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Question: In the figure A and B are two blocks of mass 4kg and 2kg respectively attached to the two ends of a ...

In the figure A and B are two blocks of mass 4kg and 2kg respectively attached to the two ends of a light string passing over a disc C of mass 40kg40kg and radius 0.1m0.1m. The disc is free to rotate about fixed horizontal axes, coinciding with its own axis. The system is released from rest and the string does not slip over the disc. Then,

(This question has multiple correct options.)
A) The linear acceleration of mass B is 1013ms2\dfrac{{10}}{{13}}\dfrac{m}{{{s^2}}}
B) The number of revolutions made by the disc at the end of 10sec10\sec . from the start is 500026π\dfrac{{5000}}{{26\pi }}.
C) The tension in the string segment supporting the block A is 48013N\dfrac{{480}}{{13N}}.
D) None of these.

Explanation

Solution

We know that weight depends on the direction because it is a vector quantity. the weight of an object is acting always toward the center of the earth.

Complete step by step answer:
Block A has a higher weight than block B, so A will have a downward acceleration and B will in the upward direction of the same magnitude because of constrained, let T1and T2{T_1}{\text{and }}{{\text{T}}_2} are tensions in the left and right hanging part of the string.
Applying Newton’s law of motion,
MA×gT1=MAa\Rightarrow {M_A} \times g - {T_1} = {M_A}a ………….. (1)
Where MA{M_A} is the mass of block A, ggis the acceleration, T1{T_1} is the tension of block A and aa is the acceleration.
T2MB×g=MBa\Rightarrow {T_2} - {M_B} \times g = {M_B}a …………… (2)
Where MB{M_B} is the mass of block B, ggis the acceleration, T2{T_2}is the tension of block B and aais the acceleration.
T1RT2R=IaR\Rightarrow {T_1}R - {T_2}R = I\dfrac{a}{R} ……………. (3)
Where II is the moment of inertia of disc C about its axis.
I=12McR2I = \dfrac{1}{2}{M_c}{R^2}
On solving above three equations and substituting given values,
4g12×40×a2g=6a\Rightarrow 4g - \dfrac{1}{2} \times 40 \times a - 2g = 6a
a=1013ms2\therefore a = \dfrac{{10}}{{13}}m{s^{ - 2}}
Angular acceleration of disc, α=aR\alpha = \dfrac{a}{R}
α=10130.1\Rightarrow \alpha = \dfrac{{\dfrac{{10}}{{13}}}}{{0.1}}
10013rads2\therefore \dfrac{{100}}{{13}}rad{s^{ - 2}}
On applying Newton motion’s equation,
θ=ω0t+12αt2\Rightarrow \theta = {\omega _0}t + \dfrac{1}{2}\alpha {t^2}
ω0=0\Rightarrow {\omega _0} = 0
θ=0+12×10013×102\Rightarrow \theta = 0 + \dfrac{1}{2} \times \dfrac{{100}}{{13}} \times {10^2}
θ=500013rad\Rightarrow \theta = \dfrac{{5000}}{{13}}rad
Number of revolutions in 10Sec10\operatorname{Sec} =5000132π = \dfrac{{\dfrac{{5000}}{{13}}}}{{2\pi }}
500026π\therefore \dfrac{{5000}}{{26\pi }}
From equation (1), 4gT1=4a4g - {T_1} = 4a
a=1013rads2\Rightarrow a = \dfrac{{10}}{{13}}rad{s^{ - 2}}
T1=48013N\therefore {T_1} = \dfrac{{480}}{{13}}N

Correct options is (A),(B),(C).

Additional information:
Suppose an object of mass MM is suspended by a string from the ceiling. The string is in a state of tension. The molecules of the string near the lower end exert force on the molecule of the object. The resultant of these electromagnetic forces is the force exerted by the string on the object which is called “tension” denoted by TT. This supports the objects and prevents it from falling. It is directed away from the object. If the string is weightless then the tension in the string is the same at each and every point of the string. If we consider a point A of the string the tension in the part below A and above A are each equal to TT and directed away from A. Similarly, the tension at the upper end pulls the ceiling down and the ceiling pulls the string upward. Here also the tension in the string is directed away from the ceiling.

Note:
In this type of question, we should know about the vector. Because here direction upward and downward can be added or subtracted based on vector direction.
Weight is the force with which the object is attracted to the earth.