(B) Let the mass of each sphere be $m$. The distance of $Q$ from $O$ is $r_Q = OP + PQ = 1\, m + 1\, m = 2\, m$. The distance of $P$ from $O$ is $r_P

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(B) Let the mass of each sphere be $m$. The distance of $Q$ from $O$ is $r_Q = OP + PQ = 1\, m + 1\, m = 2\, m$. The distance of $P$ from $O$ is $r_P = OP = 1\, m$.The tension $T_1$ in the string between $P$ and $Q$ provides the centripetal force for sphere $Q$:$T_1 = m r_Q \omega^2 = m(2)\omega^2 = 2m\omega^2$.The tension $T_2$ in the string between $O$ and $P$ must provide the centripetal force for both spheres $P$ and $Q$:$T_2 = m r_P \omega^2 + T_1 = m(1)\omega^2 + 2m\omega^2 = 3m\omega^2$.The ratio of the tensions is:$\frac{T_1}{T_2} = \frac{2m\omega^2}{3m\omega^2} = \frac{2}{3}$.

Class 11 Physics Laws of Motion Motion of Body (or Connected Bodies in horizontal or vertical) (by String or Contact)

Difficult

Two spheres $P$ and $Q$ of equal masses $m$ are attached to a string of total length $2\, m$ as shown in the figure. The string and the spheres are then whirled in a horizontal circle about $O$ at a constant angular velocity $\omega$. The distance $OP = 1\, m$ and $PQ = 1\, m$. What is the value of the ratio $\left( \frac{\text{Tension in the string between } P \text{ and } Q}{\text{Tension in the string between } O \text{ and } P} \right)?$

A
$1/2$
B
$2/3$
C
$3/2$
D
$2$

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Laws of Motion

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Motion of Body (or Connected Bodies in horizontal or vertical) (by String or Contact)

$A$ block of mass $M$ is placed on a horizontal surface and it is tied with an inextensible string to a block of mass $m$,as shown in the figure. $A$ block of mass $m_0$ is also placed on $M$. If $\mu < \mu_{min}$ (the minimum friction required to keep the block $m$ stationary),then the downward acceleration of $m$ is:

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$A$ small block of mass $m$ is placed on a wedge of mass $M$ as shown, which is initially at rest. All surfaces are frictionless. The spring attached to the other end of the wedge has a force constant $k$. If $a'$ is the acceleration of $m$ relative to the wedge as it starts sliding down and $A$ is the acceleration acquired by the wedge, then the maximum velocity of $M$ is:

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$A$ body of mass $8\,kg$ is hanging from another body of mass $12\,kg$. The combination is being pulled upwards by a string with an acceleration of $2.2\,m/s^2$. The tensions $T_1$ (in the top string) and $T_2$ (in the string between the two masses) will be respectively: (use $g = 9.8\,m/s^2$)

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$T_1$ and $T_2$ in the given figure are

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Two rectangular blocks of masses $40 \ kg$ and $60 \ kg$ are connected by a string and kept on a frictionless horizontal table. If a force of $1000 \ N$ is applied on the $60 \ kg$ block away from the $40 \ kg$ block,then the tension in the string is: (in $N$)

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$A$ body of mass $8\,kg$ is hanging from another body of mass $12\,kg$. The combination is being pulled upwards by a string with an acceleration of $2.2\,m/s^2$. The tensions $T_1$ (in the top string) and $T_2$ (in the string between the two masses) will be respectively: (use $g = 9.8\,m/s^2$)

$T_1$ and $T_2$ in the given figure are

Two rectangular blocks of masses $40 \ kg$ and $60 \ kg$ are connected by a string and kept on a frictionless horizontal table. If a force of $1000 \ N$ is applied on the $60 \ kg$ block away from the $40 \ kg$ block,then the tension in the string is: (in $N$)

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