2
1)V (Av) =v+u/2 v+u=2V (Av)v+u=2×2025+u=40v=40-25v=15m/s The initial velocity is 15m/s 2) v^2-u^2=2as a=v^2-u^2/2s =(25^2-15^2)/(2×100) =(625-225)/200 = 400/200 =2m/s^2 The acceleration of car is 2m/s^2 |
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| 2537. |
1. Two waves of the same pitch have amplitudes inthe ratio 1 3. What will be the ratio of their(i) loudness, (ii) pitch ? |
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| 2538. |
Two waves of the same pitch have amplitudes inthe ratio 1 3. What will be the ratio of their(i) loudness, (ii) pitch? |
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Answer» (i) Loudness is proportional to amplitude squareRatio of loudness= 1/3square = 1:9(ii) Frequency of the sound wave has no effect on amplitude.so, ratios of frequency remains same that is 1:1 |
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| 2539. |
8. Standing waves are produced by thesuperposition of two wavesy, -0.05 sin(3rt - 2x)andy, -0.05 sin(3xt + 2x), where x and y areexpressed in metres and t' is in seconds.Then the amplitude of the particle at x = 0.5m is(a) 2.7 cm(b) 5.4 cm(c) 9.98 cm(d) 0.10 cm |
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Answer» b. 5.4 cm x 10 -2 mresultant displacement = y1 + y2= .05 [sin(3nt-2x) + sin (3nt + 2x)]y= .05 x 2 sin (3nt) cos {(-4x) / 2} from (1), y = .1 cos 2x * sin 3nt |
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| 2540. |
. A constant force of 50N is applied to a body of mass 20kg moving initially with a speed of I5m/s How longconstant force of 50N is applied to a body of mass 20kg moving initially with a speed of 15m/s.does the body take to stop?long |
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| 2541. |
4) A lead sphere of mass 20kg has the same diameter as an aluminium sphere of mass 72kg. The spheresare simultaneously dropped from a tower. When they are 10m 10m from the ground, they have identical(neglect air resistance):a) kinetic energyb) potential energyc) momentumd) acceleration |
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Answer» option (a )is correct |
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| 2542. |
The time period of a vibrating body is 0.1 s. The waves produced by the body moves with a speed o340 ms. Find the wavelength of the wave.Ans. 34 m]find |
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| 2543. |
47. Calculate the wave length of the secondline of Lyman series in hydrogen spectra!(Given R, = 1.09737 cm-"). (t j 56. |
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Answer» 1/ λ = R[1/n₁² - 1/n₂²] For second line of Lyman seriesn₁ = 1 and n₂ = 3 1/λ = R[1/1² - 1/3²]= 8R/9 λ = 9 / (8R)= 9 / (8 × 1.097 × 10^7 m^1)= 102.5 nm ∴ Wavelength of second line of Lyman series is 102.5 nm |
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| 2544. |
amplitude is found to5. For an amplitude modulated wave, the maximumbe 10 V while the minimum amplitude is 2 V. Determine the valuemodulation index μ.what would be the value of μ if the minimum amplitude is zero volt? Whyis μ generally kept less than 1?Chennal |
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| 2545. |
12. The energy of an excited hydrogen atom is -3.4 eV. Calculate thangular momentum of the electron according to Bohr's theory |
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| 2546. |
1. The distance between two adjacent crests (or troughs) of a wave is called its2. The distance travelled by a wave in one second is called3. The order of the speed of sound is4. Loudness of sound is measured in_5. The minimum distance between an observer and the obstacle for hearing an echo (at 20°C) isunit. |
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Answer» 1. Wavelength2. Frequency3. 3×10 ^84. Decibel5. 17.2m thanks 1.Wavelength2.Frequency3.3×10^84.Decibel5.17.2m |
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| 2547. |
A sound wave had frequency 1000 Hz and wavelength 33 cm. How long will it take to movethrough 1 km.10. |
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| 2548. |
Find the value of De Broglei Wave-length of e-whose energy is 15 kev. |
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Answer» Wavelength =h/√2mEhenceWavelength =6.626*10^-34/√2*9.1*10^-31*15*10^3*1.6*10^-19as 1eV=1.6*10^-19JHencewavelength =6.626*10^-34/√273*10^-476.626*10^-34/√27.3*10^-46=6.626*10^-34/5.22*10^-23=1.27*10^-11m |
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| 2549. |
24. (a) Derive Newton's hrst law or motio(b) Find the force required to stop a car of mass 100 kg with two passengers of 50 kg eachsitting inside, if it is moving at 60 km/hr speed and take 5 s to stop |
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Answer» total weight= 200kgspeed=60km/hr=16.5m/stime to stop=5s=> aceeleration= -10/3m/s=> force=mass*acc=> force= 2000/3 N Please solve it and then send I am not able to understand in this way Thank you so much no problem at alllll |
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| 2550. |
An empty rail car of mass 100 kg is at rest on therailway track. A boy of mass 40 kg running with aspeed of 5 m/s jumps on the rail car. The speed of the(car+ boy) system is how much?. An enm |
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