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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.
| 9851. |
why is the colour of the clear sky blue |
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Answer» The molecules of air and other fine particles in the atmosphere have size smaller than the wavelength of visible light. These are more effective in scattering light of shorter wavelengths at the blue end than light of longer wavelengths at the red end. The red light has a wavelength about 1.8 times greater than blue light. Thus, when sunlight passes through the atmosphere, the fine particles in air scatter the blue colour (shorter wavelengths) more strongly than red. The scattered blue light enters our eyes. If the earth had no atmosphere, there would not have been any scattering |
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| 9852. |
The velocity time graph of the motion of a particle moving in a straight line is given in theadjoining diagram. Draw the below diagram shown. read the graph to answer the followingquestions:(L esson -2)2015ents ,)0,0-1 (s)(6) Calculate the distance travelled by the particle in 4s.(ii) Find the velocity of motion at instant t 5s(iii) At what instant the velocity of the particle is 15m s ".2.Answer any one of the following questions in about 40 to 60 words. |
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| 9853. |
text*(x*(b*y)) %2B 2*x^2 %2B 4*x^4 %2B 8*x^3 - 2 |
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| 9854. |
A rocket is moving at a speed of 200ms1 towards a stationarytarget.While moving, it emits a wave of frequency 1000Hz.Some of thesound reaching the target gets reflected back to the rocket as anecho.Calculate the frequency of the sound as detected by the target |
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| 9855. |
nd the dimensions of (i) Power, (ii) Force and) Permittivity of vacuum (Eo).[Ans : (i) [L-MT(ii) ILMT 21 |
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Answer» i) Power = Work / Time Work = Force * Distance Force = Mass * Acceleration Acceleration = Change in velocity /Time Velocity Dimension Formula = M^0 L^1 T^-1 Time Dimension Formula = T so Acceleration = M^0 L^1 T^-1 / T = M^0 L^ 1 T^-2 Force = M* M^0 L^ 1 T^-2 = M^1 L^1 T^-2 Work = M^1 L^1 T^-2 * L = M^1 L^2 T^ -2 Now Power = M^1 L^2 T^-2 / T =M^ 1 L^ 2 T ^ -3 |
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| 9856. |
reathingphotosynthesisWhich of the fLotusfollowing plants is ideal for an aquarium 20 ww(B) hydrilla (C) ceratophnylum 1D) 4d(A) |
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Answer» Hydrilla is most suitable as it is a form of algae |
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| 9857. |
6.Thedisplacement of a particle is given by, y a+ bt+ ct2- dt. The initial velocity and acceleration arerespectively(A) b,-4d(D)-b,2c(C) b,2c |
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| 9858. |
How tall isthe table?130 cm170cm |
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| 9859. |
14.For the circuit shown here, the potential difference between points A and B is :-6CB2CStoc 30 C2cmLov 2000(A) 2.5 V(B) 7.5 V(C) 10 V(D) ZeroC. TOY |
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Answer» A) 2.5V is the correct answer I think so opt A. is correct 2.5V |
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| 9860. |
28. Velocity of a particle varies with time asv t-t-2. Average speed of the particle in timeinterval from t 0 tot 4 sec is:4(1) m/s (2)2 m/s (3)3 m/s (4)5 m/s |
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| 9861. |
253lovgraw tracharacte |
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| 9862. |
(f) Why are heaters fitted near thefloor and air conditioners, near theceiling of a room? |
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| 9863. |
1.Comparison between Electric and magnetic field. |
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| 9864. |
Find the valueglamanometerofincerent throughthe given circult1002lo se152lov |
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Answer» , Jabra santulan Mein Hota Hai To Galvanometer Mein Dhara ka mann Kyun Hota Hai |
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| 9865. |
Write a program to find the largest among two numbers. |
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| 9866. |
- मामा11. (11011)2- (101)) का गान ज्ञात करने पर होगा।(1) 20(2) 21(3) 22(4) 23 |
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Answer» Ans:-(2)21.............................. |
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| 9867. |
A 2 cm diameter coin rests flat on the bottom of a bowin which the water is 20 cm deep ( μ-4/3). If the coinis viewed directly from above, what is its apparent diametera) 2cm b) 1.5 cm c) 2. 67 cm d) 1. 67 cm |
