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3-2.Motion in Plane
hard
Two spheres $P$ and $Q$ of equal masses are attached to a string of length $2\,\, m$ as shown in figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant rate. What is the value of the ratio
$\left( {\frac{{{\text{Tension in the string between P and Q}}}}{{{\text{Tension in the string between P and O}}}}} \right)?$
A$1/2$
B$2/3$
C$3/2$
D$2$
Solution
Tension between $P$ and $Q$ is
$T_1=\text { centripetal force on } Q=m r \omega^2$
$=200 \times 1 \times \omega^2\left(g \cdot m \cdot \frac{r a d^2}{s^2}\right)$
Tension between $O$ and $P$ is
$T_2=\text { centripetal force on } Q+\text { centripetal force on } P$
$=200 \times 1 \times \omega^2+200 \times \frac{1}{2} \times \omega^2$
$=300 \times 1 \times \omega^2\left(g \cdot m \cdot \frac{r a d^2}{s^2}\right)$
Ratio of tension is
$\frac{T_1}{T_2}=\frac{200 \times 1 \times \omega^2}{300 \times 1 \times \omega^2}=\frac{2}{3}$
$T_1=\text { centripetal force on } Q=m r \omega^2$
$=200 \times 1 \times \omega^2\left(g \cdot m \cdot \frac{r a d^2}{s^2}\right)$
Tension between $O$ and $P$ is
$T_2=\text { centripetal force on } Q+\text { centripetal force on } P$
$=200 \times 1 \times \omega^2+200 \times \frac{1}{2} \times \omega^2$
$=300 \times 1 \times \omega^2\left(g \cdot m \cdot \frac{r a d^2}{s^2}\right)$
Ratio of tension is
$\frac{T_1}{T_2}=\frac{200 \times 1 \times \omega^2}{300 \times 1 \times \omega^2}=\frac{2}{3}$
Standard 11
Physics
