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When a soft ferromagnetic material is placed in an external magnetic field, what happens to its magnetic domains?
In a soft ferromagnetic material placed in an external magnetic field:
- Domains whose magnetisation is aligned with the field grow in size (domain wall movement).
- Domains misaligned with the field shrink in size.
- Additionally, domain magnetisation can rotate to align with the applied field (domain rotation).
Therefore, domains may both increase or decrease in size AND change orientation. Option B is correct.
A house building advance of Rs. 2 million is paid to employees. In accounting terms, this payment is classified as a:
An advance paid to employees (such as a house-building advance) creates a receivable for the organization — the employee owes the money back. It is therefore recorded as an Asset (specifically a long-term receivable or loan to employees) on the balance sheet. It is not an expense until it is written off or adjusted, so it is neither revenue expenditure nor capital expenditure in the traditional sense.
This is a frequently misunderstood historical fact. According to the textbook, the first printing press in Muslim territory was established in Andalusia (Muslim-ruled Spain) in the 1480s. Crucially, it was operated not by Muslim scholars but by a family of Jewish merchants, who used it to print texts in the Hebrew script. After the fall of Granada to Catholic Spain in the 1490s (completing the Reconquista), this press was relocated — the textbook notes it moved to Istanbul, which had become a major destination for the Sephardic Jewish community expelled from Spain. Islamic calligraphic traditions were highly valued, and there was cultural resistance to the printing press in many parts of the Muslim world, which partly explains why the technology spread more slowly there than in Christian Europe despite the Muslim world's earlier familiarity with papermaking.
Time period: $T = \frac{5}{10} = 0.5\ \text{s}$
Formula for oscillation: $T = 2\pi\sqrt{\frac{I}{MB}}$
$B = \frac{4\pi^2 I}{MT^2} = \frac{4 \times 9.85 \times 5\times10^{-6}}{9.85\times10^{-2} \times (0.5)^2}$
$= \frac{4 \times 9.85 \times 5\times10^{-6}}{9.85\times10^{-2} \times 0.25} = \frac{20\times10^{-5}}{2.4625\times10^{-3}} = 0.01\ \text{T} = 10\ \text{mT}$
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