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Answer» But answer is 1.67cm |
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| 9868. |
Page Notes |
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| 9869. |
Q16. In the figure as shown belowS2Calculate the total electric flux of the electrostatic field through the sphere S, and S. Thewire AB of length l in the figure has linear charge density "X given by Îť=kx, where x is thedistance measured along the wire from the end A |
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| 9870. |
The distance of a particle moving on a circle of radius 12 m measured from a fixed point on the circle and measuredalong the circle is given by s 2t3 (in meters). The ratio of its tangential to centripetal acceleration at t- 2s is |
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Answer» options are not Correct , it seems like.. |
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| 9871. |
50 m(d) 25 m15. A particle is fired horizontally from an inclined plane of inclination 30 with horizontal withspeed 50 ms" lfg# 10 ms-2, the range measured along the incline is1000(b)m(e) 200 -2 nm(d) 1003 mh ot an anale of 53 ahove the horizontal with a muzzle velocity of |
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| 9872. |
The length of the seconds hand of a watch is 10 mm. What is the change in the angular speed of thewatch after 15 seconds(10z/2) ms^-1(e) (20/a) mms 1Zero10/2 mms |
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Answer» it's answer is zero as the question has asked only speed not velocity and speed is scalar quantity the change in the average speed will be zero as it is a scalar quantity |
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| 9873. |
12)4.5J(3) J22(4)9J10, A body of mass m 10 kg is moving in a medium and experiences a frictional force Fk Its initialspeed is vo E 10 ms". If after 10 s, its energy is gmv3, the value of kK will be(1) 10 Kg m's (2) 10*Kg m(3) 10 Kg s[JEE (Main) 2017; 4/120,-11(4) 10 Kg m-1 |
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| 9874. |
7. A bus starts from rest with acceleration 4 ms 2 for10 second and for next 10 second it movesuniformly then decelerating with 2 ms2 and finallycome to rest. Total travelled distance is(1) 1000 nm(3) 800 nm(2) 600 m(4) 200 m |
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Answer» tq |
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| 9875. |
26. A vehicle of mass 120 kg is moving with auniform velocity of 108 km/h. The forcerequired to stop the vehicle in 10s is:(A) 120x10.8N (B) 180 N(C) 720 N(D) 360 N |
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Answer» M=120kgU=90km/h=10m/sV=0T=5 secA=v-u/tA=-10/5A= -2m/s2F=maF= 120×(-2)F=- 240NNegative sign shows that force must be applied in opposite direction Initial velocity ( u ) = 108 Km / h => 108 × 5 / 18 => 30 m / s ⏺️ Final velocity ( v ) = 0 ⏺️ Time ( t ) = 10 sec ✔️ We know => Acc. ( a ) = ( Final velocity - initial velocity ) / time => a = ( v - u ) / t => a = ( 0 - 30 ) / 10 => a = - 30 / 10 Here Acceleration is negative be velocity is decreasing. Now , We know force is product of mass and Acceleration , i.e => Force ( F ) = mass × acc ( a ) Mass ( m ) = 120 Kg a = - 3 m/s^2 => F = m × a => F = 120 × - 3 => Here force is negative because it is in opposite direction of motion. f fufgguhihhhhhhhzadjllr |
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| 9876. |
An aeropiane is tying at conetant height of 1.96 kmwith speed 600 km/h towards a point directly over atarget. At what angle of aight should it release abomb if it is to strike the target on the ground?7,(2) 45°(3) 0 |
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Answer» Theta =30° Angle of sight=180-(90+30)=60°Hence, option (4) is correct. |
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| 9877. |
The rowing speed of a man relative to water is 5 km/h andthe speed of water flow is 3 km/h. At what angle to the riverflow should he head if he wants to reach a point on theother bank, directly opposite to starting point:(a) 127°(c) 120°(b) 143°(d) 150° |
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Answer» the answer is option (a) see the solution for a similar solved ques. |
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| 9878. |
An aeroplane is flying horizontally with a velocity of216 km/hr and at a height of 1960 m. When it isvertically above a point A on the ground, a bomb isreleased from it. The bomb strikes the ground at pointB. The distance AB is (ignoring air resistance)29.(1) 1200 m(2) 0.33 m(3) 3.33 km(4) 33 km |
