Solved: PHYSICS Question for NEET | Sushrut

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NEET PHYSICSMedium

Two stones of masses $m$ and $2m$ are whirled in horizontal circles, the heavier one in a radius $r/2$ and the lighter one in the radius $r$ . The tangential speed of lighter stone is $n$ times that of heavier stone when they experience the same centripetal forces. The value of $n$ is:

A

2

B

3

C

4

D

1

Step-by-Step Solution

Formula: The centripetal force $F_c$ acting on a body of mass $m$ moving in a circle of radius $R$ with speed $v$ is given by $F_c = \frac{mv^2}{R}$ [NCERT Class 11, Physics Part I, Sec 5.10, Eq 4.16].
Given Data:

Lighter Stone: Mass $m_1 = m$ , Radius $R_1 = r$ , Speed $v_1$ .
Heavier Stone: Mass $m_2 = 2m$ , Radius $R_2 = r/2$ , Speed $v_2$ .
Condition: Centripetal forces are equal ( $F_{c1} = F_{c2}$ ).
Relationship: $v_1 = n v_2$ .

Calculation: $\frac{m_1 v_1^2}{R_1} = \frac{m_2 v_2^2}{R_2}$ $\frac{m v_1^2}{r} = \frac{(2m) v_2^2}{(r/2)}$ $\frac{v_1^2}{r} = \frac{4 v_2^2}{r}$ $v_1^2 = 4 v_2^2$ $v_1 = 2 v_2$
Conclusion: Comparing this with $v_1 = n v_2$ , we get $n = 2$ .

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Solved: PHYSICS Question for NEET | Sushrut