When a steady current flows through a long straight wire, the magnetic field lines form concentric circles around the wire, as per Ampere’s Law. The direction of the magnetic field lines follows the right-hand rule: if you curl the fingers of your right hand around the wire in the direction of the current, your thumb points in the direction of the magnetic field.
Therefore, for a vertical current-carrying wire, the magnetic field lines will be horizontal and form concentric circles around the wire.
Two long parallel wires X and Y, separated by a distance of 6 cm, carry currents of 5 A and 4 A, respectively, in opposite directions as shown in the figure. Magnitude of the resultant magnetic field at point P at a distance of 4 cm from wire Y is \( 3 \times 10^{-5} \) T. The value of \( x \), which represents the distance of point P from wire X, is ______ cm. (Take permeability of free space as \( \mu_0 = 4\pi \times 10^{-7} \) SI units.) 
A particle of charge $ q $, mass $ m $, and kinetic energy $ E $ enters in a magnetic field perpendicular to its velocity and undergoes a circular arc of radius $ r $. Which of the following curves represents the variation of $ r $ with $ E $?
Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : If oxygen ion (O\(^{-2}\)) and Hydrogen ion (H\(^{+}\)) enter normal to the magnetic field with equal momentum, then the path of O\(^{-2}\) ion has a smaller curvature than that of H\(^{+}\).
Reason R : A proton with same linear momentum as an electron will form a path of smaller radius of curvature on entering a uniform magnetic field perpendicularly.
In the light of the above statements, choose the correct answer from the options given below
(A) Explain the following reactions and write chemical equations involved:
(a) Wolff-Kishner reduction
(b) Etard reaction
(c) Cannizzaro reaction
Following is the extract of the Balance Sheet of Vikalp Ltd. as per Schedule-III, Part-I of Companies Act as at $31^{\text {st }}$ March, 2024 along with Notes to accounts:
Vikalp Ltd.
Balance Sheet as at $31^{\text {st }}$ March, 2024
| Particulars | Note No. | $31-03-2024$ (₹) | $31-03-2023$ (₹) |
| I. Equity and Liabilities | |||
| (1) Shareholders Funds | |||
| (a) Share capital | 1 | 59,60,000 | 50,00,000 |
‘Notes to accounts’ as at $31^{\text {st }}$ March, 2023:
| Note | Particulars | $31-3-2023$ (₹) |
| No. | ||
| 1. | Share Capital : | |
| Authorised capital | ||
| 9,00,000 equity shares of ₹ 10 each | 90,00,000 | |
| Issued capital : | ||
| 5,00,000 equity shares of ₹ 10 each | 50,00,000 | |
| Subscribed capital : | ||
| Subscribed and fully paid up | ||
| 5,00,000 equity shares of ₹ 10 each | 50,00,000 | |
| Subscribed but not fully paid up | Nil | |
| 50,00,000 |
‘Notes to accounts’ as at $31^{\text {st }}$ March, 2024:
| Note | Particulars | $31-3-2024$ (₹) |
| No. | ||
| 1. | Share Capital : | |
| Authorised capital | ||
| 9,00,000 equity shares of ₹ 10 each | 90,00,000 | |
| Issued capital : | ||
| 6,00,000 equity shares of ₹ 10 each | 60,00,000 | |
| Subscribed capital : | ||
| Subscribed and fully paid up | ||
| 5,80,000 equity shares of ₹ 10 each | 58,00,000 | |
| Subscribed but not fully paid up | ||
| 20,000 equity shares of ₹ 10 each, | ||
| fully called up | 2,00,000 | |
| Less : calls in arrears | ||
| 20,000 equity shares @ ₹ 2 per share | 40,000 | |
| 59,60,000 |