Home
Class 12
PHYSICS
X and Y are two loops made from same wir...

X and Y are two loops made from same wire. The radii of X and Y are `r_(1)` and `r_(2)` and their M.I are `I_(1)` and `I_(2)` . If `I_(2)//I_(1)=4` the value of `r_(2)/r_(1)` is `(K)^(1//3)`. Find the value of K.

Text Solution

AI Generated Solution

To solve the problem, we need to find the value of \( K \) given that the moment of inertia ratio \( \frac{I_2}{I_1} = 4 \) and the relationship \( \frac{r_2}{r_1} = K^{1/3} \). ### Step-by-Step Solution 1. **Understand the Moment of Inertia**: The moment of inertia \( I \) of a loop (ring) is given by the formula: \[ I = m r^2 \] ...
Promotional Banner

Topper's Solved these Questions

  • CHAPTERWISE NUMERIC /INTEGER ANSWER QUESTIONS

    DISHA PUBLICATION|Exercise CHAPTER -7 (GRAVITATION)|15 Videos
  • CHAPTERWISE NUMERIC /INTEGER ANSWER QUESTIONS

    DISHA PUBLICATION|Exercise CHAPTER -7 (MECHANICAL PROPERTIES OF SOLIDS)|15 Videos
  • CHAPTERWISE NUMERIC /INTEGER ANSWER QUESTIONS

    DISHA PUBLICATION|Exercise CHAPTER -5 (WORK, ENERGY AND POWER)|15 Videos
  • ATOMS

    DISHA PUBLICATION|Exercise EXERCISE-2: CONCEPT APPLICATOR|30 Videos
  • COMMUNICATION SYSTEM

    DISHA PUBLICATION|Exercise EXERCISE-2 : Concept Applicator|30 Videos

Similar Questions

Explore conceptually related problems

Two loops P and Q are made from a uniform wire. The redii of P and Q are r_(1)and r_(2) respectively, and their moments of inertia are I_(1)and I_(2) respectively, If I_(2)=4I_(1),"then" r_(2)/r_(1)"equals" -

Two loops P and Q are made from a uniform wire. The radii of P and Q are R_(1) and R_(2) respectively and their moments of inertia are I_(P) and I_(Q) respectively. If (I_(P))/(I_(Q))=8 then (R_(1))/(R_(2)) is

The value of (r.i)i+(r.j)j+(r.k)k is

Two loops P and Q are made from a uniform wire. The radii of P and Q are R_(1) and R_(2) , respectively, and their moments of inertia about their axis of rotation are I_(1) and I_(2) , respectively. If (I_(1))/(I_(2))=4 , then (R_(2))/(R_(1)) is

The value of sum_(i=0)^(r)C(n_(1),r-i)C(n_(2),i) is equal to

The magnitude of gratitational field intensities at distance r_(1) and r_(2) from the centre of a uniform solid sphere of radius R and mass M are I_(1) and I_(2) respectively. Find the ratio of I_(1)//I_(2) if (a) r_(1) gt R and r_(2) gt R and (b) r_(1) lt R and r_(2) lt R (c ) r_(1) gt R and r_(2) lt R .

In the given circuit if I_(1) and I_(2) be the current in resistance R_(1) and R_(2) respectively then:

Two wires are wrapped over wooden cylinder to form two co-axial loops carrying currents i_(1) and i_(2) . If i_(2)=8i_(1) the value of x for B=0 at the origin O is: