Consider uniform electric field E= 3×10^3i N/C. Calculate flux of field through a square of area 10cm^2 when it's plane is parallel to the y-z plane?
Since flux =E.s= 3×10^3i.10×10^-4i=3 Nm^2/C
The electric potential at a point (x,y,z) is given by V=-x²y-xz³+4. What will be the electric field at that point?
Well, Electric field is nothing but negative gradient of potential. Interpreting the physical significance, “Among all the possible directions, the direction in which potential decreases at the fastest rate (w.r.t. distance) is the direction of resultant electric field (at the concerned point) and that rate of decrement equals the magnitude of the resultant field.” This relation between the two can always be exploited to obtain one from the other (i.e. from potential to field or vice-versa).Regards!
Over a certain region of space, the electric potential is V = 5x –3x^2y + 2yz^2. Find the expression for...?
Take partial derivatives with respect to each coordinate: V = 5x -3x^2y + 2yz^2 Ex = -∂V/∂x = -[5 - 6xy] = -5 + 6xy Ey = -∂V/∂y = -[-3x^2 + 2z^2] = 3x^2 - 2z^2 Ez = -∂V/∂z = -4yz
A, B, and C, can do a piece of work in 12, 15, and 20, days respectively. They started work together, but C left after 2 days. In how many days will the remaining work be completed by A and B?
The answer is 4 days.Solution:-A-15 daysB-12 daysC-20 daysLet the total work be 60u (by taking L.C.M of 15,12,20). Means,A can do 4u work per dayB can do 5u work per dayC can do 3u work per day.So, in the first two days, number of work done by A, B, C is 24u (4+5+3=12u/day)Remaining work after 2 days,=> 60u-24u=36uNow, C left the work,Work done by only A and B =9u/day (4u+5u)Number of days taken by A and B to complete the remaining work= Remaining work/work done by A and B together per day=36/9 =4 days ANSWERIf I have mistaken in solving this question, please make me correct.Thank you!
A point inside an equilateral triangle is 3 cm, 4 cm and 5cm respectively from each of its sides.What is the area of the triangle?
Let [math]\triangle[/math] ABC be the equilateral triangle, and the point P in it is such that the perpendicular distances from the sides BC, AB and AC are PD=3 cm, PE=4 cm and PF=5 cm respectively.Let the length of each side of the equilateral triangle be L cm.Now we have,[math]\Delta ABC = \Delta PBC + \Delta PAB + \Delta PAC[/math][math]= \frac{1}{2} PD.AC + \frac{1}{2} PE.AB + \frac{1}{2} PF.AC[/math][math]= \frac{L}{2} (PD+PE+PF)[/math][math]= \frac{L}{2}(3 + 4 + 5) = 6L\ cm^2 .[/math]Now, since [math]\Delta ABC[/math] is an equilateral triangle with side [math]L[/math], its area is given by the formula [math]\frac{\sqrt{3}}{4}L^2[/math].So [math]\frac{\sqrt{3}}{4}L^2 = 6L[/math].[math]\implies L = 8\sqrt{3}[/math]Hence the area of the given equilateral triangle is [math]48[/math][math]\sqrt{3}[/math] [math]cm^2[/math].Hope it helpsPeace!!
A train travels with a speed of 30km/h and returns with a speed of 50km/h. What is the the average speed of the train: (a) 36kmh, (b) 37.5 kmh, (c) 38 kmh, or (d) 40 kmh?
Lets assume the total distance covered is 2d. This is the distance covered both to and fro. Lets consider 2d to be 300 km. This should not really matter in the whole calculation.Total distance is 300. The train travels d in one direction or 150 km.Time taken to travel onwards is 150/30 = 5 hours.Time taken to return is 150/50 = 3 hours.Therefore average speed is total distance/total time = 300/(5 + 3) = 300/8 or 37.5 km/hr. Note: the formula would have worked with us not assuming specific value for d.Time taken for the onward journey would have been d/30 and the time taken for the return journey would have been d/50.Therefore the average speed is (d + d) / (d/30 + d/50) = 2d * 1500 / (30d + 50d) which is equal to 3000d/80d = 300/8 = 37.5 km/hr (d cancels out!!)
What is wrong with the Indian education system?
9th standard:PTA meetingPhysics Teacher: Your son is very smart. He is hardworking too. Dad: Thanks sir! He studies well. He is passionate about physics.Physics Teacher: Yes. He has what it takes to be an IAS Officer. Prepare him for Civil Services. He would be a great officer.Dad: That's good to hear sir. I am just a 9th standard guy. All I knew was that I loved applying physics to daily life activities. I knew I loved Maths too. But I did not know about my future. I did not know what I would be or what I wanted to be. I approached my teacher the next day.Class - the next day:Me: Sir. I like physics. What is its future.Physics Teacher: The future of physics is full of possibilities. There is infinite scope for exploring the field.Me: Then why did you want me to be an IAS Officer.Physics Teacher: Its a very stable job. You get security. You can get others to do your work for you. You will have many servants. You can earn very much. Even though the salary from government is less, you get lots of gifts (not so subtle). If you pursue physics, you can't earn much. Home:Mom: How did the meeting go. Dad: Great! His teacher told me that he has what it takes to even crack Civil Services.Mom: He's too young. Let us give him a few years before we push him towards his goals. (My mother is a professor. I am lucky. Not all are lucky to be afforded this freedom.)Conclusion:I tell my teacher I like his subject. He tells me it has good future. But money is the ultimate objective and it has less scope to earn money. He tells me to pursue another field. I feel there is something fundamentally wrong if we educate our younger generation that one field has more priority over another. All are equal. It just depends on one's passion towards a field and a drive to pursue it.As someone said: "Where would they be if Lata Mangeshkar picked up a cricket bat and Sachin Tendulkar became a singer."