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
B
T1T2
C
T1T2T1+T2
D
2T1T2T1+T2

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, $$T=\dfrac{T_1+T_2}{2}$$Physics

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