UP Board 2022 | अवकलज के अनुप्रयोग - L1 | Applications of derivatives | Hindi Medium |Arun Sir |5PM

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Elite Ltd. invited applications from public for 5,00,000 equity shares of Rs. 10 each issued at Rs. 11 per share. The payment was to be made as follos : Rs. 3 on Application, Rs. 4 on Allotment including premium, and Rs. 4 on call. Applications fr 6,50,000 shares were received. Allotment of shares was made as follows : (i) 100% shares to applicants of 4,00,000 shares, (ii) 50% shares to applicants of 2,00,000 shares, (iii) No allotment to applicants of 50,000 shares. A shareholder to whom 500 shares were allotted under category (i) paid full amount due on shares along with allotment money. Another shareholderholding 1,000 shares were subsequently re-issued as fully paid-up @ Rs. 8 per share. Pass the journal entries.

1.17 g an impure sample of oxalic acid dihydrate was dissolved and make up of 200 ml with water. 10 ml of this solution in acidic medium required 8.5 ml of a solution of potassium permanganate containing 3.16 g per litre of solution. The percentage purity of oxalic acid will be:

Two elastic wire A & B having length l_(A)=2m and l_(B)=1.5m have young's modules ratio (Y_(A))/(Y_(B))=(7)/(4) . If r_(B)=2mm then the radius of A given that due to application of the same force change in length in both A & B is same

12th UP Board 2024 | Biology के पेपर में Answers लिखने का Best तरीका ?| अब 1 भी नंबर नहीं कटेगा ?

A convex lens made up of glass of refractive index 1.5 is dippedin turn (i) in a medium of refractive index 1.65 (ii) in a medium of refractive index 1.33 (a) Will it behave as converging or diverging lens in the two cases ? (b) How will its focal length changes in the two media ?

If the measurement errors in all the independent quantities are known, then it is possible to determine the error in any dependent quantity. This is done by the use of series expansion and truncating the expansion at the first power of the error. For example, consider the relation z=x//y .If the errors in x, y and z are Delta x, Delta y and Delta z , respectively, then zpm Deltaz =(x pm Delta x)/(y pm Delta y)=(x)/(y)(1pm(Deltax)/(x))(1pm(Delta y)/(y))^(-1) . The series expansion for (1pm (Delta y)/(y))^(-1) , to first power in Delta y//y. " is " 1 pm (Delta y//y) . The relative errors in independent variables are always added. So the error in z will be Delta z =z((Delta x)/(x)+(Delta y)/(y)) . The above derivation makes the assumption that Delta x// x lt lt 1, Delta y //y lt lt 1 . Therefore, the higher powers of these quantities are neglected. In an experiment the initial number of radioactive nuclei is 3000. It is found that 1000 pm 40 nuclei decayed in the first 1.0 s. For | x| lt lt 1, "In"(1+x)=x up to first power in x. The error Delta lambda , in the determination of the decay constant lambda, " in " s^(-1) , is