For the arrangement shown in figure, the tension in the string to prevent it from sliding down, is (g = 10 m/s²)

Option 1 -

12 N

Option 2 -

21.6 7N

Option 3 -

0 N

Option 4 -

2.4 N

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Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 weeks ago

Correct Option - 4

Detailed Solution:

$\begin{array}{l} f_{l} = μ N = 0.6 \times m g c o s θ \\ f_{l} = 0.6 \times 2 \times 10 \times \frac{4}{5} = 9.6 N \end{array}$

Component of weight along incline.

$= m g s i n θ = 2 \times 10 \times \frac{3}{5} = 12 N$

$\Rightarrow m g s i n θ > f_{l} s o T = 2.4 N$

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