Manipal Entrance Test 2016 Physics Paper

Limiting friction along the plane upward
=µsmg‌cos‌θ
=08×1×10×cos‌37°
=08×10×

4

5

=6.4N
Force responsible for downward motion
=mg‌sin‌θ
=1×10×sin‌37°
=1×10×

3

5

=6N
As, downward force < limiting friction.
So, frictional force = downward force.
Hence, tension in string is zero.