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A block of mass 15 \mathrm{~kg} is held by a string on an inclined plane (angle \pi / 6). The tension T in the string is -\left(g=10 \mathrm{~m} / \mathrm{s}^2\right).

Option: 1

55N


Option: 2

60N


Option: 3

75N


Option: 4

90N


Answers (1)

best_answer

Given, 

M=15 \mathrm{~kg}, \theta=30^{\circ}, g=10 \mathrm{~m} / \mathrm{s}^2


As the block is held by the string,
 T=m g \sin \theta \\
T=15 \times 10 \times \sin 30^{\circ}

    =150 \times \frac{1}{2} \\
T=75 \mathrm{~N}

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