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Answer» When the bomb is dropped, it will have an initial horizontal velocity which is equal to the speed of the aeroplane. So the bomb fall and travel forward too.initial horizontal velocity, = 216 km/h=216×5/18=60 m/sinitial vertical velocity, = 0height, h = 1960mtime taken to fall = th=ut+1/2×gt^2........................g=9.8 1960=4.9t^2 t=√400=20sec so, AB=ux×t=60×20=1200m |
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| 9879. |
A fighter plane is flying horizontally at an altitude of 1.5 km with speed 720 km h-1. At what angle of sight (w.rt.horizontal) when the target is seen, should the pilot drop the bomb in order to attack the target? |
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| 9880. |
A drunkard is walking along a straight road. He takes5 steps forward and 3 steps backward and so on. Eachstep is 1 m long and takes 1 s. There is a pit on the road11 m away from the starting point. The drunkard willfall into the pit after(a) 21s(c) 31 s(b) 29 s(d) 37 st |
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Answer» For 8 total steps it takes 8 seconds,and moves in total 2 steps forward (2 m) Therfore,3*2 m = 6 m At 3rd cycle Time Elapsed 24 s After walking 5 m forward, he takes 5 s more.It becomes 11 m. Time = 24 + 5 = 29m thanks sonam |
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| 9881. |
rom an aeroplane when it is at a height h directly above the target. IF the aeroplanewith a speed v, the distance by which the bomb will miss the target is gven by16. A bis moving horizontally |
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Answer» The horz. velocity of the bomb = horz. Velocity of the aeroplane = v height = h time taken by ball to reach groud t = √2h/√g now for this time the range will be = (horz velocity)*(time) = v√2h√g therefore the range is √(2hv²/g). yes boss but your answer is wrong 🐍🐍🐍🐍🐍🐍🐸🐸🐸🐸🐸🐗🐗🐗🐴🐴🐴🐸🐍🐍🐼🐖🐂🐖🐖🐼🐼🐍🐸🐸🐽🐽🐽🐮🐐🐐🐤🐤🐥🐥🐺🐺🐺🐴🐸🐍🐍🐼🐯🐯🐫🐫🐪🐪🐊🐊🐊🐊🐠🐡🐠🐟🐋🐋🐳🐳🐙🐚🐬🐬🐌🐛🐛🐲🐲🐞🐝🐜🐜🐾🍸🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼☁😊☁☁😊☁😁☁☁😊☁☁😊☁☁☁☁😊😊😊😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😊☁🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼☁😊☁☁😊☁😁☁☁😊☁☁😊☁☁☁☁😊😊😊😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😊☁🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼☁😊☁☁😊☁😁☁☁😊☁☁😊☁☁☁☁😊😊😊😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😊☁💤💤💤💤💤☁☁☁💤☁☁☁💤☁☁☁💤☁☁☁💤💤💤💤💤☁ Sleepy ☁💤💤💤💤💤☁☁☁💤☁☁☁💤☁☁☁💤☁☁☁💤💤💤💤💤☁ Sleepy ☁💤💤💤💤💤☁☁☁💤☁☁☁💤☁☁☁💤☁☁☁💤💤💤💤💤☁ Sleepy ☁💤💤💤💤💤☁☁☁💤☁☁☁💤☁☁☁💤☁☁☁💤💤💤💤💤☁ Sleepy ☁💤💤💤💤💤☁☁☁💤☁☁☁💤☁☁☁💤☁☁☁💤💤💤💤💤☁ Sleepy ☁☁😊☁☁😊☁😁☁☁😊☁☁😊☁☁☁☁😊😊😊😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😁☁☁😊☁☁😊☁☁☁☁😊😊😊😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😁☁☁😊☁☁😊☁☁☁☁😊😊😊😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😁☁☁😊☁☁😊☁☁☁☁😊😊😊😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😊☁Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠☁😊☁☁😊☁😁☁☁😊☁☁😊☁☁☁☁😊😊😊😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😁☁☁😊☁☁😊☁☁☁☁😊😊😊😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😊☁🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼☁😊☁☁😊☁😁☁☁😊☁☁😊☁☁☁☁😊😊😊😊☁😊☁☁😊☁☁😊☁😊☁☁😊☁☁😊☁😊☁💤💤💤💤💤☁☁☁💤☁☁☁💤☁☁☁💤☁☁☁💤💤💤💤💤☁ Sleepy ☁🌱🍯🌱🌱🍯🌱🐝🌱🌱🍯🌱🌱🍯🌱🌱🌱🌱🍯🍯🍯🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌱🍯🌱🌱🍯🌱🍯🌱🌼🌱 Hi honey! 🌱🌼Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠Good Night 💤🌙😊 Sweet Dream🌠🌠🌠🌠🌠🌠🌠 how u can do it boss because it out of my capacity so unable to understand |
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| 9882. |
An aeroplane was flying horizontallyvelocity of 720 km/h at an altitude of 490 mWhen it is just vertically above the target abomb is dropped from it. How farhorizontally it missed the target ?(A) 1000 m(C) 100 mwith a(B) 2000 m(D) 200 m |
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Answer» h = (gt^2)/2t = sqrt(2h/g) = sqrt(2*490/9.8) = 10 sv = 720km/h = 720*1000m/3600s = 200 m/ss = vt = 200*10 = 2000 mAnswer: (b) 2000m |
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| 9883. |
ismovinghorizontally withbomb is dropped from an aeroplane when it is at a height h directly above the target. If the aeroplanes moving horizontally with a speed v, the distance by which the bomb will miss the target is given by |
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| 9884. |
ăŁc. Test Paben tror oodationar lâ State Ne, coton's larg (o grand ation. Distinguwh.between & G |
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| 9885. |
Level I (Elementary)1.. An electric heater has a resistance of 12 Ωand is operated from a 220 V power line. Ifno heat escapes from it, how much time isrequired to raise the temperature of 40 kg ofwater from 15°C to 80C? (Ans. 2703 s) |
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Answer» answer is not maching |
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| 9886. |
) sin A, oos Aa) sin C, cos C |
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| 9887. |
What is the role of cilliary muscles in the eye? Write the answer in one orsentences only |
