Question:

Two spheres PP and QQ, each of mass 200g200\, g are attached to a string of length one metre as shown in the figure. The string and the spheres are then whirled in a horizontal circle about OO at a constant angular speed. The ratio of the tension in the string between PP and QQ to that of between PP and OO is (P(P is at mid-point of the line joining OO and QQ )

Updated On: Sep 3, 2024
  • 12\frac{1}{2}
  • 23\frac{2}{3}
  • 32\frac{3}{2}
  • 21\frac{2}{1}
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The Correct Option is B

Solution and Explanation

Tension between PP and QQ is
T1=T_{1}= centripetal force on Q=mrω2Q=m r \omega^{2}
=200×1×ω2(gmrad2s2)=200 \times 1 \times \omega^{2}\left( g \cdot m \cdot \frac{ rad ^{2}}{ s ^{2}}\right)
Tension between OO and PP is
T2=T_{2}= centripetal force on Q+Q+ centripetal force on PP
=200×1×ω2+200×12×ω2=200 \times 1 \times \omega^{2}+200 \times \frac{1}{2} \times \omega^{2}
=300×1×ω2(gmrad2s2)=300 \times 1 \times \omega^{2}\left( g \cdot m \cdot \frac{ rad ^{2}}{ s ^{2}}\right)
Ratio of tension is
T1T2=200×1×ω2300×1×ω2=23\frac{T_{1}}{T_{2}}=\frac{200 \times 1 \times \omega^{2}}{300 \times 1 \times \omega^{2}}=\frac{2}{3}
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