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Question

A pendulum bob is hanging from the roof of an elevator with the help of a light string. When the elevator moves up with a uniform acceleration $$'a'$$, the  tension in the string is $$T_1$$. When the elevator moves down with the same acceleration, the tension in the string is $$T_2$$. If the elevator was stationary, the tension in the string would be


A
T1+T22
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B
T1T2
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C
T1T2T1+T2
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D
2T1T2T1+T2
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Solution

The correct option is A $$\dfrac{T_{1}+ T_{2}}{2}$$
For case-1, $$ma=T_1-mg$$
$$T_1=ma+mg .....(1)$$
For case-2, $$ ma=mg-T_2$$
$$T_2=mg-ma   ......$$(2)
when elevator is stationary, then only gravitational force of bob is the tension of the string, i.e $$T=mg$$
$$(1)+(2)$$, we get 
$$mg=\dfrac{T_1+T_2}{2}$$
Therefore, $$T=\dfrac{T_1+T_2}{2}$$

105120_3442_ans.png

Physics

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