For the arrangement shown in the figure the tension in the string is [Given: tan−1(0.8)=39∘]
A
6N
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B
6.4N
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C
0.4N
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D
Zero
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Solution
The correct option is D Zero Here tanθ=0.8 where θ is angle of repose θ=tan−1(0.8)=39∘ The given angle of inclination is equal to the angle of repose. So the 1kg block has no tendency to move. ∴mgsinθ= force of friction ⇒T=0