PrepSnap logoPrepSnap
Pass

Question

1.
2.
3.
4.
Answer (Detailed Solution Below)

Option 2 : $5\,\text{kg}\,\text{m}^2\text{/s}$ / $5\,\text{किग्रा}\,\text{मी}^2\text{/से}$

Advertisement

Detailed Solution

सही उत्तर $5\,\text{kg}\,\text{m}^2\text{/s}$ / $5\,\text{किग्रा}\,\text{मी}^2\text{/से}$ है।

Key Points
Helpful?
PrepSnapprepsnap.in • Part -A

Question:

A hoop of mass $2\,\text{kg}$ and radius $0.5\,\text{m}$ spins about its central axis at $10\,\text{rad/s}$. What is its angular momentum? / $2\,\text{किग्रा}$ द्रव्यमान और $0.5\,\text{मी}$ त्रिज्या का एक छल्ला अपनी केंद्रीय अक्ष के परितः $10\,\text{रेडियन/से}$ से घूमता है। इसका कोणीय संवेग क्या है?

1.
$2.5\,\text{kg}\,\text{m}^2\text{/s}$ / $2.5\,\text{किग्रा}\,\text{मी}^2\text{/से}$
2.
$5\,\text{kg}\,\text{m}^2\text{/s}$ / $5\,\text{किग्रा}\,\text{मी}^2\text{/से}$
3.
$10\,\text{kg}\,\text{m}^2\text{/s}$ / $10\,\text{किग्रा}\,\text{मी}^2\text{/से}$
4.
$0.5\,\text{kg}\,\text{m}^2\text{/s}$ / $0.5\,\text{किग्रा}\,\text{मी}^2\text{/से}$

Correct Answer: $5\,\text{kg}\,\text{m}^2\text{/s}$ / $5\,\text{किग्रा}\,\text{मी}^2\text{/से}$

For a hoop $I = MR^2 = 2 \times 0.25 = 0.5\,\text{kg}\,\text{m}^2$, so $L = I\omega = 0.5 \times 10 = 5\,\text{kg}\,\text{m}^2\text{/s}$. / छल्ले के लिए $I = MR^2 = 2 \times 0.25 = 0.5\,\text{किग्रा}\,\text{मी}^2$, अतः $L = I\omega = 0.5 \times 10 = 5\,\text{किग्रा}\,\text{मी}^2\text{/से}$।

Downloaded from PrepSnap (prepsnap.in)