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JEE Main Physics Class 11 Laws of Motion Part 2 Questions

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Question : 82

Total: 91

A mass of 10‌kg is suspended vertically by a rope of length 5m from the roof. A force of 30N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is θ=tan−1(x×10−1). The value of x is____(Given, g=10m∕s2 )

[27-Jun-2022-Shift-2]

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