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Answer» Ciliary Muscleis a circularmusclethat relaxes or tightens the zonules to enable the lens to change shape for focusing. The zonules are fibers that hold the lens suspended in position and enable it to change shape during accommodation |
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| 9888. |
No other thermometer is as suitable as aplatinum resistance thermometer to measuretemperature in the entire range of:-(1) 0°C to 100°C(3)-50°C to +350°C (4)-200°C to 600°c1(1(316. C6.(2) 100°C to 1500°C |
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Answer» - 200°c to 600°c is measured by a platinum resistance thermometer |
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| 9889. |
40. A person can see the area in one eye expressed indegree unit is(1) 1200(3) 1800(2) 1500(4) 360 |
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Answer» 180° is the right answer for this question |
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| 9890. |
u=20 m/sm=150 g =0.15 kgv= -25 m/st=0.01 sF=?P=? |
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Answer» p = mv-mu0.15*(-25)-0.15*(20)=-6.75kgm/s F=maa=v-u/ta=-4500m/s^2F= -675N Thanks for answer You are right F=maa=v-u/ta=-4500m/s^2F= -675N p = mv-mu0.15*(-25)-0.15*(20)=-6.75kgm/s |
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| 9891. |
2.16 Determine V, in the circuit in Fig. 2.80.16 21422WW-+10 y(+)>(+) 25 V1Figure 2.80 |
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Answer» Apply voltage node analysisso after that Vo-10/16+Vo-25/14= 0You will get Vothanks |
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| 9892. |
in SI units.10. The height of mercury column in a barometer in aCalcutta laboratory was recorded to be 75 cm. Calculatethis pressure in SI and CGS units using the followingdata Specific gravity of mercury 13.6, Density ofwater 10 kg/m, g 9.8 m/s at Calcutta. Pressure- hpg in usual symbols. |
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| 9893. |
An elevator is maving up oith an accelerotion ofht of a 60The-_apparentin the iis |
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| 9894. |
Four solid spheres each of diameter /5 cm and mass 0.5 kgare placed with their centres at the corners of a square of side4 cm. The moment of inertia of the system about the diagonalof the square is N x 10 kg-m2, then N is.9.(2011) |
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Answer» solution is not visible plz send it again |
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| 9895. |
10. The height of mercury column in a barometer in aCalcutta laboratory was recorded to be 75 cm. Calculatethis pressure in SI and CGS units using the followingdata Specific gravity of mercury 136, Density ofwater 10 kg/m', g 9.8 m/s at Calcutta. Pressure- hpg in usual symbols. |
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| 9896. |
Assuming the density of air to be 1-295 kg mfind the fall in barometric height in mm of Hg ata height of 107 m above the sea level. Take densityof mercury- 136 x 10' kg m |
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| 9897. |
A bar magnet made of steel has magneticoment of 2.5 Am* and a mass of 6.6 x 10 kg.l the density of steel is 7.9 x 10 kg/m, find theatensity of magnetization of the magnet.(Ans : 3.0 x 10 Am) |
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Answer» Like If you find it useful |
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| 9898. |
rolling fiction |
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Answer» Ans :- Rolling resistance, sometimes called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface. It is mainly caused by non-elastic effects; that is, not all the energy needed for deformation of the wheel, roadbed, etc. is recovered when the pressure is removed. |
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| 9899. |
Example 12.16. The length of each side of a cube is 0 60 m. It is completelyimmersed in a liquid of densitynsty25 x 10 kg/m in such a way that the upper face lies at a depth of 0 20 m. Calculate the thrust on each foceof the cube. |
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| 9900. |
,A constant current flows in a horizontal wire in the plane of the paper from east to west asshown in Figure. The direction of magnetic field at a point will be North to South(A) directly above the wire(B) directly below the wire(C) at a point located in the plane of the paper, on the north side of the wire(D)at a point located in the plane of the paper, on the south side of the wire |
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Answer» The magnetic field lines will move from north to south directly below the paper. |
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