કોઈ એક સ્થાને ડીપ્-એન્ગલ (કોણ) \(30^{\circ}\) અને પૃથ્વીનાં ચુંબકીય ક્ષેત્રનો સમક્ષિતિળ ઘટક \(0.5\) ઓર્સેટડ છે. પૃથ્વીનું કુલ ચુંબકીય ક્ષેત્ર (ઓર્સેટડમાં) ...................... થશે.

A \(\frac{1}{\sqrt{3}}\)
B \(\frac{1}{2}\)
C \(\sqrt{3}\)
D \(1\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1}{\sqrt{3}}\)

Step-by-step Solution

Detailed explanation

\(B _{ H }= B \cos \theta\) \(\therefore B =\frac{ B _{ H }}{\cos \theta}=\frac{0.5 G }{\cos 30^{\circ}} \Rightarrow \frac{ G }{\sqrt{3}}\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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From JEE Mains

એક યાદચ્છિક ચલ X નું સંભાવના વિતરણ નીચે આપેલ છે :

X	\(4k\)	\(\frac{30}{7} k\)	\(\frac{32}{7} k\)	\(\frac{34}{7} k\)	\(\frac{36}{7} k\)	\(\frac{38}{7} k\)	\(\frac{40}{7} k\)	\(6k\)
P(X)	\(\frac{2}{15}\)	\(\frac{1}{15}\)	\(\frac{2}{15}\)	\(\frac{1}{5}\)	\(\frac{1}{15}\)	\(\frac{2}{15}\)	\(\frac{1}{5}\)	\(\frac{1}{15}\)

જો \(E(X)=\frac{263}{15}\) હોય, તો \(P(X<20)=\) ___ .

JEE Mains 2026 Hard

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વર્ગ	\(10-20\)	\(20-30\)	\(30-40\)	\(40-50\)	\(50-60\)
આવૃતિ	\(\alpha\)	\(110\)	\(54\)	\(30\)	\(\beta\)